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Mark W. Zemansky
s» not for LOAN
HEAT FIFTH EDITION AND THERMODYNAMICS
j
Robert BoyU 1627-1691
unin Thompson Count
Rumford
*r
James
P.
Joule
1818-ISX'j
1793-1814
Nicolas Leonard 1796-1832
SadiCarnot
J. Willard Gibbs 1839-1903
lleihc Kamerlingh
Qtmes
18531926
Planck
.
Rudolf Clausius
Gustav Robert
WaltherNetmt
1822-1888
Kirchhof)
1864-19-11
Corutantin Caralhiodory 1873-1950
Peter Debye 1884-1966
1893-19?',
William Thomson
Lord Kelvin
Max
18581917
1821-1 "07
1824-1887
Clerk Maxwell 1831-1879
F. E.
Simon
Albert Einstein
18791955
HEAT AND THERMODYNAMICS
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WJ
HEAT AND THERMODYNAMICS An Fifth Edition
I
MARK
W. ZEMANSKY, Ph.D.
Professor of Physics, Emeritus
The The
City College of City University of
New
York
McGRAW-HILL BOOK COMPANY New
York
Toronto
Sydney
St.
Louis
London
San Francisco
Intermediate
Textbook
dedicated to
ADELE
C.
ZEMANSKY
PRESTON POLYTECHNIC
W^ n-^r
453S1
S^ HEAT AND THERMODYNAMICS Copyright
©
1957, 1968 by McGraw-Hill, Inc.
All Rights Reserved. Copyright 1937, 1943, 1951
by
McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Library of Congress Catalog Card Xtimber 67-26891
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•
•.-,'•'•
PREFACE
When
the
ago,
was intended
it
first
edition of "Heat for students
and Thermodynamics" appeared preparing for careers
thirty years
in physics, in chemistry,
in all branches of engineering. It was designed for sophomores, or at most juniors, and was meant to be a first introduction to the principles and subject matter of thermodynamics. To achieve this end, it started with the most elementary ideas of temperature and heat and developed the laws of thermodynamics from their experimental and engineering backgrounds, as well as presented applications of thermodynamics more or less equally divided
and
among
and engineering. purpose was maintained, but more
the disciplines of physics, chemistry,
In subsequent attention
editions, this original
was paid
to the needs of students interested in mathematics, physics,
and physical chemistry, and only small changes were made in the engineering material. Owing to the fact that in 1966 a separate engineering version of this book, entitled "Basic Engineering Thermodynamics," became available and also that the conviction was expressed by many physics teachers that it was desirable to include some kinetic theory and statistical mechanics in courses on "heat," this latest edition, the fifth, is markedly different from the first
four.
Only the
barest engineering principles involved in the operation of
heat engines and refrigerators have been retained. Details concerning convective heat transfer
and the
analysis of heat exchangers have been removed.
In their place, a chapter on the elementary principles of kinetic theory has
been substituted; in addition, a separate chapter and several other articles have been devoted to the statistical mechanics of an ideal gas, an electron gas in a metal, a vibrating lattice, and a paramagnetic subsystem in a crystal. The point of view is still preeminently thermodynamic. Thermodynamics is the fundamental discipline, and statistical methods are introduced only to
Vlll
Preface
ACKNOWLEDGMENTS
supply details that arc not included within the scope of thermodynamics, such as equations of state and temperature variation of heat capacities, and to provide greater insight with regard to processes such as the production of low temperatures by adiabatic demagnetization and the. production of negative temperatures in a nuclear magnetic subsystem. The second law of
thermodynamics
regarded as a generalization from experience with Through use of simple nonmathematical the writings of Turner, Pippard, and Landsberg, the
is still
heat engines and refrigerators.
methods suggested
in
existence of reversible adiabatic surfaces
is
shown
to follow directly
from the
Kelvin-Planck statement of the second law. The existence of an absolute temperature and of an entropy function are then deduced, without invoking the advanced mathematical techniques of Caratheodory.
Low-temperature physics is not confined to one chapter. Most of the new data in tables and graphs are recent values in the temperature range from to
300°K. As
in the previous editions, the experimental aspects of the
The parallel between the superfluidity of and the superconductivity of conductors has been accented by devoting an entire chapter to these two phenomena alone. Some of the more recent developments in the experimental and theoretical aspects of subject have been emphasized. liquid helium II
these
of
phenomena have been
what
edition
is
is
treated, in order to provide at going on in these rapidly advancing fields. As a
somewhat more
sophisticated
least
an inkling
result, this fifth
and more difficult; however, no and a year of calculus is
further preparation than a year of college physics
needed
to
understand the
text.
In the ten-year period between the fourth
and
has received from teachers throughout the world
fifth editions,
many
letters
the author
containing
and suggestions for changes. He is, of course, exceedingly grateful for this help and would like to thank these correspondents publicly and to apologize to those whose letters have gone astray and who are not included corrections
in this
list: S. E. Babb, Jr., G. R. Beacon, L. S. Castlemen, D. E. Christie, U. Condon, H. H. Denman, J. H. Horlock, N. Kurd, the late D. K. C. MacDonald, N. Pearlman, M. Sanders, F. W. Scars, H. Sepehri, I,. A. Turner, H. C. van Ness, J. H. Werntz, W. A. Woods, and H. D. Zeman. Special mention must be made of the elaborate and penetrating analysts
E.
of the deficiencies of the fourth edition and the numerous suggestions, chapter by chapter, for improvements in the fifth edition supplied by Professor B.
Kurrclmeycr of Brooklyn College. The author is tremendously grateful to Professor H. A. Boorsc, director of the Columbia University Low Temperature Laboratory, for the fruits of many years of firsthand to
him and
association with all aspects of low-tempcraturc physics. The final tribute goes to the author's wife for her diligence and patience in typing the entire
manuscript.
Mark W. Zemanskv
The
pictures of famous physicists who were pioneers in thermodynamics which adorn the end papers of this volume were provided by Professor P. T. Landsberg of University College, Cardiff, Wales, and by Mrs. Joan Warnow, Librarian of the Center for History and Philosophy of Physics of the American Institute of Physics. The author would like to express his sincerest thanks for their kindness.
CONTENTS
Preface
vn
Acknowledgments Notation
Chapter 1-1 1
-2
1-3 1-4 1-5
1-6 1-7
1-8 1-9
1-10 1-11
1-12 1-13 1-14
1
2-1
2-3
Temperature
Macroscopic Point of View
1
Microscopic Point of View Macroscopic vs. Microscopic
2
Scope of Thermodynamics Thermal Equilibrium Temperature Concept
3
Measurement of Temperature Comparison of Thermometers Gas Thermometer Ideal-gas Temperature Celsius Temperature Scale Electric Resistance Thermometry Thermocouple International Practical Temperature Scale
Chapter 2
2-2
ix
xix
2
4 6 9
12 14 17
19 21
22 23
Simple Thermodynamic Systems
Thermodynamic Equilibrium PV Diagram for a Pure Substance P8 Diagram for a Pure Substance
26 29 31
Contents
Contents
Xll
2-4
PV6
2-5
Equations of State
34
2-6
Differential
Changes of State Mathematical Theorems Stretched Wire Surface Film
36
2-7 2-8 2-9
2-10
Surface
33
40 42
Reversible Cell
43
2-12
Intensive
Chapter 3
45
and Extensive Quantities
46
Work
3-1
Work
51
3-2
Quasi-static Process
52
Work
53
3-3 3-4
•
3-5
3-6 3-7
3-8 3-9
3-10
of a Hydrostatic System
PV Diagram
55
Work Depends on the Path Work in Quasi-static Processes Work of a Wire, a Surface Film, and a Reversible Cell Work in Changing the Magnetization of a Magnetic Solid Summary Compound Systems
Chapter 4
Heat and the
First
56 57 60 62
Equation of State of a Gas
5-2
Internal Energy of a
5-7
Experimental Determination of Heat Capacities Quasi-static Adiabatic Process Clement and Desormcs Method of Measuring y Riichhardt's Method of Measuring y
5-8
Modifications of Riichhardt's
5-9
Speed of a Longitudinal
5-6
Chapter 6
65
The Microscopic Point
6-2
Equation of State of an Ideal Gas
147
6-3
Distribution of Molecular Velocities
153
6-4
Maxwellian Speeds and Temperature Equipartition of Energy
157
6-5
161
Chapter 7
Engines, Refrigerators, and the Second
76
Kelvin-Planck Statement of the Second
77
7-6
The
80
7-7
Equivalence of Kelvin-Planck and Clausius Statements
4-6
Differential
4-7
"Heat Capacity and
4-8
86
4-12
Heat Capacity of Water; The Calorie Equations for a Hydrostatic System Quasi-static Flow of Heat; Heat Reservoir Heat Conduction Thermal Conductivity
4-13 4-14
Law
145
Internal Combustion Engines
4-5
Measurement
View
7-5
Mathematical Formulation of the Concept of Heat Its
of
7-4
4-4
of the First
Kinetic Theory of an Ideal Gas
74
Internal-energy Function
Form
130 1 32
71
4-3
126
128
Method
7-3
4-2
124
Wave
Conversion of Work The Stirling Engine The Steam Engine
Work and Heat Adiabatic Work
122
6-1
7-1
Law
119
Gas
Ideal
5-4 5-5
1
11
65
Law
First
1
Gas
5-3
7-2 4-1
Ideal Gases
5-1
38
Paramagnetic Solid
2-11
Chapter 5
into Heat,
and Vice Versa
Law 166 1
68
171
173
Law
177 179
Refrigerator
185
81
82
Chapter 8
Reversibility
and
and
the Kelvin
Temperature Scale 191
88
8-1
Reversibility
89
8-2
External Mechanical Irreversibility
91
8-3
Internal Mechanical Irreversibility
192 194
93
8-4
External and Internal Thermal Irreversibility
194
Pleat Convection
96
8-5
Chemical
195
97
8-6
Conditions for Reversibility
196
4-15
Thermal Radiation; Blackbody Kirchhoff's Law; Radiated Heat
99
8-7
Existence of Reversible Adiabatic Surfaces
197
4-16
Stcfan-Boltzmann
102
8-8
Integrability of
4-9
4-10 4-11
Law
Irreversibility
Irreversibility
dQ
201
xiv
Contents
Contents
8-9
Physical Significance of X
8-10
Kelvin Temperature Scale
8-11
Equality of Ideal-gas Temperature and Kelvin Temperature
204 207 209
11-11
11-12
11-13 11-14
Heat Capacity at Constant Volume Statistical Mechanics of a Xonmetallic Crystal Frequency Spectrum of Crystals Thermal Properties of Metals
Chapt er 12 9-2 9-3
of Entropy Entropy of an Ideal Gas
The Concept 7".?
Diagram
9-4
Camot Cycle
9-5
Entropy and Reversibility Entropy and Irreversibility Entropy and Xoncquilibrium States Principle of the Increase of Entropy Engineering Applications of the Entropy Principle Entropy and Unavailable Energy
9-6 9-7 9-8 9-9
9-10 9-11
9-12 9-13
306 307 312 319
Entropy
Chapter 9 9-1
XV
Entropy and Disorder Entropy and Direction; Absolute Entropy Entropy Flow and Entropy Production
Chapter 10
Statistical
214 216 218 222 223 225 229 231
234 236 239 240 241
Mechanics
Phase Transitions; Liquid and Solid Helium
12-1
Joule-Kelvin Effect
335
12-2
Liquefaction of Gases by the Joule-Kelvin Effect
12-3
First-order Transition; Clapcyron's Equation
338 346 349 353
12-4
Sublimation; Kirchhoff's Equation
12-5
Vapor-pressure Constant
12-6
Measurement
12-7
Vaporization
361
12-8
Critical State
12-9
Fusion
of
Vapor Pressure
359
12-10
Higher-order Transitions
368 372 377
12-11
Liquid and Solid Helium
382
Chapl er 13
Special Topics
Wire
396
13-1
.Stretched
13-2
Surface Film
396
251
13-3
Reversible Cell
399
254
13-4
Fuel Cell
401
258
13-5
Dielectric in a Parallel-plate Capacitor
10-1
Fundamental
10-2
Equilibrium Distribution
10-3
Significance of
10-4
Partition Function
261
13-6
Piezoelectric Effect
403 406
10-5
an Ideal Monatomic Gas Equipartition of Energy Statistical Interpretation of Work and Heat Disorder, Entropy, and Information
263
13-7
Thermoelectric Phenomena
409
266
13-8
Simultaneous Electric and Heat Currents in a Conductor
411
268
13-9
Seebcck and Peltier Effects
269
13-10
Thomson
13-11
Thermoelectric Refrigeration
413 416 419 420 422 425 428
10-6 10-7 10-8
Principles
A and
Partition Function of
Chap ter
11
Pure Substances
11-1
Enthalpy
11-2
The Helmholtz and Gibbs
11-3
Two
275 Functions
279
11-4
Mathematical Theorems Maxwell's Equations
11-5
The T dS Equations
282 283 286
11-6
Energy Equations
291
11-7
Heat-capacity Equations
293
11-8 11-9
Heat Capacity at Constant Pressure Thermal Expansivity
11-10
Compressibility
295 297 302
Effect
and Kelvin Equations
13-12
Properties of a System of Photons
13-13
Bose-Einstein Statistics Applied to Photons
13-14
Optical Pyrometer
13-15
The Laws
13-16
Radiation Pressure; Blackbody Radiation as a
of
Wien and
of Stefan-Boltzmann
Thermodynamic System
Chap ter
14-1
14-2
14
431
Paramagnetism, Cryogenics, Negative Temperatures, and the Third Law
Atomic Magnetism Statistical Mechanics of a Magnetic-ion Subsystem
442 446
Contents
XVI
Moment
of a Magnetic-ion Subsystem
14-3
Magnetic
14-4
Thermal
14-5
Production of Millidcgree Temperatures Ionic Demagnetization
14-6
Low-temperature Thermometry
14-7
Magnetic Refrigerator Polarization of Magnetic Nuclei Production of Microdegree Temperatures
14-8 14-9
14-10 14-11
14-12 14-13
Properties of a Magnetic-ion Subsystem
Third
Law
of
456
16-11
16-12 16-13
460 470 477 479 by-
Thermodynamics
Superfluidity
451
by-
Nuclear Demagnetization Negative Kelvin Temperatures The Experiment of Pound, Purcell, and Ramsey Thermodynamics at Negative Temperatures
Chapter 15
484 487 493 496 497
1
5-2
15-3
15-4 15-5
15-6 15-7 15-8 15-9
15-10 15-11
15-12
Superfluidity of Liquid
II
Fountain Effect Second Sound Fourth Sound Creeping Film; Third Sound Other Effects of Superfluidity
Motion of He 3 through Superfluid He 4 Superconducting Transition Temperature Magnetic Properties of Type I Superconductors Heat Capacities of Type I Superconductors Energy Gap Type II Superconductors
Law
17-2
Experimental Determination of Equilibrium Constants
587
17-3
Heat of Reaction
17-4
Ncrnst's Equation
17-5
Affinity
590 594 596
17-6
Displacement of Equilibrium Heat Capacity of Reacting Gases
17-7
509
18-2
512
18-3
516
Law
Dalton's
16-2
Semipermeable Membrane
16-3
Gibbs'
16-5
16-6 16-7 16-8 16-9
16-10
Mass Action
18-4
586
in Equilibrium
600 602
Heterogeneous Systems
Thermodynamic Equations
Heterogeneous System Phase Rule without Chemical Reaction Simple Applications of the Phase Rule Phase Rule with Chemical Reaction
607
Determination of the Number of Components Displacement of Equilibrium
622 626
for a
609 613 617
18-5
522 526
18-6
531
Appendix A:
Physical Constants
635
Appendix B:
Riemann Zeta Functions
636
532 536 542 547
Bibliography
639
550 to Selected
Problems
643
Chemical Equilibrium
16-1
6-4
of
522
Index
1
580
17-1
Answers Chapter 16
575 577
Ideal-gas Reactions
Chapter 18
and Superconductivity
Helium
Chemical Potentials Degree of Reaction Equation of Reaction Equilibrium
Chapter 17
18-1 15-1
xvn
Contents
Theorem
Entropy of a Mixture of Inert Gases Gibbs Function of a Mixture of Inert Ideal Gases Chemical Equilibrium Thermodynamic Description of Noncquilibrium States Conditions for Chemical Equilibrium Condition for Mechanical Stability Thermodynamic Equations for a Phase
557
558 559 561
563 565 566 568
570 572
649
NOTATION
CAPITAL ITALIC
LOWERCASE ITALIC
B C
Area Second virial coefficient Heat capacity; Curie constant
A A
D
Debyc function
Differential sign
E
Electric intensity; energy
Naperian logarithm base; electronic
F G
Helmholtz function Gibbs function
H
Enthalpy
I
Current; nuclear
J K
Electronic
A
dimension; also
gHa^/fcT
dimension; or a constant
Molar heat capacity; speed of
light
charge
Molar Helmholtz function; variance Molar Gibbs function; Landc ^-factor; degree of degeneracy
quantum number quantum number
Thermal conductivity; equilibrium
h
Molar enthalpy; Planck's constant
i
Vapor-pressure constant
J k
Valence Boltzmann's constant
constant
Latent heat per gram or per mole
I,
Length; latent heat
I
M
Magnetization; mass
m
Mass
N
Number
n
Number number
P
Pressure; polarization
P
Partial pressure;
Q R
Heat
1
Universal gas constant; electric
r
of molecules
of a molecule or electron of gram-moles;
quantum
momentum
Heat per mole Radius; number of individual reactions
resistance; radius s
Molar entropy
T
Entropy Kelvin temperature
t
Celsius temperature; empiric
U
Internal energy
u
S
temperature
Molar energy; radiant-energy density
HEAT AND THERMODYNAMICS
Notation
Capital italic
{continued)
lowercase italic
(continued)
Molar volume Speed of a wave or
V
Volume
v
W
Work
w
X
Generalized displacement
x
a molecule Space coordinate; mole fraction
Y
Generalized force; Young's modulus
y
Space coordinate; fraction
Z
Electric charge; partition function;
z
Space coordinate
compressibility factor
SCRIPT CAPITALS /$
Magnetic induction
£
Electromotive force
J7
Tension; force
'/7l
Magnetic intensity Molecular weight Negciptcmp
% '71
GREEK LETTERS a P
Volume
7
Ratio of heat capacities; electronic
Linear expansivity expansivity,
\/kT
term in heat capacity
Thermodynamic
probability; solid
angle
Energy of a magnetic ion
A?
Radiant omittance
s
J
Surface tension
A
Finite difference
£
Degree of reaction; molecular energy Efficiency
ROMAN SYMBOLS e
Ideal-gas temperature; angle
Dcbyc temperature
cm
centimeter
e
g
gram
K
Compressibility
s
second
X
Wavelength; Lagrange multiplier;
deg
degree
penetration depth
atm
atmosphere
Joule-Kelvin coefficient; molecular
A
ampere coulomb
magnetic moment; chemical
C Hz
hertz (cps)
Molecular density; frequency;
J
joule
stoichiometric coefficient
N
newton
Peltier coefficient
Oe
oersted
Density (mass per unit volume)
V
volt
VV
watt
potential
Thomson
mann
coefficient; Stcfan-Boltz-
constant
Time; period Angle; function of temperature
SPECIAL SYMBOLS
A?A
Avogadro's number
Number
w
Coefficient of performance; angular
d
Inexact differential
Faraday's constant
X
T*
Magnetic temperature
?
of phases
speed
Magnetic susceptibility Coherence length
1. TEMPERATURE
Macroscopic Point of View
1-1
of any special branch of physics starts with a separation of a restricted region of space or a finite portion of matter from its surroundings. The portion which is set aside (in the imagination) and on which the attention
The study
is
focused
is
called the system,
and everything outside the system which has a
When a system has quantities that will be terms of been chosen, the next step is to describe it interactions with the system or its helpful in discussing the behavior of the
direct bearing
on
its
behavior
is
known
as the surroundings. in
surroundings, or both. There are in general two points of view that may be adopted, the macroscopic point of view and the microscopic point of view.
Let us take as a system the contents of a cylinder of an automobile engine. A chemical analysis would show a mixture of hydrocarbons and air before explosion, and after the mixture has been ignited there would be combustion products describable in terms of certain chemical compounds. A statement of
amounts of these substances is a description of the composition of the system. At any moment, the system whose composition has just been described occupies a certain volume, depending on the position of the piston. The volume can be easily measured and, in the laboratory, is recorded automatically by means of an appliance coupled to the piston. Another quanthe relative
tity
that
is
indispensable in the description of our system
gases in the cylinder. After explosion this pressure
is
is
the pressure of the
large; after
exhaust
it is
In the laboratory, a pressure gauge may changes of pressure and to make an automatic record as the engine operates. Finally, there is one more quantity without which we should have no adequate idea of the operation of the engine. This quantity is the temperature; as we shall see, in many instances, it can be measured just as simply as the other
be used to measure the
small.
quantities.
We
have described the materials
in a cylinder of
an automobile engine by
1-4
Temperature
3
Heat and Thermodynamics
2
specifying four quantities: composition, volume, pressure, and temperature. These quantities refer to the gross characteristics, or large-scale properties, of the system
and provide a
macroscopic description.
They
The quantities that must be
macroscopic coordinates.
are therefore called
a but
specified to provide
macroscopic description of other systems are, of course, different; macroscopic coordinates in general have the following characteristics in
common: matter. 2 3
4
In short, a macroscopic description of a system involves the specification of a few fundamental measurable properties of a system.
From
Microscopic Point of
View
—
the viewpoint of statistical mechanics or, as it is often called, statistical a system is considered to consist of an enormous number
—
N
thermodynamics
of molecules, each of
which
is
capable of existing in a
set of states
whose
molecules are assumed to interact with one another by means of collisions or by forces at a distance. The system of molecules may be imagined to be isolated or, in some cases, may be considered to energies are
ei, e 2 ,
.
.
.
•
The
be embedded in a set of similar systems, or ensemble of systems. Concepts of probability arc applied, and the equilibrium state of the system is assumed to be the state of highest probability. The fundamental problem is to find the of molecules in each of the molecular energy states (known as the populations of the states) when equilibrium is reached.
number
Since statistical thermodynamics will be treated at some length in Chap.
1
0,
not necessary to pursue the matter further at this point. It is evident, however, that a microscopic description of a system involves the following
it is
characteristics: 1
2 3
4
1-3
tity,
pressure,
is
the average rate of change of
molecular collisions is
made on
momentum due
a unit of area. Pressure,
perceived by our senses.
We
however,
feel the effects of pressure.
is
to all the
a property
Pressure
was
experienced, measured, and used long before physicists had reason to believe
molecular impacts. If the molecular theory is changed or even discarded at some time in the future, the concept of pressure will still remain and will still mean the same thing to all normal human beings. Herein lies an important distinction between the macroscopic and microscopic points of view. The few measurable macroscopic properties are as sure as our senses. They will remain unchanged as long as our senses remain the same. The microscopic point of view, however, goes much further than our senses. It postulates the existence of molecules, their motion, their energy states, their interactions, etc. It is constantly being changed, and we can in the existence of
Only a few coordinates are needed for a macroscopic description. They arc suggested more or less directly by our sense perceptions. They can in general be directly measured.
1-2
macroscopic description are really averages over a period of time of a large number of microscopic characteristics. For example, the macroscopic quan-
that
involve no special assumptions concerning the structure of
They
1
both points of view are applied to the same system, they must lead to the same conclusion. The relation between the two points of view lies in the fact that the few directly measurable properties whose specification constitutes the
based on the macroscopic point of view.
vs.
Microscopic
Although it might seem that the two points of view are hopelessly different and incompatible, there is nevertheless a relation between them; and when
Scope of Thermodynamics
1-4
It has
been emphasized that a description of the gross characteristics of a
system by means of a few of directly
its
measurable properties, suggested more or less constitutes a macroscopic description. Such
by our sense perceptions,
descriptions are the starting point of all investigations in all branches of physics.
For example,
in dealing with the
macroscopic point of view rigid
Assumptions are made concerning the structure of matter; e.g., the existence of molecules is assumed. Many quantities must be specified. The quantities specified arc not suggested by our sense perceptions These quantities cannot be measured.
Macroscopic
never be sure that the assumptions are justified until we have compared some deduction made on the basis of these assumptions with a similar deduction
is
adopted
body are considered. The
mechanics of a
rigid
body, the
in that only the external aspects of the
position of
its
center of mass
is
specified
reference to coordinate axes at a particular time. Position and time
with
and a
combination of both, such as velocity, constitute some of the macroscopic quantities used in mechanics and are called mechanical coordinates. 1 he mechanical coordinates serve to determine the potential and the kinetic energy of the rigid body with reference to the coordinate axes, i.e., the kinetic and the potential energy of the body as a whole. These two types of energy constitute the external, or mechanical, energy of the rigid body. It is the purpose of mechanics to find such relations between the position coordinates and the
time as are consistent with Newton's laws of motion.
1-5
Heat and Thermodynamics
4
of a In thermodynamics, however, the attention is directed to the interior macroscopic only those adopted, but view is point of system. A macroscopic quantities are considered which have a bearing on the internal state of a sysquantities that are tem. It is the function of experiment to determine the quantities having a Macroscopic purpose. such a sufficient for necessary and
SYSTEM
Adiabatic
All
wall
system that
may be
described in terms of thermodynamic coordinates
and air; and a vapor in contact with its liquid, such as liquid and vaporized ammonia. Chemical thermodynamics deals with the above gasoline vapor
Physical systems and, in addition, with solids, surface films, and electric cells. stretched as systems above, such addition to the includes, in thermodynamics wires, electric capacitors, thermocouples,
and magnetic substances.
values of Y.
different values of pressure and volume are possible. kept constant, the volume may vary over a wide range of values, and vice versa. In other words, the pressure and the volume are independent coordinates. Similarly, experiment shows that, for a wire of constant
constant mass, If the
pressure
many
is
mass, the tension and the length are independent coordinates, whereas, in the case of a surface film, the surface tension and the area may be varied
Diathermic
^wall
All
V,
Only restricted values of Y',
values of
(b)
(a) Fig. 1-1
Properties of adiabatic
A state of a system
X'
possible
X' possible
and
rmic walls.
Y and X have
definite values that remain conunchanged is called an equilibrium state. Experiment shows that the existence of an equilibrium state in one system depends on the proximity of other systems and on the nature of the wall separating them. Walls are said to be either adiabatic or diathermic. If a wall is adiabatic (sec Fig. 1-ltf), a state Y, X for system A and Y', X' for in
which
stant so long as the external conditions are
B may
coexist as equilibrium states for any attainable values of the
four quantities, provided only that
We have seen that a macroscopic description of a gaseous mixture may be given by specifying such quantities as the composition, the mass, the pressure, and the volume. Experiment shows that, for a given composition and for a
X
SYSTEM B SYSTEM B
system
Thermal Equilibrium
1-5
Only restricted possible
is
thermodynamic called a thermodynamic system. In engineering, the important such as mixture, steam; a vapor, such as a such as air; gas, systems arc a
SYSTEM A
A
values of
Y,X possible
coordinates. bearing on the internal state of a system are called thermodynamic It is the pura system. internal energy ol determine the serve to Such coordinates thermodynamic among the relations general to find thermodynamics pose of coordinates that arc consistent with the fundamental laws of thermodynamics.
A
Temperature
the.
wall
is
able to withstand the stress
two sets of coordinates. Thick layers of wood, concrete, asbestos, felt, etc., are good experimental approximations to adiabatic walls. If the two systems are separated by a diathermic wall (see Fig. 1-1 b), the values of Y, A" and Y', X' will change spontaneously until an equilibrium state of the combined system is attained. The two systems arc then said to be in thermal equilibrium with each other. The most common associated with the difference between the
diathermic wall
is
a thin metallic
sheet. Thermal equilibrium is the stole achieved
by two (or more) systems, characterized by restricted values of the coordinates of the systems, after they have been in communication with each other through a diathermic wall.
seem quite complicated, such as an electric cell with two different electrodes and an electrolyte, may the still be described with the aid of only two independent coordinates. On other hand, some systems composed of a number of homogeneous parts require the specification of two independent coordinates for each homogeneous part. Details of various thermodynamic systems and their thermodynamic
Imagine two systems A and B separated from each other by an adiabatic wall but each in contact with a third system C through diathermic walls, the whole assembly being surrounded by an adiabatic wall as shown in Fig. l-2a. Experiment shows that the two systems will come to thermal equilibrium with the third and that no further change will occur if the adiabatic wall
For the present, to simplify our discussion, we shall deal only with systems of constant mass and composition, each requiring only one pair of independent coordinates for its description. This involves no essential loss of generality and results in a considerable saving of words. In referring to any nonspecificd system, we shall use the symbols Y
instead of allowing both systems
independently.
Some
systems that, at
coordinates will be given in Chap.
and
X
for the pair of
first
sight,
2.
independent coordinates.
separating
A and B
is
then replaced by a diathermic wall (Fig. \-2b). If, A and B to come to equilibrium with C at
same time, we first have equilibrium between A and C and then equilibrium between B and C (the state of system C being the same in both cases), then, when A and B are brought into communication through a diathermic the
wall, they will be found to be in thermal equilibrium.
sion
"two systems are
in thermal equilibrium" to
We shall use the expres-
mean
that the
two systems
1-6
Heat and Thermodynamics
Y
A
SYSTEM
7
Temperature
SYSTEM B
SYSTEM
C
SYSTEM
SYSTEM
A
D
X Fig. 1-3
X'
Isotherms of two different systems.
ments on simple systems indicate usually that at least a portion of an isotherm If
A
and
B
are each
A andB
in
are
in
thermal
is
equilibrium with each other
thermal equilibrium with C, then
Similarly, with regard to system B,
designated by The zeroth law of thermodynamics. (Adiabatic walls are
1-2
cross
shading; diathermic walls, by heavy lines.)
diathermic wall, are in states such that, if the two were connected through a equilibrium. thermal in the combined system would be following These experimental facts may then be stated concisely in the form:
Two
of
all
(a) Fig.
a continuous curve.
thermal equilibrium systems in thermal equilibrium with a third are in Fowler, we shall call this postulate the zeroth
H.
with each other. Following R.
which are
thermal equilibrium with call If
curves
I
A
in the state
2,
of system B. Experiment shows that there exists a thermal X\; Y 2 X* Yh X,; etc., every one of which is in
X[
the original state Y[, set of states
Yh
,
zeroth equilibrium with this same state Y[, X[ of system B and which, by the that all shall suppose law, are in thermal equilibrium with one another. 1-3, in Fig. such states, when plotted on a FA" diagram, lie on a curve such as I
We
which we states at
shall call
X'2
;
and
I'
all
the states on isotherm I' of system B.
etc.,
We shall
rwo systems.
corresponding isotherms of the
the experiments outlined above are repeated with different starting
every one of which
is
A lying on curve II may be found, thermal equilibrium with every state of system B way, a family of isotherms, I, II, III, etc., of system
in
an
which a system
make no assumption
isotherm.
is in
An
still
other systems C,.D,
etc.,
may
be obtained.
All states of corresponding isotherms of all systems have something in
Y h X\ in thermal equilibrium with a sysB in the state ¥[, X[. If system A is removed and its state changed, there with be found another state Y X\ in which it is in thermal equilibrium
whole
,
a corresponding family I', II', III', etc., of system B may be found. Furthermore, by repeated applications of the zeroth law, corresponding
Consider a system
tem
of states Y[, X[; Y'2
conditions, another set of states of system
isotherms of
will
set
A and
Temperature Concept
1-6
a
find
and therefore in thermal equilibrium with one another. These states are plotted on the Y'X' diagram of Fig. 1-3 and lie on the isotherm I'. From the zeroth law, it follows that all the states on isotherm I of system A are in
lying on curve II'. In this
law of thermodynamics.
we
thermal equilibrium with one state (Y\,X\) of system A,
in
isotherm
is the locus
of all points representing
thermal equilibrium with one state of another system.
as to the continuity of the isotherm,
We
although experi-
mon, namely, that they are
in
systems themselves, in these
states,
thermal equilibrium with one another.
may
property temperature.
The temperature of a system
whether or not a system
is in
manner.
is
We
call this
a property that determines
thermal equilibrium with other systems.
of temperature
When a system A
The
be said to possess a property that
ensures their being in thermal equilibrium with one another.
The concept
com-
may be
arrived at in a
with coordinates
Y,
X
is
more
sophisticated
separated from a system
C
with coordinates Y", X", the approach to thermal equilibrium is indicated by changes in the four coordinates. The final state of thermal equilibrium is
denoted by a relation among these coordinates which
may
be written in the
general functional form f.,c(Y,X;
Y",X")
=
0.
(1-1)
8
1-7
Heat and Thermodynamics
For example, if A were a gas with coordinates P (pressure) and V (volume) and obeying Boyle's law, and if C were a similar gas with coordinates P" and V", Eq. (1-1) would be
PV - P"Y" =
0.
Thermal equilibrium between system B, with coordinates tem C is similarly denoted by the relation fBC {Y',X'; Y",X")
=
where f B c may be quite different from fAC but behaved function. Suppose Eqs. (1-1) and (1-2) are solved for
)"',
A",
and
also
}'";
assumed
to
be a well-
=
g IK (Y',X',X").
(1-3)
to the zeroth law, thermal equilibrium between and C implies thermal equilibrium between A and
relation
among
coordinates of systems
fAli {Y,X; Y',X')
=
A
and
B
A and C B,
which
only; thus,
The extraneous
coordinate
X"
in
(Y,X)
number
is
is
merely
the equation of
an isotherm of system A,
given a different numerical value, a different
to a different set of corresponding isotherms.
Measurement
1-7
of
Temperature
an empirical temperature scale, we select some system with A as a standard, which we call a thermometer, and adopt a set of rules for assigning a numerical value to the temperature associated with each of its isotherms. To every other system in thermal equilibrium with the thermometer, we assign the same number for the temperature. The simplest procedure is to choose any convenient path in the YX plane, such as that establish
Y and
by the dashed line Y = Yi which intersects the isotherms each of which has the same Y coordinate but a different X coordiThe temperature associated with each isotherm is then taken to be a
shown
in Fig. 1-4
at points nate.
convenient function of the (1-4)
0.
Since Eq. (1-3) also expresses the same two equilibrium situations, it must agree with Eq. (1-4): that is, it must reduce to a relation among Y, X; Y\ A" only.
different
coordinates
Now, according and between B is denoted by a
hA
then
g l
gAC (Y,X,X")
or
=
such as curve I curve is obtained, such as II in Fig. 1-3. The temperature of all systems in thermal equilibrium may be represented by a number. The establishment of a temperature scale is merely the adoption of a set of rules for assigning one number to one set of corresponding isotherms
To
Y" =
t
(1-2)
Y" = gAC (Y,X,X"), and
relation
of Fig. 1-3. If/
and a
0,
is
sys-
The
9
Temperature
X at
this intersection point.
The
coordinate
X
is
and the form of the thermometric function d(X) determines the temperature scale. There are six important kinds of thermometer, each with its own thermometric property, as shown in Table 1-1. called the thermometric property,
Eq. (1-3) must therefore drop out,
and the equation must reduce to h A (Y,X)
=
h D (Y',X').
Applying the same argument a second time with systems A and C in equilibrium with B, we get finally, when the three systems are in thermal equilibrium. h A (Y,X)
=
llls
(Y',X')
=
h c (Y",X").
(1-5)
In other words, a function of each set of coordinates exists, and these functions are all when the systems are in thermal equilibrium with one another. The common
equal
value
/
of these functions
I
is
common
to all the systems.
hc(Y",X").
(1-6)
the empirical temperature
= hA (J,X) =
h„(Y',X')
=
Fig.
1-4
Setting
up a temperature
scale involves assignment of numerical values to the
isotherms of an arbitrarily chosen standard system, or thermometer.
Heat and Thermodynamics
10
Table
1
Thermometers and Thermometric Properties
1-1
Gas
(const,
volume)
Symbol
Pressure
P
Electric resistor (const, tension)
Electric resistance
R'
Thermal emf
£
Pressure
1>
Paramagnetic salt Blackbody radiation
Magnetic susceptibility Radiant emittance
and solving
X
e{X)
Now
systems in thermal equilibrium with
it
=
assign
interval
where "linear
a is
X
An
Method
x
d(X2)
aV
*
(1-7)
temperature 0(A) of a system, cither of two procedures
Apply Eq.
(1-7) to
a thermometer placed
first in contact with the system whose temperature 0(A) is to be incas-" urcd, and then in contact with an arbitrarily chosen standard system
in
an
easily reproducible state
Apply Eq.
(1-7) to the
'
* $ ~ p> X
(const. Y).
an arbitrary number of "degrees" to the temperature — d(Xi). Then 0(A) can be calculated from the three .
easily reproducible stale
of an arbitrarily chosen standard system
Before 1954 there were two fixed points: ice coexisted in
(1)
is
the
equilibrium with air-
saturated water at one atmosphere pressure (the ice point); and (2)
g(Ai)
in use be/ore 1954.
2.
get
temperature at which pure
(const. Y),
-A A
e(X
measurements A, Ai, A"2
the following linear function of A':
an arbitrary constant. It follows that two temperatures on this scale" arc to each other as the ratio of the corresponding A''s, or
To determine the may be adopted: 1
= aX
Ai
0(Aj
called a fixed point.
0(A)
0(A2)
we
for 0(A'),
fcn.y.
Let A' stand for any one of the thermometric properties listed in Table 1-1, and let us arbitrarily choose, for the temperature common to the thermometer to all
-
0(AQ
0(A)
Thermocouple (const, tension) Helium vapor (saturated)
and
11
Temperature
Subtracting Eq. (1-9) from Eq. (1-8),
Thermometric property
Thermometer
-7
where the temperature
0(Xi)
Ai
0(A)
X'
thermometer
is
0(A\).
temperature
at
Ait. 1-12.
Method after 1954. With this method, only one fixed point is chosen, namely, the temperature and pressure at which ice, liquid water, and water vapor coexist in equilibrium, a state known as the triple point of water. We choose arbitrarily for the temperature at this fixed point 273.16 degrees Kelvin, abbreviated 273.16°K. (The reason for the use of Kelvin's name will be made clear later.) Thus, designating the triple point of water by the subscript 3, we have, from Eq. (1-7),
Thus
(1-8)
at the
and pure steam one atmosphere pressure (the steam point). The temperature interval between these two fixed points was chosen to be 100 deg. A critical discussion of this abandoned method will be given in the temperature of equilibrium between pure water
O(A'),
and
0(A)
A
0(A3 )
"A
'
3
0(A 3 ) = 273.1 6°K.
with
0(A)
Hence,
=
273.16°-^
(1-10)
(const. Y).
then at the temperature of another arbitrarily chosen standard system in another easily reproducible state where the temperature is
0(A2 ). Thus
The temperature
of the triple point of water
paint of thermometry.
0(A2 )
Aj
0(A)
X'
(1-9)
purity
When
is
distilled into
all air
is
the standard fixed
To achieve the triple point, water of the highest a
vessel depicted schematically in Fig. 1-5.
has been removed, the vessel
is
sealed
off.
With
the aid
12
1-8
Meat am! Thermodynamics
Temperature
13
Seal-off
Thermometer bulb
Vapor
Water layer
Pig.
1-5
Triple-point
cell.
of a freezing mixture in the inner well, a layer of ice
the well.
When
the freezing mixture
bulb, a thin layer of ice
is
and vapor phases coexist point.
The
is
is
formed around
replaced by a thermometer
melted nearby. So long
as the solid, liquid,
in equilibrium, the. system
is
at the triple
actual shape of the apparatus used by the U.S. National
Bureau of Standards
is
shown
in Fig. 1-6. Fig.
1
-6
Diagram
of the
NBS triple-point cell
[B,
D)
Dewarjlask (H). A, water vapor; C, thermometer well; E,
Applying the principles outlined in the preceding paragraphs to the first Table 1-1, we have three different ways of measuring temperature. Thus, for a gas at constant volume, three thermometers listed in
6{P)
ice
bath (G) within a
=
273.16°^-
(const. F);
Xow
imagine a series of tests in which the temperature of a given system measured simultaneously with each of the three thermometers. Such a comparison is shown in Table 1-2. The initials NBP stand for the normal boiling point, which is the temperature at which a liquid boils at atmospheric is
pressure; the letters
and
an
mantle; F, liquid water.
Comparison of Thermometers
1-8
lor
in use in ice
an
for a
NMP stand
d(R')
=
273.16'
8(6)
=
273.16' £_
thermocouple,
St
for normal melting point,
N'SP tor the normal
and TP for the triple point. The numerical values arc not meant to be exact, and 273.16 has been written simply 273. If one compares the 6 columns, it may be seen that at any fixed point, except the triple point of water, the thermometers disagree. Even the two hydrogen thermometers disagree slightly, but the variation among gas thermometers may be greatly reduced by using low pressures, so that a gas thermometer has been chosen as the standard thermometer in terms of which the empiric sublimation point,
electric resistor.
temperature scale
is
defined.
1-9
Heat and Thermodynamics
14
Fixed
(hoppercons tantan
Platinum
Constant-
Constant-
resistance
thermocouple
thermometer
volume H 2 thermometer
volume Ho thermometer
—Mercury
point
£,
mV
*()
R',
ohms
W)
P,
atm
HP)
P,
atm
reservoir
e(P)
N2
(NBP)
0.73
32.0
1.96
54.5
1.82
73
0.29
79
2
(NBP)
0.95
41.5
2.50
69.5
2.13
86
0.33
90
4.80
193
0.72
196
=
273
(NSP)
HO z
(TP)
H
2
(NBP)
3.52
£3 =
Sn (NMP)
6.65
154
6.26 273
185
R 3 = 9.83 273
P3 =
6.80 273 Pz
1.00
10.05
440
13.65
380
9.30
374
1.37
374
17.50
762
18.56
516
12.70
510
1.85
505
Gas Thermometer
1-9
A
15
Comparison of Thermometers
Table 1-2
C0 2
Temperature
is shown in and dimensions differ in the various bureaus and institutes throughout the world where these instruments are used and depend on the nature of the gas and the temperature range for which the thermometer is intended. The gas is contained in the bulb B (usually made of platinum or a platinum alloy), which communicates with through a capillary. The volume of the gas is kept the mercury column constant by adjusting the height of the mercury column until the mercury level just touches the tip of a small pointer in the space above M, known as the dead space or nuisance volume. The mercury column is adjusted by raising or lowering the reservoir. The difference in height A between the two mercury columns and M' is measured when the bulb is surrounded by the system whose temperature is to be measured, and when it is surrounded by water at
schematic diagram of a constant-volume gas thermometer
The
Fig. 1-7.
materials, construction,
M
Fig.
1
meniscus at
The
bulb, capillary,
ume,
when
3
4
If the
M
The
indicial point.
diameter of
capillary
Some
5
molecules of the gas, a pressure gradient exists
(Knudsen
gas
and nuisance volumes undergo changes of voland pressure change. the capillary is comparable with the mean free
the temperature
path, of the.
is
in the
effect).
adsorbed on the walls of the bulb and capillary, the
lower the temperature, the greater the adsorption.
There are
6
the triple point.
always touches
left
pressure.
M
M
Mercury reservoir is raised or lowered Bulb pressure equals h plus atmospheric
Simplified constant-volume gas thermometer.
-7
so that
various values of the pressure must be corrected to take account of
effects
due
to
temperature and compressibility of the mer-
cury in the manometer.
the following sources oi error:
Many 1
The gas
2
temperature different from that in the bulb. The gas in the capillary connecting the bulb with the manometer has a temperature gradient; i.e., it is not at a uniform temperature. umes)
is
present in the dead space (and in any other nuisance volat a
made
great improvements in the design of gas thermometers have been
in recent years.
Two
of these arc depicted schematically in Fig. 1-8.
Instead of the thcrmomctric gas in the bulb communicating directly with the
mercury
manometer, there arc two separate volumes of gas: the which goes as far as a diaphragm and exerts a pressure
in the
thcrmomctric
gas,
1-10
Heat and Thermodynamics
16
ft-
-
S
•:
:
-• :
-
1-10
-•-•-^^
Ideal-gas
17
Temperature
Temperature
Suppose that an amount of gas is introduced into the bulb of a constantvolume gas thermometer so that the pressure P 3 when the bulb is surrounded by water at its triple point, is equal to 1000 mm Hg. Keeping the volume V ,
-Second
To ac
capacitor plate
(J-
Bulb containing
constant, suppose the following procedures are carried out:
bridge 1
thermometric gas
Surrounding the bulb with steam condensing at determine the gas pressure Ps and calculate
1
atm
pressure,
Diaphragm Equalizing
(capacitor
gas space
Fig. 1-8
d(P.)
plate)
=
273.16°
-.
1000
Remove some of the gas so that Ps has a smaller value, say, 500 Hg. Determine the new value of P, and calculate a new value
Schematic diagram of two improvements in a gas thermometer in use at the U.S.
mm
National Bureau of Standards.
e{P.)
on one
side of
leading to the
and a manometric gas on the other side of the diaphragm manometer. The diaphragm itself is one plate of a capacitor.
=
273.1
C
500'
it,
with the other plate fixed nearby.
phragm causes
A
3
a slight motion of the diaphragm, resulting in a change of
capacitance that
is
observed with the aid of an ac bridge. At about
pressure differential of
1
part per million
is
detectable.
When
the
1
Continue reducing the amount of gas in the bulb so that P% and Ps have smaller and smaller values, Pi having values of, say, 250 Hg, 100 mm Hg, etc. At each value of Ps, calculate the corresponding
mm
difference of pressure across the dia-
atm, a
6(PS ).
diaphragm
4
shows no deflection, the manometric gas pressure is the same as that of the thermometric gas, and a reading of the manometer gives the gas pressure
Plot 6(PS ) against Pi
where
P3 =
0.
and extrapolate the the graph
resulting curve to the axis
Read from
in the bulb.
lim 0(P,).
Another improvement depicted in Fig. 1-8 is an equalizing gas space surrounding the bulb. The manometric gas is allowed to (ill this space. At the moment when a manometer reading is made, there is no net force tending to alter the dimensions of the bulb, and therefore no correction need be made for a variation of bulb volume with pressure. The greatest improvements have been made in the mercury manometer. The mercury meniscus in each tube is made very flat by widening the tubes, since the dead space docs not depend on this width as it did in the older instrument depicted in Fig. 1-7. The position of a mercury meniscus is obtained by using it as one plate of a capacitor, with the other being fixed nearby, and measuring the capacitance with an ac bridge. Gauge blocks are used to measure the difference in height of the two mercury columns. Pressures can be measured exact to a few ten-thousandths of a millimeter of
i'l-'O
The
results of
a
series of tests of this sort are plotted in Fig. 1-9 for four
different gases in order to also of
the readings of
measure 6(P) not only of condensing steam but
The graph conveys
the information that, although a constant-volume gas thermometer depend upon the nature
condensing
sulfur.
of the gas at ordinary values of Pa, all gases indicate the same temperature as Ps is
lowered and made
We
to
approach zero.
therefore define the ideal-gas temperature 9
273.16° lim (-=3 Pr-0
V
by the equation
(const. V).
(1-11)
mercury.
One gas thermometer in the
has been built with a differential pressure diaphragm
thermometer bulb
itself,
thereby eliminating the dead space entirely.
Although the ideal-gas temperature scale is independent of the properties it still depends on the properties of gases in general.
of an)' one particular gas,
18
1-11
and Thermodynamics
Jlcat
0U J )
pcrature.
be shown in Chap. 7
how
the absolute zero of tempera-
defined on the Kelvin scale. Until then, the phrase "absolute zero" will have no meaning. It should be remarked that the statement, found in so = 0, many textbooks of elementary physics, that, at the temperature ture
^0,
719.5
It will also
19
TempcraUirc
is
T
719.0
molecular activity ceases is entirely erroneous. First, such a statement involves an assumption connecting the purely macroscopic concept of temperature and the microscopic concept of molecular motion. If we want our theory to be general, this is precisely the sort of assumption that must be avoided. Second, when it is necessary in statistical mechanics to correlate
all
718.5
______ Air 718.0
N?
717.5
^e(sulfur) 717.0
!=
H2
717.75°K
1 1
250
500
1
_.
750
P3 mm ,
of
temperature to molecular activity, it is found that classical statistical mechanics must be modified with the aid of quantum mechanics and that, when this modification is carried out, the molecules of a substance at absolute zero
have a finite amount of kinetic energy, known as the
zero-point energy.
Hg
Celsius
1-11
The
Temperature Scale
Celsius temperature scale
as that of the ideal-gas scale, but
employs a degree of the same magnitude zero point
its
is
shifted so that the Celsius
temperature of the triple point of water is 0.01 degree Celsius, abbreviated 0.01 °C. 373.50 -
373.25
Thus,
if
I
denotes the Celsius temperature,
-,
1
=
6-
273.15
c
(1-12)
373.00
250
750
500
P3 mm ,
Fig.
1
of
1000
Hg
Thus, the Celsius temperature
Readings of a constant-volume gas thermometer for
-9
the temperature of condensing
sure
steam and for that of condensing sulfur, when dijerent gases are used at various values of P3.
Helium is the most useful gas for therinometric purposes for two reasons. At high temperatures helium does not diffuse through platinum, whereas
/„
at
which steam condenses at
1
atm
pres-
is /s
and reading
hydrogen does. Furthermore, helium becomes a liquid at a temperature lower than any other gas, and therefore a helium thermometer may be used to measure temperatures lower than those which can be measured with any
0,
from
=
6„
-
273.15°,
Fig. 1-9,
ts
-
=
373.15°
=
100.00°C.
273.15°
other gas thermometer.
The
lowest ideal-gas temperature that can be measured with a gas ther-
about 0.5°K, provided that low-pressure He 3 is used. The temperature 6 = remains as yet undefined. In Chap. 7 the Kelvin temperature scale, which is independent of the properties of any particular substance, will be developed. It will be shown that, in the temperature region in which a gas thermometer may be used, the ideal-gas scale and the Kelvin scale are
mometer
is
identical. In anticipation of this result,
we
write
°K
after an ideal-gas teiu-
The
accurate measurement of temperature with a gas thermometer re-
work and mathematical computaand when completed becomes an international event. Such work is published in a physical journal and eventually is listed in tables of physical constants. The temperatures of the normal boiling points (NBP) and normal melting points (XMP) of a number of materials have been measured, and the results arc tabulated in Table 1-3. The fixed points designated in the quires months of painstaking laboratory tion
20
Heat and Thermodynamics
1-12
tabic as basic are used to calibrate other
thermometers
in a
manner
that
Electric Resistance
1-12
Temperature
21
Thermometry
will be described in Art. 1-14.
When Temperatures of Fixed Points
Table 1-3
Temp., °C
Fixed points
Standard
Triple point of water
Basic
NBP of hydrogen (hydrogen point) NBP of oxygen (oxygen point)
0.01
Equilibrium of ice and air-saturated water (ice point) NBP of water (steam point)
NMP of zinc (zinc point) NMP of antimony (antimony NMP of silver (silver point) NMP of gold (gold point) Secondary
point)
NBP of helium NBP of neon NBP of nitrogen
NMP of mercury Transition point of sodium sulfate
NBP
Temp., °K 273.16
-252.88 -182.97
20.26 90.17
0.00 100.00 419.51 630.50 961.90 1064.5
273.15 373.15 692.66 903 65 1235.05 1337.65
-268.93 -246.09 -195.81 -38.86
4.22 27.09 77.35 234.29 305 53 491.11 505.00
32 38 .
NMP of tin
217.96 231.913
NBP of benzophenone
305 90
NMP of cadmium NMP of lead
320.90 327.30
of naphthalene
the resistance thermometer
.
.
.
among
form of
a long, fine wire,
it is
constructed so as to avoid excessive cooling. In special circumstances the
whose temperature is be measured. In the very-low-tcmpcrature range, resistance thermometers often consist of small carbon-composition radio resistors or a germanium crystal doped with arsenic and scaled in a helium-filled capsule. These may be bonded to the surface of the substance whose temperature is to be measin the material
to
ured or placed in a hole drilled for that purpose. It is customary to measure the resistance by maintaining a known constant current in the
thermometer and measuring the potential difference
with the aid of a very sensitive potentiometer. A typical circuit is shown in Fig. 1-10. The current is held constant by adjusting a rheostat so that the potential difference across a standard resistor in series with the theracross
it
mometer, as observed with a monitoring potentiometer, remains constant. The platinum resistance thermometer may be used for very accurate work within the range of involves the
—253
measurement
to 1200°C. of
R Fl
The
at various
calibration of the instrument
known temperatures and
the
representation of the results by an empiric formula. In a restricted range,
the following quadratic equation
579 05 594.05 600.45
is
often used:
.
R[, t
Battery
Notice that
in the
is
wound around a thin frame strains when the wire contracts upon wire may be wound on or embedded usually
m
the basic fixed points there are two normal boiling
oxygen and of water. The normal boiling point is the temits vapor remain in equilibrium when the vapor exerts a pressure of exactly 1 atm. Should the pressure depart from 1 atm by only 1 cm of mercury, the temperature change would be several tenths of a degree. It is necessary therefore to have elaborate controls on the vapor pressure to keep it constant to within a fraction of a millimeter of mercury. Such controls are available at the major bureaus of standards, points, those of
=
flj(l
+ At +
Bfi),
Rheostal
perature at which a liquid and
|
To monitoring
, J potentiometer
To main potentiometer
but not in the average laboratory.
A
normal melting (or freezing) point is defined also at 1 atm pressure. change of pressure of 1 cm of mercury on a solid, however, produces a change in melting point of only about 0.00001 deg, and therefore no precautions need be taken. As a result, there is a growing tendency to eliminate all boiling points as fixed points and to retain only melting points and triple
A
points.
Fig, 1-10
Circuit for measuring the resistance of a resistance thermometer through which a
constant current is maintained.
22
1-14
Heat and Thermodynamics
Wire
equation
B
\ 1 1
1 I
i
23
depends upon the materials of which it is composed. A platinum-platinum-rhodium couple to 1600°C. The advantage of a thermocouple is that it has a range of comes to thermal equilibrium quite rapidly with the system whose tempera-
Wire/1
Test junction
Temperature
1
is
often sufficient.
The range
of a thermocouple
i
:
I
k
ture
i
is
to be measured, because
its
mass
is
small. It therefore follows tem-
i
perature changes easily, but
.
it is
not so accurate as a platinum resistance
thermometer. 1
life
i
Copper wire
\SfeS&$8j§$7
International Practical Temperature Scale
1-14 Reference junction
Copper wire Fig. 1-11
Potentiometer
Thermocouple of wires A and B with a reference junction consisting of two junca potentiometer.
tions with copper, connected to
where R is the resistance of the platinum wire when water at the triple point and A and B are constants.
it
is
surrounded by
The
into three
correct use of a thermocouple
is
shown
in Fig. 1-11.
The thermal
measured with a potentiometer, which, as a rule, must be placed at some distance from the system whose temperature is to be measured. The reference junction, therefore, is placed near the test junction and consists of two connections with copper wire, maintained at the temperature of melting ice. This arrangement allows the use of copper electromotive force (emf)
arc
made
1
is
The
wires for connection to the potentiometer.
tiometer are usually
practical scale agrees with the Celsius scale at the basic fixed points listed in Table 1-3. At temperatures between these fixed points, the departure from the Celsius scale is small enough to be neglected in most practical work. The temperature interval from the oxygen point to the gold point is divided
Thermocouple
1-13
At the meeting of the Seventh General Conference of Weights and Meas1 927, at which 31 nations were represented, an international practical temperature scale was adopted, not to replace the Celsius or ideal-gas scales, but to provide a scale that could be easily and rapidly used to calibrate scientific and industrial instruments. Slight refinements were incorporated into the scale in revisions adopted in 1948 and in 1960. The international ures in
thermocouple
is
binding posts of the poten-
If the
two binding
posts are at the
R'
calibrated by measuring the thermal
emf
temperatures, the reference junction being kept at 0°C.
of such
measurements on most thermocouples can usually be represented by
results
as follows:
£ =
a
+
bt
+
1
cl"
R'6[\
+ At + Bt + 2
C(t
-
100)( s ].
by measurements at the oxygen steam point, and the zinc point. From the triple point of water to the antimony point. The same platinum resistance thermometer is used, and the temperature is determined from a relation between R' and t containing three constants:
The
constants are determined
point, the triple point of water, the
at various
The
=
same
known
a cubic equation
From the oxygen point to the triple point of water. A platinum resistance thermometer with a platinum wire whose diameter must lie between the limits 0.05 and 0.20 mm is used, and the temperature is determined from the following relation between R' and / containing four
of brass, and therefore at the potentiometer there
two copper-brass thermocouples.
parts, as follows:
constants:
temperature, these two copper-brass thermocouples introduce no error.
A
main
2
+ dl\
£ is the thermal emf and the constants a, h, c, and d arc different each thermocouple. Within a restricted range of temperature, a quadratic
R'
where
The
for
the steam point,
=
/?'
(1
+At +
Bt*).
constants are determined by measurements at the triple point,
and the zinc
point.
24
Heat and Thermodynamics
From
3
Temperature
the antimony point to the gold point.
The
Celsius temperature
/
is
When B and C are
found to hold.
we
in thermal equilibrium,
25
get
defined by the formula
is
£ =
nM@M'
+ bt + a",
a
where where £ is the cmf of a standard thermocouple of platinum and of an alloy of 90 percent platinum and 10 percent rhodium, when one of the junctions
is
at the triple point of water.
determined by measurements
and
at the
antimony
The
the gold point.
fully in later chapters.
are constants.
are the three functions that are equal to one another at
Set each of these functions equal to the ideal-gas temperature
(b)
and see whether any in Chap. 2. 1-3
The international practical temperature scale is not defined below the oxygen point. Temperature measurements in the liquid hydrogen and liquid helium ranges, as well as in the range above the gold point, will be. discussed
C, and
What
0,
thermal equilibrium?
constants are
point, the silver point,
R, C,
»,
(a)
+ nRC'W - M'PV =
8,
of these equations arc equations of state as discussed
In the table below, a
number
in the top
row represents the pres-
sure of a gas in the bulb of a constant-volume gas thermometer (corrected
dead space, thermal expansion of bulb, etc.) when the bulb is immersed water triple-point cell. The bottom row represents the corresponding readings of pressure when the bulb is surrounded by a material at a constant
for
in a
unknown temperature. rial.
(Use
Calculate the ideal-gas temperature
of this mate-
five significant figures.)
PROBLEMS Pa,
Systems A, B, and
1-1
P", V".
When A and C arc
in
C
found
to be satisfied.
V;
nbP
When B
- P"V" =
and
C
mm Hg
mm
/',
thermal equilibrium, the equation
PV is
are gases with coordinates P, V; P',
1000.0
Hg
1535.3
+—
nfi'P"!'" yl
=
The symbols n, b, and B' are constants. (a) What are the three functions which is
arc equal to one another at equal to /, where / is the empiric
temperature?
What
is
the relation expressing thermal equilibrium between
A
1000
—1.16 and
b
=
Systems
A and B
resistor
are paramagnetic salts with coordinates 9£,
and %', M', respectively. System C is a gas with coordinates A and C arc in thermal equilibrium, the equation
nRO¥ - MPV =
=
a
+
b
log R',
0.675.
S2.
P,
V.
is
found
to
be exactly
What is the temperature? Make a log-log graph of R'
(b) against 6 in the resistance range from 1000 to 30,000 U. 1-5 The resistance of a doped germanium crystal obeys the equation
and B? 1-2
383.95
In a liquid helium cryostat, the resistance
(a)
holds.
(b)
767.82
= with a
thermal equilibrium and each of which
250.00
are in thermal equilibrium, the
Hog R'
P'V - P"V"
1151.6
500.00
1-4 The resistance R' of a particular Allen-Bradley carbon obeys Clement's equation, namely,
relation •
750.00
log/r
M
=
4.697
-
3.917 log
6.
When (a)
What (A,)
200
In a liquid helium cryostat the resistance is
is
measured to be 218
Q.
the temperature?
Make
a log-log graph of R' against 6 in the resistance range from
to 30.000 U.
2-1
2.
some
cases,
rium
is
is
exceedingly slow.
The change
ceases
when chemical
when
Thermal equilibrium exists separated from
THERMODYNAMIC
is
SYSTEMS
rium, is
the
equilib-
reached. there
is
no spontaneous change
coordinates of a system in mechanical and chemical equilibrium
SIMPLE
27
Simple Thermodynamic Systems
its
in the
when
it
surroundings by a diathermic wall. In thermal equilib-
all parts of a system are at the same temperature, and this temperature same as that of the surroundings. When these conditions are not satisfied,
a change of state will take place until thermal equilibrium
When
the conditions for
all
reached.
is
three types of equilibrium are satisfied, the
is said to be in a state of thermodynamic equilibrium; in this condition, apparent that there will be no tendency whatever for any change of
system it is
system or of the surroundings, to occur.
state, either of the
dynamic equilibrium can be described involve the lime,
Thermodynamic Equilibrium
2-1
i.e.,
in
whatsoever, cither spontaneously or by virtue of outside influence, the
system
is
said to
undergo a change
of state. f
When
a system
is
not influenced
any way by its surroundings, it is said to be isolated. In practical applicathermodynamics, isolated systems are of little importance. We usually have to deal with a system that is influenced in some way by its surroundings. In general, the surroundings may exert forces on the system or provide contact between the system and a body at some definite temperain
tions of
ture.
When
the state of a system changes, interactions usually take place
between the system and its surroundings. When there is no unbalanced force in the interior of a system and also none between a system and its surroundings, the system is said to be in a state of mechanical equilibrium.
When
these conditions arc not satisfied, either
the system alone or both the system and
change of
state,
which
will cease only
its
surroundings will undergo a
when mechanical
equilibrium
is
restored.
When
a system in mechanical equilibrium does not tend to undergo a spontaneous change of internal structure, such as a chemical reaction, or a transfer of matter from one part of the system to another, such as diffusion or solution, however slow, then
A
it is said to be in a state of chemical equilibrium. system not in chemical equilibrium undergoes a change of state that, in
This must not be confused with the terminology of elementary physics, where the is often used to signify a transition from solid to liquid or liquid to gas, etc. Such a change in the language of thermodynamics is called a change of t
expression "change of state" phase.
thermodywith any problem involving the rate at
to deal
which, a process takes place.
way
Stales of thermo-
terms of macroscopic coordinates that do not
terms of thermodynamic coordinates. Classical
namics does not attempt
Suppose that experiments have been performed on a thermodynamic system and that the coordinates necessary and sufficient for a macroscopic description have been determined. When these coordinates change in any
in
The
investigation of such problems
carried
is
out in other branches of science, as in the kinetic theory of gases, hydro-
dynamics, and chemical kinetics.
When
the conditions for any one of the three types of equilibrium that
constitute
be
to
in
thermodynamic equilibrium are not
the interior of a ing
satisfied,
the system
is
said
Thus, when there is an unbalanced force in system or between a system and its surroundings, the follow-
a nonequilibrium
state.
phenomena may take
place: acceleration, turbulence, eddies, waves, etc.
While such phenomena arc in progress, a system passes through nonequilibrium states. If an attempt is made to give a macroscopic description of any one of these nonequilibrium states, it is found that the pressure varies from one part of a system to another. There is no single pressure that refers to the system as a whole. Similarly, in the case of a system at a different tem-
perature from
its
up and there
set
We
is
surroundings a nonuniform temperature distribution
no
single
temperature that
refers to the
is
system as a whole.
therefore conclude that, when the conditions for mechanical and thermal
equilibrium are not satisfied, the states traversed by a system cannot be described in terms o]
thermodynamic coordinates referring It
to
the system as
a whole.
must not be concluded, however, that wc arc
with such nonequilibrium
states. If
we
entirely helpless in dealing
divide the system into a large
number
thermodynamic coordinates may be found in terms of which a macroscopic description of each mass element may be approximated. There are also special methods for dealing with systems in mechanical and thermal equilibrium but not in chemical equilibrium. All these special methods will be considered later. At present we shall deal exclusively with systems in thermodynamic equilibrium. Imagine, for the sake of simplicity, a constant mass of gas in a vessel so of small mass elements, then
Simple Thermodynamic Systems
2-2
Heat and Thermodynamics
28
equipped that the pressure, volume, and temperature may be easily measured. If we fix the volume at some arbitrary value and cause the temperature to assume an arbitrarily chosen value, then we shall not be able to vary the all. Once J" and 6 are chosen by us, the value of P at equilibrium determined by nature. Similarly, if P and 9 arc chosen arbitrarily, then the value of V at equilibrium is fixed. That is, of the three thermodynamic coordinates P, V, and 6, only two are independent variables. This implies that there exists an equation of equilibrium which connects the thermodynamic coordinates and which robs one of them of its independence. Such
29
mixture of chemically active gases, a mixture of liquids, or
gases, a
a solution.
A heterogeneous mixture, such as a mixture of different gases with a mixture of different liquids.
3
pressure at
in
contact
is
Every thermodynamic system has its own equation of state, although in some cases the relation may be so complicated that it cannot be expressed in terms of simple mathematical
an equation
is
called
an
equation of stale.
functions.
equation of state expresses the individual peculiarities of one system another and must therefore be determined either
An
in contradistinction to
by experiment or by molecular theory. A general theory like thermodynamics, based on general laws of nature, is incapable of expressing the behavior of one material as opposed to another. An equation of state therefore
not a theoretical deduction from thermodynamics but
is
experimental addition to thermodynamics.
ments
in
tion
is
which the thermodynamic coordinates
of a system
of values measured.
formula-
As soon
as this
is exceeded, a different form of equation of state may be valid. equation of state exists for the states traversed by a system that is not mechanical and thermal equilibrium, since such states cannot be, described terms of thermodynamic coordinates referring to the system as a whole.
For example, if a gas in a cylinder were to expand and to impart to a piston an accelerated motion, the gas might have, at any moment, a definite volume and temperature, but the corresponding pressure could not be calculated from an equation of state. The pressure would not be a thermodynamic coordinate because it would not only depend on the velocity and the acceleration of the piston but would also perhaps vary from point to point.
Any
system of constant mass that exerts on the surroundings a uniform
hydrostatic pressure, in the absence of surface, gravitational, electric, and
magnetic
effects,
we
meter, are used in various applications of thermodynamics and will be used occasionally in this book. In the absence of any special remarks about units,
however,
2-2
it
will
be understood that cgs units are
PV Diagram
for a
to
be employed.
Pure Substance
vapor, the pressure of the
equation of its
No in
cubic centimeters; the most convenient scale of tempera-
is
were measured
range in
in
the ideal-gas scale. Other units of pressure, such as pounds per square inch, atmospheres, millimeters of mercury, and kilograms per square centi-
ture
An
therefore only as accurate as the experiments that led to
and holds only within the range
an
and the volume
If 1 g of water at about 94°C is introduced into a vessel about 2 liters in volume from which all the air has been exhausted, the water will evaporate completely and the system will be in the condition known as unsaturated
is
It expresses the results of experi-
as accurately as possible, within a limited range of values. state
usually
Experiment shows that the states of equilibrium')" of a hydrostatic system can be described with the aid of three coordinates, namely, the pressure P exerted by the system on the surroundings, the volume V, and the absolute temperature 6. The pressure is measured in dynes per square centimeter
shall call a hydrostatic system. Hydrostatic systems are
divided into the following categories:
shown
in Fig. 2-1
,
vapor being
this state is
less
than
1
atm.
On
the
PV diagram
represented by the point A. If the vapor
is
then
compressed slowly and isothcrmally, the pressure will rise until there is saturated vapor at the point B. If the compression is continued, condensation takes place, with the pressure remaining constant (isobaric process) as long as the temperature remains constant. The straight line BC represents the isothermal isobaric condensation of water vapor, the constant pressure being called the vapor pressure.
At any point between B and
in equilibrium; at the point C, there
is
Since a very large increase of pressure
C,
water and steam are
only liquid water, or saturated
liquid.
needed to compress liquid water, the line CD is almost vertical. At any point on the line CD, the water is said to be in the liquid phase; at any point on AB, in the vapor phase; and at any point on BC, there is equilibrium between the liquid and the vapor phases. ABCD is a typical isotherm of a pure substance on a PV diagram. is
At other temperatures in Fig. 2-1. It
is
the isotherms are of similar character, as shown seen that the lines representing equilibrium between liquid
and vapor phases, until a certain
or vaporization
temperature
is
lines,
reached
get shorter as the temperature
—the
critical
temperature
rises
above which
is no longer any distinction between a liquid and a vapor. The isotherm at the critical temperature is called the critical isotherm, and the point
there 1
2
A pure
one chemical constituent in the form of a solid, a liquid, a gas, a mixture of any two, or a mixture of all three. A homogeneous mixture of different constituents, such as a mixture of inert substance,
which
is
t
In the remainder of this book the word '•equilibrium," unmodified by any adjective, thermodynamic equilibrium.
will refer to
30
I
Icat
2-3
and Thermodynamics
Simple Thermodynamic Systems
31
vapor, or sublimation. There is obviously one such line that is the boundary between the liquid-vapor region and the solid-vapor region. This line is associated with the
point
triple point.
at a pressure of 4.58
is
In the case of ordinary water, the triple
mm and a temperature of 0.01°C,
extends from a volume of 1.00
206,000
cm 3 /g
2-3
Pd Diagram
cm
3
,'g
and the line volume of
(saturated liquid) to a
(saturated vapor).
for a
Pure Substance
vapor pressure of a solid is measured at various temperatures until is reached and then that of the liquid is measured until the
If the
the triple point
point
reached, the results
when
plotted on a
P6 diagram appear compressed until there is no vapor left and the pressure on the resulting mixture of liquid and solid is increased, the temperature will have to be changed for equilibrium to exist between the solid and the liquid. Measurements of these pressures and temperatures give rise to a third curve on the Pd diagram, starting critical
is
as in Fig. 2-2. If the substance at the triple point
at the triple point
and continuing
coexistence of (1) solid and vapor
vapor
lie
on the
indefinitely. lie
on the
vaporization curve; (3) liquid
is
The
points representing the
sublimation curve; (2) liquid
and
solid lie
In the particular case of water, the sublimation curve
Pressure
Vapor saturation curve-
Volume Fig. 2-1
It
is
P -Fusion curve
seen that the critical point pressure and
is
called the
a point of inflection on
at the critical point are
volume
is
known
critical point.
the critical isotherm. as the
critical pressure
and the critical volume, respectively. All points at which the liquid is saturated saturated vapor lie on the liquid saturation curve, and all points representing by dotted lie on the vapor saturation curve. The two saturation curves denoted are isotherms lines meet at the critical point. Above the critical point the the approach continuous curves that at large volumes and low pressures
Vaporization curve
isotherms of an ideal gas.
In the
PV diagram
shown
Vapor in Fig. 2-1, the low-temperature region repre-
senting the solid phase has not been shown.
The
solid region
region
and the region ^^Sublimation curve
and vapor are indicated by isotherms of the same general character as those in Fig. 2-1 The horizontal portion of one of equilibrium
between
solid
.
of these isotherms represents the transition front saturated solid to saturated
Temperature Fig. 2-2
P8
diagram for a substance such as water.
and
curve.
called the frost
Isotherms of a pure substance.
that represents the limit of the vaporization lines
The
V
is
on the fusion
line,
the vaporization curve the
called the steam
is
line,
and the fusion curve
is
called
slopes of the sublimation
stances are positive.
The
and the vaporization curves
slope of the fusion curve, however,
for all sub-
may
The
which expands upon melting. intersection of the sublimation and vaporizapoint of point is the The triple only on a PO diagram is the triple understood that curves. must be tion It it is a line. Triple-point data point. On a PV diagram represented by a point 2-1. in Table substances are given interesting for some very high pressures, Bridgman line of water at investigating the ice In
whereas the opposite
III,
Tammann
III,
these forms of ice
fusion curve of
one of the important exceptions.
and
ice,
among
be positive
most substances has a positive slope. Water When an equation known as the Clapeyron is equation is derived in Chap. 12, it will be seen that any substance, such as water, which contracts upon melting has a fusion curve with a negative slope, or negative.
points,
IV and \
33
were found to be unstable. Equilibrium and liquid give rise to six other triple which, along with the low-pressure triple point, are listed in Table 2-2.
modifications of conditions
ice line.
The
Simple Thermodynamic Systems
2-4
Heat and Thermodynamics
32
is
true for a substance
discovered five
V, VI, and VII
new
— ordinary
modifications of
ice
ice,
ice
I.
Two
Triple-point Data
Table 2-1
Temp., Substance
Helium (4) (\ point) Hydrogen (normal) Deuterium (normal)
Neon Nitrogen
Oxygen
Ammonia Carbon dioxide Sulfur dioxide
Water
II,
other
All the information that
is
represented on both the
grams can be shown on one diagram are plotted along rectangular axes.
Two
such surfaces are shown
2.172 13.84 18.63 24.57 63.18 54.36 195.40 216.55 197.68 273.16
The
result
is
and
vapor
Pressure
4. 579
Ice
I,
liquid,
Ice
I,
liquid, ice III
2,115
Ice
I,
ice II, ice III
2,170 3,510 3,530 6,380 22,400
Ice II, ice III, ice Ice III, liquid, ice
Ice V, liquid, ice
V V
VI
Ice VI, liquid, ice
VII
mm Hg kg/cm 2 kg/cm 2 kg/cm 2 kg/cm 2 kg/cm 2 kg/cm2
critical point
whole
surface.
solid-liquid region projects into the
is
critical isotherm
denoted by the is
marked
C.
letters Cr,
and the triple point by Tr. no free surface and with
A substance with
128
324 94
1.14 45.57 3880 1.256 4.58
1
4- 0.01
-22.0 -34.7 -24.3 -17.0
+ 0.16 + 81.6
PVB
a volume determined by that of the container is called a gas when its temperature is above the critical temperature; otherwise it is called a vapor.
37.80 52.8
Temp., °C
called the
finally the triple-point line projects into the triple point.
Triple Points of Water
Phases in equilibrium
and the P0 dia1', and
a PV0 surface projected on the PV plane, he will see the usual PI" diagram. Upon projecting the surface on the PO plane, the whole solid-vapor region projects into the sublimation curve, the whole liquid-vapor region projects
The The
Ilg
PV
the three coordinates P,
and 2-4: the first for a substance like H 2 that contracts upon melting, and the second for a substance like CO2 that expands upon melting. These diagrams are not drawn to scale, the volume axis being considerably foreshortened. If the student imagines
fusion curve,
Pressure,
mm
K
if
in Figs. 2-3
into the vaporization curve, the
r
Table 2-2
designated as ice
being denoted by
PV6 Surface
2-4
Fig. 2-3
Surface for a substance that contracts on melting.
34
I
Icat
2-5
and Thermodynamics
Simple Thermodynamic Systems
35
100 81.6
20.000 atm
Surface for a substance that expands on melting.
Fig. 2-4
shown on the PVl surface shown in Fig. 2-5, which was constructed by Verwiebe on the basis of measurements by All the triple points of water are
Bridgman.
land
Equations of State
2-5
II
Fig. 2-5 It
is
impossible to express the complete behavior of a substance over the
whole range of measured values of P, V, and by means of one simple equation. There have been over sixty equations of state suggested to represent only the liquid, vapor, and liquid-vapor regions, ranging from the ideal gas
Surface for water, showing all the triple points. (Constructed by Verwiebe on the basis of measurements by Bridgman.)
which, because of its five adjustable constants, represents with some accuracy the whole range above the triple point.
Some
equation
of these equations are frankly empirical, designed to represent as
measured values of P, V, and 6, while others are theohaving been calculated on the basis of the kinetic theory of gases. One of the most famous of the theoretical equations of state, based on assumptions concerning molecular behavior that are still of use today, is the van dcr closely as possible the
Pv
which holds only at low pressures
= m. in the
(2-1)
vapor and gas regions,
to the
Waals equation of
Bcattic-Bridgman equation:
p= where
A = A
m^A (s+B)
°H>
retical,
A
»-*(i-0'
(2-2)
= '
W
state:
3 (» - b) = ('+*) J
Re.
(2-3)
This equation holds fairly well in the liquid region, the vapor region, and near and above the critical point.
2-6
Heat and Thermodynamics
36 In
all these
equations,
where each
37
is itself a function of 8 and P. Both the above have an important physical meaning. The student will remember from elementary physics a quantity called the average coefficient of volume expansion, or volume expansivity. This was defined as
volume where
the gram-molar
s is
Simple Thermodynamic Systems
partial derivative
partial derivatives 1
and
7Fl is the
g mole =
R
molecular weight;
grams
T/l
a constant called the universal gas constant.
is
.
,
Av. vol. exp.
2-6
Differential
If a
Changes of
State
system undergoes a small change of state whereby
initial state
it
passes
from an
of equilibrium to another state of equilibrium very near the
=
change of vol. per unit 2-= change or£ temp.
vol.
and referred to conditions under which the pressure was constant. If the change of temperature is made infinitesimal, then the change in volume also becomes infinitesimal and we have what is known as the instantaneous volume expansivity, or just volume expansivity, which is denoted by 0. Thus,
one, then all three coordinates, in general, undergo slight changes. change of, say, V is very small in comparison with V and very large in comparison with the space occupied by a few molecules, then this change of V initial
If the
may
be written as a differential dV.
to the
volume
If
dV could
of space, then
the
Fwere a geometrical quantity referring
be used to denote a portion of that space
V
arbitrarily small. Since, however,
is
a macroscopic coordinate denoting
volume of matter, then, for dV to have a meaning, it must be large enough enough molecules to warrant the use of the macroscopic point of
to include
view. Similarly,
if
the change of
P
is
very small
in
represented by the differential dP. Every it
respect to the quantity itself
behavior of a
few
molecules.
represents
and
in a quantity
large in comparison with the
The
reason for this
is
that
it
also
may
which
effect
is
another
way
be
is
small with
produced by the
imagine the equation of of the other two. Thus,
later
in reciprocal degrees.
The effect of a change of pressure on the volume of a hydrostatic system when the temperature is kept constant is expressed by a quantity called isothermal compressibility and is represented by the symbol k (Greek kappa). Thus,
when
state solved for
any coordinate
in
terms
The unit of compressibility is The value of k for solids and pressure, so that k
V=
If the
may
infinitesimal
state of equilibrium to
equilibrium involves a dV, a dd, and a dP,
down
in the previous
in partial differential calculus
dV
atm-1 where ,
1
liquids varies but
atm
=
little
is
solved for P, then
function of (9,P).
change from one
the condition laid
1
often be regarded as constant.
equation of state
P= An
(2-5)
v \dp).
of saying that thermodynamic
coordinates are macroscopic coordinates.
We may
/3 is a function of 6 and P, but experiments to be described show that there are many substances for which is quite insensitive to a change in P and varies only slightly with 0. Consequently, within a small temperature range /3 may, as a rule, be regarded as a constant. (3 is expressed
Strictly speaking,
thermodynamic coordi-
nates such as volume, pressure, and temperature have no meaning
applied to a few molecules. This
and
thermodynamics must
infinitesimal in
a change
P
comparison with
very large in comparison with molecular fluctuations, then satisfy the requirement that
(2-4)
v \dd ),
all
of which
paragraph.
shall
dP,
function of
(9,
V)
*-(&* + (#."•
assume satisfy theorem
A fundamental
enables us to write
-(!),*+(:
we
another state of
Finally,
if
6
is
imagined as a function of P and
V,
1.01
X
10 6
dyn/cm 2
.
with temperature and
2-7
Heat and Thermodynamics
38
above equations the system was assumed to undergo an infinitesimal process from an initial state of equilibrium to another. This enabled us to use an equation of equilibrium (equation of state) and to solve it for any coordinate in terms of the other two. The differentials dP, dV, and dO In
all
and arc
therefore are differentials of actual functions If dz is
Now
the
an exact differential of a function
of, say,
called exact
of the three coordinates, only
An
infinitesimal that
is
=
^
(XL
1
not the differential of an actual function
is
called
\dyj.
If dx
=
and dz
* 0,
(2-7)
partial differential calculus that are
There are two simple theorems in a relation
among
The
and
proofs arc as follows: Suppose there exists
the three coordinates
and
x, y,
/(w) =
r; thus,
In the case of a hydrostatic system, the second theorem yields the result
0-
V3PM**/* Then
all
o.
Mathematical Theorems
in this subject.
and
(2-6)
(dy/dx)
Ua W- + w» =
clear later.
used very often
.v
follows that
it
an inexact differential and cannot be expressed by an equation of the above tvpe. Other distinctions between exact and inexact differentials will be made
2-7
=_L_
(*s\
or
'%).*
dx
are independent. Choosing
CSlOOl—
may be
written dz
two
z as the independent coordinates, the above equation must be true for and dx =^ 0, it follows that sets of values of dx and dz. Thus, if dz =
differentials.
x and y, then dz
39
Simple Thermodynamic Systems
* can be imagined as a function of y and
z,
\
and
The volume
expansivity
/3
and
the isothermal compressibility k were de-
fined as
*-(©.* + (£X*
=1 V\dd Jp Also, y can be imagined as a function of x and
z,
and
and
A * =- hf (,£
dx
I
Substituting the second equation into the
first,
V\dP)i
Therefore,
we have
An
infinitesimal
\ddjv change
K
in pressure
may now
be expressed
in
terms of these
physical quantities. Thus,
-fo m.
-a
iv
ig).
S>*
dx\ (dy
Sfh \dz
dz.
or
dP
=
®-d0 k
- ~dV. kV
(2-8)
Heat and Thermodynamics
40
2-8
At constant volume,
The tension in the wire ¥, measured in newtons (X). The length of the wire /,, measured in meters (m). The ideal-gas temperature 9.
1
cause the temperature to change a
we
stant volume, the pressure will
amount from
finite
change from Pi
i and / denote the initial and final between these two states, we get
given in terms
is
of only three coordinates:
ff-fi.
If
complete thermodynamic description of a wire
sufficiently
41
Simple Thermodynamic Systems
to Pf,
to 9/ at con-
2
where the subscripts
3
states, respectively.
9,-
Upon
integrating
The
states of
thermodynamic equilibrium are connected by an equation For a wire
of state that as a rule cannot be expressed by a simple equation. .
Pi '•
The
right-hand
member can
= ~ _.
9/0 fOtP
at constant
dB.
K
/ft
be integrated
we know
if
the
way
in
S=
which
vary with 9 at constant volume. If the temperature range 9/ — 9i is small, very little error is introduced by assuming that both are constant. With these assumptions we get
3 and
temperature within the limit of
elasticity,
Hooke's law holds;
namely,
k
where La If a
is
const. (L
—
La),
the length at zero tension.
wire undergoes an infinitesimal change from one state of equilibrium
to another, then the infinitesimal
change of length
is
an exact
differential
and can be written P,
-
ft
=
?
{9{
-
9 t ),
dl from which the final pressure may be calculated. For example, consider the following problem: A mass of mercury at a pressure of 1 atm and a temperature of 0°C is kept at constant volume. If the temperature is raised to 10°C, what will be the final pressure? From tables of physical constants, 3 and k of mercury remain practically constant within the temperature range of to 1 0°G and have the values
and
whence
P,
-
3
=
181
X
K
=
3.87
X 10-6 atm-i X 10- deg-' X 10 3.87 X 10- atrn-
Pi
=
2-8
P/
=
+ = 1
d7.
where both partial derivatives are functions of 9 and Jr These derivatives are connected with important physical quantities. We define the linear .
expansivity
a
as
IdL a= L\d9 1
(2-9)
of a will be considered later. Measurements depends only slightly on S' and varies mostly with 9. In a small temperature range, however, it may be regarded as practically conof
deg
1
a show
stant,
By
a
is
that
it
expressed in reciprocal degrees.
definition, the isothermal Young's modulus,
denoted by Y,
is
469 atm.
Y= L(6J A \dL
Stretched Wire
Experiments on stretched wires are usually performed under conditions in which the pressure remains constant at 1 atm and changes in volume arc negligible. For most practical purposes, it is found unnecessary to include the pressure and the
d9
69
The experimental measurement
;
6
468 atm,
468
(6L\
til
1
6
= and
181
lO-Megr
=
volume among the thermodynamic coordinates.
A
(2-10)
where A denotes the area of the wire. The isothermal Young's modulus is found experimentally to depend but little on J and mostly on 9. For a 2"
small temperature range, unit of
}' is
1
dyn/cm 2
.
it
may
be regarded
as practically constant.
The
Heat and Thermodynamics
42
The
study of surface films
is
an interesting branch of physical chemistry.
esting optical properties.
films:
surface of a liquid in equilibrium with
A soap bubble,
J
The
There are three important examples of such
The upper
43
— Jw is sometimes called the surface pressure. Such films difference can be compressed and expanded and, when deposited on glass, have inter-
Surface Film
2-9
Simple Thermodynamic Systems
2-10
its
vapor.
or soap film, stretched across a wire framework, con-
Reversible Cell
2-10
two surface films with a small amount of liquid between. (sometimes monomolccular) oil film on the surface of water.
sisting of
A
thin
A surface film is somewhat like a stretched membrane. The surface on one side of any imaginary line pulls perpendicular to this line with a force equal and opposite to that exerted by the surface on the other side of the line. The force acting perpendicularly to a line of unit length is called the surface tension. An adequate thermodynamic description of a surface film is given by the specifying three coordinates: The The The
1
2 3
surface tension
J, measured
in
area of the film A, measured in ideal-gas temperature
reversible cell consists of
electrolyte.
tions of the electrolytes,
.
film, the
accompanying
may
considered as part of the system. This
liquid
must always be
be done, however, without
is
this
is
surface of a pure liquid in equilibrium with
vapor has a particularly simple equation of state. Experiment shows that the surface tension of such a film docs not depend on the area but is a function of the temperature only. For most pure liquids, the equation of state can be written
J = J, where
and
In Fig. 2-6 a schematic
than the emf of the cell. Under these conditions, be described conventionally as a transfer of
from the copper electrode
to the zinc electrode.
deposited, and copper sulfate
is
is
formed, copper
used up. These changes are expressed by
the chemical reaction
Zn
+
C11SO4 -*
Cu
+
Z11SO4.
its
When
positive electricity
is
transferred in the opposite direction,
H)'
J
is the surface tension at 0°C, t' degrees of the critical temperature, and n
is is
a temperature within a few lies between
a constant that
from this equation that the surface tension decreases as / becoming zero when t = /'. The equation of state of a monomolecular oil film on water is particularly interesting. If J w denotes the surface tension of a clean water surface and 1
the temperature.
the case, zinc goes into solution, zinc sulfate
introducing the pressure and volume of the composite system because, as a rule, the pressure remains constant and volume changes are negligible.
The
in a different
the nature of the materials, the concentra-
may
positive electricity externally
6.
When In dealing with a surface
and
slightly smaller
is
the current that exists
2
two electrodes each immersed
The cmf depends on
diagram of a reversible cell, the Daniell cell, is shown. A copper electrode immersed in a saturated C11SO4 solution is separated by a porous wall from a zinc electrode immersed in a saturated solution of Z11SO4. Experiment shows that the copper electrode is positive with respect to the zinc. Suppose that the cell is connected to a potentiometer whose potential difference
dyn /cm.
cm
A
2. It is clear
increases,
J the surface tension of the water covered —
--©uSOfl-soltrtiori-
—
~
Saturafe3 ?rtSOi solution
by the monolayer, then, within a C11SO4 crystals-
restricted range of values of A,
{J
— -SafuTged^T—
J W )A
=
const.
9.
Fig. 2-6
The Daniell
-ZnS0 4 crystals
cell.
2-6
i.e.,
44
I
(cat
and Thermodynamics
externally from zinc to copper, the reaction proceeds in the reverse direction (hence the name reversible cell); thus,
Cu
+ ZnSO., -» Zn +
CuSQ 4
feature of a reversible cell
is
.
transferred in the reverse direction. Furthermore, according to one of electrolysis, the simultaneous disappearance of 1 mole of
Faraday's laws of zinc
and deposit
of
1
mole of copper are accompanied by the transfer of where) is the valence and ,VF is Faraday's
exactly jNF coulombs of electricity, constant, or 96,500 C. charge of the cell, as a
but whose change
is
positive electricity
is
We may
therefore define a quantity Z, called the absolute magnitude is of no consequence
number whose
from the positive to the negative if An moles of zinc disappear and An moles of copper are deposited, the charge of the cell changes from Z; to Z,, where
(2-11)
nates only;
A is
paramagnetic substance
not a magnet.
slightly
the cell
is
on open is
in
Upon
magnetized
in the absence of
an external magnetic field field it becomes
being introduced into a magnetic
in the direction of the field. Its permeability,
however,
whose permeability may be very large. The magnetized state of a may be described macroscopically by specifying the total magnetic moment of the substance, which we shall call the magnetization. This quantity should not be confused with the intensity of magnetization, which is the magnetic moment per unit volume. In thermodynamics it is not convenient to refer to the properties of a unit
a unit pole, placed
may be measured
volume of material. The in pole centimeters.
a magnetic field, is acted on by a force of 1 dyn, 1 Oe. Most experiments on magnetic solids are performed at constant atmospheric pressure and involve only minute volume changes. Consequently, wc may ignore the pressure and the volume and describe a paramagnetic solid with the aid of only three thermodynamic
is a tendency for diffusion to take not in equilibrium. If the cell is connected to
circuit, there
thermodynamic equilibrium. The
20°) 3 ,
very nearly unity, in contradistinction to a ferromagnetic substance
the magnetic intensity
a potentiometer, however, and the circuit is adjusted until there is no current, then the emf of the cell is balanced and the cell is in mechanical and chemical equilibrium. When thermal equilibrium is also satisfied, the cell then
+ y(t -
20°) 2
paramagnetic substance
If
place slowly and the cell
+ 0{t -
Paramagnetic Solid
2-11
magnetization of any substance
The emf S, measured in V. The charge Z, measured in C. The ideal-gas temperature 0.
is
20°)
like iron,
= — An/A'p.
Now, if wc limit ourselves to reversible cells in which no gases are liberated and which operate at constant atmospheric pressure, wc may ignore the pressure and the volume and describe the cell with the aid of three coordi-
When
-
determined.
is still
Zi
a(l
t is the Celsius temperature, £20 is the emf at 20°C, and a, /3, and y are constants depending on the materials. We shall see later that, once the equation of state of a reversible cell is known, all the quantities of interest to a chemist which refer to the chemical reaction going on in the cell can be
transferred externally
electrode. Thus,
—
+
where
numerically equal to the quantity of electricity that is transferred during the chemical reaction, the change being negative when
Z;
c? 20
that the chemical changes
accompanying the transfer of electricity in one direction take place to the same extent in the reverse direction when the same quantity of electricityis
45
tions of the electrolytes. The emf will therefore remain constant. Experiment shows that the emf of a saturated reversible cell at constant pressure is a function of the temperature only. The equation of state is usually written
€ The important
Simple Thermodynamic Systems
2-11
states
of thermodynamic
equilibrium of a reversible cell arc connected by an equation of state among the coordinates £, Z, and 6. If the electrolytes are saturated solutions, a transfer of electricity accompanying the performance of the chemical reaction at constant temperature and pressure will not alter the concentra-
is
in
said to be
coordinates:
1
The magnetic
intensity 9f,
measured
in oersteds (1
Oe =
1
dyn/
pole).
2 3
The
The magnetization M, measured The ideal-gas temperature 0. states of
in pole centimeters.
thermodynamic equilibrium
of a paramagnetic solid can
among these coordinates. many paramagnetic solids is
be represented by an equation of state
Experi-
ment shows
a func-
that the magnetization of
46
tion of the ratio of the
five intensive quantities listed in the table
of this
whereas
magnetic intensity to the temperature. For small values ratio the function reduces to a very simple form, namely,
M-CCT which
is
known
Curie constant
as Curie's equation
may be
— Cc
constant.
Intensive
and
M
in the
KM when
The
unit in
%
yfM. is measured in which the Curie constant
Unit of Cc
=
Intensive coordinate
y
erg/ cm
Hydrostatic system Stretched wire
Pressure
P
Volume
V
Tension
3
Surface film
Surface tension
J
L A
Electric cell
Emf
S
Length Area Charge
Magnetic
solid
deg 11
= cm
3
•
is,
PROBLEMS
therefore,
The equation
2-1
dcg.
(a)
=
ji
1/0,
and
An
k
(/;)
an
of state of
=
P(v
—
b)
=
Per mole
Per g
cm 3 deg
cm 3 dcg
Per
cm 3 (b)
seeMi later
i
g
thermodynamics. It will be how these are used to obtain extremely low temperatures. solids arc of particular interest in
Intensive
R6.
Show
that
+ bP/RB MP +
mm
deg
An
2-3 is
approximate equation of state of a real gas at moderate pres= A'0(1 B/v), where R is a constant and B is a function
+
given by Pv
of 6 onlv.
Show
that v
2-12
=
•
sures
Paramagnetic
Pv
1/0 1
mole
is
approximate equation of state of a real gas at moderate to take into account the finite size of the molecules, R6, where R and b are constants. Show that
Units of the Curie Constant
deg
ideal gas
\/P.
pressures, devised is
cm
M
%
expressed
is
2-2
3
Z
Magnetization
intensity Of-
is
Since the Curie constant depends upon the amount of material, its unit may be taken to be any one of the four listed in the accompanying table:
Total
Extensivc coordinate
measured in oersteds 2 ergs and in ergs per cubic
in pole centimeters,
centimeter.
and Extensive Quantities
1
next chapter that,
letters,
italic capital letters.
The
Paramagnetic
be shown
are denoted by script capital
the extensive quantities arc represented by
Simple systems
being called the Curie
M
w
all
Table 2-3
,
written
Cc = It will
47
Simple Thermodynamic Systems
Heat and Thermodynamics
and Extensive Quantities K
Imagine a system in equilibrium to be divided into two equal parts, each with equal mass. Those properties of each half of the. system which remain
2-4
A
=
+ B + 6(dB/c v + 2B 1
1
P
1
+
'
BRd/Pv 2
metal whose volume expansivity 10 -6
is
5.0
X
10~* deg-1 and
iso-
-1
which arc halved are called extensive. The intensive coordinates of a system, such as temperature and pressure, arc
thermal compressibility is 1.2 X atm is at a pressure of 1 atm and a temperature of 20 C. A thick surrounding cover of invar, of negligible.
independent of the mass; the extensive coordinates are proportional to the The thermodynamic coordinates that have been introduced in this chapter are listed in Table 2-3. The student should notice that four of the
expansivity and compressibility,
the
same are
mass.
said to be intensive; those
{a)
What will
(b)
If the
fits it
very snugly.
if the temperature is raised to 32°C? surrounding cover can withstand a maximum pressure of
be the final pressure
48
Heat and Thermodynamics
1200 atm, what
is
49
Simple Thermodynamic Systems
the highest temperature to which the system
may be
raised?
2-10
If a
wire undergoes an infinitesimal change from an
equilibrium state to a final equilibrium state,
2-5 A block of the same metal as in Prob. 2-4 at a pressure of 1 atm, a volume of 5 liters, and a temperature of 20°C undergoes a temperature rise of 12 deg and an increase in volume of 0.5 cm 3 Calculate the final
is
show that the change
initial
of tension
equal to
dJ= -aAYdd + ^-dL.
.
pressure.
2-6
A
2-11
Express the volume expansivity and the isothermal compressibility in terms of the density p and its partial derivatives. (a)
(b)
Derive the equation
The thermal
X
supports
.2
1
m
apart. If the temperature
reduced to 8°C, what
is
tension? (Assume that a and F remain constant at the values
dV 77 = pdd V 2-7
of 2
metal wire of cross-sectional area 0.0085 cm 2 under a tension at a temperature of 20°C is stretched between two rigid
10 dyn and 6
-
and 2.0 2-12
KdP.
X
a
graph showing how
is
the final
10- 5 deg_1
,
The fundamental frequency
mass m, and tension
Draw
X
10 12 dyn/cm. 2 respectively.)
3
is
of vibration of a wire of length L,
given by
expansivity and the compressibility of liquid oxygen
are given in the accompanying table.
1.5
=
/i
\J1.
1
21:V
(dP;i)0) v
depends on the temperature.
With what frequency
e,
°k
60
65
70
75
80
deg" 1
3.48
3.60
3.75
3.90
4.07
4
33
4.60
lO-'atm-'
0.95
1.06
1.20
1.35
1.54
1.78
2.06
3 ft 10-
k,
85
.
90
will the wire in Prob. 2-11 vibrate at
20°C; at 8°C? (The density of the wire is 9.0 g/cm 3 .) 2-13 If, in addition to the conditions mentioned in Prob. 2-11, the supports approach each other by 0.012 cm, what will be the final tension? 2-14 The equation of state of an ideal elastic substance is
LA 2?>
2-8 The thermal expansivity and the compressibility of water arc given in the accompanying table. Draw a graph showing how (dP/dQ) v
depends on the temperature. If water were kept at constant volume and the temperature were continually raised, would the pressure increase indefinitely?
where
K is a constant and £
(the value of/, at zero tension)
(a)
Show
that the isothermal Young's
A \L a t,
°c
50
100
150
200
250
300
0.45
0.74
1.02
1.35
1.80
2.90
0.45
0.50
0.62
0.85
1.50
3.05
(b)
ft
10-* deg-'
k,
10-' atm-'
0.45
point,
the critical point (dP/dV) T
=
0.
Show
that,
Show
a function of
given by
is
L2
that the isothermal Young's
Fo
modulus
Show
=
3K8
A
that the linear expansivity
is
given by
at the critical
both the volume expansivity and the isothermal compressibility arc
infinite.
modulus
given by
(c)
At
2-9
is
the temperature only.
=
oca
—
J -7777.
AY 6
=
«u
—
1
IJ/Ll
Tf,
/^o
-
1
at zero tension
is
50
Heat and Thermodynamics
where a
is
dL "**
1
*"5 Assume the following values
(d)
=
K=
300°K,
Calculate
Show
J
r ,
Y,
1.33
and a
X
for
10 3 dyn/deg,
A =
WORK a certain sample of rubber: 2 10-' cleg-'. 1 nun « 5
for the following values of
,
L/L B
= :
X
0.5, 1.0, 1.5, 2.0.
how J7, Y, and a depend on the ratio L/La. The equation of state of an ideal paramagnetic material
graphically
2-15
for all values of the ratio
M where
3
the value of the linear expansivity at zero tension, or
= Ngtm \(J
+
9f/0
I)
is
valid
given by Brillouin's equation, as follows:
coth (J
+
C i)<^-i* coth *
kd
hb, J, and k arc atomic constants. Find out how the hyperbolic cotangent of x behaves as x approaches
A', g,
(a)
If a
zero. (b)
Show
that Brillouin's equation reduces to Curie's equation
(c)
Show
zero.
that the Curie constant
_ NgU(J
is
given by
+
\)y.l
when
system undergoes a displacement under the action of a force, work
amount of work being equal to the product of the force and the component of the displacement parallel to the force. If a system as a whole exerts a force on its surroundings and a displacement takes place, the work that is done either by or on the system is called external work. Thus, is
9^/0 approaches
Work
3-1
said to be done, the
a gas, confined in a cylinder and at uniform pressure, while expanding and
imparting motion to a piston docs external work on its surroundings. The work done, however, by part of a system on another part is called internal work. The interactions of molecules or electrons on one another constitute internal work.
Internal
work has no place
in
thermodynamics. Only the work that
involves an interaction between a system
When
and
its
surroundings
is
significant.
a system does external work, the changes that take place can be
means of macroscopic quantities referring to the system as a which case the changes may be imagined to accompany the raising or lowering of a suspended body, the winding or unwinding of a spring, or, in general, the alteration of the position or configuration of some external
described by
whole,
in
mechanical device. This may be regarded as the ultimate criterion as to whether external work is done or not. It will often be found convenient throughout the remainder of this book to describe the performance of external in terms of or in conjunction with the operation of a mechanical device
work
such as a system of suspended weights. Unless otherwise indicated, the word work, unmodified by any adjective, will mean external work. A few examples will be found helpful. If an electric cell is on open circuit, changes that take place in the cell (such as diffusion) arc not accompanied
by the performance of work. If, however, the cell is connected to an external circuit through which electricity is transferred, the current may be imagined
' 52
I
Icat
3-3
and Thermodynamics
produce rotation of the armature of a motor, thereby lifting a weight or winding a spring. Therefore, for an electric cell to do work it must be connected to an external circuit. As another example, consider a magnet far removed from any external electric conductor. A change of magnetization within the magnet is not accompanied by the performance of work. If, however, the magnet undergoes a change of magnetization while it is surrounded by an electric conductor, eddy currents are set up in the conductor, constituting an external transfer of electricity. Hence, for a magnetic system to do work it must to
interact
with an
electric
we
In mechanics,
on by
the resultant force exerted on a mechanical system is in the same direction as the displacement of the system, the work of the force is
positive,
work
is
said to
be done on the system, and the energy of the system
increases.
In thermodynamics, however, we focus our attention on the resultant force exerted by
the system on its surroundings.
as the displacement, call this
this
work
is
done
When
work
is
is
opposite to the
this force
by the system,
work positive. Conversely, when
surroundings
may As a
2
may
forces
turbulence, waves,
etc.,
execute some sort of accelerated motion. result of this turbulence, acceleration, etc., a nonuniform tem-
perature distribution
A
is
in the same direction
and the convention
is
to
the force exerted by a system
displacement, work
is
done
on
on its the system, and
called negative.
in
thermodynamic equilibrium
satisfies
the following stringent
requirements:
1
its
finite
surroundings.
in the forces and in the temperature may produce a chemical reaction or the motion of a chemical constituent.
The sudden change
3
It follows that a finite unbalanced force may cause the system to pass through nonequilibrium states. If it is desired during a process to describe every state of a system by means of thermodynamic coordinates referring to the system as a whole, the process must not be brought about by a finite unbalanced force. We are led, therefore, to conceive of an ideal situation in which the external forces acting on a system are varied only slightly so that the unbalanced force is infinitesimal. A process performed in this ideal
way
is
said to be quasi-static. During a quasi-static process, the system
is at all
times
and all the states through can be described by means of thermodynamic
infmilesimally near a state of thermodynamic equilibrium,
which the system passes
A
An
equation of state
is
valid,
an idealization that is applicable to all thermodynamic systems, including electric and magnetic systems. The conditions for such a process can never be rigorously satisfied in the laboratory, but they can be approached with almost any degree of accuracy. In the next few articles it will be seen how approximately quasistatic processes may be performed by all the systems treated in Chap. 2. quasi-static process
is
There arc no unbalanced forces acting on on the system as a whole. Thermal equilibrium. There are no temperature differences between parts of the system or between the system and its surroundings. Chemical equilibrium. There arc no chemical reactions within the system and no motion of any chemical constituent from one part of
that the cylinder has a cross-sectional area A, that the pressure exerted
a system
by the system at the piston face
of the system or
to another part.
3-3
Work
Once a system is in thermodynamic equilibrium and the surroundings arc kept unchanged, no motion will take place and no work will be done. If, however, the sum of the external forces is changed so that there is a unbalanced force acting on the system, then the condition for mechaniis no longer satisfied and the following situations may arise:
cal equilibrium
of a Hydrostatic System
Imagine any hydrostatic system contained in a cylinder equipped with a movable piston on which the system and the surroundings may act. Suppose is P, and that the force is PA. The surroundings an opposing force on the piston. The origin of this opposing force is irrelevant; it might be due to friction or a combination of friction and the push of a spring. The system within the cylinder does not have to know how the opposing force originated. The important condition that must be satisfied is that the opposing force must differ only slightly from the force PA. If, under these conditions, the piston moves a distance dx, the system performs an
also exert
finite
and
Mechanical equilibrium.
any part
3
be brought about, as well as a
coordinates referring to the system as a whole.
Quasi-static Process
system
may
difference of temperature between the system
therefore, for all these states.
3-2
be created within the system; as a result, be set up. Also, the system as a whole
may
conductor or with other magnets.
arc concerned with the behavior of systems acted
When
external forces.
Unbalanced
1
53
Work
54
I
Icat
and Thermodynamics
infinitesimal
through
it
amount
of work
3-4
dlF
(the differential
will be explained later),
symbol with the line drawn
where
which the volume changes from
In a finite quasi-static process in
the
work
and hence
d
W=
dV,
Since the change in volume
P dV.
(3-1)
(3-2)
expressed as a function of path of integration
is
done by a system
in
process such as friction taking place
pressed as
all
these processes.
The
lack
is
performed
quasi-statically,
P is
at all times a
thermodynamic coordinate and can be expressed as a function of 8 and V by means of an equation of state. The evaluation of the integral can be accomplished once the behavior of 8 is specified, because then P can be
During this process, wc might have a chemical reaction or a transport of a component from one point to another taking place slowly enough to keep the system near mechanical equilibrium; or we might have some dissipative
—or even
Vi to Vf,
"
'
Adx =
55
is
W =-/;'P dV. But
Work
V
only. If
P
is
expressed as a function of V, the
Along a particular quasi-static path, the work going from a volume F,- to a larger volume F/ is ex-
defined.
W
of chemical equilibrium (and therefore of complete thermodynamic equilib-
= \l[Pdv-
it
dW
rium) and the presence of dissipation do not preclude writing = P dV. lack of mechanical equilibrium, however, such as exists when there is no opposing force, definitely precludes expressing dlY as P dV.
A
'
The expansion
whereas from / to i, along the same path but work absorbed by the system is
in the opposite direction, the
or contraction of a hydrostatic system need not necessarily
take place in a cylinder. In Fig. 3-1, the system has
any arbitrary boundary and expands quasi-statically to another boundary, as represented by the dashed curve. Any small clement of area LA is acted on by a force P LA and undergoes a displacement ds in a direction normal to LA. The work of P LA is therefore P LA ds, and thus the total work done by the system is
W'«n =- /; PdV. When
the path
is
quasi-static,
W
it
d
W
= P2 LA
ds
= P dV.
= ~ W,i.
approximation to a quasi-static process may be achieved in practice by having the external pressure differ from that exerted by the system by only a small finite amount. The inks unit of P is 1 X/m 2 and that of V is 1 3 The mks unit of work is therefore 1 J. It is often convenient to take 1 atm as a unit of P and 1 liter as a unit of V. The unit of work is then 1 liter atm, which is equal to 101 J. Sufficient
m
3-4
.
PV Diagram
As the volume of
a
chemical system changes by virtue of the motion of a
any moment is proportional pen whose motion along the X axis of a diagram follows exactly the motion of the piston will trace out a line every point of which represents an instantaneous value of the volume. If, at the same time, this pen is given a motion along the Y axis such that the Y coordinate is proportional to the pressure, then the pressure and volume changes of the system piston in a cylinder, the position of the piston at to the
Fig.
3-1
surface is
Work
Px AA
ds
of the infinitesimal hydrostatic force
= P dV.
P 4.-1
is
P AA
ds.
Work
over the entire
volume.
A
during expansion or compression are indicated simultaneously on the same
56
diagram. Such a device is
3-6
Heat and Thermodynamics
plotted along the
The diagram
and volume along the
X axis
is
in
p
which pressure an indicator
called
PV diagram.
diagram or
In Fig.
called an indicator.
is
Y axis
3-2(7,
are indicated
by curve
area under curve
I.
and volume changes of
The JPdV for
I.
a gas
shaded
(Isochor)
work absorbed by
is represented by the shaded area under curve II in Fig. 3-26. In conformity with the sign convention for work, the area under I is regarded
the gas
and
as positive,
drawn together is
that under II as negative. In Fig. 3-2e, curves so
brought back to
a closed figure,
is
cycle.
The
work done
II
work is work would be negative.
1
1
1
1
i
v
obviously
and therefore repre-
in the cycle. It should be noticed that the cycle
traversed in a direction such that the net reversed, the net
and
is
positive. If the direction
/
b
I
area within the closed figure
the difference between the areas under curves I sents the net
Pa
and II are that they constitute a series of processes whereby the gas its initial state. Such a series of processes, represented by
called a
a
during expansion
this process is evidently the
Similarly, for a compression, the
(Isobar)
i
2P
the pressure
Work
Fig. 3-3
2V
Work Depends on
On a
the
final
were
depends on the path.
work
the
is
the area under the line
represents another path,
the intermediate states,
the Path
PV diagram depicted
in Fig. 3-3,
for
an
initial
bf,
or
where the work
i
on the path. This
i.e.,
P
is
V(j.
and
taken from
/,
i
equilibrium state and
respectively.
to
.
is
The. straight line
f/V'o-
from
i
to /
We can sec, therefore, that
initial
and final
merely another
stales but also on
way
of saying that,
a quasi-static process, the expression
= w'-JT PdV
equilibrium state of a chemical system are represented by the two
There are many ways in which the system may be example, For the pressure may be kept constant from i to a /. (isobaric process) and then the volume kept constant from a to f {isochoric process), in which case the work done is equal to the area under the line ia, which is equal to 2PoVQ Another possibility is the path ibf, in which case points
V
is
the work done by a system depends not only on the
3-5
57
Work
cannot be integrated until
The
P dV
P
is
specified as a function of V.
an infinitesimal amount of work and has been represented by the symbol d(F. There is, however, an important distinction between an infinitesimal amount of work and the other infinitesimals we have considered up to now. An infinitesimal amount of work is an inexact expression
differential; i.e., it is not
is
the differential of an actual function of the thermo-
dynamic coordinates. There is no function of the thermodynamic coordinates representing the work in a body. The phrase "work in a body" has no meaning. To indicate that an infinitesimal amount of work is not a mathematical differential of a function and to emphasize at all times that it is an inexact differential, a line is drawn through the differential sign thus: dlt'.
W
3-6
Work
in Quasi-static Processes
The preceding
ideas will be clarified by the following examples:
Quasi-static isothermal expansion or compression of an ideal gas Pig. 3-2
PV diagram,
together constitute
a
cycle.
(a)
Curve
I, expansion; {b) curve II, compression; (c) curves
I and II
JVi
PdV,
58
but an ideal gas
is
one whose equation of
PV = where
n
and R are
state
Since the isothermal compressibility
is
1 (tE) v \dPjs
constants. Substituting for P,
we
get
we
v,nRB
and, since
is
have, at constant temperature,
,„
dV = -kV dP.
V Substituting for dV,
also constant,
W=
dV
v,
Now
= where the symbol of the
common
In
is
nR6
JVi
59
Work
3-6
Heat and Thermodynamics
nR6
In
J^
>
V and
the changes in
may be
k at
PdP.
JPi
constant temperature arc so small that they
neglected. Hence,
kV W~-*— [Pj 2
denotes the natural, or napierian, logarithm. In terms
/>?)
logarithm, denoted by log, Since the
W=
volume
m
equal to the mass
is
divided by the density
p,
2.30nR6 log pf-
W~-f (P]-Pl). rriK
g moles of gas kept at a constant temperature of 0°C and if this gas is compressed from a volume of 4 liters to 1 liter, then n = 2 moles, R = 8.31 J/g mole deg, 6 = 273 deg (using three significant figures), V{ = 4 liters, Vt = 1 liter, and
p
If there are 2
.
W=
2.30
X
2 moles
X
8.31
—
r—
-.
mole
•
deg
X
273 deg
X
For copper at 0°C, p = 8.93 g.'cm 2 k = 0.725 X 10"" 6 ater 1, = 0, and P, = 1000 atm = 1.01 X 10 9 dyn cm 2 Hence, ,
•
Pi
W=
log ° 4
-
= -6300 J
10 g
X
X X
0.725 2
7.25
cm
3
•
10- 6 atm-' 8.93
g/cm
atm = -0.407
17.8
(the
minus
sign indicating that
work was done
1
liter
atm
=
101
Quasi-static isothermal increase of pressure on a solid Suppose the pressure on 10 -2 kg of solid copper is increased quasi-statically and isothermally at
to
1000 atm. The work
W=
is
10
g,
X
10 6 atm 2
3
X
10" 3
liter
•
atm.
on the gas).
Since
0°C from
m =
.
J,
W=
-0.0411
J,
calculated as follows:
where the minus sign
[P dV,
"(Si.*
result,
indicates that
together with that of the
first
work was done on the copper. This example, indicates that
when
a gas
is
compressed we can usually neglect the work done on the material of the
dV do.
container.
Heat and Thermodynamics
60
Work
3-7 If the
(L
of a Wire, a Surface Film,
length of a wire in which there
+ dL), the
infinitesimal
amount
of
is
and a Reversible
a tension J7
work
that
is
is
done
Cell
changed from is
/.
to
equal to
dW = -JdL.
(3-3) Fig. 3-4
used because a positive value of dL means an extension of the wire, for which work must be done on the wire, i.e., negative work.
The minus
sign
For a
change of length from Li
finite
3 indicates the
force
is
siretched across
a wire framework.
cannot be evaluated
referring to the wire as a whole.
maintained
contactor.
may moment
undergoing a motion involving large
is
reversible cell of
cmf c?
to
be connected to a potentiometer so that an almost may be obtained with a sliding
continuous variation of potential difference
--£**>
wire
forces, the integral
namic coordinates
to L/,
instantaneous value of the tension at any
If the
during the process.
unbalanced
Surface film
is
\\>
where
61
Work
3-7
in
If,
terms of thermody-
however, the external
at all times only slightly different
from the tension, the
is sufficiently quasi-static to warrant the use of an equation of state, which case the integration can be carried out once ff is known as a funcwill be in tion of L. When 3F is measured in newtons and /. in meters,
process
The
circuit
be made equal
is
shown
The
or slightly
external potential difference
&
more than
by sliding the
contactor. If the external potential difference is
made
S, then, during the short time this difference a quantity of electricity
dZ through
infinitesimally smaller than exists,
on the than
work
outside. If the external potential difference
c?, electricity is
on the
cell.
there
is
is
a transfer of
amount
of
work
done by the
made
transferred in the opposite direction
In either case, the
is
the external circuit in a direction from
the positive to the negative electrode. In this case,
in
W
in Fig. 3-5.
to, slightly less,
cell
slightly larger
and work
is
done
is
joules.
dW =
Consider a double surface film with liquid in between, stretched across a is movable, as shown in Fig. 3-4. If the
-SdZ.
(3-5)
wire framework, one side of which
movable wire has a length L and the surface tension is J the force exerted by both films is 2JL. For an infinitesimal displacement dx, the work is ',
+i in Ml—
Potentiometer
dW = 2Ldx =
but
the
minus
sign?)
For a
finite
W= A
dA.
AW = -J dA.
Hence
(Why
+vwwwwv\
-IJLdx;
quasi-static process
may
change from Ai
-
/:•
(3-4)
to
/!/,
dA.
be approximated by maintaining the external
force at all times only slightly different from that exerted by the film.
The conventional description
of an electric current
from a region of higher
is
that
it is
the motion of
a region of lower potential. this is opposite to the direction of electron drift, the convention used, and it is convenient to adopt it in thermodynamics. Imagine a
positive electricity
to
Although is still
Fig. 3-5
Approximately quasi-static transfer of
electricity in
a reversible
cell.
3-8
Heat and Thermodynamics
62
When
the cell
discharging through the external circuit,
is
dZ
is
negative;
is a quantity Z connected with the slate of charge of the cell which by an amount dZ, where dZ is the actual quantity of electricity transferred. Charging the cell involves an increase in Z or a positive dZ.
i.e.,
there
decreases
If
Z changes
by a
finite
is
i,
an amount d/S. Then, by Faraday's principle of electromagnetic induction, there is induced in the winding a back cmf £, where
amount,
then in time dr the quantity
W=
-
dZ =
i
dr,
d e = -NA 4. dr
and
During the time interval dr, a quantity of electricity dZ is transferred and the work clone by the system to maintain the current
f*8iek.
circuit,
With
£
in volts
and the charge
in
coulombs, the work
will
be expressed
in the is
cal-
culated as
AW = £dZ
in joules.
3-8
63
The effect of a current in the winding is to set up a magnetic field with magnetic induction /3. If the dimensions are as shown in the figure, /3 will be nearly uniform over the cross-section of the toroid. Suppose that the current is changed and that in time dr the magnetic induction changes by
W= If the current
Work
Work
in
-NAfdZ dr
=
-A'/l
dZ dr
Consider a sample of magnetic material in the form of a ring of crosssectional area A and of mean circumference L. Suppose an insulated wire is
wound on
= Changing the Magnetization of a Magnetic Solid
= -NAidS,
top of the sample, forming a toroidal winding of A" closely spaced
shown in Fig. 3-6. A current may be maintained in the winding by a battery, and by moving the sliding contactor of a rheostat this current
where
may be changed.
given by
turns, as
equal to
i,
The magnetic
dZ
dr, is
intensity 9-f
nr
momentary value of the due to a current i in a
the
AvNi
ArrNAi
^~~L
Ring of magnetic
current.
toroidal winding
is
AirNAi
V~
AL
'
material
where
V is
the volume of magnetic material. Therefore,
Toroidal windin
V NAi = t-K,
dH/= - %-7Pt
and
If
or
M
is
the
total
magnetic moment of the material (assumed to be isotropic),
total magnetization,
we have
the relation
M Current
i
Therefore,
=
dIC Fig. 3-6
Changing the magnetization of a magnetic
solid.
(3-6)
=
-J-WdW- P^dM. 4tt
(3-7)
Heat and Thermodynamics
64
Work
3-10
no material were present within the equal 7f, and would IS If
toroidal winding,
M would be zero,
Summary
3-9
The work values of the various simple systems are summarized It should be noted that each expression for
1/
dW = - 4tt t-^£¥ This
is
(vacuum
only).
the work necessary to increase the magnetic
Wc
field in
V
a volume
etc.,
book with changes of temperature, energy, of the material only, brought about by work done on or by the material.
shall be
concerned
this
AW =
book,
-?fdM.
(3-8)
an increase of magnetization (posiIf 9£ is measured in oersteds (dynes per unit pole) and / in pole centimeters, then the work will be expressed in ergs. t to If the magnetization is caused to change a finite amount from f
The minus sign merely provides tive dM) involves negative work.
work
will
_.
'
, \ (generalized force) ,
,.
,
'
is
in
Table
3-1
the product of an intensive
an
extensive quantity.
of Simple Systems
Intensive quantity
,
dimple svstem
is
atm
Hydrostatic system
P, in
Wire
3,
Surface film
J, in dyn/cm
Extensive quantity
Work
(generalized
displacement)
P dV,
V, in liters
in liter
inN
L, in
m
A, in
cm 2
atm
-atm =
(1 liter
— ydL,
in J
— J dA,
in ergs
(10'ergs
=
-Sd7,
in J
1
101 J)
J)
that
M
the
Work
Table 3-1
in this
Consequently, for the purpose of
work
and an extensive quantity; consequently, work
The second term, —9£dM, is the work of empty space by an amount dW. done in increasing the magnetization of the material by an amount AM.
65
M
Reversible
Magnetic
e, in
cell
%,
solid
in
V
7, in
Oe
M,
C
in pole
cm
—% dM, in ergs
,
be
We
dM.
Jm Modern experiments on paramagnetic on samples
in the
form of cylinders or
the P/' field inside the material
is
materials arc usually performed
ellipsoids, not toroids.
somewhat smaller than
In these
the
7f
cases,
field
up by magnetic poles which form on the surfaces of the samples. In longitudinal magnetic fields, the demagnetizing effect either may be rendered negligible by using cylinders whose length is much larger than the diameter or may be corrected for in a simple way. In transverse magnetic fields, a correction factor must be applied. reverse field (demagnetizing field) set
shall limit ourselves to toroids or to
where the In any actual
fields
quasi-statically,
long thin cylinders
any system.
If
we
gen-
erated by the electric current in the surrounding winding because of the
We
have seen that a work diagram is obtained if any one of the intensive is plotted against its corresponding extensive coordinate. There are therefore as many work diagrams as there are systems. It is desirable at times, for the sake of argument, to formulate a work diagram which does not refer to one system in particular but which represents the behavior of coordinates
in longitudinal
and external JK fields arc the same. case, a change of magnetization is accomplished very nearly and therefore an equation of state may be used in the
generalized forces
designate the intensive quantities P,
and
their
S-,
J
corresponding extensive quantities
,
&, and 7f as
V, L, A, Z,
and
M as generalized displacements, we may represent the work done by any simple system on a generalized work diagram by plotting the generalized force
Y
against the generalized displacement X. Conclusions based on such a diagram will hold for
3-10
any simple system.
Compound
Systems
internal
integration of the expression denoting the work.
Up
to this point
we have
coordinates, one of
which
is
whose thermodynamic
dealt exclusively with simple systems,
equilibrium states are described with the aid
of three
always the temperature.
A
single equation of
3-10
Heat and Thermodynamics
66
=
Work
67
const.
V=
=
const.
A—
—
const.
^Solenoid
r
°i
"r
V=
const.
T!" Heat
Diathermic/ I
wall-^ (a) Fig. 3-7
(a)
A
composite system whose coordinates are P, V, P',
of the independent coordinates 9,
V, and
V
V, and
9.
(A)
A
graph
b
-b
reservoir
b
b^M =
.
(a)
each case, so that only two of the coordinates are independent. The laws of thermodynamics, however, which are to be developed in the next few chapters, must apply to any system no matter howcomplicated i.e., to systems having more than three coordinates and more state
was found
to exist in
than one equation of state. Consider the composite system depicted schematically in Fig. 3-7a with two different simple hydrostatic systems separated by a diathermic wall, which ensures that both parts have the same temperature. There are five
thermodynamic coordinates (P, state, one for each of the simple
V,
/",
V, and
0)
and two equations
of
systems. Consequently, only three of the
five
coordinates arc independent. In any small displacement of each piston,
the
work
is
&W = PdV + P' d\". The most convenient diagram
Fig. 3-8
(a)
A
(&)
composite system whose coordinates are P, V,
of the independent coordinates 9, V,
and
%, M,
and
9.
(b)
A
graph
M.
Consider an ideal paramagnetic gas, such as oxygen at low pressures, as
The oxygen may have its pressure P and volume V varied with the aid of a piston-cylinder combination, and it is immersed in a magnetic field whose intensity 9fi may be varied by varying the current in the surrounding solenoid. The gas is kept at a uniform temperature 9. The coordinates are P, V, 9f, M, and 9, only three of which are independent because of the two equations of state: the ideal-gas equation PV = ?iR9 and Curie's equation :% = C/9. Since the work done in any depicted schematically in Fig. 3-8a.
M
infinitesimal process
is
&W - PdV - WdM,
to use in demonstrating the features of this
V, and is a three-dimensional diagram with 9, rectangular axes, as shown in Fig. 3-76. A typical isothermal process would = const. A curve on a be a curve on a plane such as the one marked
const.
V" plotted along
system
plane such as that marked V = const, would represent a process in which no work is done by the left-hand part. The points a and h lie on a vertical The straight line ab line every point of which refers to a constant V and work done by the composite is therefore represents a process in which no
V
.
system.
Two simple systems do not have to be separated spatially by a diathermic wall in order to have two equations of state and a common temperature.
9, V, and M, which are Any vertical line would represent
the most convenient independent coordinates are plotted along rectangular axes in Fig. 3-86.
which no work is done. we shall have occasion to refer to a general five-coordinate system whose coordinates arc Y, X, 1", A", and 9 and whose work is a process in
Later on,
dll'
= YdX
The most convenient independent
+
Y'dX'.
coordinates of this system are
9,
X, and
X
.
Work
Heat and Thermodynamics
68
PROBLEMS
If this
(b)
words,
A thin-walled metal bomb of volume 7 contains a gas at high Connected to the bomb is a capillary tube and stopcock. When the stopcock is opened slightly, the gas leaks slowly into a cylinder equipped with a nonleaking, frictionless piston where the pressure remains constant 3-1
I
;;
pressure.
at the atmospheric value
Show
(a)
of
P
.
much gas as
that, after as
possible has leaked out, an
amount
work
W=
- VB
P„(Po
if
A
3-5
device
the gas
is
is
69
—
used only to produce effects in the gas or, in other what expression for work is appropriate?
the system
—
stationary vertical cylinder, closed at the top, contains a gas
whose volume may be changed with the aid of a heavy,
frictionless,
non-
leaking piston of weight w.
How much work
(n)
is
done by an outside agent
in
compressing the
gas an amount dV by raising the piston a distance dyi If this device is used only to produce temperature changes of the (b) gas, what expression for work would be appropriate?
Compare
(c)
)
this situation
with that of Prob. 3-4, and also with that
involved in increasing the magnetic induction of a ring of magnetic
has been done, where and temperature. (b)
Va
is
the
volume of
the gas at atmospheric pressure
be done
Calculate the work done by
isothermal expansion from an the equation of state
The
3-6
How much work would
if
the gas leaked directly into the
atmosphere? 3-2
material.
initial
mole of gas during a quasi-static, volume v; to a final volume V; when 1
,
(a)
Pv
3-3
b)
= R6 ( 1 -
=
R6
(R, b
\R
-J
=
=
const.;
B=
During a quasi-static adiabatic expansion of an any moment is given by the equation
K arc
where 7 and a state
(Pi, F«)
=
ideal gas, the (b)
to a state {Pi,
Show
Vs )
work done
in
PiVi
and isothermally and isothermal Young's constant, show that the work
increased quasi-statically
practically
- L 2AY
m1
in
Jt).
long and 1.0
X
10~ 7
m
2
in area
increased quasi-statically and isothermally at 0°C from 10 to 100 N. arc done? (Isothermal Young's modulus at 0°C
.)
The equation
of state of
an
ideal elastic substance
is
expanding from
*-»(c-9
- P/>'/ fV
7-1
where
and volume are 10 atm and 1 liter, the final values are 2 atm and 3.16 liters, how many joules by a gas whose 7 = 1.4? (1 liter atm of work = 101 J.)
A
is
tension in a metal wire
How many joules of work 2 2.5 X 10 n N/m
If the initial pressure
3-4
The
is
that the
in joules.
is
W=
vertical cylinder, closed at the bottom,
is
respectively,
and
of work are done
placed on a spring
The cylinder contains a gas whose volume may be changed with aid of a frictionless, nonleaking piston. The piston, is pushed down. (a) How much work is done by an outside agent in compressing gas an amount dV while the spring scale goes down a distance dyi
scale.
is
K,
work
the length, cross-sectional area,
w-
3-8 constants.
tension in a wire
/(0)].
pressure at
PV
The
J/, to J7/. If
modulus of the wire remain done is
const.).
increased quasi-statically and
3-7
from
-
is
1000 atm. Assuming the density and the isothermal remain constant at the values 10 g/cm 3 and 0.675 X to
compressibility to -6 10 atm-1 respectively, calculate the
is:
P(v
pressure on 100 g of metal
isothermally from
the
the
A' is
a constant
and L
(the value of
L
at zero tension)
is
a function
of temperature only. Calculate the work necessary to compress the substance from L = Lo\a L = La/2 quasi-statically and isothermally. 3-9 Prove that the work done during a quasi-static isothermal change of state of a paramagnetic substance obeying Curie's equation is given by
Heat and Thermodynamics
70
3-10 is
Show
that the magnetic energy per unit
volume
in
empty space
#^/8x.
A volume of 200 cm of a paramagnetic substance is maintained temperature. A magnetic field is increased quasi-statically and constant at to 15,000 Oe. Assuming Curie's equation to hold and isothermally from the Curie constant per unit volume to be 0.15 deg: 3-11
(a) (b)
rial
4.
3
FIRST
LAW
How much work would have to be done if no material were present? How much work done to change the magnetization of the mateis
when
(c)
HEAT AND THE
the temperature
How much work
is
is 300°K and when it is 1°K? done at both temperatures by the agent supply-
ing the magnetic field?
A
3-12
chamber with
rigid walls consists of
two compartments, one
containing a gas and the other evacuated; the partition between the two compartments is destroyed suddenly. Is the work done during any infinitesimal portion of this process (called a free expansion) equal to
P dV?
A reversible cell consists of an electrolyte, one solid electrode, and
3-13
one electrode involving gaseous hydrogen. (a)
What
coordixiates are
How many
needed to describe the equilibrium
states of
equations of state arc there?
What are typical examples
of these equations?
What
is the expression for d IV? Choose convenient independent coordinates. 3-14 A framework such as that shown in Fig. 3-4 is placed in a vessel where the air pressure may be varied at will. Consider the two surface films
(c)
(d)
and the liquid between the films as the system. (a)
(b) (c)
What
coordinates are needed?
How many equations of state are What the expression for AW?
there?
is
Choose convenient independent coordinates. 3-15 Devise a system consisting of an ideal nonmagnetic gas, a paramagnetic solid, and a reversible cell, all separated by diathermic walls. (d)
Draw
a diagram.
(a)
What
(b)
How many equations
(c)
What
(d)
Choose convenient independent coordinates.
are the coordinates?
is
Chap. 3 how a system could be transferred from an by means of a quasi-static process and how the work done during the process could be calculated. There arc, however, other means of changing the state of a system that do not necessarily involve the performance of work. Consider for example the three processes depicted schematically in 1'ig. 4-1. In (a) a fluid undergoes an adiabatic expansion It
was shown
initial
this system? (b)
Work and Heat
4-1
Explain.
of state are there?
the expression for d W?
in
to a final state
combination that is coupled to the surroundings with suspended body so that, as the expansion takes place, the body is lifted while the fluid remains always close to equilibrium. f In (b), a liquid in equilibrium with its vapor is in contact through a diathermic wall with the hot combustion products of a bunsen burner and undergoes vaporization, accompanied by a rise of temperature and of pressure, without the performance of work. In (c) a fluid is expanded while in contact with the flame in a cylinder-piston
a
of a bunsen burner.
What happens when two
systems at different temperatures arc placed one of the most familiar experiences of mankind. It is well known thai the final temperature reached by both systems is intermediate between the two starting temperatures. Up to the beginning of the nineteenth century, such phenomena, which comprise the subject of calorimelry, were explained by postulating the existence of a substance or form of matter termed caloric, or heat, in every body. It was believed that a body at a high temperature together
is
much
caloric and that one at a low temperature had only a little. two bodies were put together, the body rich in caloric lost some to the other, and thus the final temperature was intermediate. Although we now know that heat is not a substance whose total amount remains
contained
When
t
the
The author
is
indebted to Professor Ernst Schmidt for the clever coupling device
consisting of rack, pinion,
and cam.
4-1
Heat and Thermodynamics
72
1
Icat
and the
First
Law
73
^wwwwwww^
(a) Fig. 4-2
Whether a process
is designated
as a work or a heal interaction depends on the
choice of system.
W/W/'/i
'diWa
we ascribe the changes that take place in Fig. 4-26 "something" from the body at the higher temperature to the one at the lower, and this something we call heal. We therefore adopt as a calorimelric definition of heat that which is transferred between a system and its surroundings by virtue of a temperature difference only. It is obvious that an adiabatic wall is one which is impervious to heat, or a heat insulator, and that constant, nevertheless
and
',
w
w
',
_^_
(b)
c
to the transfer of
a diathermic wall is a heat conductor. It is important to observe that the decision as to whether a particular change of state involves the performance of work or the transfer of heat requires first an unequivocal answer to these questions: What is the system, and what are the surroundings? For example, in Fig. 4-2, a resistor immersed in water carries a current provided by an electric generator that is rotated with the aid of a descending body. If the shafts of the pulleys
and
we assume
the absence of friction in
the absence of electrical resistance in the gen-
and the connecting wires, we have a device whereby the thermodynamic state of a system composed of water and the resistor is changed by erator
purely mechanical means, the resistor
then there
is is
i.e.,
a transfer of heat from the resistor
difference between the resistor
water
by the performance of work.
If,
however,
regarded' as the system and the water as the surroundings,
by virtue of the temperature
and the water.
Also,
if
a small part of the
water being considered the surroundings, then again there is a transfer of heat. Regarding the composite system comprising both the water and the resistor, however, the (c) I'ig.
is
regarded as the system, with the
rest of the
surroundings do not contain any object whose temperature differs from that 4-1
work;
(c)
Distinction between
work and
heat.
work and
heat,
(a)
Adiabalic work; (b) heat flow without
of the system,
and
its
and hence no heat
surroundings.
is
transferred between
this composite
system
74
4-2
Heat and Thermodynamics
^>y^sSSSS^
kWXW\\\\^5^ P,
LS\V5
Heat and the
Law
75
J
PlU
V
\N\\\1
J*
ETT^
First
-AWWVI (a)
^\\\\\\\\\\\\\\^ Electric cell
Fig. 4-3
Adiabatic work,
4-2
Adiabatic
Work
Figures 4-1 a and 4-2 show that a system completely surrounded by an adiabatic envelope
may
be done.
Fig. 4-3. It
is
may
still
be coupled to the surroundings so that work
Three other simple examples an important
fact of
of adiabatic
work are shown
experience that the state of a system
be caused to change from a given formance of adiabatic work only.
initial state to a final state
Consider the composite system shown
in
may
by the per-
in Fig. 4-4<7, consisting of a
hydro-
and an immersed resistor on each side of a diathermic wall. This system can undergo an adiabatic work interaction with its surroundings in two ways. It may be done by moving one or both of the pistons in or out, static fluid
either slowly (a quasi-static process) so that
W=
jP dV,
with
P
being equal
to the equilibrium value, or very rapidly (a non-quasi-static process) so
that the pressure at the piston face
the piston
is
than the equilibrium value.
do no work on the
called a free expansion, and
To
less
If
repeat, one
it
way
will
piston at
all.
Such a
process
is
be discussed in some detail in the next
of doing work
or out, of either or both of the pistons.
by slow or rapid motion, in Another way in which work may be is
done on the system is by dissipation of electrical energy in the resistors, in which currents are maintained by generators actuated by the descending bodies. (Exactly the same effects could be produced by the dissipation of mechanical energy in the fluids by irregular churning of the fluids with paddle wheels actuated by the descending bodies.) As in Chap. 3, the most convenient independent coordinates of this system are 6, the common temperature, and the two volumes V and States i and of the system, shown on a 6VV diagram in Fig. 4-46, are arbitrarily chosen, / and it is purely a coincidence that / corresponds to a higher temperature
V
than static
i.
Fig. 4-4
(a)
Joining two stales
A i
may
composite system on which adiabatic ivork
and f by
be done in two ways. (A)
several different adiabatic paths.
pulled out at a faster rate than the velocity of the molecules
of the fluid, the fluid will
chapter.
is
In the path
iaf,
.
the dashed curve ia represents a frictionless, quasi-
adiabatic compression accomplished with one of the pistons.
It
is
drawn on
a surface cutting the two isothermal planes.
a surface will be proved in a later chapter, but point that, since ia
is
it
The
existence of such
should be noted at
accomplished only through a
frictionless
this
slow motion
of the piston, it may be performed in either the direction ia or ai. The curve af represents the adiabatic dissipation of electrical energy in conjunction with piston movements that keep the system at constant temperature. In
other words, the line af represents a process that
is
both adiabatic
and
iso-
however, an important distinction between this process and the previous one: the process af can go in only one direction. You can add energy with a current in a resistor, but you cannot extract it. The path ibf represents another adiabatic way of changing the system thermal! There
from
('
to /.
with the
is,
The curve
resistors,
and
ib
represents the dissipation process accomplished
the curve bf stands for the quasi-static process achieved
with frictionless pistons only. As before, bf could go in either direction, but ib in
only one.
There are, of course, many Other adiabatic paths joining if, such as icdj, in which the process cd is a non-quasi-static expansion accomplished by a rapid outward motion of one or both of the pistons, and the process df is performed by keeping both pistons motionless and by dissipating electrical energy in either or both of the resistors. Another possible adiabatic patli consists of rapid outward motion of the pistons, producing a non-quasistatic
expansion
ie,
followed by isochoric dissipation of electric energy
eb,
function, the difference
the system. The internal energy
between two values of which is
a function of as
bf. Although accurate measurements work along different paths between the same two states have never been made, indirect experiments indicate that the adiabatic. work is the same along all such paths. The generalization of this result is known as
since the third
the first law of thermodynamics:
It
means
only, the
to
work done
change from an
is the
same for
a final stale by adiabatic
initial state to
all adiabatic paths connecting the
two
states.
found to depend only on the initial and final states, and not on the path connecting them, an important conclusion can be drawn. The student will recall from mechanics that, in moving an object from a point in a gravitational field to another point, in the absence of friction the work done depends only on the positions of the two points and not on the path through which the body was moved. It was concluded from this that there exists a function of the space coordinates of the body whose final value minus its initial value is equal to the work done. This function
Whenever
a quantity
is
was called the
potential-energy function. Similarly, the work done in moving an electric charge from one point in an electric field to another is also independent of the path and therefore is also expressible as the value of a function (the electric potential function) at the final state minus its value at the initial state. It therefore follows from the first law of thermodynamics that there exists a function of the coordinates of a thermodynamic system whose value at
the final state minus
work
in
its
value at the
initial state is
equal to the adiabatic
going from one state to the other. This function
is
known
as the
77
is
the energy change of
many thermodynamic coordinates The equilibrium states of a means
hydrostatic system, for example, dcscribable by
of adiabatic
caused
Law
as are necessary to specify the state of a system.
followed by a quasi-static compression
is
First
should be emphasized, however, that the equation expresses more than the principle of the conservation of energy. It states that there exists an energy
dynamic coordinates P, V, and
If a system
Heat and the
4-4
Heat and Thermodynamics
76
of three thermo-
are completely determined by only two,
is fixed by the equation of state. Therefore, the internal energy thought of as a function of only two (any two) of the thermodynamic may be This is true for each of the simple systems described in Chap. 2. coordinates. is
not always possible to write this function in simple mathematical form.
Very often the exact form of the function however, that function If
is,
it is
not necessary to
so long as
we can
is
know
unknown. It must be understood, what the internal-energy
actually
be sure that such a function
exists.
the coordinates characterizing the two states differ from each other only
change of internal energy is dU, where dU is an exact differential, since, it is the differential of an actual function. In the case of a hydrostatic system, if U is regarded as a function of 6 and V, then infinitesimally, the
dU = or,
regarding
U as
a function of
dd
6
and
+
dU
dV,
dV
P,
rg).* The
student should realize that the two partial derivatives (dU/dd)v and
The first is a function of 6 and (', and the second a They arc different mathematically and also have a differ-
(dUj d6)p are not equal. function of 6 and P.
ent physical meaning.
internal-energy junction.
Denoting the internal-energy function by U, we have
— Wt-*f where the
4-3
signs are such that,
(adiabatic)
if
= U{ —
Mathematical Formulation of the First
4-4
U
the system does work,
its
energy decreases.
We
have been considering, up
to
now, processes wherein a system under-
goes a change of state through the performance of adiabatic work only. Such
experiments must be performed in order to measure the change in the energy function of a system, but they arc not the usual processes that arc carried out
Internal-energy Function
two examples
in the laboratory. In Fig. 4-5 there are depicted
The
physical interpretation of the difference U/
—
Ui
is
the change in
energy of the system. The equality, therefore, of the change of energy and the adiabatic
work
Law
(4-1
t,
expresses the principle of the conservation of energy.
It
involving changes of state that take place nonadiabatically. In
contact with a bunscn flame whose temperature
and
at the
same time
is
a gas
is
in
higher than that of the gas
is
allowed to expand. In
of processes
(a)
(b)
the magnetization of a
78
Heat and Thermodynamics
4-4
Heat and
the First
Law
79
where the convention has been adopted that Q is positive when it enters a system and negative when it leaves (just the opposite of the sign convention for W). The preceding equation is known as the mathematical formulation of the first
law.
seems a pity that the sign convention for work is opposite that for heat. Positive work ought to be work that goes into a system and raises its energy, The conventions in thermodynamics, as it is in the subject of mechanics. fit the behavior of heat engines whose normal operhowever, were devised to inflow heat and the output of work. There are so manyof ation involves the articles, and tables in which these standard conventextbooks, handbooks, It
Gas
adopted that a change would be feasible only if agreed to by engineers, chemists, and physicists. At the present time, such an accord seems out of the question. A convenient way to remember the sign convention is to tions arc
Magnet poles-
write the
first
law
in the
Uf - U = Q
(b) Fig. 4-5
t
Nonadiabatic processes.
paramagnetic solid is increased while it is in contact with liquid helium, the temperature of which is lower than that of the solid. As a matter of fact, some of the helium boils away during the magnetization. Let us now imagine two different experiments performed on the same system. In one we measure the. adiabatic work necessary to change the state of the system from i to/. This is U/ — Ui, In the other we cause the system to undergo the same change, of state, but nonadiabatically, and measure the work done. The result of all such experiments is that the nonadiabatic work is not
Uj
equal to
—
£/,-.
In order that this result shall be consistent with the principle
of the conservation of energy,
transferred transfer
by means
we
are forced to conclude that energy has been
than the performance of work. This energy,
oilier
between the system and
its
surroundings
is
whose
required by the principle
of the conservation of energy and which has taken place only by virtue of the
temperature difference between the system and
its
surroundings,
is
what wc
We
therefore give the following as our thermo-
dynamic definition of heat:
When
a system whose surroundings are at a different
temperature and on which work
may
have previously called heat.
be done undergoes a process, the energy transferred
by nonmechanical means, equal to the difference between the internal-energy change the
work done,
is called heat.
Denoting
Q =
this difference
and
by Q, we have
Ut - Ui- (-W),
Q = U, -
Ui
+
W,
form
(4-2)
W.
and to think of a system as an energy "bank," in which positive heat is regarded as a deposit and positive work as a withdrawal. It should be emphasized that the mathematical formulation of the first lawcontains three related ideas: (1) the existence of an internal-energy function; (2) the principle of the conservation of energy; (3) the definition of
heat as
by virtue of a temperature difference. It was many years before it was understood that heat is energy. The first really conclusive evidence that heat could not be a substance was given by Benjamin Thompson, an American from Woburn, Massachusetts, who later became Count Rumford of Bavaria. In 1798, Rum ford observed the temperature rise in brass chips produced during the boring of cannon and concluded energy in transit
work of boring was responsible for the flow of heat. One year later, tried to show that two pieces of ice could be melted by rubbing them together. His idea was to show that heat is a manifestation of energy, but his experiment was highly inconclusive. The idea that heat is a form of energy was put forward in 1 839 by Seguin, a French engineer, and in 1842 by Mayer, a German physician, but no conclusive experiments were performed by either. It remained for Joule, an independent investigator with a private laboratory, in the period from 1 840 to 1849, to convince the world by performing a series of admirable experiments on the relation between heat and work and to establish once and for all the equivalence of these two quantities. Von Helmholtz recognized the epoch-making importance of Joule's work and wrote a brilliant paper in 1847, in which he applied Joule's ideas to the sciences of physical chemistry and thai the Sir
Humphry Davy
physiology.
80
Heat and Thermodynamics
4-5
Concept of Heat
Heat and the
4-6
+
W+
ble to separate or divide the internal energy into a mechanical
and a thermal
part.
Law
81
£'',: is the change in energy of the composite sysiff) — {Lh + the work done by the composite system, it follows that and is tem Q' transferred by the composite system. Since the composite is the heat Q + system is surrounded by adiabatic walls,
Since (Ly
Heat is energy in transit. It flows from one part of a system to another, or from one system to another, by virtue of only a temperature difference. When this flow has ceased, there is no longer any occasion to use the word "heat." It would be just as incorrect to refer to the "heat in a body" as it would be to speak of the "work in a body." The performance of work and the flow of heat are methods whereby the internal energy of a system is changed. It is impossi-
First
W
=
Q'
Q
o,
Q = -Q'-
and
(4-3)
In other words, under adiabatic conditions, the heat
lost (or
gained) by system
A
is
equal to the heat gained (or lost) by system B.
We
have seen that, in general, the work done on or by a system is not a function of the coordinates of the system but depends on the path by which the system
was brought from
the initial to the final state. Exactly the
same
true of the heat transferred to or from a system. Q is not a function of the thermodynamic coordinates but depends on the path. An infinitesimal amount of heat, therefore, is an inexact differential and is represented by the
symbol dQ.
To
4-6
Differential
Form
A
At first sight, it seems too bad that / dQ is not independent of the path, for some such quantity would be useful. It would be pleasant to be able to say, in a given state of the system, that the system had so and so much heat energy. Starting from the absolute zero of temperature, where we could say that the heat energy was zero, we could heat the body up to the state we were interested in, find / dQ from absolute zero up to this state, and call that the heat energy. remains that we should get different answers if we heated it up in different ways. For instance, we might heat it at an arbitrary constant pressure until we reached the desired temperature, then adjust the pressure at constant temperature to the desired value; or we might raise it first to the desired pressure, then heat it at that pressure to the final temperature; or many other equally simple processes. Each would give a different answer, as we can easily verify. There is nothing to do about it. fact
Imagine a system A in thermal contact with a system B, the two systems being surrounded by adiabatic walls. For system A alone,
Q =
Uf -
Ui
+
the
known
is
as
an
B
for system
dQ = dV
If the infinitesimal process is quasi-static,
in
is
one
in
(4-4)
dU and AW can be expressed An infinitesimal quasi-static
which the system passes from an
neighboring equilibrium
initial
= W, -
u-
dQ = dU + P where
U
is
dV,
a function of any two of the three thermodynamic coordinates and
is, of course, a function of V and 6. A similar equation may be written each of the other simple systems as shown in Table 4-1.
Table 4-1
The
First
Law
for
Simple Systems
U is a
+ w.
Hydrostatic system Surface film Electric cell
=
{Ui
+
U't)
-
(Ui
+
Dj)
+ W+
Paramagnetic IV.
first
(4-5)
P
IV;
Adding, we get Q'
equilibrium state to a
state.
For an infinitesimal quasi-static process of a hydrostatic system, the law becomes
Wire
+
then
terms of thermodynamic coordinates only.
process
thermodynamic
For such a process
+ dW.
First
law
alone,
0!
Q
infinitesimal process.
law becomes
first
System
and
Law
process involving only infinitesimal changes in the
coordinates of a system
quote from Slater's "Introduction to Chemical Physics":
But the stubborn
of the First
is
solid
dQ dQ dQ dQ dQ
= dU + P dV = dU - J dL = dU - J dA = dU - £ dZ = dU - 9£dM
function of any two of
P, V, B
y,L,e J, a, e £,z, e %, m, e
for
82
To
more complicated systems, it is merely necessary to replace AW in the first law by two or more expressions. For example, in the case of a composite system consisting of two hydrostatic parts separated by a diadeal with
thermic wall,
we may
Heat and the
4-7
Heat and Thermodynamics
dQ
express
temperature from
0,
heat capacity of the
system
during the transfer of
to 0/
is
whereas
for a
(4-6)
right-hand
members
dli
+ P dV - ?{ dM.
forms,
Q
and
of Eqs. (4-5), (4-6), and (4-7) are
and the question of
their integrability
(9/
—
0,-)
heat capacity
known
units of heat, the average
-
6>
get smaller, this ratio approaches the instantaneous
C
thus:
C=
(4-7)
Q
lim It-tBi If
as Pfaffian
an interesting and important one that will be studied later. At this point, however, it is worth while to state a fundamental difference between Eq. (4-5) and Eq. (4-6), in order to justify the seemingly undue emphasis on systems of more than two independent coordinates. Since the left-hand expression in Eqs. (4-5) and (4-6) represents an infinitesimal amount of heat dQ, and since the heat transferred depends on the path, dQ is an inexact differential and the Pfaffian differentia! forms are inexact differentials. An inexact differential, however, may often be made exact by multiplying it by a function, known as an integrating factor. The Pfaffian differential form representing dQ of a simple system with two independent coordinates has the mathematical property that an integrating factor can always be found. This is not the result of a law of nature; it is a purely mathematical result of the fact that there arc only two independent differential
As both
value of the
paramagnetic gas
dQ = The
dlf+PdV+P'dV
83
Q
= Or
dQ =
La
defined as the ratio
Average heat capacity
as follows:
Q
First
dQ c=
is
(4-8)
de
The heat
capacity of a system per unit mass
called the specific heat
is
per gram-degree. The gram, however, is not a convenient unit of mass to use in most calorimetric work. Most physicists and chemists prefer the gram-mole (rather than the kilogram-mole) that is, the number of grams equal to the molecular weight. capacity, or
specific heal,
and
is
measured
in joules
—
The
heat capacity per mole, or molar heat capacity,
is
designated with a lower-
case letter thus:
_ C_
1
dQ
coordinates.
When
there are three or
situation
is
more independent
coordinates, however, the
entirely different. In general, a Pfaffian differential form con-
taining three differentials does not admit of
an integrating
factor! But,
because of the existence of a new law of nature (the second law of thermodynamics), the Pfaffian differential form representing dQ does have an integrating factor. It
is
a most remarkable circumstance that the integrating
dQ which
is
found for systems with any number of independent
factor for
variables
is
a function of the empiric temperature only, which
for all systems.
This
is
the
This quantity
The
measured
may
in joules
per mole-degree.
infinite, depending on the process the system undergoes during the heat transfer. It has a definite value only for a definite process. In the case of a hydrostatic system, the ratio dQ dO has a unique value when the pressure is kept constant. Under these conditions, C is called the heal capacity at. constant pressure and is denoted by the symbol Cp, where
be negative, zero, positive, or
same function
an absolute thermodynamic in Chap. 8.
result enables us to define
(or Kelvin) temperature, as
we
shall see
{de), In general, Cp
4-7
is
heat capacity
Heat Capacity and
Its
Measurement
When heat is absorbed by a system, a change of temperature may or may not take place, depending on the process. If a system undergoes a change of
volume
is
a function of P and
0.
(4-9)
Similarly, the heat capacity at constant
is
Ck
= dQ de
(4-10)
84
I
and Thermodynamics
Icat
and depends on both V and will be its
own
8. In general, Cp and C r are different. Both thoroughly discussed throughout the book. Each simple system has heat capacities as shown in Table 4-2.
Table 4-2
Heat capacities
System
If
&
Symbol
heating
the
At constant pressure At constant volume
T.inear
Surface
Electric
Cp Cv
At constant tension At constant length
C,7
At constant surface tension
At constant area
Cj CA
At constant emf
Ce
At constant charge
Cz
At constant field At constant magnetization
Cv
measured
is
coils,
skill
In
shape,
thermometers,
sample
is
particularly in the case of solids at low tempera-
suspended
threads of nylon or
Ci
a function of
two
The
placed in a small hole drilled for that purpose. the heater, for the current in the thermometer, across the
thermometer are made very is
and
connecting wires for
for the potential difference
much heat to The temperature when plotted as in
thin so as not to allow
be transferred between the sample and
its
surroundings.
measured as a function of the time; AB, marked "foreperiod." At the time correspond-
Fig. 4-6, this gives the line
Cm
ing to point B, a switch
same moment
is
closed
Within a small range may be regarded as practically constant. Very often, one heat capacity can be set equal to another without much error. Thus, the CV of a paramagnetic solid is at times very nearly equal to Cp. variables.
of variation of these coordinates, however, the heat capacity
the temperature
is
and a current
is
that an electric stop clock
established in the heater is
started. After a short
opened and the stop clock is stopped. Then again measured as a function of time and is plotted as the
interval of time At, the switch is
and construction of the calorimeter, depend on the nature of the material to
size,
etc.,
in a highly evacuated space by means of fine some other poorly conducting material. A heating coil is wound around the sample, and a thermocouple or a resistance thermometer (platinum, carbon, or germanium, depending on the temperature range) is tures, the
at the
Each heat capacity
is
DE, marked "afterperiod" in Fig. 4-6. As a rule, no reading of temperature or time
line
is
attempted while the stop
The measurement of the heat capacity of solids is one of the most important experimental projects of modern physics, because numerical values of heat capacity provide one of the most direct means of verifying the calculations and of deciding on the validity of the assumptions some of the modern theories. An electrical method of measure-
of theoretical physicists
constituting
ment
used almost invariably. If a resistance wire is wound around a sample of material and if both the wire and the sample are regarded as the system, then the electrical energy dissipated in the wire is interpreted as work. When the wire is not included as part of the system, is
cylindrical
however, the energy which is dissipated within the wire and which flows into the sample by virtue of the temperature difference between the wire
and the sample (however small) called a heating
across is
it is
coil.
If the
is
designated as heat.
current in the wire
is
The
wire
is
often
/ and the potential difference
£, then the heat dQ that leaves the heating
coil
over a time dr
calculated as
Time
dQ = 61 dr.
85
of an expert glass blower.
modern calorimetry,
of the sample
Magnetic
The
Lav
/ in amperes, and r in seconds, the heat will be
in volts,
expressed in joules.
First
be studied and the temperature range desired. It is impossible to describe one calorimeter that suffices for all purposes. In general, the measurement of any heat capacity is a research problem requiring all the ability of a trained physicist or physical chemist, the facilities of a good workshop, and
Heat Capacities of Simple Systems
Hydrostatic.
Heat and the
4-7
Fig. 4-6
Temperature-vs .-lime graph in the measurement of heal capacity.
r
86 clock
C
4-8
Heat and Thermodynamics
is
on, that
of the line
is,
from
B
BD, and both
extrapolated to this vertical
molar heat capacity given by
c
A
to D.
vertical line
the foreperiod
line,
is
drawn through
the center
and the aftcrperiod lines are F and G, as shown. The
giving the points
at the temperature corresponding to point
C
C
is
then
joules.
shown and it
is
in Fig. 4-6
made is
and even when the electrical method
of
IT
calorie)
was defined
as
graph
versus r but of the resistance R' versus
r,
possible to
1
IT
cal
=
~
wall
•
hr
=
4.1860
.1.
ooli
The
tiwrmochemkal
calorie,
1
on the other hand, was given the following value:
thermochcmical cal
=
4.1840
J.
Heat Capacity of Water; The Calorie and chemists today, the calorie is being dropped, and at least at very low and very high temperatures, all thermal quantities are but electrical methods arc employed exclusively where water is not used instead equivalent of heat, but is no mechanical expressed in joules. There there is the specific heat of water, expressed in J/g deg, whose temperature to 100°C is shown in Fig. 4-7. variation in the range from
Among
When
87
4.1860 J for the 15-deg calorie. They therefore defined their own calories in terms of the joule. For purposes of mechanical engineering, the International
the aid of a computer.
4-8
Law
Mechanical and chemical engineers also preferred to keep the calorie as a were not inclined to accept the conversion factor of
SI At = ~ riAd'
6
First
unit of heat, but they
as small as 0.01 deg. Strictly speaking, the
not a graph of
and the
calorimetry was used and thermal quantities were actually measured in joules, the measurements were then converted to calories.
have the entire R't curve drawn automatically with the aid of a recording potentiometer. When many values of R' and AR' arc read from recorder graphs, the corresponding 0's, AO's, and c's are obtained with is
[eat
This move proved to be somewhat premature. Physicists and chemists
preferred to think in terms of calories,
Table calorie (or
Sometimes A6
1
the subject of calorimetry was
first
presented in the middle of the
eighteenth century, measurements were confined to the temperature range
between the freezing and boiling points of water. The unit of heat found most convenient was called the calorie and was defined as the amount of heat required to raise the temperature of 1 g of water one Celsius degree. To measure the amount of heat transferred between a system and some water, it was necessary merely to make two measurements: of the mass of water and of its temperature change. Later on, as measurements became more precise and more elaborate corrections were made, it was discovered that the to 1 °C was different from the heat necessary to change 1 g of water from heat necessary to go from, say, 30 to 31 °C. The calorie was then defined to be the heat necessary to go from 14.5 to 15.5°C (the 15-deg caloric). The amount of work that had to be dissipated in water either by maintaining a current in a resistor immersed in water or by churning the water in an irregular maimer per unit mass of water in going from 14.5 to 15.5 C was called the mechanical equivalent of heal, which was measured to be 4.1860 J/cal. In the 1920s it was recognized that the measurement of this mechanical equivalent of heat was really a measurement of the specific heat of water, with the joule as the unit of heat. Since heat is a form of energy and the joule is a universal unit of energy, the calorie seemed superfluous. Therefore, in a large and important collection of tables of physical constants published at that time, which were called the International Critical Tables, all thermal quantities such as specific and molar heat capacities were expressed in terms of
physicists
—
—
—
—
40
50
60
Temperature, °C Fig. 4-7
Most probable values
of specific heat of water. {Compiled by
W.
./.
de Haas, 1950.
t
88
Heat and Thermodynamics
4-10
Equations for a Hydrostatic System
4-9
The mathematical
formulation of the
first
law
for a hydrostatic
system
P
If
(A)
is
is
+p
\dOjr
=
:
where
U
is
V,
and
0.
Choosing
"-(&•+$). Therefore the
by Hence,
we
0,
wc have
dV.
If
V
dd) P
(BU\
_
Cp and
»).+'
- Cy Cpj-
$\ m.
+F
dV df
-
also
(dV dd)p
=
f/3.
re
(4-14)
p.
(4-12)
4-10 and
It
Quasi-static
was shown
in
Flow of Heat; Heat Reservoir
Chap. 3 that a process caused by a
finite
unbalanced force
attended by phenomena such as turbulence and acceleration which cannot be handled by means of thermodynamic coordinates that refer to the system is
constant,
dV =
as a whole.
and
0,
A
similar situation exists
when
the temperature of a system and that of
'dQ\ de) v K But the ratio on the stant
{**)*
;
true for any process involving any temperature change dQ
is
89
Law
Although this equation is not important in its present form, it is a good example of an equation that relates a quantity {dU dV)t, which is ordinarily not measured, with quantities such as Cp, Cv, and fi, which may be measured.
(4-11)
any volume change dV. (a)
(6Q
CP = C v
dV.
£>
+
'
dO
is
V,
get
dQ
This equation
and
definition,
or
«»-G&* Dividing by
6
law becomes
first
First
+ P dV, But,
a function of any two of P,
and the
constant, Eq. (4-12) becomes
\de/p dQ = dU
Ileal
volume Cy\
left,
by
\deiv definition,
is
the heat capacity at con-
therefore,
Cv =
(4-13)
dejy
there
is
a finite difference between
surroundings.
A
nonuniform tem-
up in the system, and the calculation of this distribution and its variation with time is in most cases an elaborate mathematical problem. During a quasi-static process, however, the difference between the temperature of a system and that of its surroundings is infinitesimal. As a result, the temperature of the system is at any moment uniform throughout, and its changes arc infinitely slow. The flow of heat is also infinitely slow and may be calculated in a simple manner in terms of thermodynamic coordinates perature distribution
(d_U\
its
is
set
referring to the system as a whole.
Suppose that a system is in good thermal contact with a body of extremely and that a quasi-static process is performed. A finite amount of heat flow during this process will not bring about an appreciable change in the temperature of the surrounding body if the mass is large enough. For example, a cake of ice of ordinary size, if thrown into the ocean, will not prolarge mass
U is calculated mathematically by making special assumptions about the atoms of a particular material, one of the first methods of cheeking these assumptions is to differentiate U with respect to at If
constant
V and
to
compare the
resulting quantity with the experi-
mentally measured value of CV.
duce a drop
in
temperature of the ocean.
outside air will produce a
rise
No
ordinary flow of heat into the
of temperature of the air.
outside air arc approximate examples of an ideal
body
The ocean and
the
called a heat reservoir.
90
A
4-11
Heat and Thermodynamics
heat reservoir is a body of such a large mass I hat
it
may absorb
or reject
an unlimited
quantity of heat without suffering an appreciable change in temperature or in any other thermodynamic coordinate. It is not to be understood that there is no change in
thermodynamic coordinates of a heat reservoir when a finite amount of heat flows in or out. There is a change, but an extremely small one, too small to be measured. In other words: in any small unit of mass, a change in a physical property is infinitesimal, but there is an infinite number of such units the
of mass in the heat reservoir.
Any
quasi-static process of a system in contact with a heat reservoir
is
To
describe a quasi-static flow of heat involving a change of temperature, one could conceive of a system placed in contact successively with a series of reservoirs. Thus, if we imagine a series of reser-
bound
to be isothermal.
voirs ranging in temperature
from
0,
to 0/ placed successively in contact
with
a system at constant pressure of heat capacity Cp, in such a way that the difference in temperature between the system and the reservoir with which it is
in
contact
is
infinitesimal, the flow of heat will be quasi-static
and can be
C'Cpde. q p = JO:
of each small volume clement of the intervening shows a continuous distribution of temexperiment measured, substance is neighboring volume elements by between of energy transport The perature. tures
and the temperature
virtue of the temperature difference between
Qp =
them
is
known
as heat conduction.
The fundamental law of heat conduction is a generalization of the results of experiments on the linear flow of heat through a slab perpendicular to the Ax and of faces. A piece of material is made in the form of a slab of thickness Ad. the other at temperature and 6 the maintained at face is One area A. time is measured. The faces for a r perpendicular to the flows heat that The Q
+
experiment values of
is
repeated with other slabs of the same material but with different A. The results of such experiments show that, for a given for a given is proportional to the time and to the area. Also,
Ax and
value of Aft
Q
time and area,
Q
is
are small.
proportional to the ratio A0, Ax, provided that both A0
These
results
which
is
may
be written
*
A
only approximately true
rigorously true in the limit as
For example, the heat absorbed by m g of water from a series of reservoirs varying in temperature from ft to 0/ during a quasi-static isobaric process is
91
Law
When two parts of a material substance arc maintained atdiffcrcnt tempera-
r
and therefore
First
Heat Conduction
4-11
and Ax
calculated as follows: by definition,
Heat and the
—
Ax
>
when A0 and Ax
A0 and Ax approach
are finite but
zero. If
we
which
is
generalize this
is a temperaand introduce a constant of proportionality K, the fundamental law of heat conduction becomes
result for
an
infinitesimal slab of thickness dx, across
which there
ture difference dd,
CpdB
= m
(4-15)
-9/ re.
Cp dd.
I
Jet
If
the specific heat cp
is
assumed to remain practically constant,
Qp =
tncp{fi,
-
Oi).
The derivative dd/dx is called the temperature gradient. The minus sign is introduced in order that the positive direction of the flow of heat should coincide with the positive direction of x. For heat to flow in the positive must be the direction in which decreases. K is called the substance with a large thermal conductivity is known as a thermal conductor, and one with a small value of K as a thermal insulator. It will be shown in the next article that the numerical value of K depends upon a
direction of
For
a quasi-static isochoric process
Qv
=
x, this
thermal conductivity.
]g
Cv
dO.
number
A
of factors, one of
conducting material Similar considerations hold for other systems and other quasi-static processes.
may
is the temperature. Volume elements of a therefore differ in thermal conductivity. If the
which
temperature, difference between parts of a substance
is
small, however,
K can
92
I 4-12
Hrat and Thermodynamics
be considered practically constant throughout the substance. This simplification
is
usually
made
in practical
problems.
To
handle general problems in the conduction of heat, it is necessary to transform the general equation into the form of a second-order partial differential equation.
The
boundary
solution of this equation subject to given
conditions involves, as a rule, the use of functions and series beyond the scope of collegiate mathematics.
There are three simple
be handled in an elementary way. In
we
all cases,
cases,
shall
K
is
con-
Fig. 4-8
Radial flow of heat
in
a
First
Law
93
1
"
L
however, that can
assume that
Heat and the
cylinder.
stant throughout the conducting substance.
1
Linear flow of heal perpendicular
—
difference 0i
d*
to the
faces of a slab.
and the thickness
x are small,
If the it is
temperature
and an outer sphere of radius
temperature
Q\
perature
there will be a steady radial flow of heat at the constant
82,
Q = KA
and
+
r
dr,
Suppose the conducting between an inner cylinder of radius r% and an outer cylinder of radius r«, both of length L. If the inner cylinder is maintained at the constant temperature 0i and the outer at 62, there will
de ~KAdr
Radial flow of heal between two coaxial cylinders.
=
lies
be a steady radial flow of heat at the flow of this
amount
bounded by
the cylinders at r
temperature at of the shell
r,
bounded by
we have
a
Q = material
held at constant tem-
rate of Q. Considering this flow across the spherical shell
obvious that
the spheres at r
-
r%
constant rate of Q.
d6
-KA-wr
dr
Q dr_ 4wKr
and
'
2
Consider the
of heat across a cylindrical shell of material
and
6
+
and
+ dr
r
Let d be the
(see Fig. 4-8).
dd be the temperature at
r
Integrating between
r\
and
ri,
we
get
+ dr. The area
is
A =
6i
lirrL,
and the temperature gradient
is
dO/dr.
and
dd
&i
" - aIk \n 7J
(4-17)
Hence,
4-12
Q =
~
Thermal Conductivity
-KlirrL — d9 dr
When
Q
dr
of a bar,
2ttLK
r
= -
a
the substance to be investigated
and one end
is
is
a metal,
it is
made
heated electrically while the other end
into the is
form
cooled with
stream of water. The surface of the bar is thermally insulated, and the heat through the insulation is calculated by subtracting the rate at which heat
lost
Integrating between
r\
and
r«,
wc
which electrical energy is supplied. In the from the surface is very small in comparison with that which flows through the bar. The temperature is measured with suitable thermocouples at two places L cm apart, and the equation
enters the water from the rate at
get
case of most metals, the heat lost In
1-kLK
—
Radial flow of heal between two concentric spheres.
material
lies
(4-16)
r,
between an inner sphere of radius
If the r\
conducting
held at constant
K= A(0i
-
-Q 62)
94 is
Heat and Thermodynamics
4-12
Meat and the
First
Law
95
used to determine the average thermal conductivity within the given temis practically equal to the thermal con0i d-i is small,
—
perature range. If ductivity at the
When
mean
K
temperature.
the substance to be investigated
is
a nonmetal,
it is
made
into the
form of a thin disk or plate, and the same general method is used. The substance is contained between two copper blocks, one of which is heated electrically and the other cooled by running water. The thermal contact between the copper blocks and the substance is improved by smearing them with glycerin. In most cases, the rate at which heat is supplied is almost equal to the rate at which heat enters the water, showing that there is little loss of heat through the edges. Since the thermal conductivity
K=
1
is
equal to Q:A(tl6d.x),
J /sec
-
m* deg/m
=
has the unit
W 1
•
Lengths arc usually expressed
it
in centimeters,
m
deg
however, so that
the custom to express thermal conductivity in the hybrid unit
has become W.'cm deg.
it
-
1
Experiments show that the thermal conductivity of a metal is extraordinarily sensitive to impurities. The slightest trace of arsenic in copper reduces
by a factor of 3. A change in internal structure brought about by continued heating or a large increase in pressure also affects the value of A'. No appreciable change in the K of solids and liquids the thermal conductivity
takes place, however,
under moderate changes of pressure. Liquefaction
always produces a decrease in the thermal conductivity, and the thermal conductivity of a liquid usually increases as the temperature is raised. Nonrnetallic solids behave
in
a manner similar
to that of liquids.
temperatures these are poor conductors of heat; ductivity increases as the temperature
range, however, the behavior it
may
is
is
in general, the
raised.
quite different, as
be seen that the thermal conductivity of sapphire
W/cm
in Fig. 4-9,
a
•
1000°K. As a general temperature
is
rule, the
Fig. 4-9
Powell,
maximum
is
1000
Typical curves shelving temperature dependence of thermal conductivity. (R. L. McGraw-Hill, 1063.)
the thermal to the electric conductivity
was approximately the same for most metals at the same temperature. Lorcntz later showed that this ratio was not only the same for all metals at the same temperature but varied directly with the absolute temperature. If v denotes the electric conductivity in (reciprocal of resistivity), Lorentz's
—K
reached. Further reduction of
temperature causes a decrease toward zero, as shown
500
AW Handbook,
thermal conductivity of metals increases as the
lowered, until a
200
where
maximum
deg at 30°K (four times the conductivity of silver at room temperature). The thermal conductivity of some metals remains quite constant over a wide temperature range. Thus, silver, copper, and gold have thermal conductivities that remain practically constant at the values 4.2, 3.8, and 2.9 VV/cm- deg, respectively, in the temperature range from 100 to
of over 50
100
At ordinary
thermal con-
rises to
50
Temperature, °K
In the low-temperature
shown
20
10
5
in the case of
copper
=
const.
o~0
in
=
_1
cm"
1
law can be written
2.23
X
lO" 8
V
2
(4-18)
deg 2
Fig. 4-9.
Thermal and electric conductivity of metals go together. This important was first stated by Wiedemann and Franz, who showed that the ratio of
fact
The approximate to
100°C
is
validity of this law in the temperature range
shown
in
Table
4-3.
from —170
4-14
Heat and Thermodynamics
96
Values of
Table 4-3
KM
in 10 _s
VVdeg
and the main body
2
is
-170°C
-100°C
0°C
18°C
100°O
Copper
1.85
2.30
Silver
Cadmium
2.04 2.20 2.39
2.17 2,29 2.39 2.43
Tin Lead
2.48 2.55
2.32 2.33 2.43 2.39 2.47
2.32 2.37 2.33 2.44 2.49
2.51
2.51
Metal
of the fluid.
to find the value of h that
is
Heat and the
The fundamental problem
First
97
Law
of heat convection
appropriate to a particular piece of equipment.
Experiment shows that the convection
coefficient
depends on the following
factors:
Zinc
2 33 .
2.45 2.40 2.49 2.53
2.51
2.54
3
Whether the wall Whether the wall Whether the fluid
4
The
1
2
is flat is
or curved.
horizontal or vertical.
in
contact with the wall
density, viscosity, specific heat,
is
a gas or a liquid.
and thermal conductivity of
the
fluid.
Whether the
5
value,
1 is
atm. The
thermal conductivity of a gas always increases as the temperature
raised, as
is
Since the physical properties of the fluid depend upon temperature and pressure,
A
Heat Convection
current of liquid or gas that absorbs heat at one place and then
called natural convection. If the fluid
or a fan,
it is
is
made
to
move by
the action of a
an enormously complicated problem. problem good enough for practical purposes have been achieved with the aid of dimensional analysis. Such analysis yields an expression for containing the physical properties and velocity of the fluid and unknown constants and exponents. The constants and exponents arc then evaluated by experiment. only
is
in recent years that solutions of the
Thermal Radiation; Blackbody
4-14
pump
called forced convection.
A
Consider a fluid in contact with a flat or curved wall whose, temperature is higher than that of the main body of the fluid. Although the fluid may be in motion, there is a relatively thin film of stagnant fluid next to the wall, the thickness of the film depending upon the character of the motion of the main
substance
number
may
be stimulated to emit electromagnetic radiation
in a
of ways:
An
electric
conductor carrying a high-frequency alternating current
emits radio waves.
The more turbulent the motion, the thinner the film. Heat is from the wall to the fluid by a combination of conduction through the film and convection in the fluid. Neglecting the transfer of heat by radiation (which must be taken into account separately), we may define a convection coefficient A that includes the combined effect of conduction through the film and convection in the fluid. Thus, body of
clear that the rigorous calculation of a convection coefficient
/;
moves to and rejects another place, where it mixes with a cooler portion of the fluid heat, is called a convection current. If the motion of the fluid is caused by a difference in density that accompanies a temperature difference, the phenomenon is
it is
appropriate to a given wall and fluid
evident in Fig. 4-9.
small enough to give rise to
is
Whether evaporation, condensation, or formation of scale takes place.
6
It is
4-13
velocity of the fluid
laminar flow or large enough to cause turbulent flow.
above a certain
Gases are by far the poorest heat conductors. At depending on the nature of the gas and the dimensions of the containing vessel, the thermal conductivity is independent of the pressure. Under the usual laboratory conditions, this limiting pressure is considerably below pressures
A A
fluid.
transferred
hot solid or liquid emits thermal radiation. gas carrying an electric discharge
may
emit
visible or ultraviolet
radiation.
A A A
X
rays. metal plate bombarded by high-speed electrons emits substance whose atoms are radioactive may emit 7 rays.
substance exposed to radiation from an external source
may emit
fluorescent radiation.
Q = hA
AO,
(4-19) All these radiations are electromagnetic waves, differing only in
where Q is the rate of heat transfer by convection, A is the area of the wall, and A8 or At is the temperature difference between the surface of the wall
i.e.,
We
be concerned the radiation emitted by a
length.
shall
in this article
wave-
only with thermal radiation,
solid, liquid, or
gas by virtue of
its
tempera-
98 turc.
llcat
and Thermodynamics
When
thermal radiation obtained.
The
is
dispersed by a suitable prism, a continuous
distribution of energy
among
the various
wave-
spectrum
is
lengths
such that, at temperatures below about 500°C, most of the energy
is
associated with infrared waves, whereas at higher temperatures some visible radiation is emitted. In general, the higher the temperature of a body, the is
greater the total energy emitted. The loss of energy due to the emission of thermal radiation
pensated in a variety of ways. itself,
such as the sun; or there
Heat and the
4-1 5
emitting body
The
may
may
may
be com-
be a source of energy
be a constant supply of electrical energy
from the outside, as in the case of the filament of an electric light. Energy may be supplied also by heat conduction or by the performance of work on the emitting body. In the absence of these sources of supply, the only other way in which a body may receive energy is by the absorption of radiation
from surrounding bodies. In the case of a body that is surrounded by other bodies, the internal energy of the body will remain constant when the rate at which radiant energy is emitted is equal to that at which it is absorbed. Experiment shows that the rate at which a body emits thermal radiation
depends on the temperature and on the nature of the surface. The total radiant power emitted per unit area is called the radiant emittance of the body. 2 For example, the radiant emittance of tungsten at 2177°C is 50 W/cm When thermal radiation is incident upon a body equally from all directions, the radiation is said to be isotropic. Some of the radiation may be absorbed, some
and which communicates with
the outside
First
99
Law
by means of a hole having a
diameter small in comparison with the dimensions of the cavity. Any radiation entering the hole is partly absorbed and partly diffusely reflected a large number of times at the interior walls, with only a negligible fraction eventually finding
its
way
out of the hole. This
is
of which the interior walls are composed. The radiation emitted by the interior walls
diffusely reflected a large
number
true regardless of the materials
similarly absorbed
is
of times, so that the cavity
is
filled
and with
isotropic radiation. Let us define as the irradiance within the cavity the radiant falling in unit time upon unit area of any surface within the cavity. Suppose a blackbody whose temperature is the same as that of the walls is introduced into the cavity. Then, denoting the irradiance by //,
energy
Radiant power absorbed per unit area Radiant power emitted per unit area
and
= aaH = H, =
/Gp.
Since the temperature of the blackbody remains constant, the rate at which the energy
is
absorbed must equal the rate at which
it is
emitted;
whence
.
reflected,
absorbed depends on the temperature and the nature of the surface of the absorbing body. This fraction is called the absorptivity. At 2477°C the absorptivity of tungsten is approximately 0.25. To summarize:
Absorptivity
=
/3
=
= a =
or
the irradiance within
radiant emittance of a blackbody at the same temperature.
tion within
radiant power emitted per unit area.
There are some substances, such stance capable of absorbing
as
all
lampblack, whose absorptivity is very useful to conceive of an ideal sub-
it is
is
called blackbody radiation.
Such radiation
is
to
studied
it
follows that the radiant emittance of a blackbody
is
the
by
a function of the tem-
perature only.
The
Kirchhoff's
Law; Radiated Heat
radiant emittance of a non-blackbody depends as
Such a
derive as follows: Suppose that a non-blackbody at the temperature
it.
radiant emittance /S and absorptivity a,
=
much on the
indicated by the subscript
we have
interior walls are at the
1.
is
very good experimental approximation to a blackbody
provided by a which the interior walls are maintained at a uniform temperature
d,
same temperature and where the irradiance
Radiant power absorbed per unit area
=
aH,
Radiant power emitted per unit area
=
/S.
is
and
nature
we may with
introduced into a cavity whose
Then,
cavity for
is
of the surface as on the temperature, according to a simple law that
the thermal radiation falling on
called a blackbody. If a blackbody
an
A
equal
absorbed.
is
nearly unity. For theoretical purposes
B,
is
this reason, the radia-
allowing a small amount to escape from a small hole leading to the cavity. Since // is independent of the materials of which the interior walls are com-
4-15
is
a cavity
For
fraction of the total energy of isotropic radiation
that
substance
a cavity whose walls are at the temperature
is
posed, total
(4-20)
/S,,
In general, the fraction of the incident
and some transmitted.
isotropic radiation of all wavelengths that
Radiant emittance
H=
is //.
100
I
leat
4-15
and Thermodynamics
Since the non-blackbody
is
in equilibrium,
column
1.
Thus, the absorptivity of
for the long infrared
IQ
But, from Eq. (4-20),
H=
=
=
the radiant erniltance of
(4-2 1}
temperature is equal to a fraction of the
being the absorptivity radiant erniltance of a blackbody at that temperature, this fraction at that temperature.
This equation, known as Kirchhoff's law, shows that the absorptivity of a body may be determined experimentally by measuring the radiant cmittancc temperature. of the body and dividing it by that of a blackbody at the same Values of the absorptivity of various surfaces, measured in this way, arc given in
Table
be emphasized that the tabulated values of absorpthermal radiation appropriate to the temperature listed in
4-4. It should
tivity refer to the
Table 4-4 (Values at
101
0.97 not for visible radiation but
waves associated with matter
at 0°C.
an amount of internal energy equal surroundings to the energy radiated minus the energy absorbed, whereas the absorbed minus the the energy gain an amount of internal energy equal to The gain or loss of the other. energy radiated. The gain of one equals the loss in a given interval of time the
a/3,„
any body at any
Law
It
of internal energy, equal
or
is
First
should be noticed that the word "heat" has not appeared as yet. If there is a temperature difference between a body and its surroundings, then
all.
/?«; hence,
/e
ice
Heat and the
Approximate Absorptivities of Various Surfaces, as Compiled by Hottel intermediate temperatures may be obtained by
which
is
to
body
loses
the difference between the energy of the thermal radiation
absorbed and that which
is
radiated, is called heat.
This statement
is
in
agreement with the original definition of heat, since a gain or loss of energy by radiation and absorption will take place only if there is a difference in tempera-
between a body and its surroundings. If the two temperatures are the same, there is no net gain or loss of internal energy of either the body or its surroundings, and there is therefore no transfer of heat. Imagine a cavity whose interior walls are maintained at a constant temperature 0w- Suppose that a non-blackbody at a temperature 6 different from that of the walls is placed in the cavity. If the body is small compared with ture
the size of the cavity, then the character of the radiation in the cavity will not be appreciably affected by its introduction. Let H, as before, denote the irradiance within the cavity, and /& and tivity, respectively, of the
a
the radiant emittancc
and absorp-
body. Then, as before,
linear interpolation.)
Material
Temperature range, °C
Absorptivity
and
Polished metals:
Aluminum Brass
Chromium Copper Iron Nickel Zinc Filaments:
Molybdenum Platinum
Tantalum Tungsten Other materials: Asbestos
250- 600 250- 400 50- 550 100 150-1000 20- 350 250- 350
0.039-0.057 0.033-0.037 0.08 -0.26 0.018 0.05 -0.37 0.045-0.087 0.045-0.053
750-2600 30-1200 1300-3000 30-3300
0.096-0.29 0.036-0.19 0.19 -0.31 0.032-0.35
40
350
Ice (wet)
Lampblack Rubber (gray)
20- 350 25
0.93 -0.95 0.97 0.95 0.86
but
now
these
Radiant power absorbed per unit area
=
aH,
Radiant power emitted per unit area
=
F3\
two
rales are not equal.
The
difference between
transferred by radiation per second per unit area. If dQ in
time dr to the whole bodv whose area
=
dQ
is
is
them
is
the heat
the heat transferred
A, then
= A(aH -
(4-22)
/&),
; dr
where, it must be remembered, a and /S refer to the temperature the temperature B\p. Now,
and
H = /Suidw), B = a/Bnie).
Hence,
Q = A
or the rate at which heat
is
transferred by radiation
and II to
(4-23)
is
proportional to the
102
Ileal
Heat and
and Thermodynamics
difference between the radiant emittances of a blackbody at the
Electrical energy
Stefan-Boltzmann
is
Law
Si until the sphere which the rate of supply of the rate of emission of radiation. Assuming the
is
equal to
sphere to be a blackbody,
measurements of the heat transferred by radiation between a body and its surroundings were made by Tyndall. On the basis of these experiments, it was concluded by Stefan in 1 879 that the. heat radiated was proporfirst
we have
tional to the difference of the fourth
powers of the absolute temperatures. This purely experimental result was later derived thcrmodynamically by Boltzmann, who showed that the radiant emittance of a blackbody at any temperature 8 is equal to
=
/S,s(8)
This law
is
now known
?6
and a
is
4^(8*
between a body at the temperature
for the heat transferred 6
and walls
Q = Aaa{6\v where a
Two
refers to the
temperature
simple methods
may
at
4-1 6.
for the determination
A
blackened
of the
silver disk
The
is
placed in the ccnter
is covered hemisphere achieves the temperature of condensing steam; this temperature is measured with a thermocouple. Then the disk is uncovered, and its tempera-
and shielded from radiation
silver disk
until the copper
measured as a function of the time. From the resulting heating curve, the slope dB/dr is obtained. Assuming the silver disk to be a blackbody and putting dQ = CP dB, where CP is the heat capacity at constant pressure, we have is
is
2
•
(4-26)
deg 4
A
gas contained in a cylinder surrounded by a thick layer of
quickly compressed, the temperature rising several hundred degrees.
felt
Has
Has the "heat of the gas" been increased? combustion experiment is performed by burning a mixture of fuel and oxygen in a constant-volume "bomb" surrounded by a water bath. During the experiment the temperature of the water is observed to rise. If we regard the mixture of fuel and oxygen as the system: (a) Has heat been transferred? Has work been done? (/;) (c)
4-3
CF dd =
-
Ao
(a)
(c) <>
4
Cr A{8\v
-
),
dB 0')
dr
A
What
A
is
the sign of
liquid
thereby undergoes a
(b)
wher
lW
m
there been a transfer of heat?
of a large blackened copper hemisphere.
ture
56.697
PROBLEMS
(4-25)
4-2
Nonequilibrium method.
0L)
Ir .
Stefan-Boltzmann constant:
1
=
by radiation
-
be employed
-
called the Stefano-
we have
at equilibrium
where r is the radius of the sphere. The best measurements, to date, have yielded the value
Boltzmann constant. Referring to Eq. (4-23),
6 at
£i
whence
(4-24)
as the Stefan-Boltzmann law,
103
supplied at a constant rate
achieves an equilibrium temperature
energy
The
Law
A hollow blackened copper sphere is provided Equilibrium method. with an electric heater and a thermocouple and is suspended inside a vessel whose walls are maintained at a constant temperature Bw-
two tem-
peratures in question.
4-16
the First
4-4
is
AL?
irregularly stirred in a well-insulated container
rise in
temperature.
If
we regard
and
the liquid as the system:
Has heat been transferred? Has work been done?
What is the sign of Ac/? The amount of water in
a lake
may
be increased by action of
underground springs, by inflow from a river, and by by various outflows and by evaporation.
rain. It
may be decreased
Heat and the
Heat and Thermodynamics
104
correct to ask:
(a)
Is
(b)
Would
lake
is
it
due
it
How much
rain
be preferable or sensible
is
there in the lake?
to ask:
How much water
in the
to rain?
What
analogous to "rain in the lake"? and covered with asbestos is divided into two parts by a partition. One part contains a gas, and the other is evacuated. If the partition is suddenly broken, show that the initial and final internal (c)
4-5
A
concept
is
with
vessel
rigid walls
energies of the gas are equal. 4-6 A gas is enclosed within a cylinder-piston combination.
Embedded whose con-
in the gas is a junction of two dissimilar metals (a thermojunction) necting wires pass through the walls of the cylinder and lead to a reversing switch and an electric generator rotated by means of a descending weight.
As the weight descends, the current generated
may
be caused to
exist in the
thermojunction in either direction. Owing to the Peltier effect, the thermojunction undergoes a rise of temperature when the current is in one direction and a drop when in the opposite direction. The entire system is adiabatically shielded so that
all
work
interactions are adiabatic.
Assume
that
and
U refer to
the entire system, composed of gas plus thermojunction. Draw a schematic diagram of the apparatus. (a) (b)
in the (c)
What
What
Keeping the piston
What
If the piston
is
.1
at the
same
time, electrical energy
air.
dissipated in the resistor at such a rate
bomb, the
Show
capillary,
that, after as
and the water
much
has leaked out during time
r,
the change of internal energy
how
stationary,
could one produce a
At/
=
Sir
rise of
-
is
the sign of At/ ?
is
kept stationary,
1
is
is
gas as possible is
the sign of AT/?
is it
possible to
produce a drop
mole of water
is
of heat arc absorbed.
What
-P
molar volume of the gas and i is the current
vo is the
across the resistor, cell of
clcctrolyzed into hydrogen
(nv
- Vn
),
in
where
and
oxygen, 2 faradays of electricity are transferred through a seat of emf (1 faraday = 96,500 C). The energy change of the system is +286,500
and 50,000
.
kept equal to that of the outside
temperature? If so, what is the sign of At/? (e) How could one produce an adiabatic, isothermal process? 4-7 When an electric current is maintained in an electrolytic acidulated water and
105
Lavv
4-9 An exhausted chamber with nonconducting walls is connected through a valve to the atmosphere, where the pressure is P The valve is opened, and air flows into the chamber until the pressure within the chamber is Pa. Prove that u + /Vo — tij, where «o arid vo arc the molar energy and molar volume of the air at the temperature and pressure of the atmosphere and u_r is the molar energy of the air in the chamber. (Hint: Connect to the chamber a cylinder equipped with a frictionlcss nonlcaking piston. Suppose the cylinder to contain exactly the amount of atmospheric air that will enter the chamber when the. valve is opened. As soon as the first small quantity of air enters the chamber, the pressure in the cylinder is reduced a small amount below atmospheric pressure, and the outside air forces the piston in.) 4-10 A bomb of volume Vb contains n moles of gas at high pressure. Connected to the bomb is a capillary tube through which the gas may slowly leak out into the atmosphere, where the pressure is Pa. Surrounding the bomb and capillary is a water bath, in which is immersed an electrical resistor. The gas is allowed to leak slowly through the capillary into the atmosphere while, that the temperature of the gas, the
the result of allowing the piston to go out, with no current
thermojunction?
temperature? (d)
is
First
A
4-11
Pj.
It
is
atmospheric pressure,
almost empty gasholder in which the pressure
J,
value P' , very nearly atmospheric.
£? walls and covered with
asbestos is 4-8 A cylindrical tube with rigid divided into two parts by a rigid insulating wall with a small hole in it. frictionlcss insulating piston is held against the perforated partition, thus preventing the gas that is on the other side from seeping through the hole.
is
the
pd
n,
moles of
connected through a valve with a large
£
is
£
in the resistor.
metal chamber contains
thick-walled insulated
helium at high pressure
at
The
valve
is is
maintained at a constant
opened
slightly,
and the
helium flows slowly and adiabatically into the gasholder until the pressure on the two sides of the valve is equalized. Prove that
A
is maintained at a pressure P,- by another frictionlcss insulating Imagine both pistons to move simultaneously in such a way that, as the gas streams through the hole, the pressure remains at the constant value Pi on one side of the dividing wall and at a constant lower value Pf on the other side, until all the gas is forced through the hole. Prove that
h' n.
h'
-
U;
u,
The gas piston.
where
= number of moles of helium left in the chamber, m = initial molar energy of helium in the chamber,
nf
;//
h'
Ui
+ PiV^
Uf
+ Pffy.
= =
final
molar energy of helium in the chamber, and (where u' = molar energy of helium in the gasholder; molar volume of helium in the gasholder).
w'
+ P'v'
v
=
106
Heat and the
Heat and Thermodynamics
The molar
4-12
heat capacity at constant pressure of a gas varies with the temperature according to the equation
=
cP
a
+
—
bd
4-16
Derive the equations
a, b,
and
are constants.
c
isobaric process in
How much
heat
is
transferred during an
which n moles of gas undergo a temperature
rise
from
6i to 0/?
Paramagnetic
solid
obeying
Curie's equation
The molar
4-13
Law
107
table.
Heat capacity at
variable
constant intensive variable
*-®.
*-(S),—
c" =
-=(I)^f
,
Stretched wire
where
accompanying
Heat capacity at constant extensive
System -5
listed in the
First
\fe).
heat capacity of a metal at low temperature varies
with the temperature according to the equation
One mole
4-17
of a gas obeys the equation of state
(p
where
a,
("),
and
b
How much
arc constants.
where heat per mole
is
v
+ £) (v -
the molar volume,
is
and
u
Regarding the internal energy of a hydrostatic system to be a function of and P, derive the equations:
dV
~/dU
do
dP
J
a, b, c,
R
m (c)
4-15
dO ip
dlP dP
=
co
)
monatomic
of state of a
solid
is
dP.
+ J{V) =
T(u
-
uo),
PV8.
where
v
is
the molar volume,
and Y and
«o are constants.
Prove that
= PVk - (CP - Cv) cvk
Taking
V
to be a function of
P
and
where V, derive the following
k
4-19 (a)
w
m, + p au\ _ Cvk OPjv P dip
dV/P
Cp
is
the isothermal compressibility. This relation,
first
derived by
Griineisen, plays a role in the theory of the solid state.
equations:
(b)
given by
arc constants. Calculate molar heat capacities cv and Cp.
The equation
4-18
Pv
= CP -
is
v
where
p \ao) P + \de)
molar internal energy
0?
4-14
Re,
transferred dur-
ing a process in which the temperature changes from 0.01 to 0.02
(a)
its
=
b)
dV.
In the case of a paramagnetic gas: Derive the equation
$L*
d_ir
P dV
dll
dM
-%
dM.
'
(b)
-
4-20 p.
Derive expressions
and Cp,%?. outward through a cylindrical insulator of surrounding a steam pipe of outside radius r\. The tempera-
Heat flows
outside radius
r«
for Cv,u, Cv.k, Cp,it,
radially
108
Meat and Thermodynamics
ture of the inner surface of the insulator
is 0,, and that of the outer surface is 2 from the center of the pipe is the temperature exactly and 2 ? .
At what
radial distance
halfway between
Two
4-21
0i
thin concentric spherical shells of radius 5
have their annular cavity
respectively,
filled
with charcoal.
and 15
cm
When
energy is supplied at the steady rate of 10.8 to a heater at the center, a temperature difference of 50.0°C is set up between the spheres. Find the thermal con-
W
ductivity of charcoal.
4-22 A wall, maintained at a constant temperature l w is coated with a layer of insulating material of thickness x and of thermal conductivity K. ,
The outside is
of the insulation
is
in
contact with the air at temperature
transferred by conduction through the insulation
through the (a)
tA Heat and by natural convection .
air.
Show
that, in the steady state,
A overall coefficient
U(t w
-
tA ),
of heat transfer,
is
given by
and the
First
109
Law
4-25 A solid cylindrical copper rod 10 cm long has one end maintained at a temperature of 20.00°K. The other end is blackened and exposed to thermal radiation from a body at 300°K., with no energy lost or gained elsewhere. When equilibrium is reached, what is the temperature difference
between the two ends? {Mote: Refer to Fig. 4-9.) 4-26 A cylindrical metal can blackened on the outside, 10 cm high and 5 cm in diameter, contains liquid helium at its normal boiling point of 4.2°K, at which its heat of vaporization is 21 J./g. Completely surrounding the helium can are walls maintained at the temperature of liquid nitrogen
(78°K), and the intervening space
per hour? 4-27
is
How much
evacuated.
helium
lost
is
operating temperature of a tungsten filament in an incan2460°K, and its absorptivity is 0.35. Find the surface area of the filament of a 100-W lamp. 4-28 A copper wire of length 130.2 cm and diameter 0.0326 cm is blackened and placed along the axis of an evacuated glass tube. The wire is connected to a battery, a rheostat, an ammeter, and a voltmeter, and the current is increased until, at the moment the wire is about to melt, the ammeter reads 12.8 A and the voltmeter 20.2 V. Assuming that all the energy descent
Q = -
where U, the
feat
I
The
lamp
is
supplied was radiated and that the radiation of the glass tube
is
negligible,
calculate the melting temperature of copper.
4-29
4 = ^ + 1. U K
The
solar constant
is
the energy falling per unit of time on a unit
area of a surface placed at right angles to a
k
sunbeam just
outside, the earth's
atmosphere. Measurements by Abbot have yielded the value 1.35 (b)
How do
The you determine
t,
the temperature of the outer surface of
the insulation?
Show that the unit of K/
4-23
blackbody, calculate
.
4-24
area of a sphere with a radius of 93,000,000 miles
and the surface area
is
at a temperature
,
is
of ice;
and p
is
face of the ice
K
is
the thermal conductivity of ice;
the density of ice. (Hint: is
variable.
an infinitesimal thickness
Assume dy to
is
assumed constant
/ is
the heat of fusion
The temperature
in time dr.)
m
2 .
is
2.806
X
Assuming the sun
to
,
be a
A small
40 3w Aaa(0 ]v
-
6).
upper surand imagine
If the
=
—Cp
40 3w Aa
1
for
" 9w
of the
the ice to have a thickness y,
form
10 18
body remains at constant pressure, show that the time the temperature of the body to change from 0i to 0» is given by (b)
Pl
the convection coefficient per unit area and
X
.
surface temperature.
Q =
r_
while the ice forms;
6.07
2
body with temperature and absorptivity a is placed in a large evacuated cavity whose interior walls arc at a temperature 0w- When is small, show that the rate of heat transfer by radiation is dw — (a)
thence into the air by natural convection, prove that
where h
its
is
2
m
4-30
A while the water is at its freezing point 6,- (with 0.., < 0,). After a time t has elapsed, ice of thickness y has formed. Assuming that the heat which is liberated when the water freezes flows up through the ice by conduction and
2K
of the sun
kW/m 10 23
(c)
Two
one of copper, the threads within a large hole in a
small blackened spheres of identical
other of aluminum, are suspended by
silk
size,
HO
Ilcat
and Thermodynamics
block of melting
ice. It is found that it takes 10 min for the temperature of aluminum to drop from 3 to 1°C, and 14.2 min for the copper to undergo the same temperature change. What is the ratio of specific heats of aluminum and copper? (Densities of Al and Cu are 2.7 and 8.9 g cm 3
the
5. IDEAL GASES
,
respectively.)
A blackened solid copper sphere with a radius of 2 cm is placed an evacuated enclosure whose walls are kept at 100°C. In what time does its temperature change from 103 to 102°C? (c P = 3.81 J g deg- p = 4-31
in
•
8.93
g/cm 3 .)
Equation of State of a Gas
5-1
It
was emphasized
in
Chap.
1
that a gas
is
the best-behaved thcrmomctric
substance because of the fact that the ratio of the pressure
temperature to the pressure
and
P-.i
gas.
The
P3
of the
same gas
P
any both P
of a gas at
at the triple point, as
approach zero, approaches a value independent of the nature of the
by 273.16 deg, was defined temperature 6 of the system at whose temperature the gas
limiting value of this ratio, multiplied
to be the ideal-gas
exerts the pressure P.
investigating the
way
for this regular behavior may be found by which the product PV of a gas depends on the denconstant, on the reciprocal of the volume.
The reason in
mass is Suppose that the pressure P and the volume V of n moles of gas held at any constant temperature are measured over a wide range of values of the pressure, and the product Pv, where v = V/n, is plotted as a function of 1 /v. Experiments of this sort were first performed by Amagat in France in 1870 and later by Holborn and Otto in Berlin and by Kameilingh-Onnes and sity or, if the
Keesom
Nowadays, cuch measurements are made at many bureaus and universities. The relation between Pv and l/v may be expressed by means of a power series (or virial expansion) of the form in Leiden.
of standards
Pv
where A, B, C,
B
etc.,
= -'(
are called
71
71*
71
virial coefficients
-> (A being the
(5-1)
first virial coeffi-
depend on the temperature and on the nature of the gas. In the pressure range from to about 40 atm, the relation between Pv and If is practically linear, so that only the first two terms in the expancient,
the second, etc.) and
sion are significant. In general, the greater the pressure range, the larger the
number
of terms in the virial expansion.
112
Heat and Thermodynamics
The viral coefficients play an important role not only in practical thermodynamics but also in theoretical physics, where they are related to molecular properties. Except at very low temperatures, the virial coefficients arc quite small, as shown in Table 5-1, where the virial coefficients are given for nitrogen in the temperature range 80 to 273°K and in the pressure range from to 200 atm. These, data, obtained by Friedman, White, and Johnston of the Ohio State Cryogenic Laboratory, are listed in the "American Institute of Physics
Ideal Gases
5-1
Handbook" (McGraw-Hill Book Company, Xew York, "AIF Handbook."
113
o E -s
E ?
30.5
Lim (PlOsteam
= 30.621
liter-atm/mole
p~
1963)
hereafter referred to simply as
Table
5-1
Virial Coefficients for Nitrogen
T,
B,
"K
cm a/mole
80 90 100 110 120 150 200 273
-250.80 -200.50 -162.10 -131.80 -114.62
-
-
c,
D,
10 2 cm°/mole 2 10 4
10 5
cm ,2/mole
P, 10'
l
cm
1
''/mole''
-2000 -1000
210 135
-
85 65
48
71.16 34.33 9.50
R,
cm 3/mole 3
600 200 27
22
13
12
14
-123 -118
16
—
8.2
75
41
36
-16
The remarkable property of gases that makes them so valuable in thermometry is displayed in Fig. 5-1, where the product Pv is plotted against P for four different gases, all at the
graph,
all
temperature of boiling water in the top at the triple point of water in the next lower graph, and all at the
temperature of solid C() 2 in the lowest. In each case, it is seen that as the pressure approaches zero the product Pv approaches the same value for all gases at the same temperature. As the pressure of a constant mass of gas approaches zero, the volume approaches infinity, and according to the virial
expansion [Eq. Thus,
(5-1)], the
product Pv approaches the
first virial
Fig. 5-1
Fundamental property of gases is that lim (Pv) 9
gas and depends only on
so that
coefficient A.
lira
(Pv)
=A =
i'-.o
The
ideal-gas temperature
is
unct ' on °f temperature only, (independent of gas.
-«*--Sfe -»««•£$?. lim (Pv)
=
-5* 3
(const.
273.16°
e
The
bracketed term
is
called the universal gas constant
and
is
denoted by R.
Thus,
defined as
273.1 6° lim
lim (Pv)j
f5-2l
R =
=
of the nature of the
a.
and |'
is independent
V and
lim (Pv ),
(5-3)
273.16°"
n),
The accepted
value (1968) of lim (Pv),
is
22.4144
liter
atm/g mole (where
114
1
Heat and Thermodynamics
liter is exactly
equal to 10 3
R =
cm
3
Hence,
).
0.08206
liter
•
its
value
I
"
Since lim (Pv)
atm
8.3143 J/molc
Finally, substituting for v
Ideal Gases
5-2
n,
'mole
•
we may
=
R6, the virial expansion
may
therefore be written
cleg
ne
deg.
write the equation of stati
nRO.
v
v-
and is denoted by Z. It is and pressure in several reports and circulars published by the National Bureau of Standards. A condensed set of tables of Z values for the important gases is given in the "AIP Handbook."
The
ratio
Pv/Rd
tabulated for
of a gas in the limit of low pressures in the form
lim (PV)
= A =
115
The
(5-4)
is
many
called the compressibility factor
values of temperature
general behavior of the compressibility factor
—
graphically by plotting v{Pv/Rd
•0HIn the region of small values of
1
!v,
may
be displayed
1) against 1/v, since
V
the left-hand
member
should be linear
in
l/o, with B equal to the j> intercept and G equal to the slope. A few typical curves for nitrogen are shown in Fig. 5-2, where it may be seen that the
graphs are linear in the low-density region and cut the y axis at values in agreement with the B values in Table 5-1.
5-2
a. la*
Internal Energy of a Gas
Imagine a thermally insulated vessel with rigid walls, divided into two compartments by a partition. Suppose" that there is a gas in one compartment and that the other is empty. If the partition is removed, the gas will undergo what is known as a free expansion in which no work is done and no heat is transferred. From the first law, since both Q and arc zero, it follows that
-40
W
the internal energy remains unchanged during a free expansion.
The
question of
whether or not the temperature of a gas changes during a free expansion and, if it docs, of the magnitude of the temperature change has engaged the attention of physicists for over a hundred years. Starting with Joule in 1843, many attempts have been made to measure either the quantity (dd/dV)u, which may be called the Joule coefficient, or related quantities that are all a measure, in
one way or another, of the
called, the Joule
-100 0.005
0.010
Molar density Fig.
5-2
P, V, ,
56
and
d.
Considering
U as
is
(U.S.
National
—
or,
as
it is
often
a function of any two of the coordinates and V, we have
a function of 8
moles/cm 3
Graphical representation of the virial equation for nitrogen.
Bureau of Standards, Circular
In general, the energy of a gas
0.025
—
effect of a free expansion
effect.
'«ws?
dV.
116 If
5-2
Heat and Thermodynamics
no temperature change
then
it
(dd
=
0) takes place in a free
expansion (dU
=
0),
It
was
therefore concluded that
no temperature change took
liquid in the
manometer
and
of
P,
V docs
not depend on V. Considering
no temperature change,
then
it
U to
be a function
we have
*-(5W«). If
(dd
=
0) takes place in
dp.
a
free
expansion (dU
=
0),
follows that
(dU
=
0;
other words, V does not depend on P. It is apparent then that, if no temperature change takes place in a free expansion, V is independent of V or, in
and therefore U is a junction of 6 only. methods of studying the Joule effect have been employed. In the original method of Joule, two vessels connected by a short tube and stopcock were immersed in a water bath. One vessel contained air at high pressure, and the other was evacuated. The temperature of the water was measured before and after the expansion, the idea being to infer the temperature change of the gas from the temperature change of the water. Since the heat capacity of the vessels and the water was approximately one thousand times as large as the heat capacity of the air, Joule was unable to detect any temperature change of the water, although, in the light of our present knowledge, the air must have undergone a temperature change of several degrees. The second method of studying the Joule coefficient consists of an attempt to measure the temperature of the gas almost immediately after the free and
of P,
Two
during
this time, there
a transfer of heat between the gas and the walls of the vessel, which vitiates the results. Further complications arise from the conduction of heat through metal valve connections due to a temperature difference created by the rapidly streaming gas. In the experiments of Kcyes and Sears in 1924, the gas was not used as its own thermometer; instead, a platinum resistance thermometer was employed to measure the temperature immediately after expansion. Temperature changes of approximately the right order were measured, but only a few rough measurements were made. A direct measurement of the temperature change associated with a free expansion is so difficult that it seems necessary to give up the idea of a precise is
measurement of the Joule
coefficient.
Modern methods
of attacking the prob-
lem of the internal energy of a gas involve the measurement of the quantity (dufdP)t by having the gas undergo an isothermal expansion in which heat is transferred and work is done. The most extensive series of measurements of this kind was performed by Rossini and Frandsen in 1932 at the National Bureau of Standards with a method elaborated by Washburn. The apparatus is shown in Fig. 5-3. A bomb B contains n moles of gas at a pressure P and
Outlet
expansion, before the gas has had a chance to exchange heat with its surroundings, by using the gas itself as its own thermometer. In the experiments of Hirn in 1865, a vessel was divided into two equal compartments by a thin partition, which could be broken with the aid of a metal ball. Originally,
both compartments contained air at atmospheric pressure. The air from one compartment was then pumped into the other compartment, and the temperature of the compressed gas was allowed to come to its original value. The partition was then broken, and soon afterward the pressure was measured with the aid of a U-tubc manometer containing a light liquid and found to be the value that existed
when
Cazin
tube. If sufficient time elapses in order to allow the
oscillations of the inanomctric liquid to subside, then,
other words,
place.
117
repeated this experiment in 1870 with similar apparatus. The results of this method are doubtful because of the oscillations of the
follows that
$).-* or, in
Ideal Gases
the gas originally occupied the whole vessel.
Fig. 5-3
Apparatus of Rossini and Frandsen for measuring (du/dP)t of a gas.
118
Heat and Thermodynamics
5-3
communicates with the atmosphere through a long coil wrapped around the bomb. The whole apparatus is immersed in a water bath whose temperature can be maintained constant at exactly the same value as that of the surround-
The experiment
is
performed
same
electrical
coil
energy supplied to the water
W = P (nv
a
the water.
The
therefore the heat
Q absorbed by evidently
is
- rB),
P is atmospheric pressure, v a is the molar volume at atmospheric temperature and pressure, and Vs is the volume of the bomb. If u(P.d) is the molar energy at pressure P and temperature and if u(PB ,0)
where
is
the molar energy at atmospheric pressure and the same temperature, then,
from the
first
law,
-
u(P,B)
u(P
,6)
t-2
=
follows that the slope of the resulting curve at
is
clone by the gas
is
The work
it
into the air.
electric heating coil
the gas during the expansion.
any value of P is equal Within the pressure range of 1 to 40 atm, it is seen that (du, dP)s independent of the pressure, depending only on the temperature. Thus,
stant,
opened At the and the water is
the stopcock
and out
bomb, the coils, immersed in
time, the temperature of the gas, the
maintained constant by an
When
as follows:
through the long
slightly, the gas flows slowly
119
to (du/dP)e. is
ing atmosphere.
Ideal Gases
,
u
and
=
f(d)P
+ F(d),
where F(0) is another function of the temperature only. Rossini and Frandsen's experiments with air, oxygen, and mixtures of oxygen and carbon dioxide lead to the conclusion that the internal energy of a gas is a function of both temperature and pressure. They found no pressure or temperature range in which the quantity (du/dP)e was equal to zero.
Washburn's method has somewhat the same disadvantage as Joule's original method, in that the heat capacity of the gas
is
much
smaller than
To
keep the temperature of the gas constant within reasonable limits, the temperature of the water must be kept constant to within less than a thousandth of a degree. In Rossini and Fraudthat of the calorimeter
and water bath.
measurements, the final precision was estimated to be 2\ percent. Another determination of the average value of (dufdP)e of air in a pressure range of over 50 atm was made by Baker in 1938 with a somewhat different
sen's
provided that corrections have been made to take account of the energy changes due to the contraction of the walls of the bomb. In this way, the energy change was measured for various values of the initial pressure and
was plotted against the
shown
pressure, as
-400
in Fig. 5-4.
Since u(Pn ,6)
Air at 2B"(J~
<»
o
w„ =
£
X —
-300
slope
-6. 08
J,
mole atm
«of
-200
5>^
D^
1
method, capable of an accuracy of 0.1 percent. In Baker's experiment, the expanded from a thin spherical metal bomb into the space between it and a thin outer concentric spherical shell. The temperature of the outer shell was measured at small time intervals, starting from the moment after the expansion until the temperature returned to its original value; during this time, heat flowed into the gas at a rate that was a known function of the temperature difference between the outer shell and its surroundings, which were held at a constant temperature. By making the two shells very thin, the ratio of the heat capacity of the shells to that of the gas was very much smaller than in any previous work. The heat transferred to the gas could be calculated with great accuracy, and many corrections were applied with extreme care. The few measurements that have been made with this apparatus up to the present time have substantiated the results of Rossini and Frandsen. air
is
con-
K"
-100 D -^**^ J
3
5-3
Ideal Gas
1
0\
5
10
15
20
25
30
35
Pressure, atm Fig. 5-4
Dependence of internal energy of a gas on pressure.
40
45
We have seen that,
in the case of
a real gas, only
in the limit as the pressure
approaches zero docs the equation of state assume the simple form Furthermore, the internal energy of a real gas
is
PV =
nRB.
a function of pressure as well
120
5-3
Heat and Thermodvna
as of temperature. It
is
convenient at
this point to define
an
ideal
gas whose
any existing gas, arc approximately those of a real gas at low pressures. By definition, an ideal gas satisfies properties, while not corresponding to those of
partial derivative with respect to 6
=
may
dQ = Cv
all
dU\ /dP
dV
dPlAdV
PV =
+
PdV.
(5-8)
nRB,
P dV + V dP = nR is
d8.
not zero, whereas Substituting the above in Eq. (5-8),
dQ =
Finally, since
dO
and, for an infinitesimal quasi-static process,
and since (dP/dV)e = — nRO/V* = —P/V, and therefore (dU/dP)e is zero, it follows that for an ideal gas
(ideal gas).
3rX—
'
dd
equilibrium states are represented by the ideal-gas equation
be written in other ways. Thus,
dU
dU
(5-5)
Now, that (dU/dP)e
the same as the total derivative. Conse-
cv = and
The requirement
(5-6)
(CV
we
+ nR)
get
dd
- V dP,
and, dividing by dd, d(2
_ r
,„o
v
dP
both (dU/dP)e and (dll/dV), are zero.
At constant pressure, the left-hand member becomes
U=
f(0) only
(ideal gas).
Whether an
actual gas
may be
may
may
its
only a small error
For an law
Even
liquid, the ideal-gas
if
An
actual gas at pressures
be treated as an ideal gas without introducing an
error greater than a few percent.
equilibrium with
the vapor pressure
in the case of a saturated
equation of state is
may
vapor
in
be used with
We
have the
+ nR.
(5-9)
at constant
a hydrostatic system, the
an ideal gas is remaining constant and equal to nR. Since
U
is
Cy =
In the special case of an ideal gas,
is
and
given by
One more
is
it
Cy
=
Cr
— Cv + nR =
-et do
=
useful equation
dQ =
36
U
a function of 6 only,
first
+ P dV,
volume
capacity at constant pressure of
follows that
low.
infinitesimal quasi-static process of
dQ = dU
result, therefore, that the heat
always larger than that at constant volume, the difference
is
and the heat capacity
whence
be treated as an ideal gas depends upon the
tolerated in a given calculation.
below about 2 atm
C/>;
(5-7)
Cr = Cv error that
121
quently,
the equations
(ideal gas).
is
Ideal Gases
a function of
we 9 only; therefore the
get
a function of 6 only,
a function of
6 only.
can be obtained. Since (C v
+ nR)
dQ = Cpdd
d9
- V dP,
- V dP.
(5-10)
122
heat capacities of gases arc measured by the electrical method.
measure Cv, the gas wire wound around
contained in a thin-walled
is
volume, the gas
where
of time,
from
is
and
the initial (inlet)
known equivalent heat per
_£
3
2
unit
final (outlet) temperatures, the rate of
supply of heat, and the rate of flow of gas, the value of Cp results of
o=
allowed to flow at constant pressure through
receives electrically a
it
—7*
4
it. By maintaining an electric amount of heat is supplied to the gas, and the specific heat at constant volume is obtained by measuring the temperature rise of the gas. The same method is used to measure Cp except that, instead of confining the gas
a calorimeter,
is
calculated.
The
such measurements on gases at low pressures (approximately ideal 10
gases)
can be stated
a simple
in
manner
in
25
250 500 7501000 2500
50 75 100
All gases:
cv
is
a function of
(h)
cp
is
a function of
(c)
cp
—
const.
=
only,
and >cy.
R.
y = cp/cv — a function
equal to (b)
cp
y
of 6 only,
§i2.
(b)
cp
namely.
H
:! ,
and increases as the temperature
5/?, is
y
is
very nearly
is
very nearly
D->, ()», N->,
is
Polyatomic gases and gases that are chemically
is
The behavior
of hydrogen
is
quite exceptional, as
shown
in Fig. 5-5.
At
gas.
For
all
may
other diatomic gases, Cp/R
always be written
C
J1= *7 R
/(«),
where f(d)
is
often one or
more functions
of the type
raised.
is
\e)
raised.
active,
(«<"«
- \y
J,
Exact equations of the above type are
raised.
such as CO»,
NH
3,
and Br,: and cp/cv vary with the temperature, the variation being
CII.,, CI.,
cp, cv,
themselves but the ratios cy/'R and cp/R.
NO, and CO:
constant at ordinary temperatures, being equal to about
and decreases as the temperature 4
also true for
constant at ordinary temperatures, being equal to about
IR, and increases as the temperature (c)
is
constant at ordinary temperatures, being equal to about
is
is
shall specify not the heat capacities
-J
So-called permanent diatomic gases,
Cv
wc
to iR.
equal to
(a)
a very interesting consequence of theory that this
is
In the remainder of this book
r
constant over a wide temperature range and
is
of a gas. It solids.
very low temperatures, Cp/R drops to a value of -§, appropriate to a monatomic
constant over a wide temperature range and
is
equal (c)
3
=
K
logarithmic scale.
and >1. Monatomic gases, such as He, Ne, and A, and most metallic vapors, such as the vapors of Na, Cd, and Hg: (a) cv is constant over a wide temperature range and is very nearly (d)
2
cv
only.
*
Experimental values of cp/Rfor hydrogen as a function of temperature, plotted on a
Fig. 5-5
(a)
5000
terms of molar heat capacities. Temperature,
1
123
To
with a heating current in the wire, an
steel flask
equivalent
to a constant
Ideal Gases
5
Experimental Determination of Heat Capacities
5-4
The
5-4
Heat and Thermodynamics
different for each gas.
handle and arc not
suit-
approximate empiric equations arc used. Empiric equations for cp/R of some of the most important gases, compiled by H. M. Spencer, are given in Table 5-2, within the temperature range of 300 to 1500°K. In recent years, other and
These experimental results indicate that the universal gas constant R (8.31 J/mole deg) is a natural unit with which to express the molar heat capacity
difficult to
able for the practical calculations of the laboratory scientist; consequently,
more accurate empiric equations have been suggested, some of up to 3000°K.. The simple quadratic equations of Table 5-2,
these being valid
however, will serve the purposes of this book.
124
Heat and Thermodynamics
Table 5-2
=
cP/R
a
Gas
H
2
Oo Cls Br«
Ns
CO HC1 HBr
co
2
H2
NH H 2S
3
CH,
+
This equation cannot be integrated until we know something about the 7. We have seen that for monatomic gases 7 is constant, whereas for diatomic and polyatomic gases it may vary with the temperature. It requires, however, a very large change of temperature to produce an appreciable change in 7. For example, in the case of carbon monoxide, a temperature
cp/R of Important Gases b6
+
c0 a (from 300 to 1500°K)
behavior of
b,
a
10- J deg-'
3.495 3.068 3.813 4.240 3.247 3.192 3.389 3.311 3.206 3.634 3.116 3.214 1.702
-0.101
0.243
638 1.220 490 0.712 0.924 0.218 0.481 5.082 1.195 3.970 2.871 9.083
-0.512 -0.486 -0.179 -0.041 -0.141
1
10-° deg- 2
.
.
2000°C produces a decrease in 7 from 1 .4 to 1 .3. Most adiabatic we deal with do not involve such a large temperature change. We are therefore entitled, in an adiabatic process that involves only a moderate temperature change, to neglect the small accompanying change in 7. Regarding 7, therefore, as constant and integrating, we obtain from
rise
to
processes that
0.186 0.079
In
P = —7 In V +
In const.,
-1.714
PV
0.135
-0.366 -0.608 -2.164
The above equation
holds at
all
=
(5-11)
const.
equilibrium states through which an ideal
gas passes during a quasi-static adiabatic process. It
Quasi-static Adiabatic Process
When an ideal gas undergoes a
volume, and temperature change in a manner that
is described by a relation between P and V, 6 and V, or P and d. In order to derive the relation between P and V, we start with Eqs. (5-8) and (5-10) of Art. 5-4. Thus,
dd
&Q = Cpdd
and
6Q =
Since, in an adiabatic process,
-
on a
Eq. (5-11).
PV
diagram by assigning different values The slope of any adiabatic curve is
V
\dVjs
V
0,
where the subscript
3
is
used to denote an adiabatic process.
Quasi-static isothermal processes are represented
d6,
PV =
nR6. Since
P
and denoting the
_CpdV Cv V
ratio of the heat capacities
'
it
follows that an adiabatic curve has a steeper negative slope than does
p =
dV ~7 T
an
isothermal curve at the same point.
by the symbol
7,
we have
The
isothermal curves and adiabatic curves of an ideal gas
in a revealing
dP
v
vWh
P
by a family of equilateral
hyperbolas obtained by assigning different values to 9 in the equation
by the second,
dP =
be
—7 P
dP.
PdV = -CV dd. first
may
to the constant in
— 7 const. K-7-1
'dP\ Dividing the
important to under-
is
family of curves representing quasi-static adiabatic processes
plotted
+ PdV,
V dP = C P and
A
quasi-static adiabatic process, the pressure,
6Q = C v
is
an adiabatic process but is not quasi-static. It entirely fallacious to attempt to apply Eq. (5-11) to the states is therefore ideal gas during a free expansion. traversed by an stand that a free expansion
5-5
125
Ideal Gases
5-5
way on a PV6
surface. If P, V,
lar axes, the resulting surface
is
shown
and
in Fig. 5-6,
the adiabatic curves cut across the isotherms.
may
be shown
are plotted along rectangu-
where
it
may
be seen that
5-6
temperature comes back to to the final value
Pf
thus causing the pressure to rise
initial value,
its
Since the
.
and
initial
127
Ideal Gases
Heat and Thermodynamics
126
final
temperatures are equal,
we
have
=
PiVi
Eliminating
I',-
and
from the two equations, we get
Vj
i±=
(El \P,
Po
Taking the natural logarithm of both
sides
and solving
for y,
we
obtain
In (/>,//>„)
y =
In (Pi/Pf)
Surface for an ideal gas. {Isotherms are represented by dashed curves, and adiab alios
Fig. 5-6
P/Vf.
by full curves.)
The
pressures are usually measured with an open L'-tube
manometer con-
taining a convenient light liquid. Denoting the height of the barometric
column of
Clement and Desormes Method of Measuring y
5-6
The earliest and
simplest
method of measuring the y
of
an
ideal gas
is
this liquid
that of
Clement and Desormes. The gas is contained in a vessel at room temperature and at a pressure P, slightly above atmospheric pressure (see Fig. 5-7). SupBy rapidly openpose that the volume of any small constant mass of the gas is ing and closing a stopcock (flipping a poppet valve), the small amount of gas under consideration is caused to expand adiabatically until its volume is Vt and its pressure is that of the atmosphere P a with the temperature dropping slightly below room temperature. Assuming this adiabatic expansion to be I'',-.
and
where
//;
and
/(/
by
we may
ho,
=
h
(l
+ hiffa),
+ hs =
/i
(l
+ h//h
+
Pi
- k
P«
=
ho,
P/
=
ho
are very small
write
hi
compared with
ho-
),
Upon
substituting,
,
approximately quasi-static,
we may
write
pjn = p
v
7
is
In (1
4- In ho)
allowed to stand for a few minutes at constant volume until
it is
easy to show that,
if
x
is
Isochoric addition
Room temp V
Lower than
room temp
(Temp comes back to initial
Any small constant mass Fig. 5-7
of heat
Adiabatic expansion
(Temp drops) of
-
In
(1
+
h!t 6
)
compared with
small
Since the ratios Pf.V,
Room temp
/;,-
%
and
+ x)
=
are small
ftf/fta
finally
Schematic representation of the method of Clement and Desormes.
7 =
hi Ih
-
x.
compared with
value)
gas
unity,
its
In (1
Pt.Vi
+ h/ho)
V}.
Now The gas
In (1
=
hi
unity,
we
obtain
128 5-7
5-7
Heat and Thermodynamics
Riichhardt's
Method
of
Ideal Gases
129
Similarly, a small positive displacement causes a decrease in pressure which pressure P and which therefore is very small compared with the equilibrium
Measuring y
ingenious method of measuring y developed by Riichhardt in 1929, although no more accurate than the preceding method, makes use of ele-
An
mentary mechanics. The gas is contained in a large jar of volume V. Fitted to the jar (sec Fig. 5-8) is a glass tube with an accurate bore of cross-sectional area A, into which a metal ball of mass m fits snugly like a piston. Since the gas is slightly compressed by the steel ball in its equilibrium position, its pressure P is slightly larger than atmospheric pressure P Thus, neglecting .
friction,
can be denoted by dP, where dP is a negative quantity. The resultant force J7 acting on the ball is equal to A dP if wc neglect friction, or
*-§ Notice that, that
S P - p _l P-P + m-^.
is,
Now,
n
!?
is
dP
when
y is positive, a restoring force.
negative and therefore
is
J7
as the ball oscillates fairly rapidly, the variations of
is
P
negative;
and V are which
adiabatic. Since the variations arc also quite small, the states through
downward displacement and then let go, it will oscillate with a period r. Friction will cause the ball to come to rest eventually. Let the displacement of the ball from its equilibrium position at any moment be denoted by jy, where y is positive when the ball is above the equilibrium If the ball
is
given a slight
and negative below. A small positive displacement causes an increase in volume which is very small compared with the equilibrium volume V and which therefore can be denoted by dV, where
the gas passes can be considered to be approximately states of equilibrium.
Wc may therefore assume that the changes of P and V represent an
PV
position
i
Substituting for
dV = yd.
=
PyV~ dV+
and
dV and
dP,
wc
approxi-
and we may write
mately quasi-static adiabatic process,
const.
V*
dP=
0.
get
*--*£» y
positive
Equilibrium position
—
The above equation
expresses the fact that the restoring force
portional to the displacement
and
is
in the opposite direction.
is
the condition for simple harmonic motion, for which the period r
Equilibrium
volume
= V
Equilibrium
pressure
P=
=P
Po
//////////////////W Riichhardt'' s apparatus {or
The mass
measurement of y.
2ji
=
2*
is
precisely
is
-#h V PA* \jy i
4T 2mV
7 = A 2Pt"-
and
'
of the ball, the cross-sectional area of the tube, the pressure, and
the volume arc
V Fig. 5-S
Consequently,
=
directly pro-
This
to obtain y.
all
The
known beforehand, and only
the period has to be measured
values obtained by Riichhardt for air and for
CO4 were
good agreement with those obtained from calorimctric measurements.
in
130
I
feat
5-8
and Thermodynamics
Modifications of Riichhardt's
5-8
consists of holding the ball in the position
exactly atmospheric and then letting
it
131
Method
A variation of Riichhardt's method was suggested method
Ideal Gases
by Rinkel
This
in 1929.
where the gas pressure
drop, measuring the distance
L
is
S
2
that
go up again. Measuring y from this initial position, the gravitational potential energy liberated is mgL, and the work done in compressing the gas is ihc ball falls before starting to
yPoAV*
5
yPoA*L*
whence
2K
or
=
mgL,
Fig. 5-9
_ 2mgV
the advantage over
Riichhardt's that L may be meas-
an error involving the second power. Both Riichhardt's and Rinkel's methods involve errors due
(3)
that
volume changes are
second assumption
is
oscillating ball in Riichhardt's
any desired frequency by is mainexternal coils in which an the position, and friction between horizontal tained. The cylinder is kept in a weight of the piston by balancing the piston and the cylinder is reduced by
ideal, (2) that there
strictly adiabatic.
is
to three simpli-
no
friction,
and
Rinkel estimated that the
responsible for the largest error,
shown
in Fig. 5-10. It is set in
vibration at
alternating current of suitable frequency
the attraction of an electromagnet.
error in r produces
is
mass for the
A steel piston at the center of a cylindrical tube divides the gas into two equal
ured with greater accuracy than r. Moreover, an error in the measurement of /, produces an error in 7 determined by only the first power of L, whereas an
fying assumptions: (1) that the gas
the period against the
a
PoA*L
parts, as
method has
gm
experiment. (H. C. Jensen, 7963.)
7
Rinkel's
The square of
20
15
10
Mass,
amounting
to
The amplitude
of vibration of the piston
equipped with a micrometer eyepiece, at a
measured with a microscope
is
number
of values of the frequency
of the impressed alternating current, and the resonance curve
is
plotted.
about
3 percent.
The large frictional damping in the usual Riichhardt method may be avoided by using a glass tube with a slight taper, with the wider diameter at the top. A slow flow of gas maintains a steady oscillation of a ball about an equilibrium position. The flow is controlled by a throttle valve between the
-Driving coils
experimental apparatus and a high-pressure gas tank. With this method,
Hafner and Duthic found 7 to be 1.664 for argon, 1.406 for nitrogen, and 1.313 for carbon dioxide. If a small rod is attached to the underside of the oscillating ball, the total mass of the oscillating system may be varied by attaching metal washers to the rod, and the variation of period with mass may be measured. The results of H. C. Jensen are shown in Fig. 5-9, where it is seen that the line passes almost through the origin, indicating that one
^SiP
Piston
has here the equivalent of a nearly massless spring.
7,
modification of Riichhardt's experiment in which accurate account is taken of the real equation of state of the gas, the friction present, and the in 1940.
strict
adiabatic conditions
was achieved by Clark and Katz
The method was improved by Katz, Woods, and Lcverton
}/MtM
mwy/m>
A
departure from
W^777/^y/////////////:A
in
1949.
II
Fig. 5-10 pressure.
1
1
II
1
1
—— Driving
Apparatus of Clark and Katz for measuring
7
coils
of a real gas as a function of
132
1
Icat
5-9
and Thermodynamics
Volume
and very elaborate calculations not involving the assumptions made by Ruchhardt and Rinkel, y is calculated. Since friction was reduced to a great extent by the lift magnet, the corrections amounted to only about 1 percent. The authors measured y at various pressures from 1 to 25 atm and expressed the results in the form of empiric equations, as shown in Table 5-3. the resonance frequency
From
'//////
Ideal Gases
133
V
'^^S?^^^^^^^^^^^^^^?^^ w-
Pressure Variation of y (y
Table 5-3
a
°C
23.1
A H2
24.2 24.4 23.0 29.9 25.3 25.1
C0 N sO 2
CH:
+ bP + cP
b,
atm '
atm -2
c,
H
)
1.6669 1.6667 1.4045 1.4006 1.2857 1.2744 1.3029
zero pressure
-0.0002 0.00353 0.00025 0.00221 0.00629 0.00225
P
Pressure
1.667 1.667 1.405 1.401 1.286 1.274 1.303
0.000973 0.000484
-0.00105
3
extrapolated to (ideal gas)
lie
Ni
a
7
Temp.,
Gas
=
2
+
AP
•Velocity of
|=
Compressed column
T?
Wave
7Z&
front with
velocity
Propagation of a compression with constant velocity
w
P
w
by motion of a piston with
constant velocity Wq. Upper diagram at the start; lower diagram after time t.
The
AP
free
body
to the
left.
is
acted on by a force
A{P
+ AP)
to the right
and
a force
Therefore,
Speed of a Longitudinal Wave
5-9
Unbalanced
force on the compressed column
produced at one place in a substance, it will travel with a constant speed w, depending on certain properties of the substance that wc shall now proceed to determine. Directing our attention to a column of material of cross section A, let us suppose that a piston, on the left in Fig. 5-1 1 actuated by an agent exerting a force A(P AP) moves to the right with a constant velocity Wo- This sets up a compression traveling with a constant If a
compression
is
Since the unbalanced force
+
velocity w, so that in time r the compression has traveled a distance
wt while
Consider as a "free body" the compressed column whose length, pressed,
would be wt and whose uncompressed volume V
=
Awr.
if
A AP =
pAwr. At time
-r,
pAwwn,
or
w
is
the
body
is
The "uncompressed compression
AV =
pAwr
=
pAw. Therefore,
entire
free
body" of volume V
Awqt. That
is,
the
Rate of increase of mass of the compressed column compressed column has a velocity
zoo
equal to that of the piston.
AV
Awtfr
Wo
V
Awr
w
AP = pw*^,
which may be written
Therefore,
Rate of increase of momentum of the compressed column
]
J
=
pAwwt).
momentum,
uncom-
If p
density of the normal or uncompressed material, the mass of the Iree
= AAP.
equal to the rate of change of
is
the piston has traveled a distance wer.
The
Pressure
k-
Volume
Fig. 5-1
medium*
velocity of piston;
pAV/VAP
= Awr
has undergone a
1 134
Heat and Thermodynamics
This formula was
AV/V AP
who
obtained by Newton,
as the isothermal compressibility. It
that the expression is
first
5-9
regarded the quantity later by Laplace
was shown
really the adiabatic compressibility.
is
To
see
why
2K_ =
and
2
wpc v
3
this
=
one at the center of a compression and the other at the center of a X/2 apart, where X is the wavelength. Let us suppose that the temperature at the center of the compression exceeds the temperature at the center of the rarefaction by an amouiU Ad. Then the heat conducted a
In the case of a metal,
rarefaction, a distance
sated by the
distance X/2 in the time \ f 2iv (time for the
wave
to travel the distance X 2)
given by
be
still
much (3.3
Heat conducted in the time for the wave to travel a distance X/2
X
-.LB X = = KA—=—X/2
,r
—
.A6
A/1
liv
X
10 -5
QO-'J/s-cm-deg) g/cm 3 )(0.4 J/g deg)
10- 5 cm.
K would X
to raise
'
of
temperature!
mass pAK/2 by Ad\
~
X p
2
°v
U'
is
cm well
fulfilled.
would be compenand the quantity 2K/wpcv would
larger, but this p,
is
wave
We therefore conclude that,
therefore seen to be so
at ordinary frequencies are adiabatic,
Returning now
wave
in
view of the proper-
volume changes which lake place under the influence of a
the
not isothermal.
to the expression for the velocity of a longitudinal
AV/VAP
as the adiabatic compressibility ks,
we have
wave and
finally
w = From
Art. 5-5
we
have, for an ideal gas,
'
the specific heat at constant volume.
The propagation of the wave would be adiabatic if the conducted heat were much too small to raise the temperature of the mass p/lX/2 by the
and
the wavelength of ultraviolet light) that the adiabatic
is
KS is
much
cm. This quantity
10
where K is the thermal conductivity of the medium. The mass of material between the compression and rarefaction is p/lX/2, and the heat necessary to raise the temperature of this mass by the amount A6 is
Heat necessary
be
larger values of w -5
of ordinary matter,
longitudinal
w
much
smaller than 3.3
identifying
amount
X
smaller than the usual value of a wavelength of a compressional
condition ties
where cy
2
(10 4 cm/s)(10- 3
consider a column of material of cross section A, bounded by two
so, let us
planes,
is
X
3.3
X
135
Ideal Gases
and
P
=
V\dPjs
yP'
m
A9, or
where
KAAd „ .. « pA sX2 c v A6 w
7)1 is
the molecular weight and v
is
the molar volume. Hence,
.
This
may
(adiabatic condition).
«x
or (adiabatic condition).
The
usual range of wavelengths of compressional waves is from a few centimeters to a few hundred centimeters. Let us compare these values with wpcy. Taking a gas like air as a typical case,
K= w = P = cv =
=
w
=
yPv
be written
2K_ WpCy
2A
w
2 3
X X -3
1
0.4
10- 4 J/s
-cm
10 4 cm/s,
g/cm 3
J/g
•
,
deg,
-deg,
we
have, roughly,
(5-12)
Equation (5-12) enables us to calculate y from experimental measurements it; and 8. For example, the speed of sound in air at 0°C is about 1100 ft/s, or about 332 m/s. Therefore, using the values
of
w = 9 A' =
332 m/s, 273°K, 8.31 J 'mole
•
deg,
M=iX28 + lX32
=
28.8 g/molc,
wc
137
Ideal Gases
Heat and Thermodynamics
136
J.
get
7
=
raw'1
m
y
267.0
g/mole)( 3.32) 2 (10 4 m 2/s 2) 10~ 3 kg 1 g (8.31 J/mole -~deg)(273 deg)
_ (28.8
= ||| =
266.!
1.40.
s'
266.0
The speed of a sound wave in a gas can be measured roughly by means of Kundt's tube. The gas is admitted to a cylinder tube closed at one end and supplied at the other end with a movable piston capable of being set in vibration parallel to the axis of the tube. In the tube
is
a small
amount
>"
265.5
a>
>
.*?
of
Helium
265.0
powder. For a given frequency, a position of the piston can be found at which standing waves are set up. Under these conditions small heaps of powder pile up at the nodes. The distance between any two adjacent nodes is one-half a wavelength, and the speed of the waves is the product of the fre-
const, tern P-)
light
quency and the wavelength.
Much greater accuracy end of which
is
other a receiver.
frequency
is
is
achieved with an acoustic interferometer, at one
When
the distance between the
two
is
a gas
is
and CO>;
(These are correct to within about to
1.6
Pressure, atm
NBS,
(H
H
Plumb
1965.)
germanium
num-
the lack of
confined within a tube, Cronin measured the tempera-
ture variation of 7 of He, N2, air,
Between the
1.2
Speed of an ultrasonic wave in helium as a function of pressure.
kept constant and the
applying corrections for errors due to viscosity and heat conduction that are
from 14
.8
a source of waves such as a piezoelectric crystal and at the
varied, the various resonances corresponding to diflerent
when
.4
Fig. 5-13
bers of antinodes are noted. Using a frequency range of 50 to 3000 Hz, and
present
264.5
triple point of
1
his results are
shown
in Fig. 5-12.
piston forms the
percent.)
hydrogen and
resistance thermometers. Low-temperature physicists have felt an absolute standardizing instrument in this range for many years. Accordingly, an acoustic interferometer was developed by Plumb at the National Bureau of Standards, and this is now capable of a reproducibility of 0.002 deg at 2°K and of 0.007 deg at 20°K. The resonance tube has a fixed frequency of about 1 MHz, but a variable path length. A movable
the liquid helium region (roughly,
4°K), there are no fixed points with which to calibrate carbon or
upper reflecting surface, and when resonance occurs,
piston, divided
by the number
pressure of the helium 1.7
it is
detected by monitoring a small voltage applied across the crystal, with peaks appearing on a chart whenever resonance occurs. The displacement of the of peaks,
is
one-half the wavelength.
The
varied from 2 to 0.4 atm, and the speed of sound is plotted as a function of pressure as seen in Fig. 5-13. Extrapolation to zero
_o_
mm
n
pressure gives the
e
is
wave speed under
ideal-gas conditions, so that
1.6
„
1.5 1.4
r-
2
,
Air
0''
nU— °-~~ yw
1.3
0-
PROBLEMS 5.
°-C0 2
1.2
100
200
_ w 2 r/2 lu yR
300
400
1
liter-
\R
=
0.0821
atm
=
101
liter
•
atm. 'mole
•
deg
=
8.31
J/mole
•
deg;
J.]
Temperature, °K Fig.
5-12
Dependence of
(D. J. Cronin, 1963.)
7
on temperature from measurements of the speed of sound.
5-1
A
fitted
with a
thermally insulated horizontal cylinder closed at both ends jrktionless heat-conducting piston that divides the
volume
into
is
two
Heat and Thermodynamics
138
The piston is initially clamped so that the volume to the left that to the right is 3Va The left-hand volume contains an ideal and of it is I'o monatomic gas at temperature do and pressure 2/V The right-hand volume contains the same gas at temperature 0o and pressure P a The piston is now unequal
139
Ideal Gases
parts.
the following equations in the
Expand
5-7
form
.
1+- + -1 + -,
.
>
released.
(b)
What What
(c)
Describe the processes that bring the piston to
(a)
arc the final temperature and pressure on each side?
and determine the second
arc the final volumes?
Arc these
rest.
processes quasi-static?
A vertical cylindrical
5-2
end closed by a
tank of length greater than 76
(a)
cm
has
tightly fitting frictionlcss piston of negligible weight.
its
The
air (*)
to spill over the top of the cylinder?
(c)
Mercury
5-3 is
is
at
is
poured into the open end of a J-shaped
closed at the short end, trapping the air in that end.
like
an ideal
The long and due
how much mercury can
gas,
short ends are
to the curvature of the
pressure to be 75
A
5-4
cm
of
1
m and
bottom
up
may
cm
Assuming
in before
it
long, respectively,
1
5
and
in
(0
~
=
b)
(van der Waals).
ffl
c/vd 3 )
v
+B
(
Pv
= R6
+
+
B'P
C'P 2
-
4o(l
-
-
a/v)
+ D'P* +
ffl
P=
(d)
v
-
p—alR0a
(Dieterici)
b
be neglected. Take atmospheric
cm
high and 35
A card
is
cm 2 in
cross section
placed over the top and held
order that the pressure be doubled? Neglect heat conduction
An ideal gas is contained in a cylinder equipped with a frictionless, When the pressure is atmospheric Pa, the piston
nonleaking piston of area A. face
is
at a distance
/
from the closed end. The gas
the piston a distance
7/x, under (c)
(a)
x.
isothermal conditions and
In what respect
is
is
compressed by moving
Calculate the spring constant, or force constant, (/;)
under adiabatic conditions. Using
a gas cushion superior to a steel spring? (d)
Eq. (4-14), show that Cp — Cy — nR for an ideal gas. 5-9 The temperature of an ideal gas in a capillary of constant crosssectional area varies linearly
from one end
(x
=
0) to the other (x
=
L),
according to the equation
air in the capillary.
L
Prove that the volume expansivity /3 of a real gas and the compressibility factor Z are connected by the following relation: (a)
1
If the volume of the capillary is V and the pressure P is uniform throughout, show that the number of moles of gas n is given by
1 1
z\do) v PV = nR
(b)
1
effects
by a capillary of negligible volume and are initially same temperature. To what temperature must the air in the larger
bulb be raised
-
air to act
the other, are connected
through the 5-6
R6{\
)
(Another type of virial expansion)
overflows?
there while the glass is inverted. When the support is removed, what mass of water must leave the glass in order that the rest of the water should remain in the glass, if one neglects the weight of the card? (Caution: Try this over a sink.) 5-5 Two bulbs containing air, one of which has a volume three times
at the
P=
which
5-8
to the 10-cm mark.
-0
(Beattie-Bridgeman)
Hg.
cylindrical highball glass
contains water
50
be poured
glass tube
P+
top
an absolute pressure of 1 atm. The piston is depressed by pouring mercury on it slowly, so that the temperature of the air is maintained constant. What is the length of the air column when mercury starts
inside the cylinder
each ease:
virial coefficient in
Prove that the isothermal compressibility
k
of a real gas
is
0i.
—
60
In (BrJOo)'
given by
the relation 1
K
~ p
1 (b_z>
z la?
Show
that,
PV =
nR8.
when
9l
=
8
=
$,
the above equation reduces to the obvious one
Ideal Gases
Heat and Thermodynamics
140
Prove that
5-10
the.
work done by an
ideal gas with constant heat
capacities during a quasi-static adiabatic expansion
W=
(a)
CV(0i-
is
-
w=
(c)
PiVi w = 7-1
PfVf
/M"-
w
1 lv~\
(d)
Show
that the heat transferred during an infinitesimal quasi-Static
nli
pressure within the bottle
(b)
this
An
equation to an adiabatic process, show that
ideal gas of volume 0.05
3
ft
and pressure 120
PV
lb /in. 2
=
const.
undergoes
a quasi-static adiabatic expansion until the pressure drops to 15 lb/in. 2
Assuming 7
How much
is
added to the gas on the
left
side?
is
Assuming the helium to behave like an ideal show that the final temperature of the
Po-
to remain constant at the value work is done?
1.4,
what
m moles of helium at connected through a valve with a large almost empty which the pressure is maintained at a constant value Pa , very
high pressure Pi. It
gasholder
Applying
heat
An
helium in the bottle is 7#o. 5-16 A thick-walled insulated chamber contains
+ %PdV. nli
= 9l.VdP
How much
gas with constant heat capacities,
process of an ideal gas can be written
&Q
On
evacuated bottle with nonconducting walls is connected through a valve to a gasholder, where the pressure is Pq and the temperature do- The valve is opened slightly, and helium flows into the bottle until the 5-15
5-11 (a)
horizontal insulated cylinder contains a frictionless noncon-
each side of the piston are 54 liters of an inert monatomic ideal gas at 1 atm and 273°K. Heat is slowly supplied to the gas on the left side until the piston has compressed the gas on the right side to 7.59 atm. How much work is done on the gas on the right side? (a) What is the final temperature of the gas on the right side? (bj What is the final temperature of the gas on the left side? (c)
Bj).
1
A
ducting piston.
equal to
W7-1
m
5-14
141
is
the final volume?
in
is
The
opened
and the helium flows two sides of the valve are equalized. Assuming the helium to behave like an ideal gas with constant heat capacities, show that: (a) The final temperature of the gas in the chamber is nearly atmospheric.
valve
is
slightly,
slowly and adiabatically into the gasholder until the pressures on the
5-12 (a)
Derive the following formula for a quasi-static adiabatic process 7 to be constant:
of an ideal gas, assuming
8V>(b)
At about 100 ms
1
=
(b)
const.
after detonation of a
uranium
fission
The number
of moles of gas
bomb, the ft and a
n
"ball of fire" consists of a sphere of gas with a radius of about 50
temperature of 300,000°K. Making very rough assumptions, estimate at what radius its temperature would be 3000°K. 5-13
(c)
The
final
= '
<*)" \p-J
_
Derive the following formula for a quasi-static adiabatic process of an ideal gas, assuming 7 to be constant:
1
0,-
7
1
chamber
is
ni
temperature of the gas
(a)
the
left in
-
-
'
in the
gasholder
is
P,/P,
(Pf/P,)
1
'-''
(Hint: See Prob. 4-11.)
pfy-Dly
=
const.
Helium (7 = -f) at 300°K and 1 atm pressure is compressed quasiand adiabatically to a pressure of 5 atm. Assuming that the helium behaves like an ideal gas, what is the final temperature?
5-17 (a) If y is the height above sea level, show that the decrease of atmosphere pressure due to a rise dy is given by
(b)
statically
dP
Ms
,
Ideal Gases
Heat and Thermodynamics
142
where and 6
M
the molecular weight of the air, g the acceleration of gravity,
is
the absolute temperature at the height y.
decrease of pressure in
If the
(b)
show
due
(a) is
to
Assuming friction to be negligible, the air to be ideal, and the changes of volume to be adiabatic, show that the period t 2 is now
an adiabatic expansion,
that
dP
P From
(c)
and
(a)
=
dd
y
=
T-l
«'
where (c)
calculate dO/dy in degrees per kilometer.
In the Clement and Dcsormcs method, gas
5-18
2-n
2£
using some of the numerical data of Art. 5-9,
(b),
is
is
/«,
cm
2
in excess of atmospheric pressure.
When
formed. pressure
is
A
5-19
cm
1
2 .
3.5
that
admitted into a
is
cm
With what period
(b)
If the ball
A
5-20 .2
cm 2 The .
is
mass 8 g
With what period If the ball is
how
far will
go before
it
placed in a tube of cross-sectional area
is
liters
capacity, the pressure
will the ball vibrate?
is
then allowed to drop,
come up? Carbon dioxide
is
contained
how
far will
it
whose volume
is
is
go before
starts to
in a vessel
5270
cm 3 cm 2
.
mass 16.65 g, placed in a tube of cross-sectional area 2.01 vibrates with a period of 0.834 s. What is 7 when the barometer reads 72.3 cm? 5-22 Mercury is poured into a U tube open at both ends until the total length of the mercury is h. (a) If the level of mercury on one side of the tube is depressed and then the mercury is allowed to oscillate with small amplitude, show that, ball of
neglecting friction, the period
n
T\
=
is
What
wave in argon at 20°C? wave of frequency 1100 vibrations per second in a column of methane at 20°C produces nodes that are 20 cm apart. What is 7? 5-26 The speed of a longitudinal wave in a mixture of helium and neon at 300°K was found to be 758 m/s. What is the composition of the 5-24 5-25
A
is
the speed of a longitudinal
standing
mixture?
held originally at a position where the gas pressure
and
,
The atomic weight
5-27
of iodine
entrapped
air
of the
column
is
A
standing wave in iodine
Determine the temperature of the helium gas in the acoustical which the data of Fig. 5-13 were obtained. An open glass tube of uniform bore is bent into the shape of an L. immersed in a liquid of density p', and the other arm, of length L,
5-28
interferometer from
5-29
One arm
is
remains in the air in a horizontal position. The tube is rotated with constant angular speed co about the axis of the vertical arm. Prove that the height^ to
which the
liquid rises in the vertical
arm
is
equal to
given by
2-w
.1—-
U
tube
L,
and again
is
127.
is
vapor at 400°K produces nodes that are 6.77 cm apart when the frequency is 1000 vibrations per second. Is iodine vapor monatomic or diatomic?
Po(l
One end
in
is
cm Hg.
(b)
(b)
speed of a longitudinal wave
for the
V(
then allowed to drop,
is
(a)
5-21
A
and
connected to an air tank of 6
exactly atmospheric
5-23 Prove that the expression an ideal gas may be transformed to
will the ball vibrate?
held originally at a position where the gas pressure
steel ball of
of the air being 76
it
the value, of 7?
come up? tube
)
mercury.
(a)
is
\A
temperature, the is
mass 10 g is placed in a tube of cross-sectional area connected to an air tank of 5 liters capacity, the pressure,
exactly atmospheric
1
What
then per-
is
steel ball of
The tube
starts to
to its initial
in excess of atmospheric pressure.
of the air being 76
it
Ao
adiabatic expansion
come back
the gas has finally
cm
The
k yhng/L
+
the height of the barometric column.
Show
container until a manometer containing a light liquid indicates a pressure of 1
143
now
gp'
closed so that the length of the
the
.-u-tf'/n/ZRB
mercury
is
caused to
oscillate.
where Po
is
atmospheric pressure,
the acceleration of gravity.
771 is the
molecular weight of
air,
and g
is
Heat and Thermodynamics
144
One mole
5-30
of an ideal paramagnetic gas obeys Curie's law, with a
Curie constant Cc- Assume that the internal energy so that (a)
dU —
is
a function of
Cv,m dO where cv,m is a constant. that the equation of the family of adiabatic surfaces
Show
6y,U
where A (A)
U
is
only,
is
6. KINETIC THEORY OF AN IDEAL GAS
2
In 9
Ad' + In V = W6c +
In A,
a constant for one surface.
Sketch one of these surfaces on a
OVM
diagram.
6-1
The Microscopic Point
of
View
We
have emphasized that the point of view of classical thermodynamics is with the aid of their gross, or large-scale, properties. The first law of thermodynamics is a relation among the fundamental physical quantities of work, internal energy, and heat. When the first law is applied to a class of systems, a general relation is obtained which holds for any member of the class but which contains no quantities entirely macroscopic. Systems are described
or properties of a particular system that
would
distinguish
it
from another.
Equation (4-13), for example,
Cv = is
d_U\ 96 )v'
true for all hydrostatic systems whether solid, liquid, or gas. It enables one
to calculate
Cv of a hydrostatic system, provided thai one knows the internal energy and V. The heat transferred during an isochoric process which is
as a function of 6 (Art. 4-10),
Qv = f" C v
dO,
may be calculated once, the Cv of the particular system under consideration known as a function of 6. But there is nothing in classical thermodynamics
is
that provides detailed information concerning
Another example of
the limitation of classical
U
or CV.
thermodynamics
is its
failure
any desired system. To make use of any thermodynamic equation involving P, V, 0, and the derivatives (dP/dV)e, (dV/dd)p, and {d8/dP)y, one must have an equation of state. Experimental to provide the equation of state of
values are very often useful, but there arc occasions
when
perform the necessary experiments. If an experiment
is
it is
not feasible to
performed on,
let
us
146
6-2
Heat and Thermodynamics
numerical constants in the equation of state of oxygen only are obtained, and no clue is at hand concerning the values of the constants for any other gas. To obtain detailed information concerning the thermodynamic coordinates
of an ideal gas.
and thermal properties of systems without having to resort to experimental measurements, we require calculations based on the properties and behavior of the molecules of the system. There are two such microscopic theories; one
6-2
say, oxygen, the
is
called kinetic theory,
and the other
is statistical
mechanics.
Both
theories deal
with molecules, their internal and external motion, their collisions with one another and with any existing walls, and their forces of interaction. Making use of the laws of mechanics and the theory of probability, kinetic theory itself with
concerns
the details of molecular motion and impact and
is
capable
of dealing with the following nonequilibrium situations:
1
Molecules escaping from a hole
in
for
Chap.
A simplified
treatment of statistical mechanics will be reserved
10.
Equation of State of an Ideal Gas
The fundamental hypotheses
of the kinetic theory of an ideal gas are as
follows:
1
small sample of gas consists of an enormous number of molecules N. For any one chemical species, all molecules are identical. If m is the mass of each molecule, then the total mass is mN. If 771 denotes the molar mass in grams (more commonly called the molecular weight), then the number of gram-moles n is given by
Any
a container, a process known as
mN
effusion.
2
Molecules moving through a pipe under the action of a pressure difference, a motion called laminar floiv.
3
Molecules with
The number
molecules
N\, where
momentum moving across a plane and mixing with of lesser momentum a molecular process responsible for
147
Kinetic Theory of an Ideal Gas
of molecules per
mole of gas
is
called Avogadro's number
viscosity.
4
Molecules with kinetic energy moving across a plane and mixing with molecules of lesser energy a process responsible for heal
,\r
—
N 77/ = =--=!££
6.0225
X
10=
molecules
g mole
conduction.
5
Molecules of one sort moving across a plane and mixing with molecules of another sort, a process known as diffusion.
6
Chemical combination between two or more kinds of molecules, which takes place at a finite rate and is known as chemical kinetics.
1
Inequality of molecular impacts
made on
various sides of a very
Since a mole of ideal gas at 273°K and at 1 atm pressure occupies a volume of 2.24 X 1 cm 3 there are approximately 3 X 10 19 3 molecules in a volume of only 1 cm 3 and 3 X 10 lc molecules per mm and even a volume as small as a cubic micron contains as many as 1
,
,
3
X
10' molecules. ideal gas are supposed to resemble small hard
small object suspended in a fluid, a difference that give rise to a
The molecules
haphazard zigzag motion of the suspended
spheres that are in perpetual
particle that
is
known
as
ture
Brownian motion.
of
an
and pressure range of an
random motion. Within
the tempera-
ideal gas, the average distance
between
lar
neighboring molecules is large compared with the size of a molecule. The diameter of a molecule is of the order of 2 or 3 X 10 -8 cm. Under standard conditions, the average distance between molecules
kinetic theory, although
The
mechanics avoids the mechanical details concerning molecumotions and deals only with the energy aspects of the molecules. It relies heavily on the theory of probability but is mathematically simpler than Statistical
—
more
subtle conceptually.
Only equilibrium
states
can be handled but in a uniform, straightforward manner, so that once the energy levels of the molecules or of systems of molecules are understood a program of calculations may be carried out through which the equation of state, the
In
this
energy, and other thermodynamic functions
chapter
we
may
be obtained.
shall limit ourselves to a small part of the kinetic
theory
about 50 times their diameter. molecules of an ideal gas are assumed to exert no forces of attraction or repulsion on other molecules except when they collide with one another and with a wall. Between collisions, they therefore move with uniform rectilinear motion. The portion of a wall with which a molecule collides is considered to be smooth, and the collision is assumed to be perfectly elastic. is
148
Ilcat
and Thermodynamics
If it is the 1
dicular
Kinetic Theory- of an Ideal Gas
149
speed of a molecule approaching a wall, only the perpen-
component
w±
wall, from 5
6-2
to
w ± is changed upon collision with the or a total change of 2w ±
of velocity
— w±,
—
When there is no external
field of force, the
The
uniformly throughout a container.
assumed constant, so that are dN molecules, where
molecular density
any small clement
in
dN =
j'-
i)
.
molecules are distributed of
X V
is
volume dV there
'.
r
The
infinitesimal
dV must
satisfy the
same conditions
theory as in thermodynamics, namely, that
it
is
in
kinetic
small compared
V but large enough to make dN a large number. If, for example, 3 1 cm contains 10 19 molecules, then one-millionth of a cubic centimeter would still contain 10 u molecules and would with
a volume of
qualify as a differential 6
volume element. no preferred direction for the velocity of any molecule, so that at any moment there are as many molecules moving in one There
is
direction as in another.
7
Not
molecules have the same speed.
all
A
few molecules at any
moment move slowly and a few move very rapidly, so that speeds may be considered to cover the range from zero to the speed of light. Since most molecular speeds are so far below the speed of light, no error is introduced in integrating the speed from Oto «. If dNa
Kg.
6-1
w and remains constant at equilibrium, even though the molecules are perpetually colliding and changing
dA',
is
represents the
W
+ dw,
it
is
number
of molecules with speeds between
assumed that
consider an arbitrary velocity vector .
cules have velocity vectors in
quantity involves the concept of a r, 0,
and
,
w
directed from the point
in
important to know how many molethe neighborhood of w. The calculation of this
Fig. 6-1 to the elementary area dA' It
we construct
Taking
a sphere of radius
as the origin of polar r.
The
area dA' on the
two circles of latitude differing by dO and two by d, has the magnitude
surface of this sphere, formed by circles of longitude differing
dA'
The
solid angle
dP.,
=
(r
dd)(r sin
dn
=
d9.
=
2 ,
the
The will
maximum
dA'
_
{rdB){r sin 6
of
(6-1)
-\-
If r/.V u
,
is
w and w
the
dii Air,
+
number
dw, then the fraction of
angle dil is range dw, in the
solid
is
that of the entire sphere
sr (sleradians)
fraction of molecules with velocity vectors in the
have speeds between
dQ about w.
w
is 4tt
these
dw and
of molecules with speeds
so that the
number
range of dd and the
an equation expressing the
neighborhood of
=
w
directions within the solid angle
between
molecules whose directions
d 3 .\ w .o,<,
and touching the edge
d)
dO
sin
angle
solid
d
formed bylines radiating from
sin 8
Since the largest area on the surface of the sphere 47rr
is
solid angle.
dQ =
by definition
or
Since the velocity vectors of the molecules of gas have no preferred direc-
coordinates
solid angle
dNv
their speeds.
tion,
The
lie
w and
within the
of molecules within the speed
range of
~
d,
is
given by
dQ di^ a
(6-2)
fact that molecular velocities have no preferred direction.
150
Heat and Thermodynamics
6-2
y
Portion of wall
Kinetic Theorv of an Ideal
151
Gas
molecules will be contained within the cylinder. Therefore the number of range, d) striking dA in time dr is molecules (speed range, dw; B range, dd;
expressed as
No.
o[w,0,
molecules striking dA in time dr
=
d 3N Wf e^-p-
(6-4)
,
which expresses the fact that molecules have no preferred location. According to our fundamental assumptions, a molecular collision is perfectly elastic. It follows, therefore, that a molecule moving with speed w in a direction making an angle Q with the normal to a wall will undergo a change only in its perpendicular component of velocity, as shown in Fig. 6-2. Furthermore,
it
momentum
follows that the total change in
Change
of
momentum
= —2mv
per collision
per collision
cos
is
(6-5)
$.
Now, Fraction of
No. of molecules
Total change
of speed
momentum
of
w
in
striking
dQ
solid angle
in
I
y
Fig. 6-2
ponent
All the molecules in the cylinder of length w dr strike the area dA at the angle 6 to perpendicular component of velocity w cos 6 is reversed, but the parallel com-
The change
6
from
The
w sin
is
unchanged.
all
in
momentum
directions
is
dd d(p
group of molecules approaching a small area dA of the Many of these molecules will undergo collisions along the way, but if we consider only those members of the group that lie within the cylinder (Fig. 6-2) whose side is of length w dr, where dr is such a short time interval that no collisions are made, then all the d 3 .\' W{ s, a molecules within lids cylinder will collide with dA. The volume of the cylinder dV is consider
this
dPw
exerted by the
—
{-2mw
jiu dr cos 6
dA
I
(
cos 6)
— 2mw cos 6)
[Eq. (6-3)]
dNw
[Eq. (6-5)]
per unit time and per unit area due to collisions
the pressure
dPw
molecules. Reversing the sign of the
Now
per collision
4ir
[Eq. (6-1)]
the normal.
dA
.
S1 sin
in
momentum
time dr
(
w = (dN -r~
Change
these molecules
exerted by the wall on the
momentum
molecules on
change,
we
dNw
gas
get the pressure
the wall:
wall of the containing vessel.
dV = w and
if
V
is
the total
dPK =
The quantity
,
dN,
in parentheses
so that the total pressure dr cos
dA,
volume of the container, only the
V
/2
(il>i; may
due
to
cos 2 e sin
be integrated at sight and molecules of
all
(6-3)
fraction
dV/ V
of the
dO
PV = $m
X
j
w i d\'w
.
speeds
is
is
(6-6)
found
given by
to
be -5,
Heat and Thermodynamics
152
The average
of the square of the molecular speeds
<«;=>
so that
=
X
i
(
i\
JO
2
(hi ) is
defined to be
dNm
iv°-
2
quantity
%m(w3)
is
n
=
7ft,
=
),
(6-8)
PV =
or
Comparing
A'
of system
No. of g moles Mass per mole lar
%N(tm(w*)).
the average kinetic energy per molecule.
Kinetic theory
Thermodynamics
M' = Mass
we have
PV = ^Vm(w
The
Comparison of Symbols
Table 6-1 (6-7)
153
Kinetic Theory of an Ideal Gas
6-3
=
No. of molecules
m = Mass
= M'/n
Na =
(molecu-
weight)
R = Universal gas constant V = Volume p = Mass density = M'/V
=
k
of a molecule Molecules per mole = N/n (Avogadro's number) Boltzmann's constant = R/N.\
V = Volume Molecular density
=
N/ V
our theoretical equation of state with the experimental one, we sec that an association must be made between the average kinetic energy per molecule
and the
ideal gas temperature.
Thus,
Distribution of Molecular Velocities
6-3
iN$m(w
2
))
=
nR6.
In the preceding derivation, the average square of the molecular speed
was defined by the equation But
N/n
is
the
number
of molecules per mole, or Avogadro's
A',,,
so that 1
<'«•>
*«w>-§*#-!*
(6-9)
where
where k
is
Boltzmatm's constant, given by
dNw was the number
It is often
necessary to
f*
= */.
;
2 d.\' w.
of molecules with speeds
know
the expression for
dNw
between
in
iv
and
terms of w
w
+ dw.
a relation
as the Maxwellian law of distribution of velocities, since it was first derived by James Clerk Maxwell. To derive this law is merely a mathematical exercise, for it is not necessary to use any physical laws governing the behavior of molecules when colliding with one another or with the wall. These physical laws are hidden within the fundamental assumptions of chaos: that, at (or near) equilibrium, the molecules have no preferred direction of velocity and
known _fl
NA
_
8 .3143
6.0225
X 10 ergs /mo le deg = X 10 23 molecules/mole 7
•
,
In this derivation, the average energy per molecule,
10-^deg lm{w 2 ),
is
wholly
is the only kind of energy that a hard by its neighbors, can possess. Wc have therefore limited ourselves to monatomic molecules only. Diatomic and polyatomic molecules can also rotate and vibrate and may therefore be expected to possess kinetic energy of rotation and vibration as well as potential energy of vibration, even though there are no forces among neighbors. It is worth while at this point to compare the symbolism used in our treatment of kinetic theory with that used in thermodynamics. This is shown in Table 6-1. The molecular form of the ideal-gas equation has another simple form. Since PV = -§A'(^'n{a! 2 )) and ?m(w 2 } = %kd, we get
kinetic energy of translation. This
spherical molecule, uninfluenced
PV =
(fAO(f*0)
=
no preferred location within the container. If altogether there are A' molecules, a fraction of have x components of velocity lying within the range assumption of chaos allows us to say that
and
proportional to dw x or ,
dNw JN = f(w x) dw x
dNWl = Nf(wx dw x )
There are similar equations
for the y
UN*,
MM and
and
is
this fraction
and
= Nf{wv
.
these, or dNw JN, will w x and w x + dw x The is a function of w x only .
Therefore,
.
(6-11)
z directions:
)
dw,„
dN„ = Nf{w,) dw z
.
(6-12) (6-13)
2
(6-10)
Xow A* is enormously large and dw x is small compared with the range of wx
.
dwx
Nevertheless, in the range
there are very
many
molecules
—
so
many
in
fact that the fraction of the dNa, molecules whose y components of velocity dw„ (d'2 X ,<„„.J is given by the same lie in the range, between iv„ and zva
volume
iM„JN. Thus,
d 2 N Wi
wj
,
.
and from Eq.
_
dN, r „
N
=
given by
is
using Eq. (6-12),
'
Since there
J{lVy) dlUy,
SS£
-
*/*mw/w>.
Nw „u
w
= Nf(wx )f(wy) dwx
whose radius
du-y.
.
,
z components of velocity lie between w, and tion dNw ,/N given by Eq. (6-13). Hence, d3
w
Nw„ Wl/M = Nf{w x f{w y f{w ,
)
)
z
+ dw,
be constant at
the
same
.
dw x dwy dwz
when
w\
+
density of velocity points,
,
(6-14)
or the
,
Dumber
The
we
+ w\
=
(6-16)
const.
(6-17)
arc plotted
This
of molecules
same
satisfied at the
could consider the three variables
time. If Eq. (6-16)
w x wy and wz to be ,
,
were
independent.
presence, of Eq. (6-17), however, limits or constrains the values of the
variables z
u>l
const.
as the frac-
.
w x wv and w
«.'-,
Nf(w x )f(wy)f(wz ) =
alone, z)
is
points on the sphere
all
the speed w. Therefore,
is
Both of these equations must be
consider a velocity space (Fig. 6-3) in which
along rectangular axes, the
is
(6-15)
for the velocities (the velocity space
no preferred direction
isotropic), the density should
wz =
The number of molecules represented by the symbol tPJV«„w, is still large, and again we assume that the fraction of these molecules di NWt u viW whose
we
is
=
(6-11),
d2
If
be
said to
155
Gas
volume clement, divided by that
associated with points within the cubical
+
expression as the fraction
Kinetic Theory of an Ideal
6-3
Heat and Thermodynamics
154
is
and
Taking the df
an
x)
dwx
dw x +f( Wz )f( Wx )
and
d J^dw diVy
+
2wx dw x
and
of independent variables to two.
equation of constraint.
differential of Eqs. (6-16)
^
f{w„)f{w z )
number
essentially reduces the
therefore called
2w,j
we
(6-17),
y
dwu
get
d
+f{ Wx )f{ lU y)
J^tdw =
0,
2w z dw z =
0.
z
+
or
flfe**^
1
Since there are only values to
all
Therefore, is
let
Fig. 6-3 to
w+
Velocity space of the molecules. All molecules
dw
have components within the spherical shell.
whose speeds
lie in the
range from
)
0Klj
w
two independent
variables,
z
dw,
=
0,
dw,
=
0.
wc cannot
give arbitrary
and dwz
.
us multiply the second equation by the quantity mfi, where /J is
an arbitrary, unknown function. The Adding the two equa-
called a Lagrange undetermined multiplier.
then get
df(w x )
u«w r
dj(w.)
three differentials, but only to two of them, say, dwy
wi(3 is
wc
tions,
1
/(«'*)
w y dw u +
the mass of a molecule and
product
w
f{Wy)
w x dw x +
and
m
5J&U, dlVy
1
X
dlV x
f(lV x)
dw c
+ m@w x
dw x
1
d/K)
/(»•„)
BWy
-f-
+
mfiWy
dw,.
1
df(wz )
f(Wz)
dw z
+
mjiw.
dw z
=
0.
Since the two variables
zc„
differentials are arbitrary,
dw,
—
Then
0.
Choosing dw,
may be
1
df(w x)
}{w x)
dw..
and dwy
and choosing
dw,,
?± 0,
we
1
df(w„)
f(Wy)
dWy
+ mpw x =
0.
get
+
mjiwa
we
get
+
m0w,
=
0;
and dwz
5^ 0,
1
df(w.)
f(wt)
dw 2
Fig. 6-4
=
Graph
of
Maxwell
w -1 L w + dw
Let us go back for a
speed-distribution function.
moment
satisfy
would have
any one of the preceding equations,
constant or a function of the variable in that equation. that satisfies
value of satisfies
all
the equations. It
to
=
A constant
the same differential equation (as indeed
it
is
the only-
/
must) regardless of the
dw.
and
where
In
A
is
or the cubical
number
of molecules whose components of velocity lie in the small volume element shown in Fig. 6-3, divided by that volume element.
But p is constant at all points within the infinitesimal spherical shell lying between the radii w and w dw, whose volume is 4nte»s dw and which con-
component. Taking the x component, _d_
+
tains the velocity points of
\nf(w x )
= —m@w x
dNw
=
—m{i\wx dwx
In /(«;,)
=
-P(l-mwl)
P ,
+\nA,
It
~= dw
f(w,)
may
therefore be
=
dN,„ A-kw'1
dw
4TNA*w 1 tr<>a'» a, \
(6-21)
which is the equation
for the
Maxwellian distribution of speeds, plotted
in Fig. 6-4.
= &+*—*>
(6-15),
6-4
= NAh~& m
density p
= Ae-^ mw-\ (6-18)
Using Eq.
The
follows from Eq. (6-20) that
a constant of integration. Therefore,
and
molecules.
expressed also as
,
Jd In f(w x)
f(w x)
dw x dwy dw z
be cither a
also seen that the function
is
to the definition of p; namely,
0.
P
To
157
make
chosen to
wm
=
Gas
and wz have been taken to be independent, their and we may assign them the values dwu = and
the multiplier m/3
=
Kinetic Theory of an Ideal
6-4
Heat and Thermodynamics
156
***'+* •>
Maxwellian Speeds and Temperature
(6-19;
We have seen that the number of molecules d.\'a with speeds between w and w + dw is equal to
also
= NA 3e~^""°
2) .
(6-20)
dNw =
ATrNAWe-M"-** dw. 1
6-4
Heat and Thermodynamics
158 where
is
an undetermined multiplier and A
determine the value of A,
from zero to infinity.
The
it is
is
an integration constant.
merely necessary to integrate
member
integral of the left-hand
— \ttA
3
=
r
!{j
2
"'"'" )
this
To
h
n
It
is
/„
1 1
2
on the right is commonly encountered in mathematical physics. along with others in Table 6-2. Its value is j-7r ! /(£/n/3) J therefore,
2a
ff
1
3
4 \[fl 3
2« 2
;
ff
3
4
4wA
_ =
3 ,
6
4fjm/S)i (6-22)
15
fir"
16
\a'
6
- -
or
ffi
If n
meaning of the multiplier
/3,
we. recall the definition of
/
the average of the square of the molecular speeds:
2
x"e"" dx
(6-7)
=
3
3 7 a
If n
is
1
odd,
C+ *"«-* zo
24.
/
dx
=
0.
A 3 and of
in our previous equation for dNw , Eq. (6-21), the Maxwellian equation for the distribution of molecular speeds
Substituting the values of /•«
1
a
even,
is
+« find the physical
1
5
sVa
*•»
To
159
x n e-ax- dx
2-Sfe
integral listed
/;
R
dw.
Jo
1
The
=
f/„
Ga
equation
so that
is A",
Values
Table 6-2
Kinetic Theory of an Ideal
Air
becomes Substituting for dN„,
wc
'!\~v,
get
dw 4
(H) }
=
ajV**"^
The comparison, however,
=
(6-25)
die.
6-2, the right-hand integral has the value
{wt)
m^g~h ,nw t fkd
2k0
X 4irA 3 [
This
From Table
AN
«WW
fcr*
ir*
8(£m/3)3
-|ir'.
(iw/3)
! .
Therefore,
plotted for three different temperatures in Fig. 6-5. is
The
larger the
the spread of values of the speed.
(6-23)
m^
of the theoretical equation of state of
is
temperature, the wider
an
ideal
gas with the empirical one led to the relation
(6-9)
Therefore,
and
*
=
s
(6-24) Fig. 6-5
Maxuellian speed
distribution at various temperatures,
63
>
62
>
fa.
160
A simple analysis of the parabolic paths of the atoms that pass through F and C provides a relation between the deflection s and the speed w. The departure angle 8 is practically zero. The experimental results arc shown in Fig. 6-7, where the smooth curve is a plot of Maxwell's equation and the
Beam chamber
Oven chamber
161
Kinetic Theory of an Ideal Gas
6-5
Heat and Thermodynamics
dots are experimental points.
Equipartition of Energy
6-5
One
of the most important results of elementary kinetic theory, and one
been used throughout this chapter, is the relation obtained by comparing the theoretical ideal-gas equation with the empirical one, that has
Apparatus of Estermann, Simpson, ami Stem for the
Fig. 6-6
by gravity. 0, oven;
F and C,
slits;
D,
deflection
of a molecular beam
detector.
i«$9$ = Maxwell's equation for the distribution of molecular speeds has been experimentally verified directly as well as indirectly. One of the most con-
_ „,2 W2 = WX 2
Since
vincing experiments was performed by Estermann, Simpson, and Stern in
making use of the apparatus depicted schematically in Fig. 6-6, in which atoms are deflected by gravity only. Cesium atoms issue from a minute slit in an oven at the left of a long, highly evacuated chamber. Most of these atoms arc stopped by the diaphragm F, and those which go through the slit constitute a narrow, almost horizontal beam. The slit C, called the collimating slit, is halfway between F and the detector D. The atoms are detected by the method of surface ionization, in which nearly every cesium atom striking a hot tungsten wire leaves the wire as a positive ion and is collected by a negatively charged plate. The plate current is then a measure of the number of cesium atoms striking the detector wire per unit of time.
im(w 2 ) =
1947,
Since the
x, y,
(wl)
>M
and It
h
words, that
of freedom
+ all
l
...2
IV
lm(w y )
+ ?m(wl).
equivalent,
= {«$ = 3(im{wD) = (wl)
m
is
km(wl)
= ue,
im(wl)
=
-lm(wl)
= Ud,
ue,
the average kinetic energy associated with each translational degree
^k9.
The molecules we have considered up to this point have been spherically symmetrical, rigid structures capable of undergoing translational motion only. Actual molecules, however, are neither spherical nor rigid and have
\
0.2
-
...2
l
(6-9)
follows that
Or, in
V 3
+ Wy +
w l)
and z directions are
and
0.3
i-m(
Ike.
X
been found, mostly by spectroscopic methods, to rotate about two or, in some and to vibrate with many different frequencies. A molecule such as, let us say, CO has three translational degrees of freedom, two rotacases, three axes
0.10
0.20
0.30 S,
Fig. 6-7
0.40
0.50
mm
tional degrees of
Distribution of speeds of cesium atoms in atomic beam deflected by gravity. (Ester-
mann, Simpson, and Stern.)
freedom
(since
it
is
dumbbell-shaped), and, at moderate
temperatures, one frequency of vibration. Its energy
e
=
ffflst;
imiul
+
&fa.a£
+ $Ib*>b
e
may be
written
i-H\
Kinetic Theory of an Ideal Gas
Heat and Thermodynamics
162
where 7,i and In arc moments of inertia with reference to axes A and B, about which the molecule rotates with angular speeds a>, and to B respectively; J is the displacement of the C and O atoms from their equilibrium separation; (
and
Oil is
,
the derivative of this displacement with respect to the time.
The
last two terms represent the contribution to the energy of one frequency of vibration. Notice that the expression for t contains only squares of independ-
ently varying quantities.
There arc no first-power terms, and there are no if interaction existed between
If q is large,
The
7
is
nearly
1
This result
.
is in
complete disagreement
163
with experiment.
heat capacities of such molecules arc not constant but vary
markedly
with the temperature even around room temperature, and 7 is not in the neighborhood of unity. When the principle of the equipartition of energy is applied to solids and liquids, the disagreement is still worse, and this principle must be abandoned in favor of
quantum
ideas.
"cross-product" terms, which would be present
neighboring molecules.
We have seen
that, at equilibrium, the
of the three translational terms
mechanics, a beautiful theorem energy
may be proved which
?kd.
is
average energy associated with each
By
known
methods of
the
as the principle of the equipartition of
asserts that the average energy associated with any
squared term in the energy of a molecule is ^kd. It follows that, for each degree of freedom of translation and of rotation, the molecule has the average energy $£$ but for each frequency of vibration the average energy is kO, since each vibration is represented by two squared terms.
A
monatomic ideal gas has only three translational degrees therefore Ma atoms (1 mole) will have an energy u equal to
=
u
NA (U6) =
Prove that the number of molecules striking unit area of the conis equal to N(w)/4V, where, by definition,
6-1
tainer wall per unit time
6-2
dw x
%R0,
against
w and compare
cy
+ R, cp
=
|fi
and
(b)
7=7. These calculations are in agreement is
with the experimental results
The diatomic
gases
H2
,
1).,
listed in Art. 5-4.
N NO,
0>,
2,
and CO,
in the
neighborhood of
—
room temperature
where rotation takes place, but not vibration are dumbbell-shaped with two rotational degrees of freedom. Hence the molar energy is
and
cv
=
|A',
cp
8
=
=
%R,
the
6-4
In Fig. 6-4,
and
7
agreement with the results of Art. 5-4. polyatomic molecules arc "soft" and vibrate q,
=
6-5
and
with
many
=
(3
+
q)R,
cP
=
arc the two curves different?
(3
(4
w m be the value of w at which dNw idw is a maximum.
.
maximum Show
value of
that the e
and
t
= w/w m and ,
dMa/N
calculate
would require
+ q )jie +
q)R,
(c)
+ de
is
that
=
2L (l\ -)
«-«'*»<&.
defined as
is
w rms = a/Vt
v (w
times the speed of sound.
that {w), defined in Prob. 6-1
y
and
4
+
q
3+q
What
of molecules dX, with translational kinetic
given by
The root-mean-square speed w Tms Show that
Show Show
dx.
dNx /N dx?
number
fre-
(b)
cv
Why
Wrrns
easily
the principle of the equipartition of energy
=
let
IN
i»
that
u
dw z
fftfl,
When
quencies, say,
6-4.
wm Choose a new variable x
energy between
(a)
in
with Fig.
duly
6-3
Calculate
=
w dNw
/
Plot
(a)
Since cp
/"<*
1 -rt
=
{w)
of freedom:
du
and
PROBLEMS
classical statistical
{«;>
=
'8^
wm
,
is
equal to
2 ).
Kinetic Theory of an Ideal Gas
Meat and Thermodynamics
164 (d)
Calculate
(e)
Show
unit time
(1
equal
is
and compare this with l,'{w). number of molecules striking unit area
6-10
'«.')
that the
of a wall per
A
spherical glass bulb 10
in radius
is
maintained at 300°K,
except for an appendix with a cross-sectional area of liquid nitrogen, as
to
cm
shown
nally at a pressure of 0.1
in Fig. P6-1.
The bulb
mm of mercury.
165
1
cm immersed 2
in
contains water vapor origi-
Assuming that every water mole-
\/2™?A-0
cule which enters the appendix condenses on the wall and stays there, find -6 of mercury. the time required for the pressure to decrease to 10
6-6 The Doppler broadening of a spectral line increases with the rms speed of the atoms in the source of light. Which should give narrower spectral lines: a mercury 198 lamp at 300°K. or a krypton 86 lamp at 77°K? 6-7 At what temperature is the mean translational kinetic energy of a
mercury and closed except for a hole A -3 2 the liquid level of area 10 cm above is kept at 0°C in a continuously evacuated enclosure. After 30 days it is found that 0.024 g of mercury has been lost. What is the vapor pressure of mercury at 0°C?
mm
molecule equal to that of a singly charged ion of the same mass which has been accelerated from rest through a potential difference of (a) 1 V? (6) 1000 V? (c) 1,000,000 V? (Neglect relativistic effects.) 6-8 An oven contains cadmium vapor at a pressure of 1 .71 XI -2 of mercury and at a temperature of 550°K. In one wall of the oven there is a -3 cm. On the other side of the slit with a length of 1 cm and a width of 10 wall is a very high vacuum. If one assumes that all the atoms arriving at the slit pass through, what is the atomic beam current? 6-9 A vessel of volume V contains a gas that is kept at constant tem-
mm
perature.
pressure
any time
gas slowly leaks out of a small hole of area A. The outside low that no molecules leak back. Prove that the pressure P at given by
The is
r
so is
P =
P„e-»'\
where P u is the initial pressure; and calculate k in terms of V, A, and (Assume that all the molecules arriving at the hole pass through.)
-^L ,-— Liq. N
%? Fig. P6-1
:
(iv).
6-11
vessel partially filled with
Engines, Refrigerators, and the Second
7-1
7.
decreases until atmospheric pressure
ENGINES,
to
REFRIGERATORS, AND THE SECOND LAW
work by
needed
is
i.e.,
or
When two stones friction
is
into Heat,
on the system. For one complete
As soon
as the
temperature of the stones
that of the surrounding water, however, there If the
is
temabove
rise of rises
a flow of heat into the water.
the
of heat absorbed by the system,
work continuously to the outside by performing and over again. The net work in the cycle is the output, and the heat absorbed by the working substance is the input. The thermal of a heat engine
efficiency
mass of water is large enough or if the water is continually flowing, no appreciable rise of temperature, and the water can be is
the performance of
same
is
to deliver
the same cycle over
of the engine
rp,
there will be
regarded as a heat reservoir. Since the state of the stones
and
cycle, let
W
work done against
transformed into internal energy tending to produce a
perature of the stones.
brought back
Both Qu and Qc have been defined as positive numbers. If Q// is larger than is done by the system, the mechanical device by whose agency Qc and if the system is caused to undergo the cycle is called a heat engine. The purpose
and Vice Versa
are rubbed together under water, the
is
the processes that constitute a cycle
amount of heat rejected by the system, = net amount of work done by the system.
W
Work
Each of
cycle.
=
Qc
Conversion of
a
involve a flow of heat to or from the system
Qu = amount
7-1
reached, at which point the process
a scries of processes in which a system
is
its initial state,
may
167
cannot be used indefinitely.
stops. It therefore
What
is
Law
1
ri
is
defined as
rr herniall efficiency •
=
at the
end of the process as at the beginning, the net result of the process is merely the conversion of mechanical work into heat. Similarly, when an electric current is maintained in a resistor immersed either in running water or in a
—
work output, in any energv units si_ heat input, in the same energy units -.
or
-.
.
,
= W_
V
(7-1)
is also a conversion of electrical work into any change in the thermodynamic coordinates of the wire. In general, work of any kind may be done upon a system in contact with a reservoir, giving rise to a flow of heat Q without altering the state of the system. The system acts merely as an intermediary. It is apparent from the first law that the work is equal to the heat Q; or in other words, the transformation of work into heat is accomplished with 100 percent efficiency. Moreover, this transformation can be continued indefinitely.
Applying the first law to one complete cycle and remembering that there no net change of internal energy, we get
To study the converse process, namely, the conversion of heat into work, we must also have at hand a process or series of processes by means of which such a conversion may continue indefinitely without involving any resultant
or
very large mass of water, there heat, without
W
W
changes in the state of any system. At
Qi,
It is seen
from
when Q c
is
in
it
discussing the conversion of heat into work. In this case there
is
no
internal-
W
energy change since the temperature remains constant, and therefore = Q, or heat has been converted completely into work. This process, however, involves a change of state of the gas.
The volume
increases
and the pressure
~
Qc
(7-2)
Qu
might appear that the
thought,
W,
Qu
and therefore
isothermal expansion of an ideal gas might be a suitable process to consider
first
-Qc-
this
equation that
zero. In other words,
is
will be unity (efficiency of 100 percent)
-q
an engine can be built to operate in a no outflow of heat from the system, there will be 100 percent conversion of the absorbed heat into work. We shall see later under cycle in
which there
what conditions practice.
if
is
this is possible in principle
and why
it
is
not possible in
168
Ilcat
The
7-2
and Thermodynamics
Law
Engines, Refrigerators, and the Second
1
69
Hot ^Jl_ Cold
transformation of heat into work
is
mmm
usually accomplished in practice
by two general types of engine: the external combustion engine, such as the Stirling engine and the steam engine, and the internal-combustion engine, such as the gasoline engine and the dicscl engine. In both types, a gas or a mixture
i
of gases contained in a cylinder undergoes a cycle, thereby causing a piston
impart to a shaft a motion of rotation against an opposing
to
force. It
is
necessary in both engines ihat, at some time in the cycle, the gas in the cylin-
1
der be raised to a high temperature and pressure. In the Stirling and steam engines this is accomplished by an outside furnace. The high temperature Hot
and pressure achieved in the internal-combustion engine, however, are produced by a chemical reaction between a fuel and air that takes place in the
—
2—3
2
„JL
Cold
In the gasoline engine, the combustion of the gasoline and through the agency of an electric spark. The diesel engine, however, uses oil as a fuel, the combustion of which is accomplished more slowly by spraying the oil into the cylinder at a convenient rate. cylinder
itself.
air takes place explosively
3-^4
The
7-2
Stirling
(a)
4
—
1
Engine
In 1816, before the science of thermodynamics had even been begun, a Church of Scotland named Robert Stirling designed and
minister of the
patented a hot-air engine that could convert some of the energy liberated by a burning fuel into work. The Stirling engine remained useful and popular for many years but, with the development of steam engines and internal-
combustion engines, finally became obsolete. In the 1940s the Stirling engine was revived by the engineers of the Philips company in Eindhoven, lolland, and has once again become the subject of a great amount of interest and I
research. I
he steps
in the operation of a
shown schematically left
in Fig. 7-1 a.
and a compression piston on the
As the
shaft rotates, these pistons
suitable connecting linkages. gas,
somewhat
Two
The
and the left-hand portion
reservoir (burning fuel),
idealized Stirling engine are
pistons,
right, are
move
an expansion piston on the connected to the same shaft.
0„
in different phase, with the aid of
space between the two pistons
of the space
is
filled
with
kept in contact with a hot while the right-hand portion is in contact with a cold is
Between the two portions of gas is a device R, called a regenerator, wool or a series of metal baffles, whose thermal conductivity is low enough to support the temperature difference between the hot and cold ends without appreciable heat conduction. The Stirling reservoir.
consisting of a packing of steel
cycle consists of four processes depicted schematically in Fig. 7-1 a
ing pressure
on the
PV
and volume changes diagram of Fig. 7-1 b.
and
involv-
plotted (as though ideal conditions existed)
V (b) Fig. 7-1
(a) Schematic
diagram of steps in
{The numbers under each diagram
the operation of
refer to the processes
(b) Idealized Stirling engine cycle on
PV diagram.
shown on
an idealized Stirling engine.
PV
diagram
in Fig. 7-1 b.)
\
—*
2
halfway up, compressing cold gas while in contact with the cold reservoir and therefore causing heat Qc to leave. This is an approximately isothermal compression and
therefore all the input
left
piston remains at the top, the right piston
is
moves
depicted as a
no change in volume, but gas is forced through the regenerator from the cold side to the hot side and enters the
there
is
left-hand side at the higher temperature
6j/.
the regenerator supplied heat Qr to the gas.
2 >
171
Qu
could not be converted into work.
rigorously isothermal process at the temperature 6c in Fig. 7-16. The left piston moves down and the right piston up, so that
—> 3
3 —
Law
indicator diagram of an actual Stirling cycle would look much more like the diagram in Fig. 7-2. Even if these idealizations could be realized in practice, there would still be some heat Qc rejected at the lower temperature, and
While the
2
Engines, Refrigerators, and the Second
7-3
Heat and Thermodynamics
170
—
*
The
4
3 in Fig. 7-1 b
is
stationary as the left piston con-
operation of such a plant can be understood
and volume changes
is shown in by following the
of a small constant mass of water as
veyed from the condenser, through the
boiler, into the
it is
con-
expansion chamber,
sion,
as
higher pressure and temperature. In the boiler the water
ator that
and then vaporized, both processes taking place approximately at constant pressure. The steam is then superheated at the same pressure. It is then allowed to flow into a cylinder, where it expands approximately adiabatically against a piston or a set of turbine blades, until its pressure and
moving down while
in
contact with the hot reservoir,
causing the gas to undergo an approximately isothermal expan-
during which heat Qu is absorbed at the temperature On, shown in Fig. 7-16. Both pistons move in opposite directions, thereby forcing gas through the regenerator from the hot to the cold side and giving up approximately the same amount of heat Qr to the regener-
»
The
and back to the condenser. The water in the condenser is at a pressure less than atmospheric and at a temperature less than the normal boiling point. By means of a pump it is introduced into the boiler, which is at a much
tinues
4—
schematic diagram of an elementary steam power plant
Fig. 7-3«.
pressure
at constant volume.
now remains
right piston
A
To accomplish this,
Note that the process
The Steam Engine
7-3
it
absorbed in the process 2
—
»
3.
This process takes
is
the absorption of heal
heated to
its
temperature drop to that of the condenser. In the condenser, finally, the steam condenses into water at the same temperature and pressure as at the
place at practically constant volume.
The, net result of the cycle
is first
boiling point
Qu at
the high tempera-
beginning, and the cycle
is
complete.
and the delivery
In the actual operation of the steam plant there arc several processes that
no net heat transfer resultmust be emphasized that
caused by the pressure difference required to cause the flow of the steam
no leakage of gas takes place, (3) no heat is lost or gained through cylinder walls, (4) no heat is conducted through the regenerator, and (5) there is no friction. An
from one part of the apparatus to another, (2) friction, (3) conduction of heat through the walls during expansion of the steam, (4) irreversible heat transfers due to a finite temperature difference between the furnace and the
ture
the rejection of heat
Bit,
of work
W=
ing from the Fig. 7-16
is
Qu
—
Qc
Qc
at the low temperature Be,
to the surroundings, with
two constant-volume
processes. It
based on the assumptions that
(1) the
gas
is
ideal, (2)
render an exact analysis
difficult.
These
arc: (1) acceleration
and turbulence
boiler.
A
P
may in
first
approximation
to the solution of the
problem of the steam plant
made by introducing some simplifying assumptions which, although no way realizable in practice, provide at least an upper limit to the be
efficiency of such a plant
and which define a cycle
called the Rankine cycle,
terms of which the actual behavior of a steam plant may be discussed. In Fig. 7-36 three isotherms of water are shown on a PV diagram: one at 6c corresponding to the temperature of the condenser, another at On for the
in
temperature of the boiler, and a third at a still higher temperature Bn- The dashed curves arc the liquid and the vapor saturation curves, respectively. In the Rankine cycle tions that arise Fig. 7-2
Indicator diagram of an actual Stirling engine.
all processes are assumed to be well-behaved; complicafrom acceleration, turbulence, friction, and heat losses are
thus eliminated. Starting at the point
1,
representing the state of
1
lb of
172
Ileal
and Thermodynamics
7-4
lingines, Refrigerators,
173
and the Second Law
always rejected during the condensation of water, Qc cannot be made equal to zero, and therefore the input Qu cannot be converted completely is
into work.
7-4
Internal Combustion Engines
In the gasoline engine, the cycle involves the performance of six processes, four of which require motion of the piston
1
Intake stroke.
A
is
called strokes:
mixture of gasoline vapor and air
is drawn into the The pressure of the mixture by an amount sufficient to
cylinder by the suction Stroke of outside
and are
greater than that of the
the.
piston.
cause acceleration and to overcome friction. 2
Compression stroke.
pressed until
Volume
its
The mixture
pressure
of gasoline vapor
and temperature
rise
and
accomplished by the compression stroke of the piston,
V
air
is
com-
considerably. This in
which
is
fric-
and heat loss by conduction are present. Combustion of the hot mixture is caused to take place very rapidly by an electric spark. The resulting combustion products attain a very high pressure and temperature, but the volume remains unchanged. The piston does not move during this process. Power stroke. The hot combustion products expand and push the piston out, thus suffering a drop in pressure and temperature. This is the power stroke of the piston and is also accompanied by friction, acceleration, and heat conduction. Valve exhaust. The combustion products at the end of the power stroke are still at a higher pressure and temperature than the outside. An exhaust valve allows some gas to escape until the pressure drops to that of the atmosphere. The piston docs not move during this process. Exhaust stroke. The piston pushes almost all the remaining combustion products out of the cylinder by exerting a pressure sufficiently larger than that of the outside to cause acceleration and overcome tion, acceleration,
3 Bservoir
(b)
(a) Fig. 7-3
(a)
Elementary sleam power plant, (b)
PV
diagram of Ranking
cycle.
4
.saturated liquid
water at the temperature and pressure of the condenser, the
Rankine cycle comprises the following
six processes: 5
Adiabatic compression of water to the pressure of the boiler
1
(only a very small change of temperature takes place during this process).
2
-
>3
3-»4 4->5
6
Isobaric heating of water to the boiling point. Isobarie, isothermal vaporization of water into saturated steam.
Isobaric superheating
temperature
of steam
into
superheated
steam
at
Explosion.
friction.
0//.
This
is
the exhaust stroke.
Adiabatic expansion of steam into wet steam.
6->l
Isobaric, isothermal condensation of steam into saturated at the
temperature
water
Qc-
In the above processes there are several phenomena that render an exact mathematical analysis almost impossible. Among these are friction, acceler-
by conduction, and the chemical reaction between gasoline vapor and air. A drastic but useful simplification is provided by eliminating these troublesome effects. When this is done, we have a sort of idealized gasoline engine that performs a cycle known as an Otto cycle. The behavior of a gasoline engine can be approximated by assuming a set ation, loss of heat
—
—» 4,
and
—
> 3, 3 * 5, heat Qrr enters the system from a hot reservoir: whereas during the condensation process 6 > 1, heat
During the processes 2
4
—
Qc
is
rejected by the system to a reservoir at $c- This condensation process
must exist in order to bring the system back to
its initial
state 1.
Since heat
Heat and Thermodynamics
174
Pressure
7-4
P
Pressure
F
3
—
»
Engines, Refrigerators, and the Second
Law
175
4 represents a quasi-static adiabatic expansion of «j moles of air,
involving a drop in temperature from
3
according to the equation
to 0|
hV2~ = OJ •v-i l x
4
—
* 1
represents a quasi-static isochoric drop in temperature
of mi moles of air, brought about by a rejection of heat
Qc
and pressure
to a series of external
from 6\ to 0i. This process is meant to approximate the drop to atmospheric pressure upon opening the exhaust reservoirs ranging in temperature
valve. 1
—* 5
represents a quasi-static isobaric exhaust at atmospheric pressure.
The volume
V
V,
3
Volume
V
Volume
(a) Fig. 7-4
V
(6)
(a) Air-standard Otto cycle, (b) Air-standard Diesel cycle.
The working substance is at all times air, an ideal gas with constant heat capacities. (2) All processes arc quasi-static. (3) There is no friction. On the basis of these assumptions the air-standard Otto cycle is composed of six simple processes of an ideal gas, which are plotted on a PV diagram in Fig. 7 -4a. 5 —> 1 represents a quasi-static isobaric intake at atmospheric pressure. There is no friction and no acceleration. The volume varies from zero to l\ as the number of moles varies from zero to «i according to the equation
of ideal conditions as follows: (1)
which behaves
from V\ to zero as the number of moles varies from n% to temperature remaining constant at the value 6\.
varies
zero, with the
The two isobaric processes 5 —> 1 and 1 —* 5 obviously cancel each other and need not be considered further. Of the four remaining processes, only two involve a flow of heat. There occur an absorption of Qu units of heat at high temperatures from 2 —> 3 and a rejection of Qc units of heat at lower temperatures from 4 —> 1, as indicated in Fig. 7-4a. Assuming CV to be constant along the line 2 —» 3, we get
like
Qn = j C v Similarly, for process
4—»1, Qc
The thermal
efficiency
is
P V = n Re u atmospheric pressure and
temperature of the outside air. 1 —> 2 represents a quasi-static, adiabatic compression of n\ moles of air. There is no friction and no loss of heat through the cylinder wall. The tem-
where Pa
is
perature rises from
a
to 6«
—
regarding
= I" Jet
Cv
dd
Qc
=
Oi).
as a positive
CV(0 4
-
Oi).
therefore
-I _% -
!
Q/f
The two
~
Qv(0s
**
- ft
03
Si
—
adiabatic processes provide the equations
according to the equation »1*7
_
=
wr
HIT* =
1
WT4
-
temperature and presan absorption of heat Qu from a series of external reservoirs whose temperatures range from 0* to 9», If there were only one reservoir at temperature 3 the flow of heat would not be quasi-static. This process is meant to approximate the effect of the explosion in a gasoline engine. 2
,
0i is the
=
de
and
%
6tV^-
= HVf\
3 represents a quasi-static isochoric increase of
sure of «i moles of air, brought about by
which, after subtraction, yield
(04- 0i)J7-' =
,
—
"1
h -
(h
"4
(0 3
-
e2
)vr\
number,
Hcai and Thermodynamics
176
r Denoting the ratio I i 'V% by expansion ratio, we have finally
where
r,
r is
called the compression ratio or the
In practice, the compression ratio of a Diesel engine can be larger than that of a gasoline engine, because there since only air
1
J
(7-3)
rr-i
(Vi/V^"r-i
7 =
is
(for air,
y
is
is
more nearly
_
.
1
_
Taking
-
—x --
2 I/SI
1/15*
%
T5
r
called preignition.
re,
Law
177
made much
fear of preignition
=
15, rg
=
5,
and
1.5,
cannot be made greater than about 10 because, if r is larger, the rise of temperature upon compression of the mixture of gasoline and air is great enough to cause an explosion before the advent In an actual gasoline engine,
no
is
compressed. Taking, for example,
*
of the spark. This
Engines, Refrigerators, and the Second
7-5
5
=
1
=
64 percent.
5(0.0895
i
-
0.0172)
equal to 9 and y equal to 1.5
r
1.4),
The
efficiencies of actual Diesel engines are
still
lower, of course, for the
reasons mentioned in connection with the gasoline engine. 1 77
= =
In the Diesel engine just considered, four strokes of the piston are needed
y=
1
0.67
V9
for the
=
only air
67 percent.
All the troublesome effects present in an actual gasoline engine, such as acceleration, turbulence,
and heat conduction by virtue of a finite temperamake the efficiency much lower than that of
ture difference, are such as to
admitted on the intake. The air is comthe temperature is high enough to ignite oil that is
In the Diesel engine, only air
is
sprayed into the cylinder after the compression.
The
rate of supply of oil
is
adjusted so that combustion takes place approximately isobarically, the piston moving out during combustion. The rest of the cycle namely, power stroke,
—
valve exhaust, and exhaust strokeengine.
The
is
and heat
cycle
we
known
imagined It is
losses,
the
fric-
same assumptions
—
>
3 in Fig. 7 -4a
horizontal instead of vertical, the resulting cycle will is
shown
be the
is
air-
in Fig. 7-4b.
is
rB
given by
_
1
(l/rsr
7 (UrE )
—
Vi
TT
—
pressure,
A
re
=
is
a
possible to
-
(\/r c
Vi
yf Vi
=
and then, instead of using the piston itself to exhaust the remaining is blown into the cylinder, replacing the combustion products.
blower, operated by the engine
itself, is
used for
this purpose,
and thus
it
accomplishes in one simple operation what formerly required two separate piston strokes.
7-5
Kelvin-Planck Statement of the Second
Law
In the preceding pages, four different heat engines have been briefly and somewhat superficially described. There are, of course, more types of engines and a tremendous number of structural details, methods of increasing thermal
which constitute the subject matter of its origin to the attempt to convert heat into work and to develop the theory of operation of devices for this purpose. It is therefore fitting that one of the fundamental laws of thermodynamics is based upon the operation of heat engines. Reduced to efficiency,
y
(1/rc)
mathematical analyses,
(7-4)
their simplest terms, the
'
summed up
be
etc.,
important characteristics of heat-engine cycles may-
as follows:
expansion ratio 1
and
is
engineering thermodynamics. Thermodynamics owes
a simple matter to show that the efficiency of an engine operating in an
V
where
by making
as the air-Standard Diesel cycle. If the line 2
idealized Diesel cycle
in the Diesel engine, it
as in the gasoline
are left with a sort of idealized Diesel engine that performs a
standard Diesel cycle. This
compressed
gases, fresh air
take place in the Diesel engine as in the
gasoline engine. Eliminating these effects as before,
same
exactly the
usual troublesome effects, such as chemical combination,
tion, acceleration,
power stroke. Since do away with the exhaust and intake strokes and thus complete the cycle in two strokes. In the two-stroke-cycle Diesel engine, every other stroke is a power stroke, and thus the power is doubled. The principle is very simple: At the conclusion of the power stroke, when the cylinder is full of combustion products, the valve is
opens, exhaust takes place until the combustion products arc at atmospheric
the air-standard Otto cycle. pressed adiabatically until
execution of a cycle, and only one of the four
compression
ratio.
There
is
some process or
series of processes
during which there
is
an
absorption of heat from an external reservoir at a high temperature (called simply the hot reservoir).
Heat and Thermodynamics
178
Engines, Refrigerators, and the Second
7-6
flow of heat. There
is
nothing
in the first law- to
Law
179
preclude the possibility of
converting this heat completely into work. The second law, therefore, is not a deduction from the first but stands by itself as a separate law of nature, referring to
law.
The
an aspect of nature different from that contemplated by the first law denies the possibility of creating or destroying energy; the
first
in a particular way. The own energy and thus violates kind. The operation of a machine
second denies the possibility of utilizing energy continual operation of a machine that creates the
first
law
is
called perpetual motion of thefirst
its
that utilizes the internal energy of only one heat reservoir, thus violating the
second law,
^Z^i Fig. 7-5
There
2
is
some process or scries of processes during which heat is an external reservoir at a lower temperature (called simply
rejected to
the cold
reservoir).
called perpetual motion of the second kind.
The Refrigerator
7-6
Symbolic representation of heat engine.
is
We have seen that a heat engine is a device by which a system is taken through a cycle in such a direction that some heat is absorbed while the temperature is high, a smaller amount is rejected at a lower temperature, and a net amount of work is done on the outside. If we imagine a cycle performed in a direction opposite to that of an engine, the net result would be the absorption of some heat at a low temperature, the rejection of a larger amount at a higher temperature,
This
is
represented schematically in Fig. 7-5.
No
engine has ever been
developed that converts the heat extracted from one reservoir into work without rejecting some heat to a reservoir at a lower temperature. This negative statement,
which
is
the result of engineering experience, constitutes the
second law of thermodynamics and has been formulated in several ways. The original statement of Kelvin is: "It is impossible by means of inanimate material agency to derive mechanical effect from any portion of matter by
below the temperature of the coldest of the surrounding objects." In the words of Planck, "It is impossible to construct an engine which, working in a complete cycle, will produce no effect other than the raising of a weight and the cooling of a heat reservoir." We may combine these statements into one equivalent statement, to which we shall refer hereafter as the cooling
system undergoing the cycle
The rise
Stirling cycle
Stirling refrigerator
diagrams shown 1
^2
If
the second law were not true,
While the
would be air.
gives ideal
and the accompanying PV diagram of Fig.
right piston remains stationary, the left piston
7-6A.
moves and
Both pistons move the same amount simultaneouslv, forcing eas through the regenerator, giving up some heat Q K to the regenerator, and emerging cold in the right-hand space. This takes
3-^4
While
place at constant volume.
absorption of heal from a reservoir
and
possible to drive a steamship
by extracting heat from the ocean or to run a
by extracting heat from the surrounding
when reversed, it The operation of an
best be understood with the aid of the schematic
in Fig. 7-6a,
the left piston remains stationary, the right piston
down and it
called a refrigerant.
capable of being reversed and,
may
A
and the
2-*3
the conversion of this heal into work.
across the ocean
is
on the system.
called a refrigerator,
up, compressing the gas isothermally at the temperature On rejecting heat Qu to the hot reservoir.
Kelvin-Planck statement of the second law, thus: is possible w/iose sole result is the
is
is
to one of the most useful types of refrigerator.
it
No process
and a net amount of work done
device that performs a cycle in this direction
The
student should notice that
law of thermodynamics. After all, both the ocean and the surrounding air contain an enormous store of internal energy, which, in principle, may be extracted in the form of a neither of these "impossibilities" violates the
first
ture Be, during which heat
power plant
moves
causes an isothermal expansion at the low tempera-
Qc
is
absorbed by the gas from the
cold reservoir.
4->l
Both pistons move and force gas at constant volume from the cold to the hot end through the regenerator, thereby taking up approximately the same heat Qn that was supplied to the regenerator in process 2
—
>
3.
180
7-6
Heat and Thermodynamics
_—
Hot
r
,
Law
181
Cow
Hot
Cold
Engines, Refrigerators, and the Second
*
ssfflfe I
—2
1
—
Hot,-,
-3
2
—.Cold
Hot
.Cold
P.
3
—-4
4
—-1
(a)
Volume V
l*'ig.
7-7
(a)
Elementary refrigeration plant, (h)
PV
diagram of commercial refrigerator
cycle.
The
Stirling refrigeration cycle has
been
utilized in recent years
by a num-
ber of engineering firms in the construction of practical refrigerators for the
down to 12°K. As in the case of the Holland has been in the forefront in
production of low temperatures, from 90 .Stirling
engine, the Philips
Company
in
and construction of large industrial installations. (Miniature units suitable for academic laboratories are available from North American Philips, A. D. Little, Hughes Aircraft, and Malakcr Laboratories.) In order to gain a little more insight into the working of a refrigerator, let us consider some of the details of a commercial refrigeration plant that are reflected in most electric home refrigerators. The schematic diagram in the design
e„
Qc
4
C
v (&) Fig. 7-6 (/»)
(a) Schematic
diagram of steps in the operation of idealized Stirling
Idealized Stirling refrigeration cycle on
PV
diagram.
refrigerator.
Fig. 7-7a shows the path of a constant mass of refrigerant as it is conveyed from the liquid storage, where it is at the temperature and pressure of the condenser, through the throttling valve, through the evaporator, into the compressor, and finally back to the condenser. In the condenser the refrigerant is at a high pressure and at as low a temperature as can be obtained with air or water cooling. The refrigerant is always of such a nature that, at this pressure and temperature, it is a saturated
1
Heat and Thermodynamics
82
liquid.
When
a fluid passes through a narrow opening (a needle valve) from
and work
W
is
done by the
u
Joule-Tliomson or Joule-Kelvin expansion. This process will be considered in
Chap.
detail in
11. It
motor that operates the
electric
a region of constant high pressure to a region of constant lower pressure adiabatically, it is said to undergo a throating process (see Prob. 4-8), or a
some
Engines, Refrigerators, and the Second
7-6
_ Qc
W
Law
183
refrigerator, then
Qc
=
-
Q„
Qc
a property of saturated liquids (not of gases)
is
that a throttling process always produces cooling
and
partial vaporization.
In the evaporator the fluid is completely vaporized, with the heat of vaporization being supplied by the materials to be cooled. The vapor is then com-
The
may be considerably larger argument, one assumes the value 5,
coefficient of performance of a refrigerator
than unity.
If,
for the sake of
pressed adiabatically, thereby increasing
the vapor
The
is
cooled until
it
in temperature. In the condenser, condenses and becomes completely liquefied.
ideal refrigeration cycle depicted on a PI"
result of ignoring the usual difficulties
due
diagram
to turbulence, friction,
ammonia
In Fig. 7-76, two isotherms of a fluid such as
etc.
shown on a
PV diagram— one
heat
losses,
the other at 6c, the temperature of the evaporator. Starting at the point
representing the state of
lb of saturated liquid at the
1
But
"
Qc
-
W= Q,i
-
5
"
W-
Qn-W_ W
or frcon are
temperature of the condenser, and
at 6u, the
w
in Fig. 7-7 A is the
hence
1,
temperature and pres-
W
sure of the condenser, the commercial refrigeration cycle comprises the follow-
°>
ing processes:
—
1
»
Throttling process involving a drop of pressure and tempera-
2
ture.
The
states
between the
initial
and
final states of a fluid
during a throttling process cannot be described with the aid of
thermodynamic coordinates
referring to the system as a
gram. Hence the 2
—» 3
scries of
dashes between
1
and
Isothermal, isobaric vaporization in which heat
by the refrigerant at the low temperature
6c,
whole
PV
and, therefore, cannot be represented by points on a
dia-
2.
Qc
—*
absorbed
thereby cooling
4
—
than that of the condenser
6//.
Isobaric cooling and condensation at 6n.
> 1
The purpose
of any refrigerator
is
to extract as
a cold reservoir with the expenditure of as
"output," so to speak, ''input"
is
work.
refrigerator energy ratio),
is
A
in
one
much
little
heat as possible from
work
as possible.
The
and the
of performance
to
(also called the cooling
where
cycle, heat
=
— — refrigerant
heat extracted from cold reservoir
Qc
—
:
work done on is
pointed out in
The
1
852 by Lord Kelvin,
who designed a machine
device was never built, however, and
Haldane, about 75 years
later, to utilize
it
remained
for
the principle and heat his house in
air, supplemented by city water. Since about 1938, many devices known as "heat pumps" have appeared on the market for warming the house in winter by refrigerating either the ground, air,
or the water supplied in the mains.
reversing the flow of refrigerant, the heat
convenient measure, therefore, of the performance of a coefficient
first
for the purpose.
the outside
the heat extracted from the cold reservoir,
expressed by the
w If,
is
refrigerating the outdoors.
Scotland by refrigerating the outdoor
Adiabatic compression of the vapor to a temperature higher
4
;
This was is
the materials of the cold reservoir. 3
and therefore the heat liberated at the higher temperature is equal to six times the work done. If the work is supplied by an electric motor, for every joule of electrical energy supplied, 6 J of heat will be liberated whereas if 1 J of electrical energy were dissipated in a resistor, one could obtain at most 1 J of heat. Consequently, it would seem to be highly advantageous to heat a house by
zr-.
absorbed by the refrigerant from the cold reservoir
pump may
By turning
a valve and
be used to cool the house in summer, as shown in Fig. 7-8. Various commercial units have coefficients of performance ranging from about 2 to 7. The design, installation, and operation of such units arc now an important branch of engineering. The operation of a refrigerator may be symbolized by the schematic diagram shown in Fig. 7-9, which should be compared with the corresponding engine diagram of Fig. 7-5. Work is always necessary to transfer heat from a cold to a hot reservoir. In household refrigerators, this work is usually done by an electric motor, whose cost of operation appears regularly on the monthly bill. It would be a boon to mankind if no external supply of energy were needed, but it must certainly be admitted that experience indicates the also
184
Heat and Thermodynamics Air from
7-7
rooms
Heated
Engines, Refrigerators, and the Second
Law
185
rooms
air to
Condenser Cooled
pressure
Motor
v
|
I
|
^——J
-.-
Compressor
I
»ld
Fig. 7-9
(a) Air
from rooms
Cooled
air to
rooms
At
reservoir
Symbolic representation of refrigerator.
first sight,
the Kelvin-Planck and the Clausius statements appear to be
quite unconnected, but
we
shall see
immediately that they are in
all
respects
equivalent. Heated
Equivalence of Kelvin-Planck and Clausius Statements
7-7
Conde
air to
outdoors
Let us adopt the following notation: A'
High pressure
Two
(a)
Healing the house by refrigerating
truth of falsity
Kelvin-Planck statement,
the.
of the Kelvin-Planck statement,
truth of the Clausius statement, falsity of the
Clausius statement.
propositions or statements arc said to be equivalent
the outside air.
heating the outside air. (J. Partington, Jr., G.E. Educational Service
(b) Cooling the house by
News, December, 1951.)
contrary. This negative statement leads us to the Clausius] statement of Uie
K= C
KD C
/hen
and
second law:
No
process
is
possible whose sole result is the transfer of heal from a cooler to a
hotter body. T
R-
J.
Clausius (pronounced Klow'-zee-oos).
when
the truth of
one implies the truth of the second and the truth of the second implies the truth of the first. Using the symbol Z) to mean "implies" and the symbol = to denote equivalence, we have, by definition,
(6) rig. 7-8
=
— A' = C = —C =
Now,
it
may
easily be
shown that
Ki C
C
DK
also
Engines, Refrigerators, and the Second
Heat and Thermodynamics
186
-AD -C
when
Thus,
and
-CD
order to demonstrate the equivalence of
in
that a violation of
one statement implies
Law
1
87
-A'.
A'
and
C,
we have
a violation of the second,
to show-
and vice
versa.
1
To
prove that
— C Z) — A,
left-hand side of P'ig. 7-10, of heat
consider a refrigerator,
which
a cold reservoir to a hot reservoir
from
shown
requires no work to transfer
in
the
Q> units
Engine
Refrigerator
and which therefore
Suppose that a heat engine (on the between the same two reservoirs in such a way operates
violates the Clausius statement. right) also
that heat
Qj
delivered to the cold reservoir.
is
The
docs not violate any law, but the refrigerator
engine, of course,
and engine
together
constitute a self-acting device whose sole effect is to take heat Qi — Q-> from the hot reservoir and to convert all this heat into work. Therefore the refrigerator and engine together constitute a violation
Fig. 7-11
and
Proof that
—K
D
—C. The
refrigerator acting together constitute
engine on the left
is
a violation of K;
the engine
a violation of C.
of the Kelvin-Planck statement.
To
prove that
—A3
~~
C
s
consider an engine,
shown on the
left-
We
therefore arrive at the conclusion that both statements of the second
and hand side of Fig. 7-11, which rejects no statement. that Suppose which therefore violates the Kelvin-Planck a refrigerator (on the right) also operates between the same two reservoirs and uses up all the work liberated by the engine. The refrigerator violates no law, but the engine and refrigerator together constitute a self-acting device whose sole effect is to transfer heat Q< from the cold to the hot reservoir. Therefore the engine and refriger-
law are equivalent.
ator together constitute, a violation of the Clausius statement.
gas cycle. All processes are quasi-static, and
heat to the cold reservoir
It
is
a matter of indifference which one
l
Fig. 7-10 refrigerator
Proof that
and engine
—C
D
7-1
Figure P7-1 represents a simplified
PV diagram Cp
is
The
refrigerator on the left is
P
2
acting together constitute a violation of K,
a violation of C; the Fig. P7-1
Joule ideal-gas
cycle.
of the Joule ideal-
constant. Prove that the
Volume
—A.
used in a
PROBLEMS
Pressure
W=Q -Q
is
particular argument.
V
" 188
Heat and Thermodynamics
Engines, Refrigerators, and the Second
thermal efficiency of an engine performing
this cycle
Figure P7-2 represents a simplified
PV
diagram of the Sargent
ideal-gas cycle. All processes are quasi-static,
and
the heat capacities are
Pressure
189
P7-3 represents an imaginary ideal-gas engine cycle. constant heat capacities, show that the thermal efficiency is Assuming 7-3
is
Figure
,= 7-2
Law
7-4
on
P
a
PI-'"
An
(V1/V2) 1
(Pi/ Pi)
-
1
l'
ideal-gas engine operates in a cycle which,
diagram,
is
a rectangle. Call Pi, Pj the lower
when
represented
and higher
pressures,
and higher volumes, respectively. Calculate the work done in one cycle. (a) Indicate which parts of the cycle involve heat flow into the gas, and (b) calculate the amount of heat flowing into the gas in one cycle. (Assume
respectively; call Vi, V-2 the lower 3
constant heat capacities.) (c)
Show
that the efficiency of this engine
V
7-
=
Volume Fig.
P7-2
7-5
V
constant. Prove that the thermal efficiency of an engine performing this cycle is
_
d
J-Jl — 02
03
p p3
.
p2
vessel contains
600
V\
Pi
cm
3
'
Vi
-
Ft
of helium gas at
2°K and jV aim. Take
the zero of internal energy of helium to be at this point.
Sargent ideal-gas cycle.
1
A
1
yPo
P2-
is
^^
(a) The temperature is raised at constant volume to 288°K. Assuming helium to behave like an ideal monatomic gas, how much heat is absorbed, and what is the internal energy of the helium? Can this energy be regarded as stored heat or stored work? (b) The helium is now expanded adiabatically to 2°K. How much work is done, and what is the new internal energy? Has heat been converted into work without compensation, thus violating the second law? (c) The helium is now compressed isothennally to its original volume. What are the quantities of heat and work in this process? What is the efficiency of the cycle? Plot on a PV diagram. 7-6 Derive the expression for the efficiency of an ideal engine operating in an air-standard Diesel cycle, Eq. (7-4).
7-7 (a)
shown
Assume an
ideal gas to
undergo the idealized Stirling engine cycle
in Fig. 7-16 with perfect regeneration.
Show
that
2 1
1
1
9c
1
1
v. Fig.
P7-3
v,
V (b)
If this cycle
is
reversed, calculate the coefficient of performance.
Heat and Thermodynamics
190
7-8 In ihc tropics the water near the surface of the ocean is warmer than the deep water. Would an engine operating between these two levels violate the second law? 7-9 A storage battery is connected to a motor, which is used to lift a weight.
The
battery remains at constant temperature by receiving heat from
the outside air. Is this a violation of the
7-10
second law?
There are many paramagnetic
solids
Why?
whose internal energy
is
8. REVERSIBILITY AND
THE KELVIN TEMPERATURE SCALE
a
function of temperature only, like an ideal gas. In an isothermal demagnet-
absorbed from one reservoir and converted completely into
ization, heat
is
work.
a violation of the second law?
7-11
Is this
Would an atomic-energy power
Why? plant violate either the
first
law
or the second law of thermodynamics? Kxplain.
8-1
Reversibility
and
Irreversibility
In thermodynamics, work is a macroscopic concept. The performance of work may always be described in terms of the raising or lowering of an object or the winding or unwinding of a spring, i.e., by the operation of a device that serves to increase or decrease the potential energy of a mechanical sys-
tem.
Imagine, for the sake of simplicity, a suspended object coupled by
means of suitable pulleys to a system tem can be described in terms of
so that
any work done by or on the
sys-
the raising or lowering of the object.
Imagine, further, a series of reservoirs which may be put in contact with the system and in terms of which any flow of heat to or from the system may be described.
We
shall refer to the
suspended object and the
as the local surroundings of the system.
The
series of reservoirs
local surroundings are, therefore,
which interact directly with the system. Other mechanical devices and reservoirs which are accessible and which might interact with the system constitute the auxiliary surroundings of the system those parts of the surroundings
want of a better expression, the rest of the universe. The word "universe" used here in a very restricted technical sense, with no cosmic or celestial
or, for is
implications. sisting of the
The
universe merely
means a
portion of the world con-
finite
system and those surroundings which
may
interact with the
system.
Now an
suppose that a process occurs
in
which
the system proceeds from
(1)
to a final state /; (2) the suspended object is lowered to an extent that M7 units of work arc performed; and (3) a transfer of heat Q takes initial state
i
place from the system to the scries of reservoirs. process, the system its
former
level,
may be
and the
restored to
If,
its initial
at the conclusion of this
state
;'.
the object lifted to
with the same amount of any other mechanical device or
reservoirs caused to part
heat Q, without producing
any changes
in
reservoir in the universe, the original process
is
said to be reversible. In other
192
I
words, a
and Thermodynamics
[eat
reversible process is one that is
oj the process, both the system states,
not
8-2
and
performed
without producing any changes in the
fulfill
The
these stringent requirements
of the universe.
rest is
question immediately arises as to
a way that, at the conclusion may be restored to their initial
in such
the local surroundings
A
W (Work)
Thermally
%
insulated
said to
familiar processes of nature, are reversible or not.
We
shall
show
193
and the Kelvin Temperature Scale
'///////////////////////////////////A
process that docs
be irreversible. whether natural processes,
Reversibility
system
the
i.e.,
that
it is
U,
- Uf =
W
I
I '/y^^^^^w zw^/
a
r
consequence of the second law of thermodynamics that arc irreversible.
By considering
all
natural processes
representative types of natural processes
and
examining the features of these processes which are responsible for their irreversibility, we shall then be able to state the conditions under which a process
may
take place reversibly.
Adiabalic transformation of work into internal energy of a system.
Fig. 8-2
extracted from the reservoir and converted completely into work. Since
would involve a violation of the second law,
this
all
processes of the above type
set of processes involves the adiabatic
transformation of work into
are irreversible.
Another
External Mechanical Irreversibility
8-2
internal energy of a system. This illustrated
There
is
by
the following five examples:
Irregular stirring of a viscous liquid in contact with a reservoir.
Coming
by the following examples, similar
to the preceding
is
list:
1
Irregular stirring of a viscous thermally insulated liquid.
2
Coming
3
Inelastic deformation of a thermally insulated solid.
4
Transfer of electricity through a thermally insulated
5
Magnetic
to rest of a rotating or vibrating thermally insulated
1
iquic
resistor.
hysteresis of a thermally insulated material.
to rest of a rotating or vibrating liquid in contact with a
reservoir. Inelastic
A deformation of a solid
in
contact with a reservoir.
Magnetic
hysteresis of a material in contact
In order to restore the system and
its
with a reservoir.
local surroundings to their initial
without producing changes elsewhere,
Q
units of heat
would have
process of this type
to
be
is
accompanied by a
rise of
temperature of the system
In order to restore the system and
its local surroundings to producing changes elsewhere, the internal energy of the system would have to be decreased by extracting Uj — U, units of heat, thus lowering the. temperature from 6/ to 0,-, and this heat would have to be completely converted into work. Since this violates the second law, all processes of the above type are irreversible. The transformation of work into internal energy cither of a system or of a reservoir is seen to take place through the agency of such phenomena as
from, say,
Transfer of electricity through a resistor in contact with a reservoir.
states
depicted schematically in Fig. 8-2 and
a large class of processes involving the isothermal transformation of
work through a system (which remains unchanged) into internal energy of a reservoir. This type of process is depicted schematically in Fig. 8-1 and is illustrated
is
6;
to
9/.
their initial states without
and magnetic hysteresis. and the work is said to be dissiof work into internal energy are
viscosity, friction, inelasticity, electric resistance,
These Heat reservoir
effects are
known
as dissipative
effects,
pated. Processes involving the dissipation
at constant
said to exhibit external meclianical irreversibility.
temperature
experience that dissipative
moving
It
is
devices. Friction, of course,
may
Fig. 8-1
Isothermal transformation of work through a system (which remains unchanged)
into internal energy of a reservoir.
it
shown that it can never be completely elimimovable device could be kept in continual operation either of the two laws of thermodynamics. Such a continual
could, a
without violating
motion
is
known
are always present in
be reduced considerably by suitable
lubrication, but experience has
nated. If
a matter of everyday
effects, particularly friction,
as perpetual motion of the third kind.
194 8-3
Heat and Thermodynamics
8-5
Internal Mechanical Irreversibility
The. following very important natural processes involve the transformation
and then back
into
To
internal energy again: its
vacuum
1
Ideal gas rushing into a
2 3
Gas seeping through a porous plug (throttling process). Snapping of a stretched wire after it is cut.
4
Collapse of a soap film after
it is
restore, at the conclusion of
local surroundings to their
a process of this type, both the system and without producing changes else-
initial states
where, heat would have to be transferred by means of a self-acting device from a cooler to a hotter body. Since this violates the second law (Clausius
(free expansion).
statement),
all
processes of this type are irreversible.
to exhibit external thermal
A
pricked.
shall
prove the irreversibility of only the
process involving a transfer of heat between parts of the
'",
W
is
Such a process
is
said to
exhibit internal thermal irreversibility.
8-5
is
opened, there
is
crystal form, etc.
some of the internal energy into kinetic energy of "mass motion" or "streaming," and then this kinetic energy is dissipated through viscosity into internal energy again. Similarly, when a stretched wire is cut, there is first a transformation of internal energy into kinetic energy of irregular motion and of vibration and then the dissipation of this energy through inelasticity into internal energy again. In all the processes, the first energy transformation takes place as a result of mechanical instability, and the second by virtue of sonic dissipative effect. A process of this sort is said to exhibit internal mechanical irreversibility.
External and Internal Thermal Irreversibility
Consider the following processes involving a transfer of heat between a system and a reservoir by virtue of a finite temperature difference:
Some important examples
follow.
Formation of new chemical constituents: 1
All chemical reactions.
Mixing of two
different substances:
2
Diffusion of
3
Mixing
a
transformation of
Chemical Irreversibility
Some of the most interesting processes that go on in nature involve a spontaneous change of internal structure, chemical composition, density,
irreversible.
In a free expansion, immediately after the stopcock
8-4
same system by
also obviously irreversible
virtue of the Clausius statement of the second law.
I
impossible, the process
is
first.
During a free expansion, no interactions take place, and hence there are no local surroundings. The only effect produced is a change of state of an ideal gas from a volume and temperature $ to a larger volume Vf and the same temperature 8. To restore the gas to its initial state, it would have to be compressed isothermally to the volume Vi. If the compression were performed quasi-statically and there were no friction between the piston and would have to be done by some outside cylinder, an amount of work mechanical device, and an equal amount of heat would have to flow out of the gas into a reservoir at the temperature 6. If the mechanical device and the reservoir arc to be left unchanged, the heat would have to be extracted from the reservoir and converted completely into work. Since this last step is
Such processes are said
irreversibility.
because of nonuniformity of temperature
We
195
and the Kelvin Temperature Scale
Conduction or radiation of heat from a system to a cooler reservoir. Conduction or radiation of heat through a system (which remains unchanged) from a hot reservoir to a cooler one.
1
2
of internal energy of a system into mechanical energy
Reversibility
two
dissimilar inert ideal gases.
of alcohol
and water.
Sudden change of phase: 4 Freezing of supercooled liquid. 5
Condensation of supersaturated vapor.
Transport of matter between phases in contact: 6
Solution of solid in water.
7
Osmosis.
Such processes are by far the most difficult to handle and must, as a rule, be treated by special methods. Such methods constitute what is known as chemical thermodynamics and are discussed in the last few chapters of this book. It can be shown that the diffusion of two dissimilar inert ideal gases is equivalent to two independent free expansions. Since a free expansion is irreversible, it follows that diffusion is irreversible. The student will have to accept at present the statement that the above processes are irreversible. Processes that involve a spontaneous change of chemical structure, density, phase,
etc.,
are said to exhibit chemical
irreversibility.
196
Icat
I
8-7
and Thermodynamics
and
Temperature Scale
the Kelvin
mately reversible process. Similar considerations apply
Conditions for Reversibility
8-6
Reversibility
to a
197
wire and to a
surface film.
Most
processes that occur in nature are included
among
the general types
of process listed in the preceding articles. Living processes, such as cell division, tissue growth, etc., are no exception. If one takes into account all
the interactions that ble.
accompany
living processes, such processes are irreversi-
a direct consequence of the second law of thermodynamics that
It is
all natural spontaneous processes are irreversible.
A
careful inspection of the various types of natural process
shows that
all
reversible transfer of electricity takes place.
conditions for mechanical, thermal, or chemical equilibrium,
thermodynamic equilibrium, are not
quasi-static only, often
satisfied.
Dissipative effects, such as viscosity, friction, inelasticity, electric
2
caused to rotate until
thermodynamic systems, it is often necessary to invoke some sort of process in which the system passes through these states. To assume the process to be
The i.e.,
Suppose that a motor whose coils have a negligible resistance is its back cmf is only slightly different from the cmf of the cell. Suppose further that the motor is coupled cither to an object suspended from a frictionless pulley or to an clastic spring. If neither the cell itself nor the connecting wires to the motor have appreciable resistance, a as follows:
In order to arrive at conclusions concerning the equilibrium states of
involve one or both of the following features:
1
A reversible transfer of electricity through an electric cell may be imagined
resistance,
and magnetic
there
may
is
not
sufficient, for if dissipative processes
(envelopes, containers, surroundings) that
hysteresis, are present.
are present
be heat flows or internal energy changes of neighboring systems
may
limit the validity of the argu-
ment. In order to ensure that equilibrium states of the system only are con-
For a process to be reversible, it must not possess these features. If a process is performed quasi-statically, the system passes through states of thermodynamic equilibrium, which may be traversed just as well in one direction as in the opposite direction. If there are no dissipative effects, all the work done by the system during the performance of a process in one direction can be returned to the system during the reverse process.
We
the assumptions
a process
made
will
so often in mechanics, such as those
which
refer to
weightless strings, frictionless pulleys, and point masses.
A heat reservoir was defined as a body of very large mass capable of absorbing or rejecting an unlimited supply of heat without suffering appreciable
changes
in its
thermodynamic coordinates. The changes that do take place and so minute that dissipative actions never develop. There-
are so very slow fore,
when
are the
heat enters or leaves a reservoir, the changes that take place in the reservoir
same as
those which,
would
lake place if the
same quantity of heat were transferred
reversihly.
possible in the laboratory to approximate the conditions necessary performance of reversible processes. For example, if a gas is confined in a cylinder equipped with a well-lubricated piston and is allowed to expand very slowly against an opposing force provided cither by an object suspended from a frictionless pulley or by an clastic spring, the gas undergoes an approxiIt
is
for the
— without having to take account of the
the system
itself
concept of a
effect of dissipated
—
or in sonic other neighboring body
reversible process,
even though
this
it is
assumption
work
in
useful to invoke the
may
at times
seem
a bit drastic.
are led, therefore,
be reversible when (1) it is performed quasi-statically and (2) it is not accompanied by any dissipative effects. Since it is impossible to satisfy these two conditions perfectly, it is obvious that a reversible process is purely an ideal abstraction, extremely useful for theoretical calculations (as we shall see) but quite devoid of reality. In this sense, the assumption of a reversible process in thermodynamics resembles to the conclusion that
sidered
8-7
Existence of Reversible Adiabatic Surfaces
Up
to this point, the only consequence of the second law of thermo-
dynamics that has been drawn processes.
along either of two
and
is
the irreversibility of natural, spontaneous
To develop further consequences,
Clausius;
lines: the
it
has been customary to proceed
engineering method, due to Carnot, Kelvin,
and the axiomatic method, due to the Greek mathematician f The engineering method is based upon the Kelvin-Planck
C. Carathcodory.
formulation of the second law or starts
first
Carnot cycle
its
equivalent, the Clausius statement.
One
by defining a particularly simple reversible cycle called the and then proving that an engine operating in this cycle between
reservoirs at
two different temperatures
is
more efficient than any other engine
operating between the same two reservoirs. After proving that
all Carnot have the same efficiency, regardless of the substance undergoing the cycle, the Kelvin temperature scale is defined so as to be independent of the properties of any particularkind of thermometer. A theorem called the Clausius theorem is then derived, and from it the existence of the entropy function. The engineering method of
engines operating between the
same two
reservoirs
developing the consequences of the Kelvin-Planck or Clausius statements of t
C. Carathcodory, Math. Ann., 67:355 (1909) (in German).
Heat and Thermodynamics
198
8-7
Reversibility
199
and the Kelvin Temperature Scale
is rigorous and general. If one is interested in the design and manufacture of engines and refrigerators, it is essential to employ principles that hold regardless of the nature of the materials involved. If, however, one
the second law
is
interested in the behavior of systems, their coordinates, their equations of
state, their properties, their processes, etc., apart
from their use
in the cylinders
of engines and refrigerators, then it is desirable to adopt a method more closely associated with the coordinates and equations of actual systems.
decade of the twentieth century Caratheodory, to replace the Kelvin-Planck and Clausius statements of the second law, presented this In the
first
axiom: In
the.
neighbor hood (however close) of any equilibrium slate of a system of any
number of thermodynamic
coordinates, there exist states that cannot be readied (are
He showed how to derive the Kelvin from this axiom and how to derive every other consequence temperature scale method. Physicists (Born, Ehrenfest, Lande) recognized of the engineering Caratheodory's work, but since the mathematics needed to the importance of Caratheodory's axiom presented much more difficulty than the deal with involving outputs and inputs of engines and refrigersimple manipulations slow in adopting his methods. In recent years, ators, other physicists were Pippard, Turner, Landsberg, and Sears, f mainly because of the activities of has been conthe mathematical machinery of the Caratheodory method inaccessible) by reversible adiabatic processes.
siderably simplified,
and now
it
appears that
the
axiom
itself
can be dispensed
All the consequences of the Caratheodory axiom follow directly from the Kelvin-Planck statement of the second law. Consider a system described with the aid of five thermodynamic coordinates: the empiric temperature t, measured on any scale whatsoever; two
Fig. 8-3
Both J\ and fa, lying on versible adiabatic processes from i.
a
line of constant
X and A",
cannot
be reached by re-
willi entirely.
X
ments
Y and
and two corresponding generalized displaceand X' For such a system, the first law for a reversible process is
generalized forces
}";
.
dQ = d U
+
Y
dX+
Y'
dX\
cated. Let f, be an equilibrium state that the system can reach by means of a reversible adiabatic process. Through /i draw a vertical line for which the values of A" and X' arc constant at every point. Let 2 be any other equilibrium
/
state
on
We now
proceed to prove that both, states /, and /2 cannot be reached by reversible adiabatic processes from i. Assume that it is possible for the system to proceed along either of the two reversible adiabatic paths i —> /, or i -» / 2 Let the system start at i, proceed to fh then to /2 and then back to along f-> —> i, which, being a reversible path, can be traversed in this vertical line.
.
,
;'
and because of the existence
of
coordinates arc independent. At
U, X, and A". reasons: (1)
it
A
two equations of first, let
state,
only three of the
us choose these coordinates to be
system of three independent variables
is
chosen for two
enables us to use simple three-dimensional graphs, and (2) all
conclusions concerning the mathematical properties of the differential will hold equally well for all systems with more or fewer than three
dQ
independent
variables.
In Pig. 8-3, the three independent variables U, X, and X' arc plotted along three rectangular axes, t A. Press,
and an
arbitrarily chosen equilibrium state
i
is
indi-
Pippard, "Elements of Classical Thermodynamics," Cambridge University New York, 1957, p. 38; L. A. Turner, Am. J. Phys., 28:781 (1960); P. T. Landsberg,
Nature,
B.
201:485 (1964); F.
W.
Scars, Am. J.
Pliys.,
31:747 (1963).
either direction. Since
energy at constant
from the
first
/2
lies
above
/i,
the system undergoes
X and A", during which process no work
law that
heat
Q
is
an
done.
must be absorbed in the process /, ~*
increase of It
follows
/...
In the
W
reversible adiabatic processes, however,
no heat is transferred but work is done. In the entire cycle //i/V, there is no energy change, and therefore Q = IV. The system has thus performed a cycle in which the sole effect is the absorption of heat and the conversion of this heat completely into work. Since
Kelvin-Planck statement of the second law, it follows cannot be reached by reversible adiabatic processes. Only
this violates the
that both
f and x
/»
one point on the line of constant process from
X and X'
can be reached by a reversible adiabatic
i.
For a different
line (different
X, and X'f ),
there
would be another
single
1
point accessible from
such points, from
3-8
Heat and Thermodynamics
200
i
/i, /a, etc.,
and
Temperature Scale
the Kelvin
201
by a reversible adiabatic process, and so on. A few are shown in Fig. 8-4. The locus of all points accessible
by reversible adiabatic processes
i
Reversibility
in oilier words, these points lie on
is
a space of dimensionality one
less than three;
a two-dimensional surface. If the system were
described with the aid of four independent coordinates, the states accessible from any given equilibrium state i by reversible adiabatic processes would lie
on a three-dimensional hypersurface, and so on. In
what
is
to follow,
it is
more convenient
to choose as
one of the independ-
ent coordinates the empiric temperature / instead of the energy U. Since, for a given i, a reversible adiabatic surface has been shown to exist in a UXX' space, such a surface must also exist in a tXX' space, although
its
shape might
be quite different. With a system of three independent coordinates t, X, and X', the reversible adiabatic surface comprising all the equilibrium states that are accessible
from
i
by
reversible adiabatic processes
where a represents some
as yet
may
=
be expressed by the equation
const.,
(8-1)
undetermined function. Surfaces correspond-
ed Pig. 8-5 the second
If
two
reversible adiabatic surfaces could intersect,
law by performing
ing to other
it
would be
possible to violate
the cycle if\f%i.
initial states
would be represented by
different values of the
constant.
Reversible adiabatic surfaces cannot intersect, for possible, as
shown
in Fig. 8-5, to
proceed from an
if
they did
initial
it
would be
equilibrium state
i
on the curve of intersection to two different final states /j and ft, having the same { and Xf along reversible adiabatic paths. We have just shown that
X
this
X
is
,
impossible.
8-8
Integrability of dQ
It
has been emphasized that
are no functions
AW and
dQ
are inexact differentials; there
W and Q representing, respectively,
the work and heat of a can be described with the aid of two independent thermodynamic coordinates, say, a temperature t (on any scale) and a generalized displacement A', then if Y is the generalized force,
body.
Fig. 8-4 surface.
All stales that can be reached by reversible adiabatic processes starling at
i lie
on a
When
a system
dQ =
dlJ
+
y
dX.
202
Heat and Thermodynamics 8-8
Regarding
U as
a function of
/
and
.V,
Reversibility
and
the Kelvin
any point in this where the constant can take on various values a x a-i, cr along with Zand X', so the value of space may be determined by specifying function as well as Y and Y' as energy U that we may regard the internal .
,
functions of
a,
dt
dt
Solving for dt/dX,
we
dX =
0,
3-2)
...
dU
dQ =
and
must be true
Y + (dU/d X) \dXj R The
right-hand
member
is
(du/ai) x
known
known
at all points.
Equation
(8-2) has therefore a solution consisting of
family of curves, and the curve through
o(t,X)
A set of curves The
is
obtained
and X, and therefore the an adiabat on a IX diagram, is
when
=
a
any one point may be written
to the
=
dX
dX'
,
dX',
dU
dU
[7 + dX dX
dX'
dX'.
X, and X' arc independent variables,
for all values of da, dX,
=
const.)
is
this
(8-4)
equation
and dX' Suppose that two of the .
is
not.
The
differ-
provision that da
=
the condition for an adiabatic process in
form
dQ =
const.
different values are assigned to the constant.
existence of the family of curves a(t,X)
.
which dQ = 0, and therefore the coefficient of dX' must vanish. If we take da and dX' to be zero, then by the same reasoning the coefficient of dX must vanish. It follows therefore that, in order for the coordinates a, X, X' to be independent, and also for dQ to be zero whenever da is zero, the equation for dQ must reduce (or a
as a function of /
derivative dt/dX, representing the slope of
dU
,
da
da and dX, are zero and that dX'
entials,
t
dU dU = -rdo- + -~ dX
g?* + da
Since the coordinates
get
.
X, and X'. Then,
where (dU/dt)x, Y, and {dU/dX), are known functions of/ and X. A reversible adiabatic process for this system is represented by the equation
dU
203
Temperature Scale
If
we
define
(8-5)
\oa /x,x-
a function X by the equation
const., representing reversible adiabatic
from the fact that there are only two independent coordinates, and not from any law of physics. When three or more independent coordinates are needed to describe a system, the situation is quite different. The second law of thermodynamics is
processes, follows
(8-6)
we
get the result
dQ =
needed to enable us to conclude that:
Through any arbitrary surface,
and
family of
noninlerseeling surfaces.
first
law
According to its definition, given in Eq. (8-6), X is a function of cr, X, and a is a function of t, X, and A", however, we may imagine A" to be eliminated, with the result that X is a function of t, a, and A'. It is seen from Eq. (8-7) that the function 1/X is an integrating factor, such that when dQ is multiplied by 1/X the result is an exact differential da. Now, A". Since
Consider a system whose coordinates are the empiric temperature
The
Y and
Y',
for a reversible
t, two and two generalized displacements X and X' process is expressed by the equation
an
dQ = dU + Y dX +
V dX',
infinitesimal of the type
(8-3)
P where U,
Y,
(8-7)
initial-slate point, all reversible adiabatic processes lie on a
reversible adiabatics through other initial states determine a
generalized forces
X da.
and Y' are functions of t, X, and X'. Since the
f,
X, X' space
subdivided into a family of nonintersecting reversible adiabatic surfaces, a(t,X,X')
=
dx
+ Q dy + R dz +
,
is
known as a linear differential form or a or more independent variables, docs
Pfaffian expression,
when
involves three
it
not admit, in general, of
an integrating
const., factor. It
is
only because of the existence of the second
law
that the differential
form for
204
Heat and Thormoctvnan:
8-9
a
common
with
five
temperature
Reversibility
and die Kelvin Temperature Scale
and together they
/,
2
3
independent coordinates.
Composite system.
and
and the
X',
The
five
independent coordinates are
Using the equation
Two
reversible adiabatic surfaces, infmiksimally close.
dQ
referring to a physical system of any
an integrating
Two
=
X do
When
of
o,
X, X', X,
for o of the
a
by
of these independent variables.
main system, we may express X'
in
terms
and X. Similarly, using the equation for 6 of the reference system, X'
the process is rep-
t The diacritical called a circumflex.
is transferred.
mark
or accent over the symbols referring to the reference system
is
number of independent coordinates possesses
factor.
infinitesimally neighboring reversible adiabatic surfaces are
Fig. 8-6.
t,
/,
reversible adiabatic hypersurfaces are specified
different values of the function
Fig. 8-6
constitute a composite system
Main system. The three independent coordinates arc /, A', and A"', and the reversible adiabatic surfaces arc specified by different values of the function a of I, X, and A". When heat dQ is transferred, o changes by do; and dQ = \ do where X is a function of t, o, and X. The three independent coordinates are t, X, and Reference system.^ reversible adiabatic surfaces arc specified by different the X', and of /, X, and X'. When heat dQ is transferred, of the function o values = X do where X is a function of /, o, and X. changes by do, and dQ
1
resented by a curve connecting the two surfaces, heat <\Q
205
One
shown
in
characterized by a constant value of the function o, and the other by a slightly different value o -\- do-. In any process represented
by a curve
surface
is
on either of the
two surfaces dQ =
0.
When
a reversible process
represented by a curve connecting the two surfaces, however, heat is
transferred. All curves joining the
same
dQ =
X
is do-
two surfaces represent processes with the
da, but the values of X are different.
Reference system
Main system
8-9
A
Physical Significance of X t,
X, x-
t, .
The
various infinitesimal processes that
may
be chosen to connect the two
neighboring reversible adiabatic surfaces shown in Fig. 8-6 involve the same change of a but take place at different values of X, because X is a function of /, o, and X. To find the temperature dependence of X, we go back to the fundamental concept of temperature as the property of a system determining ther-
through a diathermic wall, as depicted schematically in Fig. 8-7. to be at all times in thermal equilibrium having
The two systems are assumed
A
A
o{t,X,X')
o(l.X.X')
\'t, a-.X)
A A a {t. cr.X)
A
Composite system I,
X,
x\
A
a- It, cr, X.
Fig. 8-7
from a
Two
reservoir.
(t.
A
a. X,
it. cr,
A
x, x-
A
mal equilibrium between it and another system. Let us therefore consider two systems, each of three independent coordinates (for mathematical generality), in contact
A
X, X'
X)
X. X)
systems in thermal equilibrium, constituting a composite system receiving heat
206
may may
Heat and Thermodynamics
be expressed
and X. The primed quantities A"' and A'' a of the composite sysa function of/, a, a, X, and A. For an infinitesia,
/,
may
be regarded as
process between
specified
by a and a
also a function of
,
da
I.
=
two neighboring -\-
da, the heat transferred
a, a,
da
-r-dl at
We
X, and A.
da
.
reversible adiabatic
+3da
+
,
,
da
is
dQ =
207
and the Kelvin Temperature Scale
Reversibility
following structure:
therefore be eliminated from the expression for
tem, and a
mal
terms of
in
8-10
X
- V (t)f(a)
X
=
t
3-12)
hypcrsurfaces
X da, where X
is
X =
and
v{t)g{
have
da ,, —da da
da ,„ -^ d.\ dX
+ .
^.dX
(8-8)
dX
Now
suppose that, in a reversible process, there is a transfer of heat dQ between the composite system and an external reservoir, as shown in Fig. 8-7, with heats dQ and dQ being transferred, respectively, to the main and to the reference systems. Then,
(The quantity X cannot contain A', nor can X contain A, since X/X and X/X must be functions of the
dQ =
(8-13)
is an exact differential, the quantity \/
Since f(a) da for
dQ = dQ dQ, X da = X da + X da, -!-
and
dQ.
It
is
dQ
factor exist for the temperature only
and is
of
any system, but
this integrating factor is
the same, function for all systems!
a function of
This universal character of
enables us to define an absolute temperature. The fact that a system of two independent variables has a admits an integrating factor regardless of the second law ip(l)
=
da
7^0"
K
+
-^
X
da.
(8-9)
Comparing the two expressions for da given by Eqs.
-- =
not
dX
'
dt
a does
depend on
I,
u
=
X
=
2l da
dX
X but only
=
wc
course; but
get
for
and
X
we
=
8-10 on a and
a.
That
it is
shown it is
that the
the same
systems.
Kelvin Temperature Scale
Consider a system of three independent variables t, X, and A'', for which two isothermal surfaces and reversible adiabatic surfaces arc. drawn in Fig. 8-8. Suppose there is a reversible isothermal transfer of heat Q between the system and a reservoir at the temperature /, so that the system proceeds from a state b, lying on a reversible adiabatic surface characterized by the value o\,
5-11)
da'
X/X and X/X are also independent of t, X, and X. These two ratios depend only on the
all
not established until
is
a function of temperature only and that
is
is,
sec that
d --
importance in physics
'
(8-10)
da,
its
integrating factor
always
interesting, of
is
0;
a(a,a).
Again comparing the two expressions *
**
and
'
X, or
a
two
(8-9),
function for
^= therefore
and
(8-8)
dQ which
to another state
by cm. Then,
c,
lying
Q =
ratios
on another
reversible adiabatic surface specified
since Eq. (8-13) tells us that
I
"/W
da
dQ =
(at const.
da
tve
I).
For any reversible isothermal process a —* d at the temperature same two
reversible adiabatic surfaces,
the heat
Q 3 = v «>>/,>cr)
da
Q
3
have
is
(at const. I3).
l3
between the
208
Heat and Thermodynamics
8-11
the Kelvin temperature
T= To measure
is
Reversibility
defined to be
(between the same two
273.16°K->-
re-
(8-15) versible adiabatic surfaces).
a Kelvin temperature,
the heats transferred at the
209
and the Kelvin Temperature Scale
we must
therefore measure or calculate
unknown temperature and
at the triple point of
water during reversible isothermal processes between the same two reversible adiabatic surfaces. Comparing this equation with the corresponding equation for the ideal-gas
temperature
=
273.16°
lim (PV)
lim(PF),' it is
seen that, in the Kelvin scale,
Q
plays the role of a "thcrmometric prop-
erty." This does not have the objection attached to a coordinate of an arbi-
chosen thermometer, however, inasmuch as Q/Q*
trarily
is
independent of
the nature of the system. It
follows from Eq. (8-15) that the heat transferred isothermally between
two given Two isothermal heat transfers, Q at I from
Fig. 8-8
the
same two
reversible adiabatic surfaces
eri
b to c and Qs at tsfrom a to d, between and a\\. The cycle abeda is a Carnot cycle.
reversible adiabatic surfaces decreases as the temperature decreases.
Conversely, the smaller the value of Q, the lower the corresponding T. The smallest possible value of Q is zero, and the corresponding T is absolute zero.
Thus,
(/ a system
undergoes a reversible isothermal process between two reversible
adiabatic surfaces without transfer of heat, the temperature at which this process lakes
Taking the
ratio of
Q
to
Q 3 wc
place is called absolute zero.
get
,
It
Q _
Q3
f{h)
_
a function of the temp, at which Q is transferred the same function of temp, at which Q 3 is transferred'
should be noticed that the definition of absolute zero holds for
stances
and
is
therefore independent of the peculiar properties of
arbitrarily chosen substance. Furthermore, the definition
macroscopic concepts. therefore
we
define the ratio of two Kelvin temperatures
T/ Tt by
the relation
energy. that
(between
Qi (between
en
Q
and en and cu
at
7")
at 7g)
T
we
No
reference
is
made
is
in
a
sub-
terms of purely
to molecules or to
Whether absolute zero may be achieved experimentally shall defer until
all
any one
is
molecular a question
later chapter.
(8-14)
Tj 8-11
Equality of Ideal-gas Temperature
and Kelvin Temperature Thus, two
temperatures on the Kelvin scale are to each other as the heals transferred
between the same two reversible adiabatic surfaces at these two temperatures. It
that the Kelvin temperature scale istics
of any particular substance.
is
It
is
seen
independent of the peculiar character-
therefore supplies exactly
what
is
lacking
in the ideal-gas scale. If
the temperature
T
3
(the standard fixed point)
is
taken arbitrarily to be the triple point of water
and T$
is
chosen to have the value 273.1 6°K, then
For the sake of generality, systems with three or more independent coordinates have been used in most of the discussions in this chapter.
The
systems
encountered most frequently in practical applications of thermodynamics, however, usually have no more than two independent variables. In such cases, isothermal
and
reversible adiabatic surfaces degenerate into plane
curves such as those shown on the
BV diagram
of an ideal gas in Fig. 8-9.
210
Heat and Thermodynamics
Integrating from a to I
b,
we
Isotherm
\
nR
«
\
\
\ \
\
Va
^
^ \
and we
\
V
N
\
\
«\
d
\
\
,
=
In -rr
Vb
Vd
vc
3
6 and at 6 3, between two
In
f e.
3
reversible adiabatics of
is
defined by the same sort of
iff
_
an ideal gas.
T_
T»
e,
cycle.
dQ = C v equation
found
=
we have
gas, the first
law
may
be
T refer to any temperature and
and
to
is
dd
= T3 =
+ P dV.
applied to the isothermal process b
—
=
and > c,
if
63
and 7j
refer to the triple point
of water.
written
is
Q
Q
Since, however, the Kelvin temperature scale
If 6
ferred
-^
VN
\
For any infinitesimal reversible process of an ideal
this
V,-
In
get, finally,
V
When
Vd V;
=ln
and
In ==-
M.
equation,
Carnot
> c,
-e
1
Isotherm
\ \ ate
—
f rCv dd
1
isotherms, at
Vb
Similarly, for the adiabatic process d
Therefore,
Va
.
Jos
^RJe
is a
dd
re
1 \
c\
\b
abcda
211
get
\
\
Two
Temperature Scale
the Kelvin
1
\
Fig. 8-9
and
Reversibility
273.1 6° K,
r.
(8-14)
the heat trans-
be
Q =
l
V'
PdV =
The Kelvin temperature
nRO In-^-
is
therefore numerically equal to the ideal-gas tem-
perature and, in the proper range,
Similarly, for the isothermal process a
—
>
d,
the heat transferred
may be measured with a
gas thermometer.
is
PROBLEMS =
<2«
Q__
Therefore,
Q Since the process a
—*
b is
3
nRe 3
in
& V*
8-1
3
ln(Vd /Vay
adiabatic,
we may
write for
portion
Cydd =
A
gas
is
contained within a cylinder-piston combination. In the
tell (1) whether d,W = P dV or not and (2) whether the process is reversible, quasi-static, or irreversible: (n) There is no external pressure on the piston and no friction between the piston and the cylinder wall. (b) There is no external pressure, and friction is small.
following five sets of conditions,
gin (Vc /Vb )
-PdV = -~dV.
any
infinitesimal
(c)
{d)
The The
piston
is
friction
jerked out faster than the average molecular speed.
is
adjusted to allow the gas to expand slowly.
212
Heat and Thermodynamics
There
(e)
is
no
but the external pressure
friction,
and
Reversibility
is
adjusted to allow
the gas to expand slowly.
provided that g{o,&)
toward
=
/(ff).
Continuing the reasoning at the beginning of Art. 8-8 concerning a system of two independent variables, show that the expression for dQ admits an integrating factor. 8-3 Consider the differential (or Pfaflian) expression
+ dy +
yz dx
,
s
(a)
fi
Show
i
that
r
/
=
of this problem is to lead the way dependence of g(cr,&) on a. do ;
3ir
g -— and / ,
OtX
=
g
— Off
/ with respect to
dz.
213
The purpose
a proof of the functional
8-2
the Kelvin Temperature Scale
Differentiate
Differentiate
interpret the
result.
To determine whether an
integrating factor exists,
wc
investigate the possible
solutions of the Pfaffian equation
+ dy + dz ^
yzdx
(P8-1)
Q.
Holding x constant, show that the resulting equation has a solution
(a)
A Carnot cycle, as shown in Figs. 8-8 and 8-9, consists of a reversible
8-7
Tc to a higher temperature Th, then a reversible isothermal process at 77/ in which heat Qu is transferred, then a reversible adiabatic process from Tu to Tc, and finally a reversible isothermal process at T c in which heat Qc is transferred. Draw qualitatively adiabatic process from a lower temperature
a
Carnot cycle (a)
+z=
y
F(x),
(b) (c)
but that
this cannot be a solution of Eq. (P8-1). Holding z constant, show that the resulting equation has a solution
(b)
=
G(z)r**,
A A
only, on
for the following:
ideal gas
on a
PV
diagram.
liquid in equilibrium
reversible electric cell
an
with
its
vapor on a PV diagram. is a function of temperature
whose emf
SZ diagram, assuming reversible adiabatics to have a positive
slope. (d)
y
An
A paramagnetic substance obeying Curie's law on an %M diagram,
assuming
9£fT
to be practically constant during reversible adiabatic
processes.
but that (c)
this
Do
cannot be a solution of Eq. (P8-1).
the
two "cuts"
(x
=
const,
and
z
=
const.)
8-8
produce
a
smooth
surface? (d)
8-4
+
.v
dy
-f-
2z dz
=
has a solution; and if so, find the equation of the family of surfaces. 8-5 Consider the Pfaffian expression
a'Yz 2 dx
+
bhflx* dy
+ c-xy- dz.
By
inspection and guessing, find an integrating factor. Find the equation of the family of surfaces satisfying the Pfaffian equation obtained by setting the expression equal to zero. (b)
8-6
dQ = for X,
The
definition of the Kelvin scale to any
Carnot cycle and
The efficiency of a Carnot engine. The coefficient of performance of a Carnot refrigerator. 8-9 Which is the more effective way to increase the efficiency of a Carnot engine: to increase Tu while keeping Tc constant; or to decrease Tc while keeping Tu constant? (a)
Does an integrating factor exist? Determine whether the Pfaffian equation y dx
(a)
Apply the
calculate the following:
expression for
,\
given
in
Eq. (8-12) enables one to write
which is of the same form as Eq. (8-13). The expression however, would give an equation for dQ of the same form as Eq. (8-13), v(t')f(&) d&,
(b)
8-10 Cause a gas whose equation of state is P(v — b) molar energy is a function of d only to undergo a Carnot that 6
=
T.
=
Jtd
cycle,
and whose and prove
9-1
9.
Since a
is
an actual function
exact differential, which
of/, X, X',
we may
ENTROPY dS
where the subscript reversibly.
The
R
is
quantity
.
The Concept
9-1
In a system of
from a given
any number of independent variables, all by reversible adiabatic processes .
=
.)
.
conceived as being crossed
lie
on a surface
The entire t, X, X' space by many nonintersecting surfaces of this const.
,
a.
.
.
a
+
da.
We
its
In a reversible nonadiabalic
by a on another sur-
state point lies
have seen that
dQ =
X da,
where IX, the integrating factor of dQ,
The entropy
.
process involving a transfer of heat dQ, a system in a state represented
point lying on a surface a will change until
Sf
X
and therefore
dQ =
Since the Kelvin temperature
T
is
an
(9-1)
dQ
must be transferred
and dS is an change of state from i to
called the entropy of the system,
—
Si,
where
s,
-
s<
a
finite
/dQ
=
is
(9-2)
T
of a system is a Junction of the thermodynamic coordinates whose
equal to the integral of dQ/j/
reversible
path connecting the two
an entropy change
T
between the terminal
slates.
It
is
states, integrated
change any
along
important to understand that only
an absolute entropy just as in the case of the internal energy function, whose change is defined as the adiabatic work but whose absolute value is undefined. A third relation may be obtained by integrating Eq. (9-1) around a reversible cycle, so that the initial and final entropies are the same. For a reversible cycle,
is
we
defined, not
get
dQ
T =
member
whence
written to emphasize that 6" is
given by
is
the right-hand
states accessible
initial state
kind, each corresponding to a different value of
face
is
of Entropy
(or hypcrsurface) a(t,X,X',
may be
the entropy change
,
= dQ« T
infinitesimal entropy change of the system. In /,
.
.
designate by dS;
215
Entropy
=
o.
(9-3)
f(')f(
{t)f{a) da.
defined so that
is
da being the same for both heat transfers,
T=
it
T/T'
follows that
=
dQ'dQ', with
an equation known as the Clausius theorem. The concept of entropy was first introduced into theoretical physics by R. J. Clausius about the middle of the nineteenth century. Until this time there had been much confusion concerning the relation between heat and work and their roles in the operation of a heat engine. The great French Petit, Clement, and Desormcs had little knowledge of the law of thermodynamics. Carnot believed that the work output of an engine was the result of an amount of heat leaving a hot reservoir and the
engineers Carnot, k
first
where k
is
an arbitrary constant. Therefore,
same amount
and Clement computed by calculating the work done only in the power strokc without considering, as Carnot insisted one must do. the entire cycle. In the words of Mendoza, "In the hands of Clapcyron, Kelvin, and Clausius, the.
= ~t
k^
a'
of heat entering a cold reservoir. Petit
efficiency of a heat engine
216
Ilcat
and Thermodynamics 9-2
thermodynamics began
to
make headway only when
it
Entropy
217
was divorced from Dividing by T,
engine design."
we
get
Clausius proved the existence of an entropy function by first deriving his theorem |Eq. (9-3); see Prob. 9-1] and then applying it to a cycle consisting of a reversible path Ri between two equilibrium states i and /, followed by another reversible path R-2 bringing the system back to i. For this cycle
rT T
nji
T +
njf
T
U'
r/dQ R
\.
-j
rfdQ
~ I{
/,-
~f
=
now
Let us trarily
or
.
chosen
AS of the gas between an arbiwith coordinates Tr , Pr and any other state with Integrating between these two states, wc get
calculate the entropy change
A,.9
It
follows that there exists a function
S whose change
i?
reference slate
coordinates T, P.
,
independent of path.
dT dP = r Cp-=r — nK -p-
70 dS
or
dT
=
I
It,
is
Suppose we ascribe
to the reference state
p, P -» Rln pan entropy Sr and chose any
arbitrary
Then an entropy S may be associated with = AS. To make the discussion simpler, let Cp be
numerical value for this quantity.
—
where S Then,
the other state
The
constant.
derivation of the Clausius theorem, the properties of Garnet engines on is based, and Clausius' derivation of the existence of an
S
which the theorem
entropy function are in every way equivalent to and as general as the methods of Caratheodory. The only superiority of the Caratheodory approach is that it
focuses the attention on the system,
its
coordinates,
its states, etc.,
>SV
—
which may be rewritten
St
=
CV In
T —
nR
Tp
P In
-5-
>
thus:
whereas
these are apt to be overlooked in the engineering approach. Physicists
and
S
=
Cp
In
T - nR In P +
(Sr
- CP In
T,
+ nR In P
r).
engineers should appreciate both points of view.
Denoting the quantity
Entropy of an Ideal Gas
9-2 If
a system absorbs an infinitesimal amount of heat dQ R during a reversible change of the system is equal to
=
ft
dQ K
is
expressed as a
nates, then,
sum
= C P 4T - V dP.
In
P
get finally
+ .SV
entropy change.
now
return to the original differential equation,
of differentials involving thermodynamic coordiT, the expression may be integrated and the
dQfi
T - nR
wc
T and P thousands of different values, wc may calculate thousands of corresponding values of S which, after tabulation, constitute an entropy table. Any one value from this table, taken alone, will have no meaning. The difference between two values, however, will be an actual
dT
dS^Cpjr-
upon dividing by
entropy of the system obtained. As an example of this procedure, consider one of the expressions for dQ K of an ideal gas, namely,
In
,
Substituting for
Let us If
parentheses by the constant S
S = Cp
process, the entropy
dS
in
"R
dP -
Again, for simplicity, assuming Cp to be constant, integral
we may
and obtain
S = CP
In
T - nR
In
P
+
St,,
take the indefinite
9-3
Heat and Thermodynamics
218
where So
is
the constant of integration. Since this
is
precisely the equation
Entropy
219
This integral can be interpreted graphically as the area under a curve on a diagram in which T is plotted along the Y axis and along the X axis. The nature of the curve on the TS diagram is determined by the kind of reversible process that the system undergoes. Obviously, an isothermal process is a .$'
obtained previously, we see that, in taking the indefinite integral of r/.S', we do not obtain an "absolute entropy,"' but merely an entropy referred to a nonspecified reference state whose coordinates are contained within the con-
an
stant of integration. Thus, for
horizontal line.
ideal gas,
In the case of a reversible adiabatic process,
dT .-/*?
P
nil In
4- So.
dS=^,
(9-4)
dQ« =
and
To
calculate the entropy of an ideal gas as a function of
the other expression for
dQ«
V,
wc
use
whence,
if
T
is
rp
d.S
— =
dT
dS
=
P
.
^V Tp T-7p-aV, 4-
Lv-?p
0;
not zero,
of an ideal gas. Thus,
dQ„
and
T and
we have
and S
constant. Therefore, during a reversible adiabatic process, the
is
entropy of a system remains constant; or in other words, the system undergoes an isentropic process. An isen tropic process on a TS diagram is obviously a
nR -y
vertical line.
way
Proceeding in the same
as before,
we
get for the entropy, referred to
an
If
two equilibrium
states are infinitesimally near,
unspecified reference state, the expression
S=
f
Cv =p
4-
then
dQ = T dS,
nR
In
V
4- S„,
and (9-5)
dQ _
dS_
dT
if
At constant volume. which becomes,
if
Cv
is
constant,
$ = Cv
9-3
In
T+
nil In
V
+S
$),-*-
.
TS Diagram
and
For each infinitesimal amount of heat that enters a system during an is an equation
8),
process
is
amount
-
* -(£),
(9-7)
dS.
If the It follows therefore that the total
(9-6)
at constant pressure,
infinitesimal portion of a reversible process, there
dQ„ = T
of heat transferred in a reversible
temperature variation of Cv is known, the entropy change during may be calculated from the equation
an isochoric process
given by
/Cv
fc-JP
TdS.
*,-*,-{;% dT
(isochoric).
(9-8)
220
Heat and Thermodynamics
9-3
lintropy
221
Similarly, for an isobaric process,
'^Ut *-£¥
(isobaric).
(9-9)
The above equations provide a general method for calculating an entropy change but no way of calculating the absolute entropy of a system in a given state. If a set of tables is required that is to be used to obtain entropy differences and not absolute entropy, then
an arbitrary standard
state
from, this standard state to all
standard state
is
a convenient procedure to choose change of the system other states. Thus, in the case of water, the
and
it is
calculate the entropy
chosen to be that of saturated water at 0.01 °C and
mm, and all entropies are referred curve on a TS diagram representing a
vapor pressure 4.58
The
slope of a
process,
from Eq.
and from Eq.
(9-6),
its
own
to this state. reversible isochoric
is
BT
T_
es
Cy'
(9-7), the slope of
a reversible isobar
= (-) \dSj P
is
T_
CP
Entropy, cal/gm-deg Fig. 9-2
7 'S diagram for
C0
2.
The two dashed
(
lines
bounding the
solid-li,
region are a
auess.)
T
Curves representing various types of processes of a hydrostatic system are a TS diagram The TS diagram for a
shown on
curve from A
senthalp
processes in
to
/''is
which
in Fig. 9-1.
solid
is
TS
diagram.
is
shown
AB = BC = CD =
isobaric heating of liquid to
DM =
isobaric isothermal vaporization.
EF =
isobaric heating of vapor (superheating).
The Curves representing reversible processes of a hydrostatic system on a
2
in Fig. 9-2.
The
transformed finally into vapor. Thus,
isobaric heating of solid to
its
melting point.
isobaric isothermal melting.
area under the line
BC
tion. Similarly, the heat of
its
boiling point.
represents the heat of fusion at the particular
temperature, and the area under the line Fig. 9-1
C0
substance such as
a typical isobar representing a series of reversible isobaric
sublimation
is
DE represents the heat of vaporizarepresented by the area under any
Heat and Thermodynamics
222
sublimation
line. It is
9-5
equal to the
sum
of the heat of fusion
and the heat of vaporization
223
T
obvious from the diagram that the heat of vaporization
becomes zero at the critical point and, also, that the heat of sublimation
Entropy
is
at the
triple point.
Carnot Cycle
9-4
During a part of the cycle performed by the system in an engine, some is absorbed from a hot reservoir; during another part of the cycle, a
heat
smaller
amount
of heat
is
rejected to a cooler reservoir.
The
engine
is
Tc
there-
between these two reservoirs. Since it is a fact of experiis always rejected to the cooler reservoir, the efficiency of an actual engine is never 100 percent. If we assume that wc have at our disposal two reservoirs at given temperatures, it is important to answer the following questions: (1) What is the maximum efficiency that can be achieved by an engine operating between these two reservoirs? (2) What are the characteristics of such an engine? (3) Of what effect is the nature of the substance undergoing the cycle? The importance of these questions was recognized by Nicolas Leonard Sadi Carnot, a brilliant young French engineer who in the year 1824, before the first law of thermodynamics was firmly established, described in a paper fore said to operate
ence that some heat
entitled "Reflexions sur la puissance motrice
ating in a particularly simple cycle
Carnot cycle
is
du
known today
as the Carnot
cycle.
in a
heat that is absorbed is absorbed at a constant high temperature, namely, that of the hot
that of the cold reservoir.
Since
all
?)
surfaces,
Qc =
Tji
Qn
r„
(Carnot)
=
1
—
TV* t~--
(9-10)
For a Carnot engine to have an efficiency of 100 percent, Tc must be zero. Since nature does not provide us with a reservoir at absolute zero, a heat engine with 100 percent efficiency is a practical impossibility. A temperature-entropy diagram is particularly suited to display the char-
a constant lower temperature,
acteristics of
vertical lines,
a reversible cycle.
If an engine is to operate between only two reservoirs and still operate in a reversible cycle, then it must be a Carnot engine. For example, if an Otto cycle were performed between only two reservoirs, the heat transfers in the two isochoric processes would involve finite temperature differences and, therefore, could not be reversible. Conversely, if the Otto cycle were performed reversibly, it would require a series of reservoirs, not merely two. The expression "Carnot engine," therefore, means "a reversible engine operating between only two reservoirs."
Carnot engine absorbing heat
Qc
between the same two iscntropic
four processes are reversible, the Carnot cycle
reservoir. Also, all the heal that is rejected is rejected at
A
Carnot cycle of any system of any number of independent coordinates when repreis a rectangle.
diagram
depicted in Fig. 8-8, and one executed by an ideal gas with
engine operating
ing heat
Since,
TS
A general
Carnot cycle is' called a Camol engine. A Carnot engine operates between two reservoirs in a particularly simple way. All the
is
sented on a
feu" an ideal engine oper-
only two independent variables in Fig. 8-9.
An
A
Fig. 9-3
to a cooler reservoir at
Qn from
7~<-
a hot reservoir at
Tb and
has an efficiency v equal to
1
—
reject-
Qc/Qir-
a Carnot
cycle.
The two
reversible adiabatic processes arc
and the two reversible isothermal processes arc horizontal lines lying between the two vertical lines, so that the Carnot cycle is represented by a rectangle, as shown in Fig. 9-3. This is true regardless of the nature of the system and of the number of independent thermodynamic coordinates.
9-5
Entropy and Reversibility
In order
to
understand the physical meaning of entropy and
in the world of science,
it is
necessary to study
all
its
significance
the entropy changes that
when a system undergoes a process. If we calculate the entropy change of the system and add to this the entropy change of the local surroundtake place
224 ings,
1
we
leat
9-6
and Thermodynamics
obtain a quantity thai
is
the
sum
of
all
the entropy changes
brought
about by this particular process. We may call this the entropy change of Ike universe due to the process in question. When a finite amount of heal: is absorbed or rejected by a reservoir, extremely small changes in the coordinates take place in every unit of mass. The entropy change of a unit of mass is therefore very small. Since the total mass of a reservoir is large, however, the total entropy change is finite. Suppose that a reservoir is in contact with a system and that heat Q is absorbed by the reservoir at the temperature T. The reservoir undergoes nondissipativc changes determined entirely by the quantity of heat absorbed. Exactly the same changes in the reservoir would take place if the same amount of heat Q were transferred reversibly. Hence the entropy change of the reservoir is
Q/T. Therefore,
Q
whenever a reservoir absorbs heal
at the temperature
system during any hind of process, the entropy change of the reservoir
Consider now the entropy change of the universe that the performance of any reversible process. The process
is
is
Q
T from
any
brought about by
will, in general,
be
accompanied by a flow of heat between a system and a set of reservoirs ranging in temperature from 7',- to '/'/. During any infinitesimal portion of the process, an amount of heat dQn is transferred between the system and one of the reservoirs at the temperature 7'. Let dQu be a positive number, if dQn is absorbed by the system, then
dS of the system
Entropy and Irreversibility
9-6
When is
equal to
AS
=
(system)
S{
—
.5,-
fdQ
=
/
JiJi
R
where
may
gration process
is is
-?
T 1
indicates any reversible process arbitrarily chosen
be brought from the given
initial state
by which
the system
to the given final state.
No
inte-
performed over the original irreversible path. The irreversible replaced by a reversible one. This can easily be done when the
and the
final state of the system are
the initial or the final state
be used. At
first,
involve initial
we
and
is
equilibrium
states.
When
either
a nonequilibrunn state, special methods must
shall limit ourselves to irreversible processes all of
final states
which
of equilibrium.
(a) Those involving work through a system (which remains un-
Processes exhibiting external mechanical irreversibility
the isothermal dissipation of
changed) into internal energy of a reservoir, such
as:
Irregular stirring of a viscous liquid in contact with a reservoir.
dQji
=
a system undergoes an irreversible process between an initial equi-
librium state and a final equilibrium state, the entropy change of the system
initial
T,
225
Entropy
r
Coming
to rest of a rotating or vibrating liquid in contact with
a
reservoir.
dS of the
reservoir
= —
^,-
Inelastic deformation of
>
a
solid in
contact with a reservoir.
Transfer of electricity through a resistor
and the entropy change
of the universe
2 dS is
Magnetic zero. If dQit
is
rejected
by the
system, then obviously
in
contact with a reservoir.
hysteresis of a material in contact
with a reservoir.
In the case of any process involving the isothermal transformation of work is no entropy change of the system because the thermodynamic coordinates do not change. There is a flow of heat Q into the reservoir where Q = W. Since the reservoir IV through a system into internal energy of a reservoir, there
dS of the system
dS of the
reservoir
= =
~-
,
absorbs
T
Q
units of heat at the temperature T,
\-WfT. The entropy change
its
of the universe
is
entropy change therefore
is
-{-Q/T or is a
W/T, which
positive quantity.
and the entropy change of the universe 2 dS is again zero. If dQ K is zero, then neither the system nor the reservoir will have an entropy change, and the entropy change of the universe is still zero. Since this is true for any infinitesimal portion of the reversible process,
therefore
wc may conclude
that,
of the universe remains unchanged.
when a
(/;)
Those involving the adiabatic dissipation of work into
internal,
true for all such portions;
1
Irregular stirring of a viscous thermally insulated liquid.
reversible process is performed, the entropy
2
Coming
3
Inelastic deformation of
it is
energy
of a system, such as:
to rest of a rotating or vibrating thermally insulated liquid.
a thermally insulated
solid.
226
Y
Meat and Thermodynamics
Transfer of electricity through a thermally insulated resistor. Magnetic hysteresis of a thermally insulated material.
4 5
9-6
the
volume
Vj.
The
In the case of any process involving the adiabatie transformation of work into internal energy of a system whose temperature rises from 7", to T/ at is
no flow of heat to or from the surroundings, and
therefore the entropy change of the local surroundings
is
zero.
To
entropy change of the system
«fe»») =
W
constant pressure, there
For an isothermal process of an ideal
pressure P) to the final state (temperature 77, pressure P). Let us replace the irreversible performance of work by a reversible iso'/'„
to Tj.
The
in
v,
(system)
=
/
temperature from Ti
dV
_
(system)
= nR
i/
AS
whence
In -=/•
Vi
T T,dQ T'
positive
For an isobaric process,
of the universe
is
nR
therefore
AS
(system)
av
=
„
which
is
a
Those involving a
transfer
of heat by virtue of a finite temperature difference, such as:
dT
1
C,p
/
In {Vj/Vt),
number.
Processes exhibiting external thermal irreversibility
dQ R = CP dT, and
dQ
gas,
dQft
and
The entropy change A ,9
Vt to
X?
entropy change of the system will then be t
from a volume
then
dQ„ = PdV,
must be
replaced by a reversible one that will take the system from the given initial
baric flow of heat from a series of reservoirs ranging
is
227
calculate
the entropy change of the system, the original irreversible process state (temperature
T
reversible isothermal expansion at the temperature
Entropy
Conduction or radiation of heat from a system
to
its
cooler sur-
roundings. Finally,
if
Cp
is
AS and
2
assumed constant, (system)
= CP
the entropy change of the universe
is
In
Cp
Tf Tt
Conduction or radiation of heat through a system (which remains unchanged) from a hot reservoir to a cooler one.
'
In the case of the conduction of Q units of heat through a system (which remains unchanged) from a hot reservoir at T\ to a cooler reservoir at Tt,
In (T/fTi),
which
is
a positive
the following steps are obvious:
quantity.
AS Those involving the
Processes exhibiting internal mechanical irreversibility
AS
transformation of internal energy of a system into mechanical energy and then back into internal energy again, such as:
vacuum
it is
In the case of a free expansion of an local surroundings
is
zero.
To calculate
'
./,
= —•=->
(cold reservoir)
i
= + -^ 1
process).
pricked.
S AS = AS
and
>
2
ideal gas, the
(universe)
=-$---$ 1-2.
Processes exhibiting chemical irreversibility
change of internal
'
\
Those involving a spontaneous
structure, chemical composition, density, etc., such as:
entropy change of the
the entropy change of the system, the
must be replaced by a reversible process that will take the gas lrom its original state (volume F,-, temperature T) to the final state (volume temperature T). Evidently, the most convenient reversible process is a
free expansion
0,
(free expansion).
Gas seeping through a porous plug (throttling Snapping of a stretched wire after it is cut. Collapse of a soap film after
(hot reservoir)
=
1
AS Ideal gas rushing into a
(system)
1
A
2
Diffusion of two dissimilar inert ideal gases.
chemical reaction.
3
Mixing; of alcohol and water.
4
Freezing of supercooled liquid.
228
Heat and Thermodynamics
9-7
5
Condensation of a supersaturated vapor.
6
Solution of a solid in water.
7
Osmosis.
Entropy and Nonequilibrium
9-7
The
diffusion of
two
dissimilar inert ideal gases to be equivalent free expansions, for one of which
States
calculation of the entropy changes associated with the irreversible
processes discussed in Art. 9-5 presented
Assuming the to two separate
229
Entropy
all cases,
no
special difficulties because, in
the system cither did not change at all (in
which case only the
entropy changes of reservoirs had to be calculated) or the terminal states of a system were equilibrium states that could be connected by a suitable reversi-
AS = nR
In
,
A thermally conducting bar, brought to a nonuniform temperature distribution by contact at one end with a hot reservoir and at the other end with a cold reservoir, is removed from the reservoirs and then thermal
and taking a mole
of each gas with Vi
=
v
Consider, however, the following process involving internal
ble process.
^/
and V,
=
we
2v,
obtain
irreversibility.
thermally insulated and kept at constant pressure.
2 AS =
211 In 2,
will finally bring the
from an
which Table
is
a positive
number. All the
results of this article are
summarized
in
9-1
initial
An
internal flow of heat
bar to a uniform temperature, but the transition will be
nonequilibrium state to a
final
equilibrium state. It
ously impossible to find one reversible process by which the system
brought from the same
initial to the
same
final state.
is
obvi-
may
What meaning,
be
there-
may be attached to the entropy change associated with this process? Let us consider the bar to be composed of an infinite number of infinitesimally thin sections, each of which has a different initial temperature but all
fore,
Table 9-1
Entropy Change of the Universe Due
to
Natural Processes
final temperature. Suppose we imagine all the secfrom one another and kept at the same pressure and then each section to be put in contact successively with a series of reservoirs ranging in temperature from the initial temperature of the particular section to the common final temperature. This defines an infinite number of reversible isobaric processes, which may be used to take the system from its initial
of
Entropy change
Type of Irreversible process
irreversibility
of the
system
AS
Entropy change of the local
surroundings
Entropy change of the universe -2
(syst.)
A.S(surr.)
Isothermal dissipation of work through a system into internal energy of a reservoir
External
AS
W
W
T
T
nonequilibrium state to its final equilibrium state. We shall now define the entropy change as the result of integrating &Q/T over all these reversible processes. In other words, in the absence of
system from
mechanical irreversibility
Adiabatic dissipation of work into internal energy of a system
Cln^
which have the same
tions to be insulated
Up
In
— *
i
i
to/,
we
conceive of an infinite
one reversible process
number
to take the
of reversible processes
one for each volume element. As an example, consider the uniform bar of length L depicted A typical volume element at -v has a mass
in Fig. 9-4.
Internal
mechanical
Free expansion of an ideal gas
V,
dm = pA
nRln%
dx,
irreversibility
External thermal irreversibility
|
Transfer of heat through a medium from a hot to cooler reservoir
where p a
T
2
Q _ Q T,
t
2
is
the section
the density and
A
the cross-sectional area.
irreversibility
Diffusion of two dissimilar inert ideal gases
2R\n2
2R
heat capacity of
r, cp
Chemical
The
is
dm = cppA
dx.
In 2
Let us suppose that the
initial
temperature distribution
is
linear, so that the
230
Heat and Thermodvnai
Entropy
9-8
Upon
integrating over the whole bar, the total entropy change
which, after integration
2 AS = CP
(\
and
f
+ In
Tj
simplification,
+
\
To show
Tl r in— ,
that the entropy change
„
In
is
is
becomes
TL -
I
7° J
a
—
In 1
t\
1.
f
a convenient 200°K; whence I> = 300°K.
positive, let us take
= 400% Tt =
numerical case such as T„
231
Then,
2 AS Fig. 9-4
Process exhibiting internal thermal irreversibility.
section at x has
an
initial
=
temperature
n-
2.30CV
(— +
2.477
+
2.301
-
^2.30
2
X
2.602J
0.0196V.
The same method may be used to compute the entropy change of a system during a process from an initial noncquilibrium state characterized by a nonuniform pressure distribution to a final equilibrium state where the pressure is uniform. Examples of such processes arc given in the problems at
j-—x.
Jo
=
the end of this chapter. If
no heat
is
lost
and
if
conductivity, density,
we assume
for the sake of simplicity that the
and heat capacity
thermal
of all sections remain constant, then
the final temperature will be
Principle of the Increase of Entropy
9-8
Ti
= T +
T,
The entropy change
of the universe associated with each of the irreversible
up to now was found to be positive. We are led to believe, therefore, that whenever an irreversible process takes place the entropy of the universe increases. To establish this proposition, known as the entropy processes treated
Integrating
dQ/T
over a reversible isobaric transfer of heat between the series of reservoirs ranging in temperature from '1\ to e get, for the entropy change of this one volume element,
volume clement and a 2/>
w
cppA dx
I
Jr<
-= =
cj-pA ax In
1
in a
general manner,
adiabatic processes only, since is
T, ~
it
is
sufficient to confine
we have already seen
T — TL
r
r
-"""""
'>
:
(
--«--')•
We
start
by considering the special case of an adiabatic irreversible process between- two equilibrium states of a system. 1 Let the system, as usual, have three independent coordinates T, X, and X' and let the initial state be represented by point i on the diagram shown in Fig. 9-5. Suppose that the system undergoes an irreversible adiabatic process to .
CppA dx In
our attention to
that the entropy principle
true for all processes involving the irreversible transfer of heat.
the proof
7, Ti
=
principle,
t
/
In
+
bx) dx
= -
(a
+
bx) In (a
+
bx)
-
x.
232
Ileal
and Thermodynamics
T
isothermal process k
—>j,
233
Entropy
9-8
where
Rev. isothermal
Q„= T'(Sj-Sk A
net
amount of work
W
(net) has
).
been done in the
W (net)
=
cycle,
where
Q„.
It is clear from the second law of thermodynamics that the heat Qg cannot have entered the system that is, Qn cannot be positive for then we would have a cyclic process in which no effect has been produced other than the extraction of heat from a reservoir and the performance of an equivalent amount of work. Therefore Q« < 0, and
—
—
T'(Sj
and 2.
AS >
finally,
If
- Sk) <
we assume
0,
0.
that the original irreversible adiabatic process took place
it would be possible to bring the system by means of one reversible adiabatic process. Moreover, since the net heat transferred in this cycle is zero, the net work would also be zero. Therefore, under these circumstances, the system and its surroundings would have been restored to their initial states without producing changes elsewhere,
without any change in entropy, then Fig. 9-5
The
process
second law unless Sf
>
i
'.—*
f
is irreversible
and
adiabatic.
The
cycle ifkji contradicts the
Si.
the state /; then the entropy change
is
back
to i
which implies that the original process was
= Sf -
A..9
to
&.
reversible. Since this
is
contrary
our original assertion, the entropy of the system cannot remain unchanged.
Therefore,
A
temperature change may or may not have taken place. Whether or not, let us cause the system to undergo a reversible adiabatic process /—»i£ in such a direction as to bring its temperature to that of any arbitrarily chosen reservoir, say, at 7".
reservoir
entropy will
is
now
and caused the same as
to
that the system
undergo a
i
is
zero,
—»/ and
AS
A
its
k—*j
until its
final reversible adiabatic process j
original state.
The
—
>
i
net entropy change
and entropy changes take place only during the two
St)
+
- &) =
csj
= Sf - S ), t
it
only heat transfer
If
we assume
tesimal,
that the system
necessary)
if
-
it is
and that
during the
sum
of the entropies
AS
of the
it
follows that
should be emphasized that (1) that the
infini-
each part
shall
have a
we may define the entropy of its parts. If we now assume
coordinates, then
above
namely, is
its
take place.
possible to ascribe a definite temperature,
reversible processes described
It
Q R that has taken place in the cycle
it is
depending on
may
be subdivided into parts (each one
possible to take each part back to
part, then
S^
may
pressure, composition, etc., to each part, so that
that
follows that
AS = Sk
adiabatic process in which mixing and chemical reaction
of the whole system as the o.
(9-11)
3. Let us now suppose that the system is not homogeneous and not of uniform temperature and pressure and that it undergoes an irreversible
definite entropy
denotes the entropy change associated with the irreversible part of
the cycle (AS
The
brought into contact with the
k ~*j. Consequently
%If
is
reversible isothermal process
at the beginning.
bring the system back to
for the cycle
processes
Now suppose
AS>0.
entropy of a
in (1),
its
initial state
by means of the
using the same reservoir for each
whole system
we have had system may be
is
to
positive.
make two assumptions,
defined by subdividing the
1 234
Entropy
9-9
Heat and Thermodynamics
235
summing the entropies of these parts and (2) that may be found or imagined by which mixtures may be unmixed and reactions may be caused to proceed in the opposite direction. The justification for these assumptions rests to a small extent on experimental system into parts and reversible processes
Q+W
grounds. Thus, in a later chapter, there will be described a device involving semipermeable membranes whereby a mixture of two different inert ideal gases
may be separated revcrsibly. A similar device through which a chemimay be caused to proceed reversibly in any desired direction
W
cal reaction
may
also be conceived. Nevertheless, the
tions,
main justification
for these
assump-
and therefore
results
in
for the entropy principle, lies in the fact that they lead to complete agreement with experiment; for the experimental
physicist, this
As the
is
sufficient.
our argument, let us consider an assemblage of systems and reservoirs in an adiabatic enclosure. All heat transfers involving finite temperature differences involve net entropy increases, and all adiabatic 4.
last step in
Body lowered
to in
be
temp
from T, to Jj
processes involving irreversible state changes, mixing, chemical reactions, etc.,
the behavior of the
includes
it
all
entrojry of the universe as a result
be represented
in
adiabatic enclosure
systems
under consideration.
interact during the process
now
The
are also attended by net entropy increases.
constitutes the "universe" since
and
reservoirs that
(a)
(b)
It follows, therefore, that
of any kind of process
may
Fig. 9-6
(a)
Engine operating between
between a reservoir at
T\and a finite
reservoirs at
Th
and Tc-
(b) Refrigerator operating
body and lowering the temperature of the body from 77.
to
2*
the following succinct manner:
entropy principle,
>
2 AS
(9-12)
0,
n
2 AS where the equality
sign refers to reversible processes
and
the.
(universe)
inequality sign
nu =
whence
Engineering Applications of the Entropy Principle
9-9
Since
Whenever
irreversible processes take place, the
increases. In the actual operation of ator,
it is
a device such
entropy of the universe
as
an engine or a
refriger-
sum of all the entropy changes. The to draw useful conclusions concerning
often possible to calculate the
fact that this
sum
is
positive enables us
the behavior of the device.
mechanical engineering will
Two
important examples from the
illustrate the
power and
field
of
simplicity of the entropy
principle. 1 .
9-6r/,
Consider a heat engine undergoing any arbitrary cycle, as shown in Fig. extracting heat Q from a reservoir at 77/ delivering an amount of work
W, and rejecting heat Q
— W to a
colder reservoir at
Tc
.
According
to the
w
g~ >
0,
W
or
to irreversible processes.
—
= *-=
Wmxx./Q
is
the
maximum
Q
(>-ft>
efficiency of
an engine extracting Q from a and since 1 — Tc/Th
reservoir at 77/
and
was shown,
Eq. (9-10), to be the efficiency of a Carnot engine,
in
rejecting heat to a reservoir at Tc,
the result that the maximum is that
efficiency
we have
of any engine operating between two reservoirs
of a Carnot engine operating between the same two reservoirs.
Suppose it is desired to freeze water or to liquefy air, i.e., to lower the temperature of a body of finite mass from the. temperature Tj of its surroundings to any desired temperature Tj. A refrigerator operating in a cycle between a reservoir at T\ and the body itself is utilized, and after a finite number of complete cycles has been traversed, a quantity of heat Q has been removed from the body, a quantity of work has been supplied to the 2.
W
236
Heat and Thermodynamics
refrigerator,
as
shown
and a quantity
of heat
Q
+
Whas
been rejected to the reservoir,
in Fig. 9-6(6). Listing the entropy changes,
we have
is
AS AS
AS
and
body = S - S u
of the reservoir
on the universe
a form
=
in
W
W>
T^Si
It follows that the smallest possible
W (min)
=
-
desired to
same as
that
which would be produced
which
it is
it
was completely
if a certain quantity
available for
work
completely unavailable for work. This amount of energy
E is
into
To
somewhat
is
abstract, let us
namely, the irreversible conduction of heat under a finite temperature gradient. Suppose that heat Q is conducted along a bar from a region at temperature T\ to a region at temperature 7V After conduction has taken place, wc have heat Q available at the lower tempera-
S2)
-
,
first
of
a special case,
which the following amount
-
Si)
Max
Q.
W -
is
If
work
after
is
available for work:
Q.
conduction had not taken place, heat
been obtained from
Max
W
is
(
t
—
1
Q would have been
maximum amount
of
available at the
work that could have
this is
= Q
work before conduction
amount
Evidently, the
cost of operation of the refrigeration plant.
= Q
conduction
higher temperature 7\, and the
If tables of the thermodynamic properties of the material are available, a knowledge of the initial and final states is all that is needed to read from the tables the values of Si — S3 and, if the body undergoes an isobaric process, of Q. The calculated value of (min) is used to provide an estimate of the
minimum
It is
1
value for
r,0Si
the
Since the general proof of this proposition consider
l
ture 7"2
whence
completely available for work.
times the entropy ciuinge of the universe brought about by the irreversible process.
Applying the entropy principle,
1
is
0,
= Q + T
is
it
of energy were converted from a form in which
2
of the refrigerant
which
in
establish the proposition that, whenever an irreversible process lakes place, the effect
of the
form
in a
237
Entropy
9-10
of energy
E
that has
(
1
—
$
become unavailable
for
work
the difference
*-q(i-£)-q(i-£) 9-10
Entropy and Unavailable Energy
Suppose that a quantity of heat Q may be extracted from a reservoir at the temperature T and that it is desired to convert as much as possible of this heat into work. If Tu is the temperature of the coldest reservoir at hand, then
The
W (max)
= Q{1 -
proposition
duction. Since
§
simple manner,
To
=
To AS
{n~T]) (universe).
therefore seen to be true for the special case of heat con-
is
it
=
is
we
not possible to handle shall
have to adopt
all irreversible
a
more
processes in this
abstract point of view in
order to establish the proposition generally.
which represents the
Q
maximum amount
units of heat are extracted
of energy available for
from a reservoir at T.
work when
It is obvious, therefore,
any energy which resides within the reservoir at Ta and which may be extracted only in the form of heat is in a form in which it is completely that
unavailable for work.
The
potential energy,
however, of a
frictionless
mechanical device (as measured from the position of lowest potential energy)
Consider a mechanical device such as a suspended object or a compressed work on a system. Suppose the system is in contact with a reservoir at the temperature T. The mechanical device and the spring capable of doing reservoir at
T
Suppose an which the mechanical device docs work on the system, the internal energy of the system changes from £/, to Uj, and constitute the local surroundings of the system.
irreversible process takes place in
W
238 heat
Heat and Thermodynamics
Q
is
demands
9-1
transferred between the system and the reservoir.
Then
the
first
law
that
same
roundings
Q = U, and the second law
U;
+
If,
is
rejection of
Sf
Now
Si
suppose that
(system
it is
and
local surroundings)
desired to produce exactly the
>
units of heat at the temperature
same changes
is
zero,
This would require, in general, the services of Carnot engines and refrigerators, which, in turn, would have to be operated in conjunction with an auxiliary mechanical device and an auxiliary reservoir. The auxiliary mechanical device may be considered, as usual, to be either a suspended object or a compressed spring. For the auxiliary reservoir let us choose the one whose temperature is the lowest at hand, say, T These constitute the auxiliary surroundings. With the aid of suitable Carnot engines and refrigerators all operating in cycles, in .
it is
now
possible to produce in
and the local surroundings, by reversible processes only, the same changes that were formerly brought about by the irreversible process. If this is done, the entropy change of the system and the local surroundings is the same as before, since they have gone from the same initial states to the same final states. The auxiliary surroundings, however, must undergo an equal and the system
opposite entropy change, because the net entropy change of the universe during reversible processes is zero.
Since the entropy change of the system and local surroundings the entropy change of the auxiliary surroundings
is
Y'o,
that
is,
—E/Tn. Since
the
we have
1
in the
reversible processes only.
conjunction with the auxiliary surroundings,
of the auxiliary sur-
0.
system and the local surroundings which resulted from the performance of the irreversible process, but by
The entropy change
merely the entropy change of the auxiliary reservoir due to the
E
surroundings
—
St.
239
Entropy
of the entropy changes of the system, local surroundings, and auxiliary
sum
that
—
as before, namely, S;
1
is
E = T (S, -
whence
Therefore, is
To
becomes unavailable for work during an irreversible process
the energy that
times the entropy change of the universe that
process.
(9-13)
Si).
is
brought about by the irreversible
Since no energy becomes unavailable during a reversible process,
follows that the
maximum amount
when
of work is obtained
it
a process lakes place
reversibly.
Since irreversible processes arc continually going on in nature, energy continually becoming unavailable for work. This conclusion, principle of the degradation of energy
and
first
known
is
as the
developed by Kelvin, provides an
important physical interpretation of the entropy change of the universe. It
must be understood that energy which becomes unavailable for work is not energy which is lost. The first law is obeyed at all times. Energy is merely transformed from one form into another. In picturesque language, one may say that energy is "running downhill."
positive,
negative. Therefore the
Entropy and Disorder
9-11
T
must have rejected a certain amount of heat, say, E. Since no extra energy has appeared in the system and local surroundings, the energy E must have been transformed into work on the auxiliary mechanical device. We have the result therefore that, when the same changes which were formerly
It has been emphasized that work, as it is used in thermodynamics, is a macroscopic concept. There must be changes that are describablc by macroscopic coordinates. Haphazard motions of individual molecules against
produced in a system and local surroundings by an irreversible process are brought about
intcrmolecular forces
constitute work.
motion.
dissipated into internal energy, the disorderly
reservoir at
reversibly,
an amount of energy
and appears
energy for
E
work
D
is
E leaves an auxiliary
reservoir at
T
form of heal In other words,
in the
form of work on an auxiliary mechanical device. converted from a form in which it was completely unavailable
in the
which it is completely available for work. Since the original process was not performed reversibly, the energy E was not converted into work, and therefore E is the energy that is rendered unavailable for ivork because of the performance of the irreversible process. It is a simple matter to calculate the energy that becomes unavailable during an irreversible process. If the same changes are brought about reversibly, the entropy change of the system and local surroundings is the into a form in
do not Whenever work is
motion of molecules
work
into internal energy, the disorderly
the molecules of either a reservoir or a system therefore involve a transition
from order
is
to disorder.
possible to regard all natural processes
cases the result obtained
toward a
The
is
that there
is
from
motion of
Such processes Similarly, two gases that
increased.
arc mixed represent a higher degree of disorder than It is
involves order or orderly
increased. Thus, during cither the isothermal or
is
adiabatic dissipation of
Work
when they
this
are separated.
point of view, and in
all
a tendency on the part of nature to proceed
state of greater disorder.
increase of entropy of the universe during natural processes
is
an
1 240
iicat
and Thermodynamics
9-13
expression of this transition. In other words, entropy of a system or of a reservoir
To
existing in the system or reservoir.
we may
state
roughly that
the
a measure of the degree of molecular disorder put these ideas on a firm foundation the
is
concept of disorder must be properly defined. chapter that the disorder of a system
may be
be shown in the next calculated by the theory of
It will
and expressed by a quantity 12 known as the thermodynamic probaThe relation between entropy and disorder is then shown to be
probability bility.
S =
const. In
(9-14)
S2.
By means of this
equation a meaning may be given to the entropy of a system a nonequilibrium state. That is, a noncquilibrium state corresponds to a certain degree of disorder and, therefore, to a definite entropy. in
Entropy and Direction; Absolute Entropy
9-12
Entropy
241
very interesting and also a very important question in physics as to whether there exists an absolute standard state of a system in which the entropy is really zero, so that the number obtained by calculating the entropy It is a
change from the zero the system. It
was
state to
first
any other represents the "absolute entropy" of
suggested by Planck that the entropy of a single
element at the absolute zero of temperature should be taken Zero entropy, however, has statistical implications implying, in a rough way, the absence of all molecular, atomic, electronic, and nuclear disorder. Before any meaning can be attached to the idea of zero entropy, one must know all the factors that contribute to the disorder of a system. An adequate discussion requires the application of quantum ideas to statistical crystal of a pure to be zero.
mechanics.
Fowler and Guggenheim, who have considered the subject exhaustively, summarize the situation as follows:
We may assign, if we
please, the value zero to the entropy of all perfect crystals
of a single pure isotope of a single element in
its
idealized state at the absolute
zero of temperature, but even this has no theoretical significance on account of
The second law
of thermodynamics provides an answer to the question not contained within the scope of the first law: In what direction does a process take place? The answer is that a process always takes place in such a direction as to cause an increase in the entropy of the universe. In
that
is
the case of an isolated system,
it is the entropy of the system that tends to find out, therefore, the equilibrium state of an isolated system, necessary merely to express the entropy as a function of certain coordi-
increase. it is
To
nates and to apply the usual rules of calculus to render the function a maximum. When the system is not isolated but instead, let us say, is maintained at
constant temperature and pressure, there are other entropy changes to be taken into account. It will be shown later, however, that there exists another function, referring to the system alone and known as the Gibbs function, whose behavior determines equilibrium under these conditions. In practical applications of thermodynamics, one is interested only in the the entropy of a system changes in going from an initial to
amount by which a
final state. In cases
with the
where
it is
—
minimum of effort
necessary to perform
example,
many
such calculations
steam engineering, in problems in refrigeration and gas liquefaction, etc.- it is found expedient to set up an entropy table in which the "entropy" of the system in thousands of different states is represented by appropriate numbers. This is done by assigning the value zero to the entropy of the system in an arbitrarily chosen standard state
and calculating
states.
When
for
in
the entropy change from this standard state to all other
done,
it is understood that one value of what is listed as "the entropy" has no meaning but that the difference between two values is
this
is
actually the entropy change.
nuclear spin weights. For the purpose of calculating experimental results, some
conventional zero must be chosen, and the above choice or a similar one is thus often convenient. But its conventional character will no longer be so likely to be overlooked that any importance will in the future be attached to absolute entropy, an idea which has caused much confusion and been of very little assistance in the development of the subject.
9-13
Entropy Flow and Entropy Production
Consider the conduction of heat along a copper wire that
lies
between a
hot reservoir at temperature Tx and a cooler reservoir at T%. Suppose the heat current or rate of flow of heat
is
represented by the symbol
Iq.
In unit
time, the hot reservoir undergoes a decrease of entropy lq/Ti, the copper
wire suffers no entropy change because, once in the steady state,
its
coordi-
do not change, and the cooler reservoir undergoes an entropy increase Iq/T«. The entropy change of the universe per unit time is Iq/T* — Iq/Tx, which is of course positive. This process may be considered, however, from a point of view in which the attention is focused on the wire, rather than on the universe. Since the hot reservoir underwent an entropy decrease, we may say that it lost entropy to the wire, or that there was a flow of entropy into the wire equal to Iq/Ti P er unit time. Since the cooler reservoir underwent an entropy increase, we may say that the reservoir gained entropy from the wire, or that there was a flow of entropy out of the wire equal to Iq To per unit time. But Iq/ T-> is greater than Iq Tl, and hence this point of view leads us to a situation in which the flow nates
242
Heat and Thermodynamics 9-13
of entropy out of the wire exceeds the flow
quantity that can flow,
it is
in.
If
entropy
is
to
be regarded as a is produced or
necessary to assume that entropy
generated inside the wire at a rate sufficient to compensate for the difference between the rate of outflow and the rate of inflow. If the rate of production of entropy within the wire is written dS/dr, we have
and an
IL now, both a heat current
wc
taneously,
rnay say that entropy
and
if
TiTs
Tt
the temperatures of the reservoirs arc
T+
AY" and
T, so that only a
It is
an
AT _IqAT t r
T
^T
+I Ac? T r
,
(9-15)
interesting fact of experimental physics that, in the absence of a
potential difference, a heat current
ence; but
dS
Is
'
small temperature difference exists across the wire,
dr
being generated within the wire by
is
AT
.
=
dr-
lQ
'A
electric current exist in the wire simul-
virtue of both processes at a rate given by
<£S"
dr
243
lintropy
when
there
is
depends only on the temperature
differ-
a potential difference as well, the heat current (also
the entropy current) depends
on
potential difference. Similarly,
when both temperature and
both the
temperature difference and the potential differ-
ences exist across a wire, the electric current also depends on both of these
Since IQ stands for a heat current,
we may
interpret IQ /T as
an entropy
current Is, or j
differences.
The
heat flow (and entropy flow) and the electricity flow are
irreversible coup/ledflows,
-h
which
exist
by virtue of a departure from equilibrium
conditions in the wire. If the departure from equilibrium
may
be assumed that both Is and I are
and potential
is
linear functions of
not too great,
it
the temperature
differences. Thus,
Wc
have therefore the result that, when heat is conducted along a wire across which there is a temperature difference AT, entropy flows through the wire at a rate Is and entropy is produced within the wire at a rate dS
.
dr
AT
change. time,
same
rate /Ac?, since the wire itself undergoes
no energy
reservoir undergoes an increase of entropy IAS/7' per unit there is no entropy change of the wire. Hence the entropy change
of the universe per unit time is I AS/T, which is positive. Changing our point of view, as before, to a consideration of the wire, we may say that there was no flow of entropy into the wire, but that entropy flowed out at the rate
To provide for this outflow of entropy, duction inside the wire at the rate
/ Ac?, T.
dS dr
I
AT
and
The
and
-
I
Suppose now that an electric current / is maintained in this same copper wire by virtue of a difference of potential Ac? across its ends, while the wire is in contact with a reservoir at the temperature T. Electrical energy of amount / Ac? is dissipated in the wire per unit time, and heat flows out of the wire at the
1
=
AS T'
we assume an entropy
u
AS
u
AS
(9-16)
(9-17)
are the famous Onsager equations, which express the linearity between the
and the generalized forces AT/T and AS/T. The L's are connected with electric resistance, thermal conductivity, and the
flows (or currents) coefficients
thermoelectric properties of the wire.
if
Only three
of the four L's arc independ-
can be proved rigorously by means of the departure from equilibrium is small,
ent, for
it
L« =
L»
statistical
mechanics
that,
(9-18)
pro-
which is known as Onsager's reciprocal relation. By means of this strange point of view involving entropy flow and entropy production, and with the aid of Onsager's equations and reciprocal relation, the famous equations for a thermocouple will be derived in Chap. 13.
244
Meat and Thermodynamics
F.niropy
PROBLEMS
left
unchanged. Let
devices 9-1
Fig. P9-1 represents schematically a system
cycle during
which heat
transfers
Q h Q2
,
.
.
.
,
undergoing a reversible take place between it and a
Q u Q2
.
,
and the common
.
.
,
245
be the heats exchanged between the Carnot
reservoir.
By invoking
the second law of thermo-
dynamics, derive Clausius' theorem,
9-2 (a)
Derive the expression for the efficiency of a Carnot engine directly
from a TS diagram. (b)
Compare
the efficiencies of cycles
T
A and B
of Fig. P9-2.
T r,
A
jr
X
'
T
T2
2
Fig.
B
P9-2
9-3
Draw rough TS diagrams
the following ideal gas cycles:
for
and Diesel; a rectangle on a PL diagram: a "right triangle" on a PV diagram in which the base is an isobaric, the altitude an isoehoric, and the "hypotenuse" an adiabatic. 9-4 Given a system whose coordinates arc the temperature T, any number of generalized forces Y, V, and their corresponding generalStirling, Otto,
.
ized displacements X, A", lig. P9-1
System undergoing a reversible
cycle.
(a)
T L T->, where any T is the temperature of the system at the moment it is exchanging heat with the reservoir at T. (If the cycle were not reversible, the temperature of a reservoir and that of the system would not necessarily be the same.) Some Q's are positive and some are negaset of reservoirs at
tive.
Let &,
C-2,
cycle cither as
.
an
.
.
.
.
.
.
,
Prove that 1_
,
represent devices each of which operates in a Carnot engine or as a refrigerator, between one of the reservoirs .
.
....
T (b) (c)
BVh.
What is the expression for 17" Of a paramagnetic gas?
of a fluid?
,
and a common reservoir at T'. Suppose each device is arranged so that in one or more complete cycles it exchanges the same amount of heat with its reservoir that the reservoir exchanges with the system, so that each reservoir
is
9-5 (a)
For an
The
ideal gas with constant heat capacities,
entropy
.S"
is
=
given by
C'v In
P+
Cp
In
V
+
const.
show that
The
(b)
adiabatic compressibility
KB
247
Entropy
Meat and Thermodynamics
246
1 (il\ =
—
is the entropy change in units of R of a diamond of 1 .2 g mass when it is heated at constant volume from 10 to 350°K? Atomic weight of carbon is 12, and is 2230°K. 9-10 Calculate the entropy change of the universe as a result of each
What
is
-1
V\dPja
yP
of the following processes: If
(c)
a gas
both ideal and paramagnetic, obeying Curie's law, show
is
that the entropy
is
(a)
given by
S = Cv n
In
T+
nR
In
V-
(b)
^- + const.,
25
Of
electric current of
10
A
is
maintained
while the temperature of the resistor
What What
(a) (b)
is is
is
for
1
s
(c)
(d)
(a)
A
the entropy change of the universe?
the entropy change of the universe?
A kilogram
of water at
273°K
is
brought into contact with a heat
When the water has reached
373°K, what is the entropy change of the water? Of the heat reservoir? Of the universe? (b) If the water had been heated from 273 to 373°K by first bringing it in contact with a reservoir at 323°K and then with a reservoir at 373°K, what would have been the entropy change of the universe? (c) Explain how the water might be heated from 273 to 373°K with almost no change of entropy of the universe. 9-8 A body of constant heat capacity Cp and at a temperature V, is put in contact with a reservoir at a higher temperature '/'/. The pressure remains constant while the body comes to equilibrium with the reservoir. Show that the entropy change of the universe is equal to
capacitor of capacitance
1
/jF is
connected to a 100-V reversible
capacitor, after being charged to 100 V,
a resistor
where x
-
In (1
+ x)\
= —(T/~
Tf)/Tf. Prove that this entropy change is positive. According to Debye's law, the molar heat capacity at constant volume of a diamond varies with the temperature as follows:
9-9
4
cv
= 3R -£
7V 0,
is
discharged
kept at 0°C.
grams of water at a temperature of 20°C is converted atmospheric pressure. Assuming the heat capacity per gram of liquid water to remain practically constant at 4.2 J/g deg and the heat of vaporization at 100°C to be 2260 J/g, and using Table 5-2, calculate the entropy change of the system. 9-13 Ten grams of water at 20°C is converted into ice at — 10°C at constant atmospheric pressure. Assuming the heat capacity per gram of liquid water to remain constant at 4.2 J/g deg and that of ice to be one-half of this value, and taking the heat of fusion of ice at 0°C to be 335 J/g, calculate the total entropy change of the system. 9-12
Thirty-six
into steam at
250°C
at constant
•
9-14 frictionless
A thermally insulated cylinder closed at both ends is fitted with a heat-conducting piston which divides the cylinder into two parts. clamped in the center with 1 liter of air at 300°K and and 1 liter of air at 300°K at 1 atm pressure on The piston is released and reaches equilibrium in pressure and
Initially the piston
2
atm
is
pressure on one side
the other side.
temperature at a new position. Compute the
and the 9-15
total increase of entropy.
A
What
final pressure
and temperature
irreversible process has taken place?
cylinder closed at both ends, with adiabatic walls,
is
divided
two parts by a movable frictionless adiabatic piston. Originally the pressure, volume, and temperature are the same on both sides of the piston (Po,Vo, ). The gas is ideal, with CV independent of T and j = 1.5. By means of a heating coil in the gas on the left-hand side, heat is slowly supplied to the gas on the left until the pressure reaches 27P In terms of nR, Vt>, and To: (a) What is the final right-hand volume? into
C,{x
m into
such blocks, at 100 and 0°C, are joined together. is the entropy change of the universe as a result of each of
The same
(b)
9-7 (a)
dropped from a height of 100
battery at 0°C.
through
reservoir at 373°K.
is
heat capacity at
What
the entropy change of the resistor?
•
is
block, at 10°C,
total
placed in a lake at 10°C.
the following processes?
kept constant at 27°C.
has a mass of 10 g and c P = 0.84 J/g deg: is the entropy change of the resistor?
What What
Two
9-11
in a resistor of
The same current is maintained for the same time in the same resistor, but now thermally insulated, whose initial temperature is 27°C. If the resistor
The same
is
the lake.
where Cv,.\t is the heat capacity at constant volume and magnetization, assumed constant, and Cc is the Curie constant.
An
copper block of 400 g mass and with a
constant pressure of 150 J/deg at 100°C
(c)
9-6
A
.
,
(b)
What
is
the final right-hand temperature?
'
Heat and Thermodynamics
248 (c)
What
(d)
How much
is
Entropy
the final left-hand temperature?
of the body from
heat must be supplied to the^tfj on the
left?
(Ignore the
docs work W, then
How much work
w
What What What
(/) (g) (A)
A
9-16
is is
is
done on the gas on the right? the entropy change of the gas on the right? the entropy change of the gas on the left? the entropy change of the universe?
engine
is
TV)
is
immersed
in a cooler liquid (m',cP ,Tf
adiabatically
)
Prove that the condition of equilibrium, namely,
Ts = Tf
where Si
—
T\ and remain
7V
subject to the condition that the heat lost by the metal equals the heat gained
amount
of
by the liquid. 9-17 A mass m of water with an equal mass of water at verse
at
T\
isobarically
is
TV Show
,
In
9-18 let (7"o
is
+ r )/2
work obtainable H/
W
2
by drawing
positive
and undergo no change
= CP (Ti+
a
is
Show
that S/
—S = t
Cp
a semicircle of
1
-2
r-
-
In
when
identical bodies of constant heat capacity are at the
same
Two
9-22 initial
(c)
(d)
Solve the problem of Art. 9-7
that S/ that S/
In (1
-
temperature
one body
is
7",-.
A
is
when
perature
a. is
re-
Tu
—
Tj,
To
,
is
T
show that the minimum amount of
Ci
0l + r8 -2Tv\
and suppose that
medium
at the constant tem-
system may exchange Let the system undergo a process involving the absorption the only reservoir
with which
lite
work W, and a change
of entropy S/
—
S).
that:
\
-
Uj
-
W (max)
lit
+ W
S/
T
-
Si
>
0.
(Ur - T S,). (max) — (actual). (*) 7"o AS (universe) 9-24 When both a heat current and an electric current are maintained in the same wire simultaneously by a temperature difference AT" and a potential difference AS, prove that (b)
Using the entropy principle, given in Eq. (9-12), prove: (a) The Kelvin-Planck statement of the second law. (b) The Clausius statement of the second law. 9-20 A body of finite mass is originally at a temperature T"i, which is higher than that of a reservoir at temperature 7*. Suppose an engine operates in a cycle between the body and the reservoir until it lowers the temperature
=
of heat Q, the performance of
(a)
9-19
of phase,
Consider any system immersed in a 7"n
heat is this medium.
Show i
between these two bodies until the bodies remain at constant pres-
refrigerator operates
W (min) 9-23
of
only the hot reservoir
moved, and show that the entropy change of the universe /,
= VtytI-
cooled to temperature T«. If
a)
{
.S't-
Show
that,
the final temperature attained by both bodies.
is
and undergo no change work needed to do this is
- a)
(1
- S = 0.3070 when a .— 1, — positive for all permissible values
Show Show
Ti-2Tf),
a
(b)
show that the
maximum,
sure (a)
of phase,
is
T,
In the expression for the entropy
—
S%),
respectively, are used as reservoirs for a heat engine. If the bodies
and adiabatically mixed change of the uni-
diameter T\ -f- TV change of the bar in Art. 9-7, Tl)/To be designated by the symbol a. that this
= Q - T2{Si -
identical bodies of constant heat capacity at temperatures
VTiT, and prove
obtainable from the
the entropy decrease of the body.
is
at constant pressure
where 7/ 2mcp
maximum work
that the entropy
is
(TV
St
Two
9-21
may be obtained by rendering the entropy change of the universe a maximum
isobarically.
the body. If the engine
Q — W to the reservoir at TV Applying
is
W (max)
piece of hot metal (mass m, specific heat at instant pressure
temperature
and
will reject heat
it
Q from
thus extracting heat
the entropy principle, prove that the
coil!)
cp,
7~i to 7~»,
249
=
Vi
- T Si =
W
w (ralr"(rarL'
W
250
Heat and Thermodynamics
(6)
L\\
=
KA/Ax, where
A' is the
thermal conductivity and
A and, Ax
arc the area and length, respectively, of the wire. (c)
L12
=
T/R', where A"
Show
9-25
is
the electric resistance of the wire.
that, in the case of irreversible
coupled flows of heat and
d
=
(b)
\ (0
Is
STATISTICAL
MECHANICS
electricity,
1=0
10.
Show
dr/iT
that,
involves a
(d)
Show
=
involves a
with
2/
d
AT
minimum
that,
with
and
=2,,.
the equilibrium state obtained
when
rate of entropy production.
AS
minimum
fixed,
AT rf)
fixed, the equilibrium state
obtained
when
rate of entropy production.
9-26 Three identical finite bodies of constant heat capacity are at temperatures 300, 300, and 100°K. If no work or heat is supplied from the outside, what is the highest temperature to which any one of the bodies can be raised by the operation of heat engines or refrigerators?
Fundamental Principles
10-1
In the treatment of kinetic theory given in Chap. ideal gas could not be regarded as completely
6,
the molecules of
an
independent of one another,
then they could not arrive at an equilibrium distribution of velocities. was therefore assumed that interaction did take place, but only during collisions with other molecules and with the walls. To describe this limited form of interaction we refer to the molecules as "weakly interacting" or for It
"quasi-independent."
beyond
The treatment
of strongly interacting particles
is
the scope of the present discussion.
The molecules
of
an
ideal gas
quasi-independence. They are
have another characteristic besides
indistinguishable,
their
because they are not localized
Chap. 6 that the molecules have neither a preferred location nor a preferred velocity. The particles occupying regular lattice sites in a crystal are distinguishable, however, because they are constrained to oscillate about fixed positions; therefore one particle can be distinguished from its neighbors by its location. The statistical treatment of an in space. It
was emphasized
ideal crystal as a
number
in
of distinguishable, quasi-independent particles
will be given in the next chapter. In this chapter
we
shall confine
our atten-
tion to the indistinguishable, quasi-independent particles of an ideal gas. particles, where N, as Suppose that a monatomic ideal gas consists of
N
—
an enormous number say, about 10 20 Let the gas be contained in a cubical enclosure whose edge has a length L, and let the energy e of any particle, as a first step, be entirely kinetic energy of translation. In the x usual,
is
.
direction,
= where p x
is
the x
*mx 2 i
=
(mxY
m J| 2m
component of the momentum.
If the particle
is
assumed to
252
Ileal
and Thermodynamics
10-1
back and forth between two planes a distance /, apart, the simplest form of quantum mechanics provides that, in a complete cycle (from one wall to the other and back again, or a total distance of 2L), the constant momentum px multiplied by the total path 2L is an integer n x times Planck's
move
freely
constant
h.
2L
=
by values of the ri's such have the same energy. To use an example + + given by Guggenheim,! the states corresponding to the values of nx n y and n z in Table 10-1 all have the energy e = 66A 2 8mL 2 There are twelve quanquantum
state
of a particle. All states characterized
»„
that n x
nx
The allowed
— j v8me x
values of kinetic energy
when
we
,
get
,
.
tx
are
\
ni
discrete,
300°K, whose edge is, say, 10 cm. It was shown in Chap. 6 that the average energy associated with each translational degree of freedom is \kT. Then,
1.4
=
10
= 10
X
10.5
6.6
10 9
cm
10-"
X
6.6
X
2.1
X
X
10-"|3I deg
x
66
1
2
3
4
5
6
7
8
9
10
11
12
nx
8
1
1
7
1
4
7
4
1
5
5
4
"y
1
8
1
4
7
1
1
7
4
5
4
5
>h
1
1
8
1
4
7
4
1
7
4
5
5
same
energy
level,
turn states associated with the
However
close they
may
be, there
is still
molecules of an ideal gas.
mechanics to determine,
+
is
is
number of energy fundamental problem of
only a discrete
It is the
at equilibrium, the populations of these
number of particles Ni having energy ei, the number A; 2 having energy e s and so on. It is a simple matter to show that (see Prob. 10-2) the number of quantum states g corresponding to an energy level i (the degeneracy of the level) is very much larger than the number of energy levels— that
10-'" erg,
refer to this
+
statistical
300 deg
and we therefore
energy level as having a degeneracy of 12. In any actual case, n\ n n\ y an enormous number, so that the degeneracy of an actual energy level extremely large. levels for the
-
=
n;
corresponding to integer
nx
= UT<= i X
+ n; +
.
changes by unity, the corresponding change in ex is very small, because n x itself is exceedingly large. To see that a typical value of n x is very large, consider a cubical box containing gaseous helium at values of n x but
«
const, will
,
A2
,
n,
=
nz
Table 10-1
n x h.
Substituting this result into the previous equation,
and
253
Mechanics
Thus, px
e*
Statistical
is,
the
,
;
— s/8 X erg s
X
6.6
X 10-"
g
X
2.1
X
10" H erg
particles
occupying that
level.
Thus,
10~ 19
10-*'
Mi
» Nu
(10-2)
.
is very unlikely, therefore, that more than one particle will occupy the same quantum state at any one time. At any one moment, some particles are moving rapidly and some slowly,
It
when nx changes by unity is so small that, most practical purposes, the energy may be assumed to vary continuously. This will be of advantage later on, when it will be found useful to replace a sum by an integral. Taking into account the three components of momentum, we get for the
Therefore, the change of energy for
total kinetic
energy of a particle
=
Pi
+ Pi + Pi _ 2m
The
a«tf w +"S+<»-
among a large number of different quanAs time goes on, the particles collide with one another and with the walls, or emit and absorb photons, so that each particle undergoes many changes from one quantum state to another. The fundamental assumption of statistical mechanics is that all quantum stales have equal likelihood of being so that the particles arc distributed
tum
states.
The
occupied.
(10-1)
state
the
is
t E. A.
specification of
an integer
for
each
n x , n„,
and
nz
is
a specification of a
New
probability that a particle
same
may
find itself in a given
Guggenheim, "Boltzmann's Distribution Law," Interscience
York, 1955.
quantum
for all states. Publishers, Inc.,
254
I
It-at
4
3
2
1
and Thermodynamics 6
5
10-2
7
8
9
10
i
12
1
13
A
B
c
A
C
B
B
A
C
B
C
A
C
A
B
C
B
A
14
15
16
any one
'
any one time. The
level at
Ni
in
which
255
any one moment, that
specification, at
three,
distinguishable particles A, B, and
any
of the g,
C
energy
level
ei
with degeneracy gi degeneracy £2
particles in energy level et with
Ni particles
There are six ways
Mechanics
there are A'i particles in
Fig. 10-1
Statistical
in
energy
level
e,
with degeneracy
gi
can
occupy three given quantum slates.
Now
consider the
N
f
particles in
quantum
in a container of volume V when the gas has a total number of particles and an energy V is a description of a macrostate of the gas. The number ways Q in which this macrostate may be achieved is given by a product
states associated
would have $ choices in occupying g< different quantum states. A second particle would have the same gt choices, and so on. The total number of ways in which A',- distinguishable particles could be distributed among g< quantum states would therefore bcg t But the quan1 tity g* is much too large, since it holds for distinguishable particles such as A, B, and C in Fig. 10-1. This figure shows six different ways in which three with the energy
«,-.
Any one
particle
N of of
terms of the type of Eq. (10-3), or
•'.
'
had no
particular
number gfi
is
quantum
states.
of permutations of
A",
distinguishable objects
is
A';! If
The
the quantity
divided by this factor, the resulting expression will then hold for indis-
No. of ways that Ni
quantum
of microstates is,
V,
indistinguishable
U are
N, and
}
among g.
that macrostate in
(10-3)
Nil \
should be pointed out that the A" indistinguishable, quasi-independent particles were assumed to be contained within a cubical box only for the rectangular box with three different dimensions could sake of simplicity.
A
system of
render
which
Q
is
—or more simply,
if
to
find the equilibrium populations of
we
look for the values of the individual A's that
In
—a
fi
approximation, which
maximum.
natural logarithm of factorial x
may
it
is
convenient to use
be derived in the following way: the
is
If
we draw
=
In 2
+
that, in the case of
an ideal
gas, there are
many quantum
corresponding to the same energy level and that the degeneracy of each level is much larger than the number of particles which would be found in
each step
is
+
In
and
In x
shown
x.
in Fig. 10-2,
along thc^
axis,
where the
the area under
exactly equal to the natural logarithm, since the width of each x is therefore The area under the steps from x 1 to x
=
step equals unity. In (xl).
In 3
a series of steps on a diagram, as
integers arc plotted along the x axis
states
that,
which case Eq. (10-3) would be unchanged.
Equilibrium Distribution have seen
assumed
gas will correspond
this state. It is
contains factorials of large numbers,
In (x!)
We
To
and the num-
the greater the proba-
the equilibrium state of the
a maximum.
52
Since In Stirling's
N particles in
kept constant,
the energy levels, therefore,
states
in
called the thermodynamic probability of the particular macro-
V. is
Other names for this quantity are the number of complexions. Whatever its name, the larger
ber
It
have been chosen,
quantity
state.
bility of finding the
particles can be distributed
10-2
(10-4) l
identity, there
tinguishable particles. Therefore,
easily
Nt
Nx'.
quantum states 2, 7, and 13. would be only one way to occupy these That is, one must divide by 6, which is 3! The
distinguishable particles can occupy particles
=
If the
When
x
is
large,
we may
=
replace the steps by a smooth curve,
shown
256
Heat and Thermodynamics
10-2
x
In
where we have used the
In 7
In In
6
In
5
In
4
V.
a
maximum
In
3
we proceed
const.,
(10-7)
=
II
=
const.
(10-8)
2
d
In 1
{In
2
as a
The
10-2
dashed curve
dashed curve approximates
under the
area
+ In 3 + In 4 +
•
•
+ In x)
when x
the
area
Q
= y L>
d
dN =
Niln iL\
N
under the steps
is large.
when
x
is
large,
gi
dNi |f A
+
dNx
+ +
Integrating by parts,
we
In x dx.
/
and
get
In
~
(.v!)
—
x In x
x
+
we
and Then,
if
we
neglect
1
compared with
«i
dNi
—0
g-i UdNs
In
JV 2
Using
is
x In x
—
(10-5)
x.
9.
A'i In
approximation
Eq. (10-4),
in
we
+••+
u dNi
A
|r A'2
A dNi +
In
A dN 2
—
/3
dN%
—
+
+
In
+ «W» —
+ -
In
fie*
«f#s
+ +
and the third by
we
dN § A;
•
—
0,
=
0,
=
0.
/?,
=
dN + 0e; dN -
= =
A
where
In
A
get
+
t
If we add these equations, the coefficient of each
get
g
x
—
Ni
In
h\
4- A'i
+
Nt
In
g2
—A
2
In
N2 +
#8
ith
o,
o, 0.
Ni In
gi
-Y
A",-
In
A',-
-•-
V
In
Q
= y Zw
A',-
In
+ # A;
4 + ?-
Ni
.V,-,
A',
(10-6)
dN may be set equal
to zero.
term,
+ or
or
dNt
In
In
= V
dNi
•
=
Stirling's approximation.
Stirling's
=
+
&- dNi +
In
Taking the In
+
x,
and This formula
gi +••••+ hi #£ Mi + A,-
dNi e2
Ai
«
equal to zero and taking the differential of
arc Lagrange multipliers (sec Art. 6-3),
In
In (x\)
U
(10-9)
get
Multiplying the second equation by In
1.
is
=
Setting the differential of In
i
~
Q
I B|^i.
In In (xl)
mind
t
Eqs. (10-7) and (10-8),
in Fig. 10-2; therefore, approximately,
in
the differential of In
0,
X
8 Fig.
In
problem by the method of Lagrange multithat the e's and g's are constants. The populations of the energy levels, and their sum A' is
to solve this
only variables are the
In
render
=
important to bear
constant. Since
to
A*
SA',«,-
it is
is
2A = ,-
Before
N. Our problem now
subject to the conditions that r
pliers,
=
fact that 2A'f
257
Mechanics
Statistical
In
A - 0a =
In^i-
l
n
A =
N = t
0.
-0% Agie-e'<.
(10-10)
258
10-3
Heat and Thermodynamics
Mechanics
Statistical
259
The population
of any energy level at equilibrium is therefore seen to be proportional to the degeneracy of the level and to vary exponentially with the energy of the level.
The next multipliers
step
to
is
A and
A
>
1
=
A
«J
A
N ,Ns SAT,
=
*}i
A
N„ »„...
Mt
t
determine the physical significance of the Lagrange
11
A
N
Wj
=
.
.
.
A
Nj, A
.
.
.
iV
^
/?.
//-/V/y/W///Y/////y///z/^^^^^ Fig. 1
0-3
The
Significance of population
A",
of the
A and z'th
energy level
Ni
=
1
wall.
j3
is
given by
over
all
the energy
Agie-f':
levels,
we
so that the logarithm
A =
The sum It
was
in the
first
N
Each sample has a constant number
denominator plays a fundamental role
"sum over states."
who
called
in statistical it
sum
is
2A7 =
A'
=
const.
=
Jv
=
const.,
,-
and
2A'j
the partition
?»^nr*i
(10-12)
-f-
(10-13)
but the energy of each sample is not composite system is constant; thus,
JV** Substituting this result into Eq. (10-10),
we
=
,gm
A'
constant.
+ ^% = vU =
Only
the total energy of the
const.
get
Vo Ni
of molecules, so that
mechanics.
the Zuslandsumme, or
We retain the first letter of Zuslandsumme as a mathematical
symbol, but the accepted English expression for this junction. Thus.
and
+ N + y^y In Jf + N. L, U N In % Ni fy
= Y
(10-11)
H*r»
introduced by Boltzmann,
is
get
In ii
and
of two samples of ideal gas separated by a diathermic
total energy is constant.
second sample of gas, the symbols expressing energy levels, populations, etc., are distinguished with a circumflex. The thermodynamic probability of the composite system Q is the product of the separate thermodynamic probabilities,
Summing
An isolated composite system
0-3
The
find equilibrium conditions,
we proceed
as before
and get
(10-14)
It will be shown later that Z is proportional to the volume of the container. Since the properties of a gas depend on temperature as well as on volume,
In
one would expect a relation between /3 and the temperature. To introduce the concept of temperature into statistical mechanics, we must go back to the fundamental idea of thermal equilibrium between two systems, just like the procedure in Chap. 8 for relating the quantity X to temperature. Conse-
A
y dNi
= In
A V dNj =
-££«4ft-lT%d$-
0.
quently,
let us consider an isolated composite system consisting of two samples of ideal gas separated by a diathermic wall, as shown in Fig. 10-3. For the
Adding and
setting
each
coefficient of
dN
and
dN
equal to zero,
we
get two
10-4
Hrat and Thermodynamics
260
have the important link between thermodynamics and
Nj =
and all
statistical
I
",
we
mechanics:
Ag,-e-^,
1
where
261
Mechanics
specify that the reversible process take place at constant
we now
If
sets of equations,
Statistical
quantities are different, except 0.
When two systems
(?£ \dU
=
T
separated by a
(10-16)
0's are the
diathermic wall come to equilibrium, the temperatures are the same and the same. The conclusion that is connected with the temperature is
Since both S and
inescapable.
(dS/(ll/)v gives the reciprocal of the Kelvin temperature. This
was shown in Chap. 9 that the entropy of an isolated system increases when the system undergoes a spontaneous, irreversible process. At the conclusion of such a process, when equilibrium is reached, the entropy has the maximum value consistent with its energy and volume. The thermodynamic probability also increases and approaches a maximum as equilibrium is approached. We therefore look for some correlation between .9 and £2. Conone with entropy sider two similar systems A and B in thermal contact Sa and thermodynamic probability il A the other with values Sb and Ob-
macroscopic concept of temperature
It
—
U may
be calculated by
statistical
mechanics,
is injected into statistical
is the
way
the derivative in
which
the
mechanics.
In employing the Lagrange method to find the equilibrium values of the wc found that
energy level populations [Eqs. (10-9) and (10-10)],
'
and
In
il
|i
In
= y =
In -0- dN,:
-
/3
In A.
,
Since entropy
system
is
an extensive variable, the
The thermodynamic
In
Q =
7
/3e,:
dNi
-
In
A^ dN
t
= QA Q B
is
the product, or
.
where
U
is
the total energy of the system. Therefore,
let
s = then.
The
'
+ Ss,
probability, however,
£2
we
entropy of the composite Therefore,
S = Sa
If
total
is
f(QjiB)
only function that
arbitrary constant
k',
m),
=/(a A )
0+f(ih<).
satisfies this relation
we may
is
the logarithm. Introducing an
Since (dS/dlf)v
=
d
In
2 =
d
1
dU
k>du
k
'
lnn = k'
V^'> we Sct tnc beautiful
\dUj,
result
write fi
S=
k' In
S2
dl!
+ P dV.
the process takes place between two neighboring equilibrium be performed reversibly, in which case dQ = T dS, and If
dU = T dS - P dV.
it
will
10-4 states,
it
may
(10-17)
Yt
W hen the actual values of the ;
between entropy and thermodynamic probability. The first law of thermodynamics applied to any infinitesimal process of any hydrostatic system is for the relation
dQ =
~
(10-15)
Wc
be seen that
k' is
e's appropriate to an ideal gas are introduced, none other than Boltzmann's constant k.
Partition Function have seen that the population Ni
=
jV,-
of the
Agie-H'i.
;'th
energy level
is
10-5
Heat and Thermodynamics
262
Substituting \/k'T for
and
/3
N/Z
for A,
and,
get
(T
p->ilVT
Mechanics
263
finally,
a =
N =—
/V-
we
Statistical
Nk
7
In
(10-18)
rr
N+ r
-
Nk',
(10-21)
once In Z is known. which provides us with a method of calculating One more equation that is of value is the relation between pressure and .S"
Z = 2^-'<'*T
vhere
(10-19)
the partition function. Since
The
partition function
Z
TdS = dU
contains the heart of the statistical information
about the particles of the system, so that it is worth while to express other properties of the system, such as U, S, and P, in terms of Z. If wc differentiate Z with respect to T, holding V constant, wc get
~
+ PdV,
Apr-H)
fri™!*-*""
(&-Z=
_-
V
.
From the we get
f-a-p-'tili'T
relation
between S and the partition function, given
U=
It
v
€( '
TS = -Nk'T
In
in Eq. (10-21),
- Nk'T; 4 TV
''
z,
ze/
P = Nk'T
therefore,
"
d In
z\ (10-22)
/V/fcT 2 It follows that
so that again the pressure
V-«T>(!jA.
(10-20)
of
T
and
This
is
I
the
E7
may
be calculated once In
,9
Z known is
=
as a function of
T and
.9
=
-£'
Y U
In
10-5
— + k'N. gi
Substituting for /Vj/gs the value given in Eq. (10-18),
Z is known
as a function
It
provides us with a simple
obtaining the properties of a system:
Use quantum mechanics to find the e values of the quantum states. Find the partition function Z in terms of T and V. Calculate the energy by differentiating In Z with respect to 7". Calculate the pressure by differentiating In Z with respect to V. Calculate the entropy from Z and U.
k' In 0,
-V,'
In
V. Also,
where, according to Eq. (10-6),
Hence,
be calculated once
advantage of statistical mechanics.
set of rules for
and
may
'.
To we
get
must
Partition Function of an Ideal apply the rules laid first
down
in the
Monatomic Gas
preceding article
to
calculate the appropriate partition function. This
levels
an ideal gas, wc was defined to be
264
where the summation was over is
10-5
Heal and Thermodynamics
obtained
if
we sum
energy
all the
levels.
Exactly the same result
and
the expression
Z=
In
r
f In
In
fin (
Statistical
Mechanics
265
2Trmk' \
(10-24)
Pressure of an ideal monalornic gas
over
all
quantum
states.
We at first take
of translation of the particles confined to a rectangular box sides are, respectively, a,
by Eq. (10-1)
and
l>,
n t , n,„
states.
The
and
n z arc
" 8m
71x=
2=y
1
('
V
V
«y= 1
tli = 1
whose
quantum
x, y,
state;
is
and z
-««(;)
given
,
b*-
n,„
= _
+ ?)' specifying the various
quantum
therefore a threefold sum: thus,
is
Comparing
this result
x r f
and
y
-(h , ISmk'T)(n, , Ht ) e
V
with the expression for the pressure
e
k'
-(li , ;S»a-'r)(ri,I ..'c ! >_
n z that give rise to appreciable values of the
r tfis.*'rit,,w d„
i r
/*
ra>n«w)(«,w
rf
g-oto**?nwt*>
2
is
of the type listed in
Table
=
k
= R
( \
2
=
&t\
g-«*
is
6-2:
ISirrnk'T
2
\
* « i„ 2\a
4
A
2
.2"\
Nk
In
=
h (10-27)
|-
come
to statistical equilibrium, the
k T.
£+¥+m
A
In
and since abc
Z\
Entropy of an ideal monatomic gas
^=
8jrm*'r
l&irm/c'T
2\
A
1
w
same result which was obtained by the kinetic theory of monatomic ideal gas and shows that, when particles each having
energy per particle equals 3
In
exactly the
gases for a
Therefore,
Z =
d
INkT.
three translational degrees of freedom
f
(10-26)
Energy of an ideal monatomic gas
U = NkT
This integral
P = NkT/V
„ i
[ J°°
Each
(10-25)
obtained with the kinetic theory of gases, given in Eq. (6-10), where k is Boltzmann's constant, we sec that the arbitrary constant k' introduced in the equation S = k' In Q, is none other than the Boltzmann constant, or
energy are very large and since a change of n x or n y or n. by unity produces a change of energy that is exceedingly small, no error is introduced by replacing each sum with an integral and by writing
z=
i-"T
gHh*t«mkfTHnJt
C|! .'8»'*'n("^/"!)
Since the values of n r
+
\a*
quantum numbers
partition function
Z= V or
of any
as *'
where
The energy
c.
dlnZ\ bV ) T
P = Nk'T
into account only the kinetic energy
Z + | la
T
(2irmk In
V,
Z=
F
2wmk' T\i h2
(10-23)
= Nk ||
In
T
*5A
.
/2armk\i
+ $M + Nk
266 If
we
Heat and Thermodynamics
take
1
mole of
N=
gas,
10-6
ATA and
A'\k
=
and rewrite
R. Therefore,
it
s
=
T+R
c v In
-R
In v
In
/
W
h^i
+ R s
- U-
is
compared with Eq.
to he
=
s
cv In
(9-5),
d(\/kT)
(10-28)
R\n
-\-
v
-\-
Z
d In
=
(10-30)
dp
namely,
Suppose
T
267
Z
din
N
<«>
This expression
Mechanics
thus:
l\ ^ 2lrm ':
Statistical
su,
e
to consist of
terms representing translational kinetic energy of
the type fymw*, those representing rotational kinetic energy of the type i/co 2 ,
+
those representing vibrational energy gJKj8
and we sec that not only were we able to arrive at this equation by the methods of statistical mechanics but also we were able to calculate the constant s<,. Equation (10-28), which was first obtained by Sackur and Tetrode,
$&£*, etc. All these
energy are expressed as squared terms of the type
forms of
Let there be / such
bip\.
terms, or
+b
= hp\
e
+
2 p\
+
b,p).
usually bears their names.
Then, since the partition function
the product of the separate partition
is
functions,
10-6
Equipartition of Energy
Z= Both kinetic theory and statistical mechanics, when applied to the molecules of an ideal gas (each having three translational degrees of freedom), yield the result that at equilibrium the energy per particle associated with each degree of translational freedom is %kT. The methods of kinetic theory
*-**«»>
dp x
J
Let
yt
then
=
J*
e-1*** dpi
J
and
fflpi
dyi
f+rt dp, =
p
=
(S-i
could not be applied to rotational and vibrational degrees of freedom, but the simple statistical method just developed is capable of dealing with all
rw
J
=
dp,.
& dpf,
g-»ji.f ie
dyi
r*an J
dyi
types of molecular energy, not just translational kinetic energy.
The property
of the partition function which makes
whenever the energy of a molecule
expressed as a
is
it
so useful
sum
is
that,
of independent
terms each referring to a different degree of freedom,
6
=
t
+
e
+
t
where Ki does
not
contain p.
+•;
Z= =
then.
Z= \
e
~' lkT
-''' kT
ye
partition function
0-iKi
/3-»/T,
p-z^JCff,
•
•
•
•
now becomes
0-lKf
,
Kf,
£-(«'+«"+«"'+• ••)/*r
Li
i-i
=
= V
The
where none
of the K's contains 0.
Since
(e)
= — 3(ln
Kl
+ ln K +
Z)/dp,
y e-<"i kT y r**** to
= Z'Z"Z'"
(10-29;
If the various types of energy are calculated with classical physics, it is a simple matter to derive the classical principle of the equipartition of energy. take Eq. (10-20), namely,
We
U = NkT*
d In
dT
Z
- - 35 (- 2
ln
"
+
ln
*
'
'
'
In
K
2/3'
and
since
/?
=
1/kT, <<>
= {hT.
(10-31)
268
Heat and Thermodynamics
It has therefore been
proved
thai,
10-8
when a
large
number of nondistinguishable, quasisum of f squared terms come to
independent particles whose energy is expressed as the equilibrium, the average energy per particle is
f
times
Now,
it
has been shown both by kinetic theory and by an ideal gas is given by
^k T.
badly when applied to polyatomic molecules which have many vibrational is a simple matter to introduce quantum mechanical expressions for the energy of rotation and vibration in the partition function
and this
is
referred to
thermodynamic properties. For discussion of any of the books on statistical mechanics listed
Since the energy per particle
is
We
It
p
Work and Heat
Statistical Interpretation of
have been considering the
statistical
volume
lational
V.
The
$MsT.
follows that
2
m
U
equilibrium of a large
number
quasi-independent particles in a cubical container energy levels of individual particles undergoing trans-
(10-33)
we
Substituting this result into Eq. (10-32),
get
S»
N of nondistinguishable, of
translational kinetic energy only, with three
U=
in the appendix.
10-7
mechanics
degrees of freedom,
to calculate the resulting
the student
statistical
NkT P=
is
degrees of freedom. It
269
Mechanics
that the pressure of
the famous principle of the equipartition of energy that was mentioned but not proved in Art. 6-5. It was stated that the principle broke down
This
Statistical
(10-34)
e,-
motion only were given by
A
change of volume, therefore, causes changes in the energy values of the levels, without producing changes in the populations of the levels.
energy A2
8m LI Since L 3 = V, then l. 1 = V', and quantum numbers appropriate to
(»i
+K+ B
letting
the
z'th
When
nl).
the
JV;
d
be the sum of the squares of the energy level, we get Since kd In
quantum numbers
that determines
corresponding energy et depends on volume only.
In
The this
effect of
a
=
— + 8m It
In
2
a small change of V on
B we may f,
Taking the logarithm
B -
In
=
S«; dNi.
t
=
dS,
k
2e,-
dNi
=
dS,
and
setting 1:0 equal to
l/T, we
get finally,
BiV-i. U-£8m set of
In it
t
h2
Given the
change and the u remain constant, we have from page 261,
£3
2 £i dNi
say that the of
We see
e,-,
= T
dS.
that a reversible heat transfer produces changes in the populations
of the energy levels without changes in the energy values of the levels them-
In V.
Thus, the equation dU = 2e,- dN{ -\- 2JV< de,- expresses the first and second laws of thermodynamics, with 2e, dNi = T dS and 2A'; de{ = —PdV. selves.
e
is
given by taking the differential of
equation; whence dti ti
=
2dV 10-8
3 V'
Therefore,
Disorder, Entropy,
Whenever work
and Information
or kinetic energy
is
dissipated within a system because of
friction, viscosity, inelasticity, electric resistance, or
Kde^-^dV,
disorderly motions of molecules arc increased.
magnetic
Whenever
hysteresis,
the
different substances
arc mixed or dissolved or diffused with one another, the spatial positions of
and
lN u=-\pV. id
(10-32)
more disorderly arrangement. Rocks crumble, iron some metals corrode, wood rots, leather disintegrates, paint peels, and people age. All these processes involve the transition from some sort of the molecules constitute a rusts,
270
Heat and Thermodynamics
"orderliness"
10-8
to a greater disorder. This transition
is
expressed in the
thermodynamics by the statement that the entropy of the universe increases. Molecular disorder and entropy go together, and if we measure disorder by the number of ways a particular macrostate may be achieved, the thermodynamic probability S2 is a measure of disorder. language of
Then
classical
the equation
$ =
k In
U
is
the simple relation between entropy
and
ways
Statistical
Mechanics
271
of achieving this state, because there arc fewer microstatcs with position
coordinates in the smaller volume. Before the compression, each molecule is
known
to
could occupy
is
Vo/AV, where
compression, each molecule
number
Vu The number
be in the volume
of locations each molecule
.
AV to
is
some arbitrary small volume. After the be found in volume V\, with a smaller
is
of possible locations Fj/AF. It follows that
disorder.
The number of ways in which a particular macrostate may be achieved can be given another interpretation. Suppose that you are called upon to guess a person's first name. The number of choices of names of men and
women
is staggeringly large. With no hint or clue, the number of ways in which one can arrive at a name is very large, and the information at one's disposal is small. Suppose, now, that we are told the person is a man. Immediately the number of choices of names is reduced, whereas the information is
Information
increased.
is
increased further
if
we
is
the information.
convenient measure of the information conveyed choices is reduced from Q n to ily is given by r
/
=
,
,
tln
when
the
number
of
^o
n7
bigger the reduction, the bigger the information. Since k In S, then
9.
is
the
entropy
I or
Si
which can be interpreted the
to
mean
- & - Su =
Biillouin,
S -I,
that the entropy of a system
amount of information about
is
r
for the entire gas of
kln
V /AV
w^ =
,
kln
V Yi
,
N molecules, I
= Nk
In
\y v
in
,
\
agreement with the
reduced by words of
the state of a system. In the
"Entropy measures the lack of information about the exact
state
of a system."
result of classical
thermodynamics. The increase of is seen to be identical with the
information as a result of the compression
corresponding entropy reduction.
The
A
The
and
0, = kln-
name
arc told that the man's
with H, for then the number of choices (or ways of picking a man's name) is reduced very greatly. It is clear that the fewer the number of ways a particular situation or a particular state of a system may be achieved, the starts
greater
I,
connection between entropy and information can be applied to the
problem of MaxwelPs demon. Maxwell imagined a small creature stationed near a trap door separating two compartments of a vessel containing a gas. Suppose that the demon opened the trap door only when fast molecules approached, thereby allowing the fast molecules to collect in one compartment and slow ones in the other. This would obviously result in a transition from disorder to order thus violating the second law. According to Biillouin, the demon could not tell the difference between one kind of molecule and another because he and the molecules are in an enclosure at a uniform temperature and all are bathed in isotropic blackbody radiation. The demon could not sec the individual atoms. However, suppose that we allow the demon, according to the analysis of Rodd, to use a flashlight whose radiation is not in equilibrium with the enclosure. Then the demon can get information about the molecules and thereby decrease the entropy of the system. But
—
other
phenomena come
into the discussion: (1) the filament of the
lamp
the flashlight undergoes an increase of entropy; (2) a photon scattered
As an example of the connection between entropy and information, con(N molecules) from a volume ''ci to a volume Vj, We know that the reduction of entropy is equal to
sider the isothermal compression of an ideal gas
& - Si m
Nk
In
&
a molecule
is
absorbed by the
demon
demon and
in
by
serves to increase his entropy;
opening the trap door reduces the number of microstatcs available to the molecules. (The entropy change of the battery of the flashlight can be ignored.) If all these processes are taken into account and the corresponding entropy changes are calculated from the standpoint (3) the action of the
in
Rodd was able to demonThe second law is not violated.
of the increase or decrease of information, then
But,
when we
strate that the total
decrease the volume of the gas,
we
decrease the
number
of
Much
entropy change
is
positive.
has been written about reversibility and irreversibility, order and
272
Heat and Thermodynamics
and supposed violations of the second law. It would be hard to any language to compare with the exposition given in Feynman's "Lectures on Physics" (Chap. 46). This chapter is recommended wholeheartedly to all students whether naive or sophisticated in their level of attainment for its brilliance, its depth, and its human warmth. disorder,
Prove that the number of quantum
(b)
find anything in
i
dek(2ir//i )V(2m)h
Show that quantum states is
number
^
erg
X
lO" 16 erg/deg and h
=
6.63
X
10'
"
Eq. (6-25). 10-5 In the case of
s.)
•
in
A mercury atom moves
box whose edge is 1 m long. Its kinetic energy is equal to the average kinetic energy of a molecule of an ideal gas at 1000°K. If the quantum numbers n x n and n. arc all equal y 10-1
in a cubical
The quantum
states
available for gas molecules of energy
e
be achieved
a cubical box of length L correspond to integer values for each n x , «„, and n z according to Eq. (10-1). In a three-dimensional Euclidean space with
when
coordinates n x , nv , and n z each unit volume will contain one ,
The
total
number
of
quantum
quantum
state.
states g'
with energy less than «' is equal to the volume of the positive octant of a sphere of radius r = Li^mt^ih. (a)
is
g,-
Show
that
»A
(b)
Render
N
=
that g'
atm
A' of
» N,
10-3
1
e' is
helium atoms,
(d)
about
Show
Given
=f(ii A )
+/(Q
ll
ei;
A'2 particles
e%,
(2,
•
9..
S
should be
•
quasi-independent particles capable
with degeneracies
,
•
gi, gz,
.
.
.
,
which there are A'i particles in with energy etl and so on, assume the thermoin
by the Bose-Einstein expression,
to be given
o liE
= (gi+Ni)\(g2+N2)\ giWiig*m\
).
and then with respect
to
il A
.
Integrate twice to show that /(12)
•
r
any given macrostate
dynamic probability
Using First differentiate partially with respect to Q, , ;
!
a
A' indistinguishable,
Consider a function / defined by the relation /(0.,fi ;( )
Sl_>
Ar i!A' 2
should be the same as for indistinguishable particles but
energy level pressure,
so on, may-
maximum subject to the equations of constraint and 2A ,e; = U = const., and explain why U and P
il
const,
respectively. In
In a volume of 1 cm' of helium gas at 300°K and 10~ 12 erg. Calculate: (b) g' and (c) the number
Si
Stirling approximation, calculate In
In
of existing in energy levels
3A»
and
different
10-6
S
62;
,-.
Using the
2A =
number of ways quantum states with
the
with energy
states
is,
7
(«)
r j
occupying these
-
A"i particles in gi
quantum
Q = AM
in
(
given by
to n, calculate n.
10-2
~' lkTdi
iie
N distinguishable particles,
particles in gi
«jj A" 2
,
,
Sjm
which a macrostate defined by
energy
dN
Derive the Maxwellian law of the distribution of speeds, that
(d)
1.38
of ideal-gas molecules
2
=
PROBLEMS =
energy interval
given by
d
(Values of constants: k
states dg, in the
d6.
the
(c)
—
i
273
Mechanics
Statistical
Stirling's
maximum
approximation and the Lagrange method, render In
subject
to
2A = r
,-
A'
=
const,
and 2A
r
,e,-
= U=
f2 U
const.,
E a
and
show that
=
const. In
Q
+
Nt
const.
Take the expression for the kinetic energy of a particle in a and imagine a space defined by the cartesian coordinates n x n u and n z Note that a single quantum state occupies unit volume in this space. (a) Setting n 2 = n x + n\ + n\, show that the number of quantum 10-4
cubical box
,
,
10-7
=
Si
Ae-?»
-
1
Given the same system as in Prob. 10-6, except that the thermois. given by the Fermi-Dirac expression
dynamic probability
.
states in the small interval dn is !(4tt« 2 dn).
Bra
=
gi\g*i
Nil(gt- N{)W t l(g t
- N )\ 2
274
Heat and Thermodynamics
Using
Stirling's
maximum
approximation and the Lagrange method, render In Qfd a r 2iV,- = A = const, and 2A",e; = U = const., and show
subject to
11.
that
Ni
10-8
=
PURE SUBSTANCES
gi
AeS«
+
1
Given a gaseous system of jVa indistinguishable, weakly
inter-
acting diatomic molecules: (a)
Each molecule may vibrate with
an energy
e,-,
U = Show
the
same frequency
v
but with
given by
(k
+ t}hv
(i
=
0,l,2,
that the vibrational partition function
.
Z„
.
.).
is
Enthalpy
11-1 -lh,lkT
e
z = t
1
-
«-*'/«
The
laws of thermodynamics were stated and their consequences were
developed
Each molecule may rotate, and the rotational partition function ZT has the same form as that for translation, except that the volume V is replaced by the total solid angle Air, the mass is replaced by the moment of inertia /, and the exponent § (referring to three translational degrees of freedom) is replaced by since there are only two rotational degrees of (b)
-§-,
freedom. Write the rotational partition function. (V)
Taking
into account translation, vibration,
in a sufficiently
When
of coordinates.
and
rotation, calculate
manner to apply to systems of any number more independent coordinates, one
speaks of isothermal surfaces and isentropic (adiabatic reversible) surfaces. as
If,
is
often the case, there are only
faces reduce to simple plane curves.
pendent coordinates of constant mass.
the Helmholtz function.
general
there are three or
system, for
we
shall
is
two independent coordinates, these surof two inde-
The most important system
a hydrostatic one, consisting of a single pure substance
Once the thermodynamic equations arc developed for this see how simple it is to write down the analogous equations
any other two-coordinate system.
(d)
Calculate the pressure.
(e)
Calculate the energy.
PV appeared
(/)
Calculate the molar heat capacity at constant volume.
very useful to define a
In discussing some of the properties of gases in Chap. 4, the several times (see Probs. 4-8, 4-9,
new
and
sum
of
U and
4-11). It has been found
function H, called the enthalpy,
1 !
by the
H = U + PV.
relation
(11-1)
In order to study the properties of this function, consider the change in
enthalpy that takes place
from an
initial
when
dH = but
dQ =
Therefore,
dH =
t
Pronounced
a system undergoes an infinitesimal process
equilibrium state to a
en-thal'-pi.
final
equilibrium
dU+ P dV + dli + P dV. dQ + V dP.
state.
We
have
V dP;
(11-2)
276
1
and Thermodynamics
Icat
Dividing both sides by dT,
~". '.".'"
r~i':_.m dT +
dT
dT
'
and, at constant P,
[w)P -
Cp \df) P -
',:///////'/' ((')
transferred.
That
an
isobaric process
is
equal to the heat that
////////a
r~^~ ;:.:—
(11-3)
-
dH = dQ + V dP,
the change in enthalpy during
277
———
-"-
_-
Pi.ViWm
i.
Since
Pure Substances
11-1
Initial
\
state
mzzzzm
|v ///////'.'/s//-v/////fy
is
;'-'
is,
ES
-
Hi
=
II,
-
Hi
= j' CP
or
H,
Q, (isobaric)
Eft. V,
'%%zm
m
(11-4)
dT. (/) Final state
Since isobaric processes are
much more important in
try than isochoric processes, the
enthalpy
is
engineering and chemis-
Fig. 11-1
of science. If
may
A
a pure substance undergoes
an
infinitesimal reversible process, Eq. (11-2)
be written
= 'T
(§),=
and
<&"
(11-5)
throttling process
is
obviously an irreversible one, since the gas passes
on its way from the initial equilibrium state These nonequilibrium states cannot be described by thermodynamic coordinates, but an interesting conclusion can be drawn about the initial and final equilibrium states. Applying the first law to the through nonequilibrium to
dH= TdS+V dP, showing that
Throttling process.
of greatest use in these branches
its
final
equilibrium
states
state.
throttling process
Q = U,The
relations given in Eqs. (11-5) suggest that the properties of a
pure sub-
stance could be displayed to advantage on a diagram in which His plotted as a function of S and P. This three-dimensional graph would be a surface, and Tand V would be indicated at any point by the two slopes that determine a
plane tangent to the surface at the point.
One
wc have
Q =
and
W=
+
PdV
I
W,
fVi
PdV.
Since both pressures remain constant,
most interesting properties of the enthalpy function is in connection with a throttling process. Imagine a cylinder thermally insulated and equipped with two nonconducting pistons on opposite sides of a porous wall, of the
shown in Fig. 11-1;'. The wall, shaded in horizontal lines, is a porous plug, a narrow constriction, or a series of small holes. Between the left-hand piston as
and the wall there right-hand piston
a gas at
is
is
and a volume F,-; and since the any gas being thus prevented from the gas is an equilibrium state. Now
The above the
expression
work necessary
is
to
known
initial state of
=
Such a process
is
a throttling process.
or
And,
in
engineering as/low work, since
it
represents
keep the gas flowing. Therefore,
a pressure P;
imagine moving both pistons simultaneously in such a way that a constant pressure P( is maintained on the left-hand side of the wall and a constant lower pressure Ps is maintained on the right-hand side. After all the gas has seeped through the porous wall, the final equilibrium state of the system will be as in Fig. 11-1/.
W = PfV, - P^.
against the wall, with
seeping through, the
shown
U<
Ui
U,
+
-
Ui
/>,!/;•
+ P,VS - PM,
=
u,
+
pf Vj.
finally,
Hi
= H,
(throttling process)
(11-6)
In a throttling process, therefore, the initial and final enthalpies arc equal.
One is not entitled to say that the enthalpy remains constant, since one cannot
278
Low
High pressure,
pressure. 1
Pump
Constant high
speak of the enthalpy of a system that ~
/{//////////////////)//////# //// '//// 'Y////, -
-
is
279
passing through such nonequilibrium
In plotting a throttling process on any diagram, the initial and final equilibrium states may be represented by points. The intermediate states, states.
however, cannot be plotted. A continuous throttling process may be achieved by a pump that maintains a constant high pressure on one side of a constriction or porous wall and a constant lower pressure on the other side, as shown in Fig. 11-2. For every
Constant low pressure.
pressure^
mole of
fluid that
undergoes the throttling process,
'//////. hi
Fig. 11-2
Pure Substances
11-2
Heat and Thermodynamics
=
kf
we may
write
,
Apparatus for performing a continuous throttling process.
where the lower-case
The Table 11-1
Comparison of
Internal energy
letters indicate molar enthalpy.
properties of the enthalpy function must be clearly understood by the
U and H
student, for they will be used continually throughout the remainder of this
book.
The comparison
Enthalpy /f
Table
11-1 will help the student to
U
In general
and the enthalpy given
of the internal energy
remember
in
these properties.
In general
dU = dQ - PdV
#= dQ
+ VdP
The Helmholtz and Gibbs Functions
11-2
The
Helmholtz function (sometimes called the Helmholtz free energy)
is
defined
as Isochoric process
Uj- Uc= j'CydT Adiabatic process
Ui-
Ui
= - VpdV
Free expansion
W=
F= U-
Isobaric process
-Q
fl>
-
11,
-Hi = f'CpdT
fli
Hf
V V dP
Throttling process
dF = dU
U=
f
S dT,
dF =
Hence
-S dT - P dV.
(11-8)
Hi = H,
Uj
ideal gas
Cv dT +
- TdS-
T dS = dU+ P dV.
and
From For an
(11-7)
For an infinitesimal reversible process,
Adiabatic process
- Ht=
TS.
const.
this
it
follows: (1)
For
a reversible isothermal process,
For an ideal gas
dF = -PdV,
H= j CP dT+ const. /•>-
or Reversible process
Reversible process
dU = TdS - PdV
dH= TdS- VdP
-(if).
Hence the change
Fi
= - j'.PdV.
(11-9)
Helmholtz function during a reversible isothermal process equals the work done on the system. (2) For a reversible isothermal and of the
isochoric process,
dF =
-OS.
and
F=
0.
const.
(11-10)
1 280
Heat and Thermodynamics 1
These properties are of interest in chemistry and arc useful in considering chemical reactions that take place isothermally and isochorically. The main importance of the Helmholtz function, however, is in statistical mechanics, where it is closely associated with the partition function Z defined by the equation
-2
281
Pure Substances
For an infinitesimal reversible process,
AG = It will
Z=
1
dll
- T dS -
S dT.
be recalled, however, that
Zgte-xw
dH = T dS + V dP; where
and g, represent the energy values and degeneracies,
e, respectively, of the various energy levels of the system of particles. In the case of indistinguishable particles, it was shown in Chap. 10, that the entropy is given by
S=Nk\n^+^ + Nk.
dG
whence In the case of a
= -S dT +
reversible isothermal
dG =
(10-21)
G =
and Therefore,
F=
U-
| - NkT N = ~kT{N In Z - N In N +
and
V
dP.
(11-12)
isobaric process,
0,
const.
TS = -NkT In
This A'),
is
a particularly
important result
in
connection with processes involv-
ing a change of phase. Sublimation, fusion,
F = -kT{N\nZ -InNi).
and vaporization take place
iso-
thermally and isobarically and can be conceived of as occurring rcvcrsibly.
The
relation between Helmholtz function and partition function is even simpler in the case of distinguishable particles (such as those of a crystal, where the particles are localized) namely,
Hence, during such processes, the Gibbs function of the system remains constant. If we denote by the symbols g', g", and g'" the molar Gibbs functions of a saturated solid, saturated liquid, and saturated vapor, respectively, then the equation of the fusion curve
is
;
F = - NkT In Whether the particles are distinguishable
Z. the equation of the vaporization curve
function leads directly to the Helmholtz function,
and
since
g"
dF= -SdT-PdV, he entropy and the pressure th simple differentiations:
may
and the equation of
the.
is
the triple point two equations hold simultaneously, namely,
=
Gibbs Junction (also called the Gibbs free energy)
is
H-
TS.
=8 =
g
P and T only, and hence the two P and T of the triple point uniquely.
All the g's can be regarded as functions of
defined as
equations above serve to determine the
The Gibbs
G =
g'",
sublimation curve
g
The
=
then be calculated by performing the
At S
is
or not, the logarithm of the partition
(11-11)
cal reactions
also of
function is of the utmost importance in chemistry, since chemican be conceived of as taking place at constant P and T. It is
some use
in engineering.
282
Two
11-3
If a relation exists
among
expressed as a function of x and y;
whence
Theorem
Substituting this expression for
Mathematical Theorems
7
x, y,
and
we may imagine
z,
z
rfy
in the
283
Pure Substances
11-4
Heat and Thermodynamics
preceding equation,
'(!. [(iHlUt)JH(l)
we
get
(fe.
But
-(!).*+(§) tE
If
we
Equating the
let
=
«fe
where
z,
A/,
and
with respect to
A'"
arc
all
M
and A with respect
/dM\ \dy) x
+A
d 2z
dx by
two equations, wc get
last
\dy), \dz) f
r
«fe
functions of x
r
y,
terms of the
- *-(&
—GO. then
rfz
and
to x,
\dz), (11-14)
^jv.
y. Partially differentiating
wc
\dy),\dz) l
M
L
\dx)r
get
Maxwell's Equations
11-4
g£
and
\9*).
We Since the two second derivatives of the right-hand terms arc equal,
it
have seen that the properties of a pure substance are conveniently
represented in terms of these four functions'.
follows
that
Internal energy U,
H
= U + PV, Enthalpy /*' = function Hclmholtz U — TS, = 'TS. Gibbs function G
(11-13)
H
This
is
known
Theorem 2
as the condition for an exact differentia/.
/ is a function of x, y, and z, and a relation exists among x, y, and z, then / may be regarded as a function of any two of x,y, and z. Similarly, any one of x,y, and z may be considered to be a function of /and one other of x,y, and z. Thus, regarding x to be a function of /and .y,
ji
£ -(&* =
/ and
\
I),*-
may
be regarded as a function of any two of P,
equation
we may go
P, V, T, S, U, II, T,
and
.$'
(I
=
function of
(I",
T in
the
T.
V
\T), T).
to be solved for
equation,
function of
further
G may
first
and
V,
are expressed as functions of
function of
U=
z,
-(*).*«),*
and
may be imagined
substituting this value of
Consequently,
U
U= S
and
The second
i
to be a function of
of these
Suppose for example that both and T, thus:
If a quantity
& Considering
Any one
(S,
we
T in terms of
.S'
and
V;
should then have
V)
and say that any one
of the eight quantities
be expressed as a function of any two
others.
V 284
Applying
Now
imagine a hydrostatic system undergoing an infinitesimal reversible process from one equilibrium state to another:
dll=
1
1
The
internal energy changes
M=
2
P are
imagined
to be functions of
.S'
and
V.
The enthalpy changes by an amount
dH = =
3
all
hence
TdS +VdP;
hence
(g)
dU, dH, dF, and dG:
= -
(^.
=
(^
?
(§) ?
;
01-15)
= TdS - P dV, where U, T, and
TdS - PdV,
by an amount
dU = dQ - P dV
2
this result to the four exact differentials
285
Pure Substances
11-4
Heat and Thermodynamics
dU+ PdV + TdS+ V dP,
dF
= -SdT - PdV;
dG
= -SdT
hence
dS\
dT
dVjr
'**-(sX-(&-
The
four equations on the right are known as Maxwell's equations. These do not refer to a process but express relations which hold at any equilibrium state of a hydrostatic system.
V dP
where H, T, and Fare all imagined to be functions of S and The Hclmholtz function changes by an amount
Maxwell's equations are enormously useful, because they provide relationbetween measurable quantities and those which either cannot be measured or are difficult to measure. For example, the fourth Maxwell
P.
ships
equation,
dF = dU - TdS
-
S
dT \dPj T
= -SdT - PdV,
\dTjp
may be combined
4
where F, S, and P are all imagined to be functions of The Gibbs function changes by an amount
T and
with the statistical interpretation of entropy in order to provide information concerning the volume expansivity of a pure substance in the following way. If a substance is compressed isothermally and if no
V.
unusual molecular rearrangements take place (such as association or dissocimerely occupy a smaller volume and arc therefore in a
ation), the molecules
T dS - S dT - -SdT + V dP,
dG = dH -
where G,
S,
and V
Since U, H, F, and
G
are all imagined to be functions of
T
and
more orderly state. In the language of information theory, our knowledge about these molecules is increased. The entropy is therefore decreased, and the derivative (dS/dP) T is negative. It follows that (dV/dT) P is positive and that the substance must have a positive expansivity.
P.
The Maxwell equations will be used extensively in the remaining portions They need not be memorized, but they must be readily available.
of this text. are actual functions, their differentials are exact
i
For
differentials of the type
no less than ten mnemonic schemes have been suggested the pages of the American Journal of Physics, f The diagram in Fig. 11-3 based on a sentence due to J. J. Gilvarry, a former student of the author: this purpose,
all in
dz
where
z,
M, and Ar
=
M dx + N
is
dy,
are all functions of x and
y.
namely, "Good Physicists Have Studied 6'nder Fery Fine 7'eachers." The upper-right and lower-left corners of a square are slightly cut away to provide Therefore,
eight points. Starting at the upper-left point first letter
tThe
\dy) x
{dx),'
(1957).
of each
word
of Gilvarry's
and proceeding
magic sentence
is
clockwise, the
placed successively at
ten references will be found in a note by D. E. Christie, Am. J. Phys., 25:486
Heat and Thermodynamics
286
11-5
G®
Since
T dS = dQ
for a reversible process,
it
Pure Substances
287
follows that
H<> And, from Maxwell's
third equation,
(2\ \dVj T = (-) \dTjv' ffiF
T dS = C v
whence
d
'"(ft
dV.
We shall call the above equation the first T dS equation.
Fig. 11-3
Mnemonic for
obtaining
dU,
dll,
dp, and
dG
by using the sentence ''Good
1
Tds =
automatically by noting what the adjacent letters are and where they
lie.
cv
dT + T
Using the van dcr Waals equation of
RT P= v
)
dS
-
(
)
dV.
(dP\ \dTj v ~
and 1
k**
The
student
is
assumed
exercise to discover
to
how
be able to
the
fill
the parentheses. It
Maxwell equations arc hidden
is
an
interesting
in the
diagram.
hence
11-5
The T dS Equations
The entropy
Since Tis constant, cy
of a pure substance can be imagined as a function of 7"
and
'-*•(#
>
T
R v
-
b'
dT =
0;
and
+ RT v -
„ = RT
dv
("i Jvi
v
—
=-,
b
finally
-
q
b
since the process
V;
and
T [w) dV
a b
Tds = c 7 dT
q
-
dv.
state,
Therefore,
whence
and
m>
and V lies
(positive)
to the left (negative). Therefore,
dU = + (
useful in a variety
mole.
each point, and the points corresponding to the four functions U, H, F, and G are emphasized. The equations for dU, dH, db\ and dG may now be obtained
U and see that S lies above
is
1
been transferred?
For
physicists have studied under very fine teachers."
Thus, to find dU, we look at
It
mole of a van dcr Waals gas undergoes a reversible isothermal expansion from a volume o% to a volume v;. How much heat has of ways. For example,
*>V
V
(11-16)
= RT\n
ly
—
b
W^b
is
reversible, q
= fTds.
1 288
Heat and Thermodynamics
If the
11-5
entropy of a pure substance
is
regarded as a function of
T and P,
then
-(#,«•+(&*
289
Pure Substances
percent; and the volume expansivity changed from 181 X 10~ 6 deg-1 to 174 X 10 -6 deg -1 , a 4 percent change. The volume and the expansivity of most solids and liquids behave similarly; therefore V and /3 may be taken out from under the integral sign and replaced by average values, V and /3. (A bar over a quantity indicates an average value.) We then have 3-
and
Q « -Tf§j**dP, But
Q= - rflBfcP, - Pi).
or
seen from this result that, as the pressure
It is
And, from Maxwell's fourth equation,
will flow out
_ (dV\
(dS\ VSPy/r
if
^
is
positive
(such as water between .
an absorption of
\dTjp'
is increased isothermally, heat substance with a negative expansivity and 4°C), an isothermal increase of pressure causes
but that,
for a
heat.
on 0.015 liter of mercury at 0°C is increased reversibly and isothermally from zero to 1000 atm, the heat transferred will be If the pressure
whence
T dS = CpdT- T QyJ
(11-17)
<'P.
This is the second T dS equation. A third will be found among the problems at the end of the chapter. Two important applications of the. second T dS equa-
where Pi
=
T=
273 deg,
V=
0.015
=
liter,
178
X 1CH deg-',
Pi
=
0,
and
1000 atm. Hence,
tion follow.
Q = -273 1
.
When T
Reversible isothermal change of pressure
is
= -0.730
constant,
= -73.7
TdS = "*(%),*. Q =
and
Remembering
that the
-T
dT
volume expansivity
liter
0.015 •
atm
liter
=
X 10-° deg-' X X 101 J
178
-0.730
10 3 atm
compare
the heat liberated with the
work done during
the compression.
dP.
\
W= and
is
fP dV;
at constant temperature,
W= we
X
J.
It is interesting to
5F>
X
deg
&'*
obtain
Q =
-T f
Recalling the isothermal compressibility,
V0 dP,
which can be integrated when the dependence of V and /3 on the pressure is known. In the case of a solid or liquid, neither Fnor /3 is very sensitive to a change in pressure. For example, in the case of mercury, Bridgman found That as the pressure was increased from zero to 1000 atm at 0°C the volume of 1 mole of mercury changed from 14.72 to 14.67 cm 3 a change of only ,
k
= — {\ /V){dV/dP)r, we
get
= W'--/> kP dP. The sure.
isothermal compressibility
is
also fairly insensitive to a
Bridgman showed
from 3.92
X
change of pres-
that the compressibility of mercury at 0°C changed -6 10" 6 to 3.83 10 atm" (a 2 percent change) as the pressure
X
1
290
Heat and Thermodynamics
11-6
was increased from zero to 1000 atm. by average values and obtain
£
=
W
=
-|(0.015
=
-0.0291
=
-2.94
X
3.88
10
°
atm
liter liter
-
(Pj
X
1
for
3.88
X
10'
be
AT =
6
we
Therefore, is
it is
X
atm"
1
10*
atm
X
178
10~« deg- 1
X
10 3 atm
28.6 J/deg
-^-.-^ =
1
=
0.0255 dcg
=
2.58 deg.
X
0.0255
101 deg
Energy Equations
11-6
seen that,
when
the pressure on 0.015 is
liter
of
mercury at
pure substance undergoes an infinitesimal reversible process between states, the change of internal energy is
two equilibrium
liberated but only
dU = TdS - P dV.
is
we
Dividing by dV,
get
d
Jl=T d-?--P dV
dV 2.9
= -70.8
where U,
J.
S,
and
'
P are regarded as functions of T and
V. If
T
is
held constant,
then the derivatives become partial derivatives, and is obtained in the case of any substance with a positive exFor a substance with a negative expansivity, heat is absorbed and
similar result
pansivity.
the internal energy 2.
Reversible
is
(3U\ \dVj T
increased.
adiabatic
change of pressure
Since the entropy remains
n—
constant,
TdS= Q^CpdT-Tl or
X
J.
AU = Q - W = -73.7 + A
liter
)
done! The extra amount of heat comes, of course, from the store of internal energy, which has changed by an amount
work
0.015
2
-atm
increased from zero to 1000 atm, 73.7 J of heat
2.94 J of
X
get
If a
0°C
273 deg
P!):
mercury,
291
increased isentropically from zero to 1000 atm, the temperature change will
Vk
W= and taking
We may therefore again replace V and
Pure Substances
dT =
CP
®,*-
TV0 Cp
=
(dP/dT) v we get ,
dU
I - *•($,-'•
(11-18)
dV
dP.
hardly changes even for an increase of 1 0,000 atm. The above equation, applied to a solid or a liquid, may therefore be written
LP
Using Maxwell's third equation, (dS/dV) T
dP,
We shall call
In the case of a solid or liquid, an increase of pressure of as much as 1000 atm produces only a small temperature change. Also, experiment shows that Cp
AT = l^l (Pf -
_
/ 6S\
\dVjr
when
this
equation the first energy
1
.
Two examples
of
Ideal gas
nRT P= V dP\
Pi ).
It is clear from the above that an adiabatic increase of pressure will produce an increase of temperature in any substance with a positive expansivity and a decrease in temperature in a substance with a negative expansivity. If the pressure on 0.0015 liter of mercury (C P = 28.6 J/deg) at 0°C is
equation.
ness follow.
_
nR
(ft Therefore
U does not
depend on V but
is
a function of
T
only.
its
useful-
Heat and Thermodynamics
292 2.
Van
der
Wools gas
Equating the
and Consequently,
-
v
dV:
b
«
=
ce <£T
+ -f dv,
u
=
I
cy
solving for dT,
J2 v —
rv * —
dT
—
dT =
energy of a van der Waals gas increases
—
'
P
Gp
(\dP fy
Gy
—
Cy
dP
It
and Fare imagined to be functions of T and P. become partial derivatives, and
If
T
is
was shown
result that
^Mr
t(Z\ (£\
^
dU _ -4S _ p dV
S,
dP. C\
dV,
and divide by dP. Then,
where U,
—
'd)
T
Both the above equations yield the
dP
—
C*jp
dT\
and
dP
Cp
(er\
Therefore
increases,
dU = T dS - P
dV
C\
dT =
But
+ const.
-
Cp
with the temperature remaining constant. The second energy equation shows the dependence of energy on pressure. We start, as usual, with the equation
volume
equations,
'
It follows, therefore, that the internal
as the
T dS
and second
b
fe)\8v/t
and
first
R
f«*\ -
293
Heat-capacity Equations
11-7
(7 mole)
\dTjv
Pure Substances
11-7
in Art. 2-7 that
(ap\ = _ / ev\ \0Tjv \dTjp
held con-
stant, then the derivatives
.
(Sfc
and therefore d
Jl\ ( \dPj
= T (™)
\dPj T
T
A
d -p(\dPji
Using Maxwell's fourth equation, (dS/dP)r
= — (dV/d'T)p, we
CP - Cv =
is
shows
-T
w)P
-p
\w\'
0V\ 2 (dP\ rrjp \dv)r
one of the most important equations of thermodynamics, and
Since (dP/dl")r
is
always negative for
(dl'/d'Tjp must be positive, then is
the second energy equation.
it
that:
(11-19) 1
which
(11-20)
get
This
\w)T -
-T
or
Cp
can never be
less
than Cv.
Cp
—
all
known
substances and
Cv can never be
negative;
1
294
11-8
Heat and Thermodynamics
T —> 0, Cp —
2
As
3
CP = Cy when (pV/BT), =
>
density of water
Cy; or at
is
the absolute zero the
Remembering
Cv
for the difference in the
in terms of
arc
TJS=CpdT-T(-f
Laboratory measurements of the heat capacity of solids and liquids usually and therefore yield values of Cp. It would be extremely difficult to measure with any degree of accuracy the CV of a solid or liquid. Values of Cy, however, must be known for purposes of comparison useful in calculating
T dS equations
Cy.
take place at constant pressure
with theory. The equation
The two
295
For example, at 4°C, at which the
0.
maximum, Cp —
a
two heat capacities are equal.
Pure Substances
Cp and
heat capacities
is
7"
and At constant
= C v dT
dP, )p
r ''(ft
dS
tts
-r
dV.
.S',
njk)9 *~
very
other measurable quantities.
that
=-T(g) v dVs
CvdTs
and
.
V \dTjp Cp
K=
and
1
Dividing,
/dV\
-v\Jp) T But the quantity
we may
~(dV/dT) P (dP/dT) v
Cv in
square brackets
is
~\
equal to
/dP\
— (dV/dP)r-
Therefore,
write the equation in the form
Cp
—
Cy
Cp Cy
TV vi far SET), —
(dP/dVh \dP/dV) T
(11-22)
-(-) The
CP
-Cv =
TO
adiabatic compressibility
is
defined as
(11-21)
1
/dV\
i
/an
and, as usual.
As an example, let us calculate the molar heat capacity at constant volume of mercury at 0°C and 1 atm pressure. From experiment we have cp = 28.0 J/mole deg, T = 273 deg, v = 0.0147 liter/mole, (3 = 181 X 10-" deg-', and k = 3.94 X 10" 6 atm-'. Hence. •
cr>
_
=
273 deg
X
0.0147
and Finallv,
cv
= =
7
=
liter
0.0336
liter/ mole
mole
-
3.4
—=
28.0
cv
24.6
cP
jrj-7
•
= =
X
have, therefore.
Cp
X
(181)
2
X
=
=
K_
(11-23) KS
10~' 2 deg~ 2
10- 6 atm- 2
11-8
atm
•
3.39 J/mole
28.0
We
Cv 3.94
=
V \dP)r
Heat Capacity
at Constant Pressure
deg deg, 24.6 J/mole 1.14.
•
deg.
The experimental measurement of cp has already been discussed in Art. 4-7, and the general features of one type of calorimeter suitable for such measurements were described, as well as some details of technique. Data on the heat capacities of elements, alloys, compounds, plastics, etc., taken over as wide a temperature interval as possible, are of great importance in pure science and
296
Heat and Thermodynamics
11-9
297
Pure Substai
then bends and begins to flatten out in the neighborhood of room temperature. In none of the three crystals, however, does the cp curve actually BC
become 1
ǩȣ-- "
9I{
—
horizontal.
In the Ge
crystal, lattice sites are occupied by single germanium atoms, and mole of Gc consists of A'a vibrating particles, where A"A is Avogadro's number. The value of cp at room temperature is very nearly equal to 3R. In NaCl, the lattice sites are occupied by sodium ions and by chlorine ions in a face-centered cubic arrangement. Therefore 1 mole of NaCl consists of A a sodium ions in addition to A a chlorine ions, so that altogether there are
therefore
70
60
2Aa NaCl
50
no
=
0)
ft JI r? V
—
o
1
vibrating particles.
The
value of cp at room temperature
equal to 6R. In NiSe 2 the lattice ,
sites
is
very nearly
are occupied by nickel atoms
and by
selenium atoms, with the center of the line joining the two selenium atoms
and the nickel atom forming a face-centered cubic arrangement similar to that of NaCl. In 1 mole of NiSe 2 there arc 3A'A vibrating particles, and the ,
40
room temperature value of cp is very nearly equal to 9R. In all cases, the cp at room temperature of exactly N\ atoms or ions is in the neighborhood of 3R, about 25 J mole deg or about 6 cal/mole deg. •
30
The (
as—
5e
curves in Fig. 11-4 correspond to crystals that were deliberately
chosen to
room temperature. There is no magic in Not all crystals have values of cp whose 300°K. The cp of diamond, for example, rises
illustrate a regularity at
the temperature 300°K, however. 20
rapid increase,
is
tapering off at
300°K it is still quite far from the value 2>R. Furthermore, c r never approaches any value asymptotically but continues to rise at all temperatures. The laws governing the temperature variation of heat capacity cannot be stated simply in terms of cp. To express experimental results in a so slowly that at
10
neat form and also to appreciate the relation between experiment and theory, 50
100
200
150
Temperature, Fig. 11-4 Seidel;
NaCl,
250 *
300
it is
K
necessary to study the temperature variation of cy. For this purpose,
use the
Heat capacity at constant pressure of crystalline nonmetals. (Ce, Kecsom and and Perlick; NiSe 2 Gronvold and Westrum.) (R = 8.31
Clusius, Goldrnann,
,
J/mole deg.)
cP
in engineering. To the physicist the most important temperature region is from absolute zero to about room temperature (300°K), so that we shall emphasize this region more than any other. In this range, most materials arc
in the solid phase.
We
shall limit ourselves to solids in the
cither a single crystal or a rod or
powder
consisting of a large
form of
crystals,
number
of small
we
thermodynamic formula
cv
=
Tv{S 2
(11-21)
which requires a knowledge of the complete temperature dependence of /3, k, and v, as given, for example, for NaCl in Table 11-2, compiled by Meincke and Graham. Before we use this equation, let us first learn something about thermal expansivity and compressibility.
crystals.
The behavior of three different crystalline nonmetals is shown in Fig. 11-4. (Metals exhibit a special behavior because of the effect of free electrons; such behavior will be discussed later in this chapter.) The Cr of all materials approaches zero as
T approaches zero.
In the
first
20
to
40 deg, cp
rises
rapidly but
11-9
Thermal Expansivity
In modern experiments on the expansion of solids, the linear expansivity a is
usually measured. If the three rectangular dimensions of a solid are L\, Li,
1
298
11-9
Heat and Thermodynamics
Table 11-2
Thermal Properties of NaCl
{Compiled by P. P.
M. Meincke and
r.
deg
40 50
60 70 80 90
100 125 150 175 250 290
40.1
43.3 45.4 48.6 49,2
Ls,
o
Graham.)
A
K,
V,
Ai(atm-')
liter/mole
3.94
0.0264 0.0264 0.0264 0.0264 0.0264 0.0264 0264 0.0265 0.0265 0.0265 0.0266 0.0266 0267 0.0269 0.0270
0.171 1.72 7.44 17.2 29.3 41.4 52.2 61.5 69.5 75.8 88.2 96.3
0.151 1.30 4.76 9.98 15.7 21.0 25.5 29.3 32.3 35.0
10
20 30
and
•
M.
M(deg-')
cr,
J/molc
°K
G.
3.94 3.94 3.94 3.95 3.97 3.99 4.01
.
4.03 4.05 4.09 4.12 4.16 4.29 4.36
103
114 118
.
mole
•
0.151 1.30
4.76 9.97 15.7
20.9 25.3 29.1 32.0 34.7 39.5 42.4 44.2 46.6 46.7
deg
M(atm-')
3.94 3.94 3.94 3.94 3.94 3.96 3.97 3.98 3.99
M Pi
4.01
4.03 4.04 4.05 4.11 4.13
then
V=
Mi
Ks,
SJ-,
J,
299
Pure Substances
uuu,
P;
Modern
Fig. 11-5
WK^
version of Fizeau interjerotnelric dilatometer.
(James and Yates.)
a rough schematic diagram of an interferometer which is a modification due to James and Yates of a device originally designed by Abbe and Pulfrich. Highly monochromatic light, such as the red light from a low-pressure cadmium lamp, after it has passed through a filter F is concentrated by lens /.i and reflected from mirror Mi onto plates Pi and P2 which In Fig. 11-5 there
is
,
dV _ -L 2L clU
SgT
jT
VdT~ =
and
where
«i,
« 2 and a* arc ,
In the case of quartz, the
LiclT «i
i
r
cti
4- i r
dL *
+ Lidf
1
dL a
L,dt
as,
the linear expansivities along the three directions.
two
are separated by a ring or cylinder
linear coefficients perpendicular to the z axis
= 2a x + aj. If the solid is = a = a* = a and 8 = 3
isotropic, as in the case of a
regular intervals. If
N fringes
travel across the field of
Z
N\/2.
If
at T,
then
is
the length of the specimen at temperature To and
L-U u
In the X-ray diffraction method, the lattice parameter of the specimen of
is
view while the temperature changes
r from T to T, then the optical path difference has changed by A X, where X is the thickness of the air space has changed by the wavelength of the light, and
physical principles are easily understood. crystal
whose expansivitythe bottom of a cryostat,
of the material
to
2
There arc four absolute methods of measuring the linear expansivity of solids: X-ray diffraction determinations, interference fringes of visible light, variation of electric capacitance, and variation of intensity of light. Modem methods, particularly those which are used to make measurements in the to 50°K range, involve a lot of auxiliary cryogenic equipment, but the
R made
be studied. The ring and plates arc placed at where liquid hydrogen or liquid helium is used to provide the low temperatures at which measurements are often made. In Fig. 11-5 all details of the cryostat, heater, thermometer, venting tubes, electric leads, etc., have been omitted. Interference takes place between the rays of light reflected from the bottom of Pi and the top of P2 , and a camera C is used to photograph the interference fringes. The temperature is varied very slowly from, let us say, 4°K up to room temperature, and the fringe system is photographed at is
are equal, so that 8
cubic crystal,
dU
L
is
the length
NX
1U
measured as a function of the temperature, using an X-ray beam
known wavelength.
If,
therefore, A"X/2Z.o
is
plotted against
T and the slope of the resulting curve
300 is
11-9
Meat and Thermodynamics
taken at various temperatures, the linear expansivity
is
301
Pure Substances
obtained. Thus, 125
d
(N\\
In order to avoid the delay involved in photographic processing, R. K. his staff at the U.S. National Bureau of Standards have developed
Kirhy and
a photoelectric interferometer in which the is
detected by a photomultiplicr tube
and
movement number
the
of interference fringes of fringes
is
100
automati-
on a recorder against the measured temperature of the specimen. hand operations are eliminated, and the data are presented on a
cally plotted
Thus,
all
chart in a form suitable for immediate determination of expansivities. In the electric method, the expansion of the specimen is communicated to one of the plates of a capacitor, whose other plate is fixed nearby. The change in capacitance is measured by an extremely sensitive bridge; and in the hands of the Australian physicists
headed by G. K. White,
this
«1
method has proved
capable of detecting length changes of the order of 10 -8 cm. The optical grid method of R. V. Jones, which has been used with great success by Andres in Switzerland, is extraordinarily sensitive. The specimen is
connected to one of the grids shown
The
in Fig. 11-6, the other
remaining
fixed.
and the one on the right has an irregularity in the middle. In position (a), no light can pass the grids in the lower half; whereas in position (A), in which the right-hand grid has been lowered with respect to the other by one line width, no light can pass the grids in the upper half. In position (c), the two halves are equally transparent, and as a result, left-hand grid
is
regular,
the net output of two oppositely connected photocells (one behind the upper half,
and the other behind the lower
half)
is
zero. Thus, a
change
in length
of the specimen produces a change in the net output of the two oppositely
connected photocells.
The temperature dependence
of the
volume expansivity of many sub200
100
II
300
Temperature, °K Fig. 11-7
Temperature dependence of fi and of cp are almost Goldmann, and Perlick.)
Graham;
Cp. Cltisius.
sfences
is
identical.
(/3,
Meincke and
Cthe same as that of NaCl,
shown
in Fig. 11-7:
absolute zero, rises rapidly in the interval from
namely,
/3
is
zero at
to 50°K, then bends
and
out without actually becoming horizontal. Thus, the temperature behavior of ft is almost identical with that of cp, as shown in Fig. 11-7. flattens
(o) l'ig.
11-6
(b)
(c)
Optical grid method of measuring thermal expansion. (R. V.Jones.)
Another similarity between changes of pressure.
|3
and
cp
is
the insensitivity of both quantities to
Heat and Thermodynamics
302
11-10
Pure Substances
303
Compressibility
11-10
made
Compressibility measurements are
in
two ways and for two different enormous hydrostatic
reasons. Installations capable of subjecting solids to
and capable of providing numerical values up to about a million atmospheres are used to study phase transitions, changes of crystal structure, and other internal changes of solids and liquids. These arc sialic measurements. Measurements of the speed of longitudinal waves in liquids and both longitudinal and transverse waves in solids, at atmospheric or moderate pressures, arc dynamic in character and provide numerical values of the adiabalic compressibility ks, where pressures at constant temperature
of the isothermal compressibility at pressures
=
KS
4.3
NaCl
y~ isothermal
4.1
-^-
Adiabatic
V \dPji 4.0
was shown given by It
is
in
Chap. 5 that the speed of a longitudinal wave
w
in a fluid
w =
3.9
50
!p k s'
Measurements of w and p are sufficient to provide ks of a fluid, but the measurement of ks of a crystalline solid is more difficult. It is necessary to measure the speed of transverse waves as well as that of longitudinal waves and, from the two measurements, to calculate two different clastic constants. In the case of NaCl, these quantities are designated cu and cii, and it is shown in books on elasticity that where p
is
the density.
Ks
ks
is
+ 2ci
obtained, the isothermal compressibility
and Substituting for cy in the
first
—
may be
by
calculated
—
Tvfi
Cy CyK
=
CpK S
2
(11-21)
-
(11-23).
equation the value found from the second,
get Cp
—
250
300
Temperature variation of isothermal and adiabatic compressibilities of NaCl.
and Swim.)
which reduces to k
—
Ks
=
(11-24) Cp
CpKS
Tvf}
of measuring wave velocities that is most suitable for crystalline one which incorporates the familiar radar technique. Short ultrasonic pulses of about 1 p.s duration are sent through the crystal by a quartz crystal transducer. After reflection from an end of the crystal, the quartz emitter is used as a receiver. The pulse and the echo arc observed on an oscilloscope, solids
using the two thermodynamic equations derived in this chapter, namely,
Cp
11-8
(Overton
200
K
The method
= c ii
Once
Fig.
150 Temperature.
100
2
we
is
and the wave speed is calculated from the dimensions of the crystal and the time delay between pulse and echo. This is the method used by Overton and Swim to obtain the values of Kg listed in Tabic 11-2. The temperature variation of ks and k of XaCl are shown in Fig. 11-8, where it may be seen that both ks and k unlike cp and ft do not approach zero as T approaches zero. From zero to 40°K, the adiabatic and isothermal compressibilities are nearly equal. At higher temperatures, K is larger than Ks, as required by Eq. (11-24). The molar volume of XaCl is shown in Fig. 11-9, where the temperature variation may be seen to be very similar to that of k. In a rough sort of way,
—
—
304
Heat and Thermodynamics
11-10
305
Pure Substances
0.0270
50
\ Water
48 a.
£ o
0.0265
46 Isothernlal, k
E
o
O
44
^^ Adiabatic,* s 42 0.0260
200
300
Temperature, "K Fig. 11-9
20
Temperature variation of molar volume of NaCl, similar
compressibility. {Heinglen.)
cp and
/3
v
and
k are also similar,
waves
Table 11-3 T,
IV,
A
km/s
/i(deg-i)
1.404 1.448 1.483 1.510 1.530 1.544 1.552 1.555 1.555
-67
10 20 30
40 50 60
70 80
in
pressibility of
in Fig. 11-10.
The minimum achieved by
water at about 50°C
89 208 304 390 465 522 586 643
•
deg
4.2177 4.1922 4.1819 4.1785 4.1786 4.1807 4.1844 4.1896 4.1964
where k and
P,
«>
g/cm 3
^(atm -1)
0.99986 0.99973 0.99823 0.99568 0.99225 0.98807 0.98324 0.97781 0.97183
51.5 48.8
51.5 48.6
46.4 45.2 44.4 44.3 44.3 44.7 45.4
44.5 43.3 42.6 42.0 41.8 41.8
us,
M(atm
b arc constants.
The
-1 )
quite anomalous.
the isothermal
As a
com-
rule, the isois
raised
and follows a simple exponential equation quite well:
k
Of,
J/g
is
thermal compressibility of most liquids increases as the temperature
Properties of Water
°C
80
Isothermal and adiabatic compressibilities of water.
and arc shown
a liquid is usually measured with the aid of an acoustic interferometer such as the one described in Art. 5-9, which was used with helium gas to serve as an absolute thermometer in the range from 4 to 10°K. Results of measurements on water are listed in Table 11-3 of longitudinal
11-10
70
60
°
but with an
entirely different kind of temperature variation.
The speed
50
Temperature, Fig.
vary in the same manner, while
40
30
to that of isothermal
All liquids, including water,
=
(11-25)
K e°
constant
become
b for
less
mercury
is 1
.37
X
1
-3
deg-1
.
compressible the more they are
compressed; the reciprocal of the isothermal compressibility increases linearly with respect to the pressure, where
46.1
i-i-*. K
where for
Kg
is
the compressibility at zero pressure
mercury.
(11-26)
/Co
and
c is 6.7 for
water and 8.2
306
Heat and Thermodynamics
Heat Capacity
11-11
11-12
at Constant
Volume Rbl
The measurement of
and
cp, @,
is
to use these
and
a constant challenge to
experimental physicists throughout the world. This field one, for each quantity is of theoretical interest in its own is
N aCI
1.0
k of crystalline solids both metallic
nonmctallic, particularly at low temperatures,
at present
307
Pure Substances
measurements
is still
/
/
a very active
i*-
Fe
Our purpose with the thermo-
right.
in conjunction
/|V|<.n \
J
dynamic equation,
cp-cv='^->
Diamond
(11-21)
0.2
complete temperature dependence of Cr. All the measurements on NaCl are listed in Table 1 1 -2, along with the calculated values of cv, and to find the
both cp and cv are plotted against of
NaCl
Tup
to
1000°K
in Fig. 11-11. Since 1
consists of 2.Va ions, the heat capacities refer to
-J-
/^__
mole
200
100
400
300
mole, or A'a ions.
At low temperatures, below 100°K, Cp and c v are practically the same. At
Fig. 11-12
Rbl, NaCl,
all
35
30
600
800
1000
Temperature variation of cvfiR of nonmelals. (7 mole of diamond, and | mole of FeS2 .)
§
mole of
MgO,
higher temperatures, while cp continues to increase,
stant value 3R, cP
500
Temperature, °K
which
physicists
who first
perature.
We
see
is
observed that cp came near
now
C\-
approaches a con-
called the Dulong and Petit value,
that this value
is
named
after the
value at about room tem-
this
actually approached
by cv and
is
exceeded only in special situations. The temperature dependence of cy/3R of cV
25
shown in Fig. 11-12, where it may be seen Dulong and Petit value even before room temperature, whereas diamond has reached only one-fifth this value. As a matter of fact, it requires a temperature greater than 2000°K to bring the cv of diamond near 3R. five representative
3R
20
that
Rbl
nonmctals
is
practically reaches the
Although the five curves in Fig. 11-12 differ markedly in the temperature which c\- —> 3R, the curves are still very similar in shape. An experienced experimenter would be led to suspect that there existed a parameter 0, let us say small for Rbl and large for diamond, such that cv is a universal function at
15
—
—
NaCl
10
of the ratio T/(r).
we
shall sec that
Such a variation
is
called a law of corresponding
an approximate law of
this sort is
slates,
and
provided by theory.
S
11-12 100
300
400
500
500
700
800
1000
Temperature, °K tig.
1
1-11
proaches
3R
Temperature variation of cP and cv of \ mole of NaCl. The value of cv apas a limit.
The that c v
Statistical
Mechanics of a Nonmetallic Crystal
reason that the cv of a solid
=
(du/dT')v,
where
u
is
is
of
more
theoretical interest than cp
the molar internal energy, which
calculated with the aid of statistical mechanics. In general, this calculation
extremely complicated, because
many
different
is
may be
phenomena contribute
is
to the
308
Heat and Thermodynamics
Pure Substances
11-12
internal energy of the solid. Suppose, for example, that the solid
is
a crystal
having a lattice composed of molecules, each of which consists of several atoms; and furthermore, suppose that there is about one free electron per molecule. Then the total internal energy may be due to:
A
309
undergoing simple harmonic motion in the x direction has an energy equal to %kx 2 + (1 /2m)p'x The fact that the energy of A" vibrating r lattice points is given by 3A expressions of this type enables us to conclude that, when A' lattice points undergo small displacements, the motion of the particle
.
may be described as that oj 3N independent simple harmonic oscillators. These harmonic oscillators (or normal modes) are not associated with individual lattice points; each one involves motion of the entire crystal. This conclusion is independent of the type of crystal lattice and is true only when displacements from equilibrium positions are small. When the vibrations get large enough, anharmonic effects take place and the oscillators are no longer independent. Actually, to determine the normal coordinates of a given crystal and to calcu-
crystal 1
Translational motions of the free electrons.
2
Vibrations of the molecules about their equilibrium positions, called briefly lattice vibrations.
3
Internal vibrations of atoms within each molecule.
4
Partial rotation of the molecules.
5
Excitation of upper energy levels of the molecules.
6
Anomalous
It is fortunate that all these effects
do not take place
in all solids.
For exam-
and in the whose component parts do not rotate or vibrate. Furthermore, all effects do not take place in all temperature ranges. Thus, the motions of the free electrons of metals have an appreciable effect on the heat capacity only at very low temperatures, below
ple, in the case of nonmetals,
motions of free electrons do not
exist,
case of metals the lattice consists of single atoms
about 20°K. Above this temperature, they may be ignored. Similarly, excitation of upper energy levels takes place only at very high temperatures and can therefore be ignored at moderate temperatures. In this article, let us limit ourselves to a nonmetallic crystal in which the lattice sites are occupied either by a single atom or ion or by a rigid molecule
whose internal
vibrations, rotations, excitations, etc.,
may
be ignored.
We
occupying lattice sites as lattice points and shall assume there are A* of them. These lattice points arc localized in space and are therefore distinguishable by their positions; furthermore, they arc closely packed and interact very strongly with their neighbors. It would seem, at first glance, that the statistical methods described in Chap. 10 as appropriate to indistinguishable, weakly interacting particles are useless here. But this is not the case. Since each lattice point has three coordinates x, y, and z, the r system of A: lattice points has 3A coordinates. If each lattice point undergoes a displacement from its equilibrium position that is small compared with the space between lattice points, the change in potential energy of the crystal would involve many terms involving not only squares of these displacements but product terms as well. In the theory of small oscillations, it is shown that
shall refer to the particles
there always exists a
new
set of 3A7
coordinates (linear functions of the original
coordinates) such that the potential energy terms. Corresponding to these
are 3 A'
momenta, and the
contains the square of a
normal modes is a very complicated problem in mechanics. It is a fortunate circumstance that considerable information may be obtained by applying statistical mechanics to these normal modes, along with either of two simplifying assumptions- -one late the various frequencies of vibration of the 3A"
effects.
new
is
the
sum
of exactly 3JV squared
coordinates, called normal coordinates, there
kinetic energy
momentum.
is
the
sum
of
3Ar terms
that each
due
and the other due
to Dcbyc. mechanics has now been reduced to that of 3 A" independent (weakly interacting) but distinguishable simple harmonic oscillators. Suppose we have N y such oscillators each vibrating with the same frequency v. According to the quantum theory, the energy e of any such oscillator may take on only discrete values:
to Einstein
Our problem
in statistical
«.
where h
is
=
[i
®hv
Planck's constant, 6.63
a macrostate of the crystal
is
(«
X
10
= 34
0, 1, 2,
J
•
s.
.
.
.),
Suppose, that at any
with energy
a,
=
\hv
A'i oscillators
with energy
«i
=
\hv
A'j oscillators
with energy
ti
=
\hv
r
moment
specified by:
oscillators
Ao
(11-27)
These energy states arc nondcgcncratc: that is, not more than one quantum state has the same energy. The number of ways in which A'„ vibrators may be distributed
among
the energy states according to the macrostate specified
is
same as the number of ways in which AT, distinguishable objects (colored balls, marked objects, etc.) can be distributed in boxes so that there are Nt> objects in box 0, A', objects in box 1, etc. To fix our ideas, suppose that we have only four objects a, b, c, and d to be distributed between two boxes so that one object will be in one box and three in the other. The number of different ways in which four lettered objects may be arranged in sequence is the
310
11-12
Heat and Thermodynamics
a
bed
cda
b
dab
c
d
which must be maximized subject to the usual conditions,
abc
SA = A „ = r
r
const.,
;
a
bde
b
cad
c
dba
d
acb
a
cdb
b
dac
c
abd
d
bca
SA
and
The
cbd
dbc
a
dca
b
acd
b
adb
c
d
d
bda
c
deb
Fig. 11-13
adc
b
bad
c
d
4!
=
24, as
shown
The
total
number
The
etc.,
(11-29)
,
=
l,e-" lkT
(11-30)
.
P
expressions for energy U, and pressure
same
are the
as for indis-
(11-31)
of different arrangements
is
there-
P = NJcT
and
But the expression for the entropy
4 !/3 = 4. In the general case where N, vibrators are distributed among energy states with A'o vibrators in the state with energy eo, A'i in the state with energy «i,
fore
—-ulkT j—
the partition function
left
line) arc superfluous, since they involve
box, and so on.
,
tinguishable particles, namely,
box (depicted merely a shift of position within a box. Of the six arrangements on the left, only one-sixth need be counted or 1/3!. The same is true of the arrangements involving b in the left
and only
is
are in the right one. All the other microstales with a in the
beneath the dashed
10,
arrangement (microstate) dein the left box and /;, c, and d
in Fig. 11-13. Consider the
picted in the upper left-hand corner, where a
Chap.
e
7
Z,
below are superfluous.
line; those
is
cba
!
above the dashed
= A
Ni
The number of different microstales corresponding to the macrostate in which one and three vibrators are in the other energy state is equal to 4 !/3 = 4.
different microstales lie
const.
v
cab
vibrator is in one energy state
The four
,e,-
We get
bac
where Z„ a
= U =
r
details of this calculation are identical with those of
the results will be given.
a
311
Pure Substances
is
/d
In Z, \
\
dV )t
(11-32)
simpler:
!
the total
number
of microstatcs, or the thermodynamic probability
The
which
is
1
Nl \N t
k In
is
--
(11 " 28)
•
l
evaluation of the partition function
Z,
expressed as
= =
or
z,
-l"IMT
e
e~
= =
k(N, In k(M, In
-
ffl,
A'„
In #i!
— N, -2
—
A',-
In 2Va JV'i
In
In
A',-),
!
-
Nt
-f
Ari
-A
r
2
In A'2
+ iVs -
u.
(11-33)
particularly simple, since
is
less
than
1
and an
infinite
it is
number
and
h ' nkT
_|_
{\
-3hrl2kT
e
+
e->"'kT
_|_
+
g-SkrlWt e-*-'"i kT
_|_
.
.
.
+••),
g-hvllkT
= 1
k{\n N,\
+
of terms:
was included.
Q
=
Z„
merely a geometric progression with a ratio
the expression given in Prob. 10-5, where degeneracy
The entropy S S=
=
y
S2, is
given by
Q
N k In
Sy =
_
lnZ = -
(11-34)
-h*lkT
e
~ln'~
ln(1
- e~"" tkT
)-
(11-35)
312
Heat and Thermodynamics
The energy U,
of
U,
11-13
N, simple harmonic
= NJc'F—=ln a
vibrators
I
I
I
I
g(-)
Z,
-r*'i a {.hv/k 'n)
— = —2 .V,
4-
"**
3N-
- «-»*'*r
1
therefore the average energy per vibrator
=
I
I
1
v" tr- Tl il \ZkT*
M w
gW
is
313
Pure Substances
is
(o)
^ *W«P
_
(11-36)
!'
gW
gO)
r
3A equivalent simple harmonic oscillators do not have the same frequency. Let dN, be the number of oscillators whose frequency lies between v and v + dv. Then In general,
all
the
dN,
=
g(v) dv
(11-37)
where g(v), the number of vibrators per unit frequency band, must be determined for a given crystal or class of crystals and must satisfy the condition \dh\
=
=
}g( v ) dv
3A\
(11-38)
(c)
The energy
of A" particles of the crystal
is
then
Fig. 11-14
approximation;
'hv
u-
v
at constant
_du _ ~ df
f J
volume of this amount
{\/k){hv/T)yt" («*>/"
-
\y
of crystal
1
l-14a.
Equation (11-40) then reduces
r
calculation.
,
...
.
5(v) dv
-
to the following simple forr
.. r k(hv E /kT)'e k ,kv,,/kT
...
(e
(11 " 40)
we
define the Einstein
@E The
simplest assumption concerning the vibration characteristics of a
crystal
was that of Einstein, namely, that all the 3A' equivalent harmonic had the same frequency ve (subscript E for Einstein), as depicted
and
let
N=
A' A
_
^ kT
1122 I)
characteristic temperature
Frequency Spectrum of Crystals
oscillators
more rigorous
(d)
is
If
11-13
Blackman approximation;
.
in Fig.
and the heat capacity
Frequency spectra of lattice vibrations, (a) Einstein approximation; (b) Debye (c)
Ge by
the expression
—
(Avogadro's number)
—remembering that
(11-41)
NA k = R
(the
r
Heat and Thermodynamics
314
universal gas constant)
—we get
T —>
As
°o
,
the upper limit of the integral becomes small,
and the integrand
when
integrated, yields
evaluated at small values of x reduces to x 2 and, cv_
/OeV
3/e
\t)
e/t
_
e /T (e E
(11-42) !
1)
315
PureSubstar:
11-13
^i@/T)*. Therefore cy/"5R —*\, as required by the Dulong and Petit law. To find the limiting value of Cvfilt as T —» 0, it is convenient to express Eq. (11-45) in a different form by integrating by parts.
This was the first attempt to apply quantum theory to the specific heat of solids, and although the assumption of equal frequencies for all equivalent harmonic oscillators is far from justified, the Einstein expression has the same general shape as the curves in Fig. 11-12. As T —* , then cr/$R—* 1, in
agreement with the Dulong and Petit law. As T—*0, then ey/3/2 approaches zero, in agreement with experiment; but it approaches zero exponentially, which is faster than experiment indicates. The next approximation was made by Debye, who calculated a frequency distribution on the assumption that a crystal was a continuous medium sup-
e/r
and
assumption
it
was a simple matter
transverse waves.
With
show that a continuous spectrum
to
quencies was present, starting with zero and terminating at a frequency vm, according to the simple relation
dx
Si")
"
(e/ry
cv
=
It
may
mation
is
As
T-*
Cv_
(3 v
3Nk
i
3(©/T) (11-46)
L
eir
_
t
(11-43)
3A'.
h
_
e-
l
ex
jo
-
J,fW
we
is
a Riemann zeta function, f equal to
get
r(4)
=
i+I + I
90"
-yy-
= l ( T\ ~ 5 \®) '
il
Hence It is
convenient to define a hv_
kT and the Debye
new
variable of integration
x,
*»
Since
kT
3R
where
hv„
and
T'
4ir
Debye' s
l
T
/5 3
—
law
77.9,
and
(0/T)
at
3
10 90
7r
low temperatures Cv two forms:
1
=
cp
=
c,
we may
write
in either of these
6-™ (?'
characteristic temperature
=
'
i
The Dcbyc approxi-
into Eq. (11-40),
k,ikT [e
eir
3(9/72
/v 3m )(hv/kry-e l"i kT dv
/;
e
0,
where f (4)
Debye approximation
Q/ry _ I
(
1
of fre-
depicted in Fig. 11-14/;.
Substituting the
—
e*
e/T x 3 dx
4
3J2
this
maximum
•
equal to
is
1
therefore,
and be seen by inspection that jg(v) dv
-
1
'
i/T 4x 3 dx
e
= TT
-1
*d
(Q/TyJo Jo
(e*-\y-
;
porting standing (or stationary) longitudinal
s
x' e"
/.'
(Q/T)
hv„
(11-44)
(12 5)8
mJ
/yy
for
£<
0.04
(11-47)
|
mole deg \0, •
With
these
new
quantities
cv _
and with 3
Jr " (erfy
N= {&/T
h
Na, x*e" {e*
-
The entire course of cv/3R given by Eq. (11-45) or (11-46) cannot be reduced to a simple form but must be evaluated numerically. The results of
dx
iy
(11-45) t Sec
Appendix.
316
Heat and Thermodynamics
11-13
Dulong ar d
with theoretically derived functions such as that of Blackman shown in Fig. ll-14c. It is quite astonishing that the Debye curve in Fig. 11-15 is in such
Petit
1.0
good agreement with experiment. Only when heat capacities are measured with great accuracy do departures from Debye's curve show up. These are obtained as follows: The accurate experimental value oic v fiR at a known T
0.9
/
\
0.8
/
o;
o
is
of
1
0.7
compared with
the
770 is obtained;
Debye curve
from
ber of temperatures. If
this
all
is
0-6
and the corresponding value is done at a numobtained were the same, the Debye
of Fig. 11-15,
derived the value of 0. This
values of
theory would hold perfectly. Such mental points in Fig. 11-16 show.
so is
not the case, however, as the experi-
The
exact calculation of the frequency a very complicated and tedious task. The use of modern computers has helped a great deal, and physicists are having
\. -o
317
Pure Substances
spectrum of
0.5
1
0.4
lattice vibrations
is
reasonable success in explaining experimentally observed heat capacities.
Debye's model, although crude in the /
0.3
medium temperature
range,
rigorous at very low temperatures, below 0/100, well within the
The reason may be
seen in curves
(b), (c),
and
(d)
of Fig. 11-14.
T3
is
quite
region.
At very low
0.2
temperatures, only the small frequencies of lattice vibration are excited, so
/
that only the beginning segments of the curves play a role.
r 3 law
rigorously the calculation of g(v)
ir
^-
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
g(v)
2.0
is
always quadratic.
is
One would
No
matter
how
carried out, the beginning segment of
expect therefore that, at very low tcm-
T/0 Fig. 11-15
Debye's heal capacity.
Table 11-4 and are plotted in be found in the "AIP Handbook."
this calculation are listed in
extensive table will
Fig. 11-15.
A
340
more
320
Table 11-4
T
0.025 0.050 0.075 0.100 0.125 0.150
Debye's Heat Capacity Cv
T
3R
e
0.00122 0.00974 0.0328 0.0758 0.138 0.213
T
cv
3fl
0.175 0.20 0.25 0.3 0.4 0.5 0.6
cv
300
©
3R
0.293 0.369 0.503 0.613 0.746 0.825 0.869
0.7 0.8 0.9 1.0 1.2 1.5 2.0
0.916 0.926 0.942 0.952 0.963 0.980 0.988
280
260
240 40 60 Temperature. °K
The fundamental assumption
of the
Debye
theory, leading to a simple
quadratic function for^(j'), as shown in Fig. 11-146,
is
quite crude compared
Fig,
11-16
Variation of
Tenerz, and Waller.)
Q
with
T for
NaCI. (Curve calculated by Lundquist, Lundstrom,
318
11-14
Heat and Thermodynamics
and dividing both
sides
6.0
cv
^i
4.0
The graph
,/
in Fig.
_ /125V
E
iJ
mole
\C& )
•
deg 2
how well the law is obeyed between and near absolute zero are given for a number of
11-17 shows
4.5°K for KCl. Values of nonmetals in Table 11-5.
Eh"
319
by T, we get
f |
Pure Substances
_KCI
10
20
15
T 2 ,(°K) 2 Fig. 11-17
Verification of
Dcbyc
T
Debye
T
3
taw for KCl. (Keesom and Pear Iman.)
/125\ 3 cv
Table 11-5 Nonmetal
He Ne
A Kr Xe Se
Te As
Ge Si
C C
(graphite)
(diamond) LiF LiCl
NaF NaCl NaBr Nal
In a metal crystal the lattice sites are occupied by single metal atoms whose vibrations obey the same laws that have been found to hold for nonmetals. In addition to the lattice vibrations, however, there are free electrons whose of the same order of magnitude as the number of atoms and whose resemble those of the molecules of a gas. If the classical theory of an motions valid for the electron gas in a metal, the molar heat capacity ideal gas were
number is 3
law should be. strictly obeyed and also that the value obtained from such measurements should be the correct value of of at absolute zero. Using the T 3 law in the form pcratures, the
Thermal Properties of Metals
11-14
.oo
T
°K 26 67 93 72
55 90 153
volume should be augmented
at all temperatures
by the constant
3
Values 0,
at constant
mole
for
deg
Nonmetals
Nonmetal
KF KCl KBr KI RbCl RbBr Rbl
282 363 636 420 2230 730 422 492
InSb
321
NiSe 2
225 164
FeSe.
CaF 2 SiOs Fc,0,
TiOs
MgO ZnS FeSs
Bi 2 Tc 3
0, °K 336 233 174 132 165
a
20
131
103
200 510 470 660 760 946 315 637 297 366 155
100
200
300
400
500
600
700 800
900 1000 1100 1200
Temperature, °K Fig. 11-18
(Compare
Temperature variation of cp and
with Fig. 11-7.)
/3
of copper
is
similar to that of nonmetals.
320
Heat and Thermodynamics
11-14
321
Pure Substances
value fit Hence at high temperatures, instead of reaching the Dulong and Petit value 3R, it should reach the value $R, or about 37 J/mole deg; and •
low temperatures, cv should not approach zero as T approaches zero but should approach %R. The fact that the temperature variation of c v of metals had the same general features as that of nonmetals remained a puzzle until Sommerfeld applied quantum statistics (Fermi-Dirac statistics) to the free at
electrons in a metal
and showed
heat capacity of the metal
when
peratures
the linear term
is
is
that the contribution of the electrons to the
linear in
T and
Copper
appreciable only at low temterm becomes small. At higher temperatures,
Debye T 3 small compared with the
the
•
1.0
shall return to this point later, after
we have
Isotherma
is
effect of lattice vibrations.
0.9
We
--^Adia batic
studied other thermal properties
of metals.
The
0.8
E
and
of metals are insensitive to moderate changes of pressure and vary with temperature in the same way as nonmetals. The graphs for copper cp
(as typical of all metals) in Fig. 11-18,
which should be compared with those show that both c P and /3 are zero at absolute zero and that, from to 50°K, they rise rapidly. At higher temperatures the curves flatten out but do not approach asymptotic values. The temperature dependof
NaCl
0.7
in Fig. 11-7,
ence of the
of metals shows a curious regularity, as
shown
800
600
400
200
in Fig. 11-19,
Temperature,
it
1000
1200
K
Variation of isothermal and adiabatic compressibilities of copper with temperature.
Fig. 1 1 -20
where
s
may be seen
that the higher the melting point, the lower the
volume
expansivity. As a result, in the temperature interval from absolute zero to the melting point, all metals expand approximately the same fraction of their original volumes. It seems almost as if a metal like tin, realizing it is going to melt soon, expands rapidly with the temperature, whereas platinum, with a
high melting point, slows
down
its
rate.
be seen that the isothermal and adiabatic compressibilities of copper vary with the temperature like those of NaCl in Fig. 11-8. The complete temperature dependence of the thermal properties of copper is In Fig. 11-20
Q.
2 E
cv
O
may
and c v is compared with c P in Fig. 11-21. Notice that the of copper goes somewhat beyond the Dulong and Petit value above 700°K.
given in Table
CD
it
1 1
-6,
We shall see later that this additional heat capacity may be attributed to the
>
free-electron gas inside the metal.
The
free-electron gas in a metal differs
from an ordinary gas in two main
quantum states accessible to the A ,), number of molecules (gi gas molecules is very much electrons number of quantum states and the whereas in a metal the number of are comparable. Let us see how this comes about. An electron moving more or less freely in a cubical metal of length L and volume V = I? has a kinetic energy e given by exactly the same expression as that developed in Chap. 10 respects. In
an ordinary
gas, the
number
of
larger than the
200
400
600
800 1000 1200 1400 1600 1800 2000 Temperature, °K
Fig. 11-19 melting point.)
Normal
temperature variation of volume expansivity of metals.
(NMP,
normal
»
7
11-14
Heat and Thermodynamics
322
Thermal Properties of Copper
Table 11-6
Each
T,
Cp>
°K
J/mole
deg
6.25
50 100 150
16.1
20.5 22.8 24.0 24.5 25.8 27.7 30.2
200 250 300 500 800 1200
K,
ft •
M(deg
')
/i(atm
11.4 31.5 40.7 45.3 48.3 50.4 54.9 60.0 70.2
er,
»)
_1 )
cm 3/mole
J/mole
7.00 7.01 7.02 7.03 7.04 7.06 7.12 7.26 7.45
0.722 0.731 0.744 0.759 0.773 0.788 0.850 0.935 1.045
KS, •
deg
6.24 16.0 20.3 22.4 23.5 23.8 24.5 25.4 26.0
M(atm
_1
where
m
= todS
{nl
states referring to
a space where n x n y
r in
,
0.721 0.728 0.737 0.744 0.754 0.764 0.806 0.860 0.901
The quantum
the mass of
the volume
and
states is
of the positive part of this spherical shell. Since
k2
and the volume the
of the spherical shell
number
of
this
quantum
states
'
8mL * A
=
9 dr A 2r
,
de,
2
between
and
r
r
+ dr
is 4irr
g de with energy between
e
=
dr.
and
e
There-
+ de
is
volume, or
n z are very large integers.
g
=
de
}
4irr
30
2
= t?
dr
•
2r dr
8mL ,SmL 2
,
e>
de
2
c onsta nt pre ssure
4\
Co nslan volu me
-a
+ de lie
an energy interval between e and e and r + dr. and the number of these
r
8mL 2
fore,
n x, n m
and
n z are cartesian coordinates
states referring to
then
+nl+ *•
an electron and
and
between the spheres of radius
one-eighth of is
,
n„, and n z refers to one quantum state, and all the one energy e lie on the surface of a sphere of radius
)
for a molecule moving in a cubical box, namely,
6
values of n x,
triplet of
quantum
323
Pure Substances
2
,„
Ir
li
3R (11-48)
20
The number of quantum on
E
states per unit
energy interval, g,
Since an electron has a mass only about 1/10,000 that
JK*.
•v.
atom,
15
it
follows that, for a given energy, there
quantum
may
is
seen to depend
of, say,
be (10
-4
)',
a helium or 10
-6 ,
an electron gas than for an ordinary gas. Also, in a given volume (say, 1 cm 3 ), there may be 10 19 atoms of helium, but in a rnetal of the same volume there might be about 10 23 free electrons, assuming about one free electron per atom of metal. With 10 4 more electrons and about 10 -6 fewer
CJoppe r
10
states for
quantum states, it is clear that the number quantum states may be comparable.
fewer
/
of
of electrons
and the number
-Now, according to the Pauli exclusion principle, only two electrons (with opposite spin) can occupy the same 100
200
300
400
500
600
700
800
Temperature, °K Fig. 11-21
Temperature variation of cp and cv of copper.
900 1000 1100 1200
absolute zero, all
all the
energy states arc
that a
volume
V
quantum
state.
At a temperature of
electrons cannot be in the lowest energy state; instead,
filled to
an energy
of metal contains
ep,
known
N free
as the Fermi energy. Suppose
electrons.
At absolute
zero, the
324
number
of states occupied by these electrons
A'/2,
is
where from Kq. (11-48)
U of
and the energy
the entire system of A' particles
3N/V /•» f» _ ~Jf/V
?-" fflh-** N
or
V
8w /2m
k
Id, 2«i
evaluation of this integral
The
i
(11-49)
grand the
may be expanded
first
in
a
is
is
t^e <
e<'-r )lkT
_|_
-j"
When T/T?
difficult.
is
small, the inte-
and integrated term by term.
series
325
Pure Substances
11-14
Heat and Thermodynamics
When
only
retained,
two terms are
To appreciate how
large the Fermi energy is, it is instructive to calculate the temperature of an ordinary gas at which a molecule would have the energy e r This temperature, called the Fermi temperature, is defined as
U=
.
+
Uv CV = Nk^r — +
4A'e F
1
12
'kT
and
2
€W
(11-50)
Calling the constant
where k
is
Boltzmann's constant. From Eq. (11-49),
wc
y',
we may
get c.
""
The molar volume
(11-51)
2mk\V)'
\8irJ
The entire course of c„
T of copper
every two copper atoms,
7.2
is
N=
3.0
cm
X
*»
TV, then
ce
~
=
2
F»
or
2
(6.6)
0.24
and assuming one free electron for 10 23 electrons. Hence,
X
9.1
X X
- y'T
shown
is
|i?.
in Fig.
1
(for
T« r„).
1-22,
where
it
(11-52)
may be seen that, when
In the linear region of the curve.
3
;
e.
=
3 f^31 ^
Tv
write for the molar heat capacity
10-" lO-
2^
X X
X 10 4 x 10 -i6'
•
/r
(
for
T/Tw <
0.2).
14
12 j
— J—r-l Smole deg/
It
was mentioned previously that the
cv of copper at
1200°K exceeds the
~ 50,000°K.
This means that the closely packed electrons in a metal at absolute zero have the energy that an ordinary gas would have at 50,000°K This energy, known !
as the zero-point energy of the electron gas, joules per mole.
When
may
be several hundred thousand
the temperature of the metal
is raised to a value T, only electrons can be raised to higher states, so that the increase of energy per unit temperature rise dU/dT, which is the heat capacity, is
whose energies are near small.
To calculate
the metal, c e ,
it is
ev
this electronic
contribution to the molar heat capacity of
necessary to develop a
statistics
of indistinguishable particles
obeying the Pauli exclusion principle where { g
Fermi and Dirac,
this is called
the Fermi-Dirac
«
N
{.
First
formulated by
Prob. 10-7). a system obeying this type of statistics comes to equilibrium, it is a simple matter to show that the number of particles A',- with energy e< is given statistics (see
When by
N = t
-
Si
elv-t^m
+
i
Fig.
1 1
-22
Temperature variation of heat capacity of an electron gas.
Dulong and
Petit value, as
at this temperature
shown
in Fig. 11-21.
The
The
electronic contribution at
is
low
1200
=
-
mole
•
0.75 J/mole
deg/ 50,000
mj/mole deg
deg,
•
and when
this
J/molc
shown
is
added to the lattice contribution of 25 J/mole deg, the deg is in good agreement with the value 26 J/mole deg a relation first derived
t
0)
13
l£
Ji TT* -r*-
5
j
W
0.8
se"
iy
u tt"
r^-
tfL
+
y'T,
6
10
12
(ii_53)
y;
T
plotted in Fig. 11-23 verify Eq. (11-53) very well and yield reliable values of and 7' of other metals are listed in Table 11-7. and 7'. Values of
0.4
4
+
1 should yield a straight line with a graph of c/T plotted against 3 slope equal to (125/G) and an intercept 7'. The experimental measurements on copper, silver, and gold made at the Westinghouse laboratories and
I
2
ji
o?y
therefore a
prff r
poc
=
o-c
Es
O
'-
by Sommerfeld. Dividing by T, we get
1
1
(f
•
in Fig. 11-21.
o E
given by
is
)'
result 25.8
327
metal is region of greatest interest in the study of the electron gas in a this region, law holds. In temperatures (T < 0/25), where Debye's T«
the total heat capacity in
\
Pure Substances
11-14
Heat and Thermodynamics
326
14
16
18
Debye Temperatures and
Table 11-7
Electronic Constants
of Metals {Compiled by N. Pearlman, 1966.) -,°- c>
o
3.0
%E
o-"~
Metal
1
-o-
2.0
rvO
1.0
0,0
&> o£
Silv
3 Li
10
6
14
12
16
It
o
o ^^
,o-
-,o
6
E O'
Y»°
o-°
r
,°"
n-
11-23
19
K
20
Ca
21 Sc
Mn
p-^ r^ Off
26 Fc 27 Co
Gc»ld-
28 29
30 6
8
10
T*, deg I'ig.
Mg
V
1
4
Na
12 13 Al
24 Cr 25
' 2
Be
22 Ti 23
-C J
.&>
°K
Metal
mole deg 2
0,
°K
•
er
10
^
y'j
mJ
mJ mole deg 2
cfi-
4
E
Metal
©,
•
11
4
mJ mole deg 2
9*
2
0,
n-°
>-"
e-;
"
SO
•0=
7',
7',
o-"
E
12
14
16
18
2
Heat-capacity measurements of Corak, Garjunket, Satterthwaite, and Wexler.
31
m
Cu Zn Ga
33 As
344 1440 158 400 428 91
230 400 420 380 630 410 467 445 450 343 310 320 282
1.63 0.17 1.4 1.3 1.35 2.1 2.9 11
3.5 9.8 1.40 14
5.0 4.7 7.1 0.688 0.65
0.60 0.20
37
Rb
38 Sr 39
Y
40 Zr 41
Nb
42
Mo
44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In
56 147 280 291 275 450 600
480 274 225
Sb
209 108 199 210
55 Cs 56 Ba 57 La 66 Dy 69
38 110 142 210 200
50 Sn 51
Tm
2.4 3.6 10.2 2.80 7.79 2.0 3.3 4.9 9.42 0.650 0.69 1.6 1.78 0.105 3.2 2.7
10.5
79
Au
80
Hg
n
81
11
/y
82 Pb 83 Bi
105 119 1/0 182
71
72 Hf 73 la 74 75 Re
W
76
Os
77 Ir 78 Pt
,
90
Th
V
1
10
Yb Lu
120 210 252 240 400 430 500 420 240 164
70
92 94 Pu
160
2.9 11.3 2.16 5.9 1.21 2.3 2.4 3.1
6.8 0.75 1.8 1.5
3.0 0.021 4.7 10.3 13
328
Pure Substances
Heat and Thermodynamics
329
PROBLEMS 24
11-1 22
that, for
an
ideal gas:
dT - T
f
y dT - nRT
In
I
"
-
const.
T +
const.
CP dT - T
f
§ dT + nRT
hi
P
-
const.
T
const.
a)
p_
f
(/,)
G =
[
(c)
Apply the above equations
(
20
Show
11-2
cv
to
Defining the Massieu function
\-
idea! monatomic equation the by Fu
mole of an
1
gas.
(a) S-
Fn =
16
12
f + S,
show that
/ s
14
-
Palladium
dF«
(b)
=
P ±dT =dV. + TT Tl
Defining the Planck function
I-'v
by the equation
10
20
10
30
40
T5 Fig.
11-24
50
(°K)
,
Thermal expansion of palladium
to
60
70
80
Fv
2
show
the effect of conduction electrons.
show that
an interesting
fact that
conduction electrons
in a
metal contribute to
the thermal expansion as well as to the heat capacity and, according to the
same
| + S,
V H dFv = -^dr--^dP.
(G. K. White, 7967.)
It is
- -
relation,
11-3
From
the fact that
dV/V is an
exact differential, derive the rela-
tion
p
= aT+b'P',
fa
(11-54)
QPJt
dT/,.
where the
linear term represents the electronic contribution and the cubic term that due to the lattice. A verification of this equation in the case of palladium is shown in Fig. 11-24, and numerical values of a and b are listed in Tabic 11-8.
Table 11-8 03
Expansion of Metals
= aT + bTz;
at
11-4
Low Temperatures
"(Jf
compiled by K. Andres, 7063.) (c)
Metal
Mo
W
Re Pt
Cd
n(deg" 2 )
n(deg-')
1.29 0.09 2,7 6.6 0.6
0.0114 0.0135 0.0204 0.177 1.14
Metal
n(deg-
Al
3.3
Pb
3
Mg
3.6
Ta
3.1
Pd
10.8
2 )
Derive the following equations:
= -s-r/^ H=G -"(*&), dT/p
n(deg-0 0.072 4.20 0.126 0.096 0.129
(IS 11-5
Derive the third
T dS equation,
(Gibbs-Hclmholtz equation).
330
Pure Substances
Heat and Thermodvnamics
and show
that the three
T (IS equations may
(P (1)
T4S= C v dT +
—
(2)
T dS = CP dT -
V0 T dP,
Show
11-10
be written as follows:
+ a /v )(v 2
—
cv
dV,
T(o
-
The
pressure on 500 g of copper
atm
thcrmally from zero to 5000
Using the
expansion
virial
(Assume the
at 100°K.
and heat capacity constant. The values are given in Table 11-6.)
How much How much
heat
and isodensity, volume
to
(c)
(d)
11-3.)
(b) (c)
Calculate the change in internal energy.
The
reversibly
and
pressure on
1
g of water
Using
adiabatically. Calculate the temperature
(a)
and
limit as v
—>
°°. its
and bearing
and its limit as v —* », (Comof Rossini and Frandsen given in Art. 5-2.)
in
mind
=
RT+ B'P+
that B'
=
+
C'P 2
B, calculate (du/dP)r directly from Eq.
(11-19).
Show H, and
atm
that the differentials of the three
F may
thermodynamic func-
be written
-
(CV
dT
PV&)
+
V(kP
-
,87')
dP.
+ F(l - 0T) dP. dF = -{PV0 + S) dT + PVk dP.
dH = CP dT
change when the
temperature has the three different values given below: 11-13 1
Temperature, Specific volume
°C
cmVg
5
1.000 1.000 1.012
v,
{a)
cr,
ft
/t(dcg
>)
J/g
•
Derive the equation
deg
T ( av\
(dCv\
50
-
67
+
15
+465
\9VJt
4.220 4.200 4.180
(b) (c)
A
11-9
gas obeys the equation P(v
and has a constant (a)
u
is
(b)
y
is
(<-)
A
cy-
Show
a function of
T
—
b)
= RT, where
b is
\dT*Ji
Prove that Cv of an ideal gas is a function of T only. In the case of a gas obeying the equation of state
constant
Q. Rl^-
that:
4. 1 l+ 7'
1
only.
where
constant.
relation that holds during
P(v
—
an adiabatic process
B
is
a function of
is
cv *=
b)i
=
const.
limit as
calculate (du/dP)r
{b),
Pv
dU = to 1000
increased from
its
pare the solution with the results Using the virial expansion (d)
tions U,
is
>
00
11-12
How much heat is transferred? How much work is done?
11-8
>
(c)
been subjected to a reversible adiabatic compression? 11-7 The pressure on 200 g of water is increased reversibly and isothermally from 1 to 3000 atm at 0°C. (Numerical values are given in Table (a)
—
and
C
,
Using the same expansion, calculate (dP/dv) T and
(b)
V
work is done during the compression? Determine the change of internal energy. What would have been the rise of temperature if the copper had
(b)
calculate (du/dv)r
remain practi-
transferred during the compression?
is
B
= RT(\
increased reversibly
is
expansivity, isothermal compressibility,
(a)
const.
11-11
Pv 11-6
=
b)"
TdS=CfdP+^dV.
(3)
initial
with
is
K
(a)
cally
van der Waals equation a function of T only, an equation for an
that for a gas obeying the
= RT,
b)
adiabatic process
331
T
only,
show that
—
d2
RT
jp_
(BT)
+
(cr)*
where
(ci-)o is
11-17
the value at very large volumes.
11-14
Derive the following equations:
to
CP = T
(A)
dP\ r) s
dPjr
\dr-Jr
(c)
Pv
where
B
a function of
is
T
RT+
=
show
only,
Cp_
vpf
(OP/d'Os (0P/dT) v
w
Prove that Cp of an ideal gas is a function of T only. In the case of a gas obeying the equation of state
(i)
\9T)b
l)f)p
Derive the equation
((?)
y
-
1
11-18 (a)
BP,
Joule
A
measure of the
coefficient
rj
=
result of a Joule free
(dT/dV)u. Show
Cr
(cp)o
is
the value at very low pressures.
In the accompanying table arc listed the thermal properties of liquid neon, compiled by Gladun. Calculate and plot against temperature
11-15
(a) c v
\
ks
to
;
and
P,
ft
K,
°K
10~ 2 dog-'
10- 3 atm-'
25
6.15 5.99
1.33 1.46
5.81
1
5.62 5.40 5.17
2, 12
33 35 37 39
43
63
2.52 3.14 4.24 6.8
4.61 4.23 4.00 3.68
42
.
1.84
4.91
41
is
provided by the
is
-pY
measure of the result of a Joule-Kelvin expansion (throttling provided by the Joule-Kelvin coefficient p. = (dT/dP),f Show .
that:
Up
10- 2 mole/cm 3
31
Cv
(c) y.
T,
27 29
A
process)
fpT
1
=
-T-^P + W». (b)
where
expansion
that:
that
V
=
333
Pure Substances
Heat and Thermodynamics
332
10 18
0.43 0.50 0.62 0.79 1.03 1.40 2.04 3.4 6.9
11-19
cp,
J/ mole
-
deg
It
Einstein crystal
was shown
36.6 37.6 39.2 41.2 43.9 47.7
F; (b) the pressure P; and
53
in
11
62 82 100
26
160
11-12 that the partition function of an
in Art.
is
g-hrlikT
Z= 1
If the crystal consists of
U
-
«r-*w*r
NA lattice points, calculate (a) the Helmholtz function
(c) the entropy S. (d) Express the zero-point energy terms of (-)].;. 11-20 Using the Debye approximation, show that the total energy of ArA lattice points (3AA independent harmonic oscillators) is
U=
9RT rwr x dx (@/TyJo e*-l" i
$R®
Derive the following equations:
11-16
What to
is
the interpretation of the term •£#©?
11-21 ,
dV\
(dV/dT)
W WW)*s
as the
CVK _
11-22 A system consists of A:A distinguishable, independent particles, each of which is capable of existing in only two nondegenerate energy states
1
i
-
Plot the Einstein c v/2>R curve against T/Q E on the same graph Debye CvfiR curve against T/B, and then compare the two curves.
and
y
e.
Hcai and Thermodynamics
334
12.
(a)
What
(b)
Calculate the energy.
(c)
Calculate
(d)
Plot c v/R as a function of kT/e from
11-23 function of
is
the partition function?
cy.
kT/e =
to 1.
Given a crystal obeying Debyc's approximation with (-) a only, assume the entropy S to be a function of T/Q. If F is
V
PHASE TRANSITIONS; LIQUID AND SOLID HELIUM
defined by the equation
r = (a)
Show
(b)
d\n V
that
r
[Hint:
- d In 8 =
-
C VK
Use Maxwell's third equation.] NaCl at a few temperatures using Table
Calculate V of
Show
that
transition
7 = 11-24
(a)
The
Show
1+
partition function of a
Debye
9
,>,„,.
is
Calculate the Helmholtz function.
(c)
Show
o
that the equation of state of the crystal
PV
+ fO') = rp7- U&
the zero-point energy.
-_.
ro/T x s dx
+WT? /
(b)
f/o is
crystal
we
pure substances during the best-known phase transitions,
shall study the behavior of
from one phase
to another. In the
namely, the melting of ice and the vaporization of water, the regions of temperature and pressure are easily accessible without special apparatus. Some of the most interesting materials, however, such as nitrogen, hydrogen,
TfiT.
that
lnZ= _3i n( i-,-en)
where
and In this chapter
plot F against T. (c)
Joule-Kelvin Effect
12-1 11-2,
is
given by
and helium, whose phase transitions contain still unsolved problems, exist only at low temperatures. It is important, therefore, to learn how these low temperatures are achieved and maintained. The first step is to liquefy air, and the most economical way to accomplish this is with the aid of the Joule-Kelvin effect or, as it was formerly called, the porous-plug experiment. In the porous-plug experiment a gas is made to undergo a continuous throttling process. By means of a pump, a constant pressure is maintained on one side of a porous plug and a constant lower pressure on the other side. In the original experiments of Joule and Kelvin a cotton plug was used, and the gas flowed through it parallel to the axis of the pipe. In modern measurements a cup of a strong porous material capable of withstanding great force allows the gas to seep through in a radial direction. Rigid precautions are taken to provide adequate thermal insulation for the plug and the portion of the pipe near the plug. Suitable manometers and thermometers are used to measure the pressure and temperature of the gas on both sides
of the plug.
The experiment temperature
'/',•
is
on
performed
in the following
way: The pressure Pi and
the high-pressure side of the plug are chosen arbi-
pressure Ps on the other side of the plug is then set at any value P„ and the temperature of the gas Ts is measured. P,- and Ti are kept less than same; P,is changed to another value, and the corresponding Tj is the This is done for a number of different values of Pj, and the measured. trarily.
The
336
Heat and Thermodynamics
1
corresponding 7/ is measured in each case. P; is the independent variable of the experiment, and 77 is the dependent variable. The results provide a set of discrete points on a TP diagram, one point being P,7' and the others being the various P/s and T/s indicated in Fig. 12-1 by numbers (1) to (7).
337
Phase Transitions; Liquid and Solid Helium
2-1
700
1
600
Although the points shown in the figure do not refer to any particular gas, they are typical of most gases. It can be seen that, if a throttling process takes place between the states Pi T{ and PjTf (3), there is a rise of temperature. Between P/Ti and PjTj (7), however, there is a drop of temperature. In general, the temperature change of a gas upon seeping through a porous plug depends on the three quantities Pi, 7',, and Pf and may be an increase or a decrease, or there may be no change whatever. According to the principles developed in Art. 11-1, the eight points
500
400
300
plotted in Fig. 12-1 represent equilibrium states of
some constant mass of which the gas has the same enthalpy. All equilibrium states of the gas corresponding to this enthalpy must lie on some curve, and it is reasonable to assume that this curve can be obtained by drawing a smooth curve through the discrete points. Such a curve is called an isentlmlpic curve. The student must understand that an isentlmlpic curve is not the graph of a throttling process. No such graph can be drawn because in any throttling process the intermediate states traversed by a gas cannot be described by means of thermodynamic coordinates. An isenthalpic curve is the locus of all points representing equilibrium states of the same enthalpy. The porousplug experiment is performed to provide a few of these points, and the rest arc obtained by interpolation. The temperature T{ on the high-pressure side is now changed to another value, with P being kept the same. Pf is again varied, and the corresponding T/s are measured. Upon plotting the new PiTt and the new P/s and T/s, another discrete set of points is obtained, which determines another isenthe gas (say,
1
200
g) at
100 -Crit. pt.
-Liq-vap. equil. curve
of isenthalpic curves
The numerical
R Q.
E
/(3)
+
at
any point
curve
f
frm
+
+
/(7)
+
shown
in Fig. 12-2 for
is
is
value of the slope of an isenthalpic curve on a TP diagram coefficient and will be denoted by u- Thus,
locus of
all
and
is
points at
maxima
shown
=
whereas outside, where is
/j
which the Joule-Kelvin
of the isenthalpic curves,
is
where n
is
negative,
drawn
intersect the isenthalpic
(12-1)
BP)k
for A' 2 in Fig. 12-2 as
If a vertical line
Ha
diagram prepared by
obtained. Such a series
is
inside the inversion curve,
®
TS
called the Joule-Kelvin
the locus of the +
(From
nitrogen.
The /(5)
700
F. Din, 1958.)
M
/<«>*
600
inversion curve for nitrogen.
thalpic curve corresponding to a different enthalpy. In this way, a series
2
500
Pressure, atm Isenthalps and
Fig. 12-2
t
E-H
400
300
200
100
known
is
zero,
i.e.,
as the inversion
a heavy closed curve. The region
positive, is
coefficient is
is
called the region of cooling;
the region of
lieating.
some arbitrarily chosen pressure, it will curves at a number of points at which /j may be at
We
obtained by measuring the slopes of the isenthalps at these points. should then have a set of values of /* referring to the same pressure but to Pressure Fig. 12-1
Isenthalpic states of a gas.
P
different temperatures.
This can then be repeated at another pressure.
Since the Joule-Kelvin coefficient involves T, P, and
h,
we
seek a relation
338
Heat and Thermodynamics
among
the differentials of T, P, and
12-2
/;.
and, according to the second
= Tds +
Tds =
Substituting for
T ds, wc
T ds
Helium
339
is
vdP,
equation,
cpdT- T
dp.
get
Mh«•-£['(*), -]* + T{
or
Solid
In general, the difference in molar
enthalpy between two neighboring equilibrium states dk
and
Phase. Transitions; Liquid
dP,
dh.
Cp
Since
/x
=
(dT/dP) h
,
20
>-k
T
40
60
(m,-'
100
80
(12-2)
120
160
140
180
200
Pressure, atm Fig. 12-3
Isenthalps and inversion curve for hydrogen.
(
Wooliey, Scott, and Brickwedde
1948.)
This
is
the
thermodynamic equation
for the Joule-Kelvin coefficient. It
is
evident that, for an ideal gas, throttling to a pressure of if
Cp
The most important
P
1
atm,
helium originally at 200 atm and 52°C
application of the Joule-Kelvin effect
is
in the lique-
in is
used in most laboratories for
Table 12-1 gives the
Liquefaction of Gases by the Joule-Kelvin Effect inspection of the isenthalpic curves
and the inversion curve
of Fig.
12-2 shows that, for the Joule-Kelvin effect to give rise to cooling, the
initial
temperature of the gas must be below the point where the inversion curve cuts the temperature axis,
For
many
gases,
i.e.,
below the
room temperature
sion temperature, so that
is
maximum
inversion temperature.
already below the
maximum
inver-
no prccooling is necessary. Thus, if air is comatm and a temperature of 52°C, then, after
pressed to a pressure of 200
is
to 23°C.
On the
throttled to 1
atm,
other hand, its
tempera-
Figure 12-3 shows that, for the Joule-Kelvin effect to produce cooling hydrogen, the hydrogen must be cooled below 200°K. Liquid nitrogen
cooling in helium, the helium
An
be cooled
will
ture will rise to 64°C.
faction of gases.
12-2
it
this
is first
maximum
purpose.
cooled to a temperature lower than the
optimum
produce Joule-Kelvin
inversion temperatures of a few gases
monly used in low-temperature work. The shown in Fig. 12-4. It is clear from Figs. 12-2, 12-3, and 12-4 the
To
cooled with the aid of liquid hydrogen.
com-
inversion curve for helium
that,
is
once a gas has been pre-
maximum
inversion temperature,
pressure from which to start throttling corresponds to a point
on the inversion curve. Starting at this pressure and ending at atmospheric pressure, the largest temperature drop is produced. This, however, is not large enough to produce liquefaction. Consequently, the gas that has been cooled by throttling is used to cool the incoming gas, which after throttling
12-2
Heat and Thermodynamics
340
Maximum
Table 12-1
Phase Transitions; Liquid and Solid Helium
^^r^^^^^:,v.^-^!-m^^^r^
Inversion
fr^
Temperatures -
Maximum
Gas
temp.,
gas
inversion
°K
Aii-
780 764 659
Nitrogen
621
Neon
231
Hydrogen Helium still
202
~40 many
cooler. After
repetitions of these successive coolings, the
lowered to such a temperature that after throttling
is
liquefied.
shown
The
The
-i
~1500
Carbon dioxide Argon Oxygen
becomes
341
it
M
>-Hlgh
11/
pressure
becomes partly
device used for this ptirpose, a countercarrent heal exchanger,
i
U
is
"Low
in Fig. 12-5.
pressure
gas, after prccooling,
is
sent through the middle tube of a long coil
of double-walled pipe. After throttling,
it
flows back through the outer
Compressor
,
fiO
Fig. 12-5
'SO
Liquefaction of a gas by means of Joule-Kelvin
effect.
annular space surrounding the middle pipe. For the heat exchanger to be the temperature of the gas as it leaves must differ only slightly from the temperature at which it entered. To accomplish this, the heat exchanger must be quite long and well insulated, and the gas must flow through it with sufficient speed to cause turbulent flow, so that there is good thermal contact between the opposing streams of gas. When the steady state is finally reached, liquid is formed at a constant efficient,
'*"**> Helium
N
—\
30
\
*
\
+
rate: for
?n
—
and the
/
every mass unit of gas supplied, a certain fraction y is liquefied, y is returned to the pump. Considering only the heat
fraction 1
—
exchanger and throttling valve completely insulated, as shown in Fig. 12-6, we have a process in which the enthalpy of the entering gas is equal to the enthalpy of y units of emerging liquid plus the enthalpy of 1 — y units of emerging gas. If
r'
—r 10
s 20
40
60
80
100
hi
=
enthalpy of entering gas at (T,,Pi)
Iil
=
enthalpy of emerging liquid at (Ti.,Pr,),
h]
=
enthalpy of emerging gas at (T/,P/),
Pressure, atm tig. 12-4 I960.)
Isenthalps
and
inversion curve for helium. (R.
W.
}
Hill and 0. V. Lounasmaa,
and
only by varying the pressure, the condition that
(.hi.Ti.Pi)
-
(n) But
or the point (Ti,Pi) must
is
lie
=
at
T = Th
on the inversion curve.
TS diagram showing
obtained directly from such a diagram.
The
isobars
particularly useful. For example, to calculate the fraction
liquefied in the steady state y, the three enthalpies
helium are shown
that
°-
In the design of a gas-liquefaction unit, a isenthalps
is
maximum, M
and
be a minimum
it
= ~ cm = " (£)* (S)*
(§),
hence, for y to be a
343
Phase Transitions; Liquid and Solid Helium
12-2
Heat and Thermodynamics
342
in Figs. 12-7
and
and hi. may be hydrogen and for
hi, hf,
TS diagrams
for
12-8.
use of the Joule-Kelvin effect to produce liquefaction of gases has
two advantages: (1) There are no moving parts at low temperature that would be difficult to lubricate. (2) The lower the temperature, the larger the drop in temperature for a given pressure drop, as shown by the isenthalps in Figs. 12-2 and 12-3. For the purpose of liquefying hydrogen and helium, Fig.
1
2-6
Throttling valve
and heat exchanger
in steady slate.
however,
it
cooling that
h = yh L +
then
or
y
h,
=
hi
(1
and the helium must be precooled with makes the liquefaction of these gases expensive.
nitrogen,
— y)hj,
An
-h —
(12-3)
hi.
determined by the pressure on the liquid, which fixes the temperature, and hence is constant. hf is determined by the pressure drop in the return tube and the temperature at C, which is only a little below that at A; hence, it remains constant, hi refers to a temin the steady state, hi
perature Tt that
is
fixed,
is
but at a pressure that may be chosen at will. may be varied only by varying In. Since
Therefore, the fraction liquefied y
liquid hydrogen,
which
approximately reversible adiabatic expansion against a piston or a
turbine blade always produces a decrease in temperature, no matter the original temperature. If, therefore, a gas like helium could be to
Now,
has a serious disadvantage: namely, the large amount of preis necessary. The hydrogen must be precoolcd with liquid
do
external
work
adiabatically through the
medium
what
made
of an engine or a
turbine, then, with the aid of a heat exchanger, the helium could be liquefied
without precooling. But this method has the disadvantage that the temperalure drop on adiabatic expansion decreases as the temperature decreases. A combination of both methods has been used successfully. Thus, adiabatic reversible expansion
curve,
is
used to achieve a temperature within the inversion effect completes the liquefaction. Kapitza
and then the Joule-Kelvin
first to liquefy helium in this way, with the aid of a small expansion engine that was lubricated by the helium itself. Later he liquefied air with the aid of a centrifugal turbine only a trifle larger than a watch. The most significant development in the field of gas liquefaction is the
was the y
y will
be a
maximum when
hf
=
hi is
hf
a
-k
-
hL
minimum; and
since
hi
may be
varied
Collins helium liquefier, in
which helium undergoes adiabatic expansion
in
344
12-2
Heat and Thermodynamics
o O o p essure, atm lO o c o o t> — -.o lO
•
if
150
345
—
rv>
r~,
Phase Transitions; Liquid and Solid Helium
//7/
140
1
1
130
44( )
r
120
«TT
80
AC
1
1
A
/
ny J( UgOl
420
110
400H
IU
100-
80
*£
^ 50
I
90
^
r 80
_
;
-320
to
Q.
E
H
==-4
70
-300
/ yT?
to
60
280
/ /
i
—f 260
50
//S/ ^\
c ritical
y^X^V
^ (\ l\\ ^ ?0
30
4^V
20
\
\.
A
\Jr
point^^-**^
40
A.
\
3<
V >*
w*C
v.
\ XV«^ i-^- *~ -
fc^_
A
y^
240 t>o
7~
220
- 200
s
-4 .*?
<<>
180
\ 70 80 90 100
10
1
140
20
1
50
^ 3
4
5
6
7
8
9
10
12
11
13
14
15
Entropy, cal/g deg fig.
12-7
Temperature-entropy diagram for hydrogen. (Woolley, Scott, and Brickwedde
1948.)
Entropy. J/mole-deg
a reciprocating engine. The expanded gas
is
then used to cool the incom-
ing gas in the usual countercurrcnt heat exchanger.
When
the temperature
low enough, the gas passes through a throttling valve, and Joule-Kelvin is used to complete the liquefaction. The unit consists of a four-stage compressor, a gasholder, a purifier, and a cryostat containing the engines, is
cooling
heat exchangers,
Dewar
flasks,
vacuum pumps, and
all
switches
and gauges,
=
Temperature-entropy diagram of helium. Lines of constant pressure (P 4 to atm), and constant enthalpy (H 40 to 440 J/mole) are shown. (R. W. Hill Fig.
1
2-8
— H =
0. V. Lounasmaa, I960.)
00 and
1
346
12-3
Heat and Thermodynamics final
and psig)
S =
«o(l
V=
B»(l
— x)s V) + noxs - A> W + BOXPW, )
V are seen to be linear functions of .v. phase transition takes place reversibly, the heat (commonly known as a latent heat) transferred per mole is given by
Hydraulic hoist
|
for
Low pressure helium gas return (0.5 psig)
gas holder
—
and S and
(0.5 psig)
Low-pressure
347
phase at any moment. Then the entropy and volume are given by of the mixture at any moment S and V, respectively
formed into the
Four-stage compressor
(200
Phase Transitions; Liquid and Solid Helium
If the
maintenance
Experimental
chamber
/
=
r(jt/5
-
5 «>).
Pressure
u
regulator
© ©
01
B
©
ooo
o
o
existence of a latent heat, therefore,
that there
is
a change of
= -sdT + vdP,
Expansion j
means
entropy. Since
dg
OOOOD oo o ooo O oooooo o
The
Liquid helium \ drawoff tube
9g
engines
dT
Joule-Thompson
and
valve
Helium
Helium cylinder
(2000
purifier
we may
in Fig. 12-9.
This unit
is
manufactured by the A. D.
Little
Co. of
Massachusetts.
the Stirling refrigerators,
which were described
characterize the familiar phase transitions by either of the following
1
There arc changes of entropy and
2
The
of volume.
first-order derivatives of the
Gibbs function change discon-
tinuously.
simplest laboratory devices for producing small amounts of liquid
The air are
ox
equivalent statements:
Schematic diagram of Collins helium-liquefying plant.
shown
-
Heat exchanger
psig)
Fig. 12-9
as
•
Liquid helium container
in Art. 7-6.
Any phase change
that satisfies these requirements
is
known
as a phase
For such a phase change, the temperature variations and Cp are shown by four crude graphs in Fig. 12-10. The phase
change of the first order.
First-order Transition; Clapeyron's Equation
12-3
of G, S, V, transition
In the familiar phase transitions
—as well as
tion
in
some
less
— melting,
vaporization,
familiar transitions, such as from one crystal
modification to another, the temperature
and
pressure remain constant
while the entropy and volume change. Consider «
phase
i
with molar entropy
T and P and
and sublima-
***•
moles of material
and molar volume v u K Both
s
(t)
and
v
M
in
that the is
may be
regarded as accomplished reversibly
fourth graph showing the behavior of
Cp of a
Cp
0; or
in either direction.
particularly significant in
mixture of two phases during the phase transition
true because the transition occurs at constant
dT =
is
when
T is
constant
dP =
0.
is
infinite.
This
T and P. When P is constant,
Therefore,
are
hence remain constant during the phase transition which ends with the material in phase / with molar entropy s u) and molar volume v u) (The different phases are indicated by superscripts in order to reserve subscripts for specifying different states of the same phase or different substances.) Let x equal the fraction of the initial phase that has been trans-
functions of
The
CP = T
dS
dT
V \dTjp
*
= ~ 1 V dPjr
.
It should be noticed, however, that these statements are true only when both phases arc present. As shown in Fig. 12-1CV, the Cp of phase (i) remains
Heat and Thermodynamics
348
12-4
G
may be
integrated with the understanding that the various P's
which a phase I
and 7"'s at obey a relation in which P is a function of T so that (dP/d'T)v — dP/dT. Hence,
transition occurs
only, independent of V,
Gibbs function
TtsV> -j<») Phase (0
\
Phase
=
T%(v^ dl
left-hand
member
of this equation
is
dP _
It
Phase
is
equation, applies to
Phase
first-order
change
P.
instructive to derive Clapeyron's equation in another way.
It
was
Chap. 11 that the Gibbs function remains constant during a reversible process taking place at constant temperature and pressure. Hence, for a change of phase at T and P,
Phase (/)
(/)
any
T and
(/")
in
g
and
for a
phase change at
(i)
=
g
(/) ;
T + dT and P +
dP,
(d)
(C)
(c)
(12-4)
of phase or transition that takes place at constant
shown
Fig. 12-10
I
This equation, known as Clapeyron's
Tooo
Volume
(/)
and hence
Heat capacity
C,
Phase
the latent heat per mole,
T(vV> -»«>)'
dT
(b)
(a)
) T
*«).
(/")
The
dP
349
Phase Transitions; Liquid and Solid Helium
Characteristics oj a first-order phase transition, (a)
Gibbs function; (b) entropy;
Subtracting,
we
get
volume; (d) heat capacity.
finite right
up
to the transition temperature. It does not "anticipate" the
onset of a phase transition by starting to rise before this temperature
always true of a first-order transition, but not of all transitions. second T dS equation provides an indeterminate result when applied
reached. This
The
is
to a first-order phase transition.
For any small portion,
TdS= CP dT where Cp
=
first
«>
and dT
=
0; also,
/3
=
oo
Therefore,
(i>
dT
+ 0® dP = -s^ dT + dP = dT ~
j(/)
- j(Q
»>
-
dP
and,
finally,
dT
m<« dP.
'
o«> I
T(v">
-
»W)
Tl'pdP,
and dP
=
0.
12-4
Sublimation; KirchhoflF's Equation
T
dS equation, however, may be integrated through the phase transition. When 1 mole of substance is converted rcversibly, isothermally, and isobarically from phase (i) to phase (/), the first TdS equation,
The
-s
or
is
Tds =
cv
dT
'(£.*•
In dealing with phase transitions,
way The
it is necessary to indicate in a simple the initial and final phases and the corresponding heat of transition. notation used in this book is as follows: symbol representing any
A
property of the solid phase will be distinguished with a prime; for the liquid phase it will have a double prime, and for the vapor phase a triple prime.
1 350
Thus
12-4
Heat and Thermodynamics
v'
vapor.
stands for the molar volume of a
The
solid,
heat of sublimation per mole will be
(boiling) lv,
and the heat
of fusion (melting)
If-
v" of a liquid, and Is,
v"
of a
Phase Transitions; Liquid and Solid Helium
In the case of zinc between 575 and 630°K, the vapor pressure
the heat of vaporization
Using
this
system of notation,
log
P = -
v')
Sublimation usually takes place at a low pressure, where the vapor regarded as an ideal
„«'
1
Since
P is small,
of the solid that
v'" v'
may
be
gas, so that
is
large
may be
so
magnesium
is
is
= =
too small to measure. In the following pages
we
shall derive
any desired temperature. An infinitesimal change of molar enthalpy between two states of equilibrium of a chemical system is given by KirchhojPs equation for the heat of sublimation at
P'
—indeed,
7527 6787
of a solid
~ RT
for
144 kJ/mole; for zinc between 575 and 630°K, ls = X 130 kJ/mole. At other temperatures, the heat of sublima2.30/* X tion is different. If reliable vapor-pressure data existed over other temperature ranges, the temperature variation of lg could be obtained. As a rule, however, this is impossible because at low temperatures the vapor pressure 2.30/*
7V" -
dT
5.972.
T
739°K the heat of sublimation
Therefore, from 700 to
h
given by
6787 -
Clapeyron's equation for sublimation becomes
dP
is
351
much
larger than the molar
volume
neglected, or
dh Introducing the second
Clapeyron's equation can then be written
dh
T ds
= Tds
+ vdP.
equation,
we
dT
+
- T
= c P dT
+
=
c,.
v
v{\
-
get do
dP
&T)dP.
dT/T-
A
= -R d\aP
finite
change of enthalpy between the two
states Pi7~,
and PjTj
is
d{\/T)
= from which
it
solids are usually
when
log
h{
dlogP
P
Is is is
equal to —2.30/? times the slope of
plotted against 1/7".
Vapor
pressures of
measured over only a small range of temperature. Within
graph of log
log
P
against 1/7"
practically a straight line, or
is
{' cP
dT
+
j'.v(\
-
(ST) dP.
Let us apply this equation to a solid whose initial state j is at zero pressure and at a temperature of absolute zero and whose final state /' is that of a saturated solid (solid about to sublime) represented
by a point on the These two states are shown on a PVT surface in Fig. 12-11. To calculate the enthalpy change from i' to /', we may integrate along any reversible path from i to /'. The most solid-saturation curve
const. first
7527 '-¥±
triple point.
—
>/', the is the path represented by the two steps i —> A and A being isothermal at absolute zero and the second isobaric at pressure P.
Denoting the
P= -
below the
convenient
const. P= -- T
For example, within the temperature interval from 700 to 739°K, the vapor pressure of magnesium satisfies with reasonable accuracy the equation
log
-h=
d(l/T)
can be seen that
the curve obtained
this range, the
-2.307*
+
8.589.
final
enthalpy by
h'
-
),'
Q
= =
(j
h'
„(l
'
V
Io
and the
-
initial
pT) dP
by h
+
/*'
dP+ hTc dT '"
>
,
cP
dT
I
Heat and Thermodynamics
352
P
Consider
now
353
Phase Transitions; Liquid and Solid Helium
12-5
the reversible sublimation of 1 mole of a solid at temperature
T and pressure P corresponding to the transition from/' to/'" in Fig.
12-11.
We have ls
=
=
h'"
c'p
]
A'
dT -
c',,
jQ
Since the two integrals approach zero as
h—
* A'o"
and Hence, li
—
h
is the heat
dT
T
— K
Fig. 12-1
where
Portion of
PVT surface
below the
the sublimation curve ice it
ranges from
fore,
if
we
is
Now
very small for most
to about 5
mm;
for
T—
as
>
/i'
.
it
follows that
0,
of sublimation at absolute zero,
= f*c"P 'dTJo
-
"
which
P dT+l
c'
is
denoted by
la-
(12-7)
.
triple point.
the pressure at
solids.
to 0.1
only an approximate
the molar
This equation
points on
equation, being subject to the restrictions that the pressure
For example,
cadmium, from
known
cP all
the molar volume of the solid at absolute zero and
v' is
heat capacity at a constant pressure P.
h'
approaches zero,
T
ls
+
for
ordinary
mm.
is
as Kirchhojf's equation. It
the saturated vapor behaves like
an
is
is
low and that
ideal gas.
There-
limit the application of this formula to solids at temperatures
where the vapor pressure
is
very small,
h =
r
cP
we may
dT+
ignore
//„.
/
v'
Vapor-pressure Constant
12-5
dP, and
(12-5)
vapor in equilibrium with a solid is assumed to behave like an ideal and if the volume of the solid is neglected in comparison with that of the vapor, Clapcyron's equation becomes If the
gas,
Since the c P of a solid does not vary appreciably with the pressure, the value of c P at atmospheric pressure may be used in the above integral.
a low pressure behaves like an ideal gas. Going back to the general = (dh/dT)p and remembering that the enthalpy of an ideal gas a function of the temperature only, we have
h
dP
The enthalpy of the saturated vapor indicated by point/'" in Fig. 12-11 may be calculated on the basis of the assumption that the saturated vapor
RT-
dT.
at such
equation cp is
dti"
=
Integrating from absolute zero to T,
el!
If,
in
dT.
ls
where A
is
=
ln
may
suppose that the vapor pressure
is
be used. Thus,
+ fJcp'dT-
JJc'pdT.
we have The molar
h'"
we
addition to these assumptions,
very small, KirchhofPs equation
= f'4'dT
+ C,
(12-6)
the molar enthalpy of a saturated vapor at absolute zero.
heat capacity of an ideal gas can be represented as the
constant term and a term that
is
cp
sum
of a
a function of the temperature. Thus,
—
///
ca
,
///
tf;
,
(12-8)
Heat and Thermodynamics
354
where / "
equal to f # for all monatomic gases and to Jfi for all diatomic has the property that it approaches o-ases except hydrogen. The factor c t Kirchhoff's equation may therefore be zero rapidly as T approaches zero.
The
is
two terms are known
last
f
and
f\? dT -
- h + cfT +
Clapeyron's equation,
after substitution in
if.
Thus,
- 23q ~ loS 1.013,600
m
=
fJc'pdT;
we
as the practical vapor-pressure constant
355
f
written
Is
Phase Transitions; Liquid and Solid Helium
12-5
-
6.006.
Finally, expressing the pressure in millimeters, introducing the numerical
get
values
p df
.
,
j„ „ + ^dT T
h dT
RT
c
,
,
f>' dT
J
,
RT
T [ ,,
r
Jo - J
c'P
dT .n-
^T~ dT
and Integrating this equation,
we
=
log 760
and
-
2.881,
calling
get finally
B= the equation
i is a constant of integration. This relation is not rigorous, but it is accurate enough to be used in conjunction with experimental measurements of the vapor pressure of solids. Such measurements are usually attended
where
by errors that are much greater than those brought about by the simplifying assumptions introduced in the derivation. If the vapor in equilibrium with a solid is monatomic, c" has the value of IR and c' " is zero. The equation of the sublimation curve then becomes {
This
is
chemist
The
T-
2.30/e /„
dT dT,
becomes
P= -
log
(12-9)
!1±
the form in
y~f + \ log T-B+ + 2.881. i'
which the equation
is
most useful to the physicist or
in the laboratory.
sublimation equation
is
used in two ways:
compared with
theoretical calculations of
pressure of a substance at temperatures at
In both cases, the integral
B
and which
i';
to obtain experi-
(1)
mental measurements of the vapor-pressure constant
i ,
which are
(2) to calculate the
P
is
must be evaluated on the
to be vapor
too small to measure. basis cither of experi-
mental measurements of c P or of theoretical values for c P In order to do this, c'P is plotted against T from absolute zero to as high a temperature as .
,
In
P = -
5.
RT
2
_,
lnf
1
L**
T
-R
Changing to common logarithms and expressing
we
iT
dT +
i.
(12-10)
the pressure in atmospheres,
necessary.
tained.
<*
get
p
T dT
measured
is
is
thus ob-
These values are now divided by T- and plotted on another graph The area under this new curve at various values of T provides,
against T.
finally, the
l
The
area under the curve at various values of r r , with a planimetcr, and the temperature variation of / c P is
/
'-,
1
rrTc'pdT
" " 33KT + 1 log T ~ 2MR I
temperature variation of B.
measurements of the vapor pressure exist over a wide temperature range, the numerical value of log P — § log T -\- B is plotted against \/T. Since If
Sr- dT 2.30
log 1,013,600.
log
P-
± log
T+B-
^1 +
*'+
2.881,
356
the resulting graph
is
The vapor
a straight line whose
Slope Intercept
and
Phase Transitions; Liquid and Solid
12-5
Heat and Thermodynamics
= -
m
=
+
i'
pressure constant
To show
the vapor.
is
i'
this correlation,
-
//"
1
357
Iclium
connected with the entropy constant of we write the equation
h'
2.881.
/here
The data for cadmium are listed in Table 12-2. The vapor-pressure measurements were made by Egerton and Raleigh, and the measurements of B by Data
Table 12-2 T,
°K
for the
log?
#
)
Determination of i' of Cadmium log
logP£ log T + B
B
T
=
/VjJ'
dT +
r-fSfL-Mb P +
-12.00 -11.14 -10.36
1.82 1.88 1.94 2.08 2.20 2.32 2.41
6.38 6.45 6.50 6.63 6.75 6.85 6.94
-
'
dT +
h'
and
T i = f
c'P
f
h'a ,
+ sL
\/T subject to the three assumptions
-7.44 -6.57 -5.80 -4.17 -2.86 -1.77 -0.99
360 380 400 450 500 550 594
s'D
=
and
h
0.00278 0.00263 0.00250 0.00222 0.00200 0.00182 0.00168
8.72 7.41
6.30 5.52
made
previously. Substituting,
*J#iT-j f4ir + l}-f£££-Mi*p+4r-J
we c'P
In P,
we
=
cP
-|/?;
therefore, dividing
R and
by
dT
T
i
For a monatomic vapor
get
in Fig. 12-12,
lo
is
lnP=
found to be
-RT + 2 lnT -R
The two .
—
^ 4
.
solving for
get Jll
Lange and Simon. From the graph shown 112 kj/molc, and i' to be 1.50.
-s'a
l*f-tf*"
integrals within the bracket
r
S
'
R
may be combined
into
one double
integral, because
= 4.38 = ( + 2.881, whenc e t' = 1 .50
Intercept
\i\ by integration by
+ In
E-i
-4
=
parts. (Let u
P= - h
RT
, '
§c P
„
5 ,.
1
inI
(
-r]
2
and by comparison with Eq.
dT and
(12-10),
dv
dT
= dT/T'
2
/4 dT ~W
it is
,_
.
s'
"
.)
R
Therefore, s'a
5
V
seen that
a.
\
—6 Cad mium
\\
i '
_
£n
—8
S
\N
— 10 —t
Since ^ is
0.0020
the entropy of a solid at
Graph for determination of vapor-pressure
an
0.0030
1/T
f
constant of cadmium.
- «P R
•>
2
(12-11)
arbitrarily chosen standard state,
conventionally set equal to zero, so that
'
Fig. 12-12
_
^-
O
0.0010
is
£o
R
1
5 2
it
358
Heat and Thermodynamics
In Chap. 10, we calculated the entropy of an ideal monatomic gas and obtained Eq. (10-28), as follows:
=
cv
T+R
In
In v
+ R In
(Zirmk/H?) 3 '*
N I
Dividing by
R and
replacing v
U
Measurement of Vapor Pressure
12-6
The
359
Phase Transitions; Liquid and Solid Helium
12-6
determination of the heat of sublimation at absolute zero U and the i' require accurate measurements of the vapor pres-
vapor pressure constant
sure of solids. It will be seen in Art. 12-7 that the most convenient method of determining the heat of vaporization of a liquid also requires measuring the
5
no wonder that such measurements have engaged the many years and still constitute a very active branch of modern thermophysics. Let us consider just a few methods
Z
that are
by RT/P, we get
vapor pressure.
It
is
attention of physicists and chemists for
UK^ =
lnr _ lnP+ nA
> (
^y/^ +
2
i
J\ a
commonly
used:
therefore,
(2wmy k w
1
m
Since the mass of a molecule
is
In
k3
depends on
V/l
3'
2 .
7/2,
we
get finally
= -1.589
+
§ log
(12-13)
T/l.
melting-point solids.
comparison between these theoretical values and the experimental values
is
given in Table 12-3.
be measured It is
He Ne
Mg A Zn Kr Cd Xe
Hg
m
4.00 20.2 24.3 39.9 65.4 83.7 112 131 201
solid
and
whose vapor pressure its
surface area
A
is
is
to
deter-
off.
The assumption
is
made
that the rate of evaporation
equal to the rate at which vapor molecules would strike the solid there were equilibrium between the solid and the vapor. This rate
Chap. 6 (Prob. 6-5
was shown Calculated and Measured Values of the Vapor-pressure Constant Molecular weight
carefully weighed,
then placed in an evacuated enclosure and raised to the
being drawn
if
Substance
is
desired temperature. It evaporates at a constant rate, with the vapor
is
Table 12-3
The
Langmuif's evaporation method.
mined.
A
within the range from
McLeod gauge, Bourdon gauge, hot-wire, ionization), -8 which enable smaller pressures down to about 10 mm of Hg to be measured. The static method usually is quite adequate for liquids but often is of little value in measuring the vapor pressure of high2
i'
is
sensitive (e.g.,
A more refined calcula-
tion shows that this result is true provided the atom in both the solid and the vapor phases has zero spin and zero angular momentum in its ground state. Limiting ourselves to such substances and evaluating the atomic constants,
we
the vapor pressure
mm
-3
of Hg, the vessel containing the solid or liquid about lO to 10 3 is connected to a liquid column manometer, and the pressure is obtained directly. There arc at least a dozen manometers that are more
(12-12)
proportional to the molecular weight
see that the vapor-pressure constant
When
Sialic method.
2
=
i'
i'
[from Eq. (12-12)]
(measured)
-0.69
-0.68
0.37 0.49
0.39 0.47
0.81 1.13 1.30 1.49 1.59 1.87
0.81 1.21 1.29 1.50 1.60 1.85
in
f-' where
T/l is
lirRT
or
P=
the molecular weight. This
M A
\2wRT
m
method
is
very useful
when
a high-melting-point wire. This is a variation of the Langmuir method in which the weight and area of the solid do not have to be measured. Instead, the evaporating vapor is allowed to pass through an opening of known area and is then condensed in a cold trap. A measurement of the mass of condensed vapor after a time interval provides the
the substance
is
Knudsen's effusion method.
quantity
M.
360
12-7
Heat and Thermodynamics
There are many very ingenious methods involving
optical absorption,
rotation of the direction of vibration of linearly polarized light,
0.8
measurement
o o
of radioactivity, isotope exchange, etc., but these arc of interest only to
research workers in this
sound, and
A
critical
all
field.
None
Ki.l
of the methods
are attended by errors that
account of
all this
361
Phase Transitions; Liquid and Solid Helium
may go
as simple as
is
it
may
as high as 5 or 10 percent.
work, along with extensive tables,
is
U.b
given in a
book by A. N. Nesmeyanov ("Vapor Pressure of the Elements," Academic Press, New York, 1963). The complete temperature dependence of vapor pressure requires a formula with four adjustable constants. Many formulas have been suggested, but the one found most satisfactory by Nesmeyanov is
O.b
fine
o
fo c
o
U.4
o
U.d
O"
C/ o
log i>
=A-
CT + D
(12-14)
log T.
c
0.1
^ 0.2
0.1
Values of the constants and of
/o
Table
metallic elements.
0.5
arc given in his book.
A curious regularity has been found the ratio c\>/&. In
0.4
0.3
between
12-4, values of these
The graph drawn
and the
l
two
0.6
(
limit (as
T —>
0) of
Fig. 12-13
Heat of sublimation
0.7
0.8
0.9
1.0
1.2
1.1
1.3
MJ/mole
at absolute zero is proportional to (cy/ji)a.
quantities are given for 20
in Fig. 12-13 shows that
12-7
3/
Vaporization
(12-15)
The
5
heat of vaporization of liquids with normal boiling points from, let us 250 to about 550°K is generally measured directly with a calorimeter resembling that shown in Fig. 12-14. The sample liquid L% is contained in a small vessel and has immersed in it a small heating coil R%. Completely surrounding this vessel is a temperature bath consisting of a mixture of air and the vapor of another liquid L\. By choosing a suitable liquid Li and keeping it at its boiling point by means of the heating coil R\ in the presence of air at the proper pressure, the temperature bath may be maintained at any say,
Table 12-4 [(«v//3)o
Metal
h
Ratio of Heat Capacity to Expansivity and Heat of Sublimation of Metals
due to R. K. Kirby; l
^v//3)o,
kJ/mole
kJ/mole
is
due
to
Metal
A. N. Nesmeyanov.]
br/0h kJ/mole
kJ/mole
desired temperature.
rium with
Ag Al
Au Be
Cd Cu In
Fe Li
Mg
470 374 632
284 312 367
513 302 500 343 700 177 340
321
112 338 238 436 160 146
Na
Nb Pb Pd Pt
Sn
Ta Th Ti
Zn
120 1150 332 725 964 507 1280 835 943 288
its
vapor.
At
this
The
chosen temperature, the liquid Li
is
in equilib-
small vessel containing Li communicates with
may
be main-
108 722 197 382 555 302 780 470 472
another vessel on the outside (not shown in the figure), which
131.
container, with the heat of vaporization being supplied by the heating coil
tained at
any desired temperature by a separately controlled heating or
cooling device. If the
temperature of the outside container
of Li, a pressure gradient
is
is
maintained at
produced, and some of L%
less
than that
will distill over.
By
temperature of Z-2 is kept equal to that of its surroundings, and the energy necessary to vaporize it is thus supplied. There is therefore a steady distillation of L% into the outside maintaining a small current I in the heating
coil Ri, the
I 362
363
Phase Transitions; Liquid and Solid Helium
12-7
Heat and Thermodynamics
Vaporization Dataf
Table 12-5
T
Substance
°K
T/Tc
lv,
P,
lv/Tc,
f/mole J/mole
-
v'"
deg atm
-
P(v"'-v)" v",
T
liter/mole
'
J/mole deg •
N2
63.2 0.501 77. 4|0.612 94.1 0.745 104 0.823 111 0.879 116 0.918 120 0.950 124 0.981
6200 5580 4980 4430 3880 3290 2630 1840
49.1
0.124
44.2 39.4
1
35.1
10 15 20 25
83.8 87.3 106 117 124 130 135
0.556 0.580 0.704 0.777 0.823 0.863 0.896 0.924 0.970
6610 6520 5860 5290 4820 4350 3880 3420 2510
43.8 43.2 38.8 35.0 31.9 28.8 25.7 22.6 16.6
0.679
72.4 0.544 81.6 0.614 34.5 atm 99.0 0.745 0.820 109 115 0.865 0.902 120 126 0.948 0.977 130
6480 6050 5130 4500 4130 3750 2800 1990
48.7 45.5 38.6 33.8
0.3
Tc =
126.2°K
Pc =
33.5
atm
A Te = 151°K Pc = 48 atm
Fig. 12-14
Apparatus for measuring heat of vaporization.
139
and the heat of condensation being withdrawn by the surroundings of the outside container. Moreover, all the energy supplied by the heater Ri is used to vaporize Z, 2 since there is no heat loss between the inner tube and its sur/?2
146
CO
,
roundings. Consequently,
vaporization per mole
if
n moles are vaporized in a time
the heat of
Tc = 133°K
=
Pc is
lv
Of much more
t,
= Gh n
interest are the cryogenic liquids
points in the neighborhood of
100°K or
less.
For these
with normal boiling liquids,
one needs the
—
t
Taken from
F.
5
30.7 26.0 20.8 14.6
30
1
5
10 15 25
30 40
1
D
10
14
28.2 21.0 15.0
25
8.31 7.93 7.16 6.38 5.45
9.93 6.97 1.55 0.769 494 0.347 0.255 0.192 0.107
8.18 8.10
19.0 6.32 1.30 0.650 0.449 0.326 0.185 0.109
8.00 7.85 6.65 6.04 5.54 4.96
18
30
entific Publications,
temperatures from the triple point to the critical point. A few such tables now available, and the heat of vaporization may be obtained by performing the subtraction A'" — ft". In Table 12-5, vaporization data for some
phase. Determining the slope of the line in Fig. 12-15 to be 5.4,
arc
simple liquids, obtained from thermodynamic tables prepared by F. Din, are listed. In Fig. 12-15, the heat of vaporization l v divided by the critical tem-
—
v")/Thom about perature Te has been plotted against the quantity P(v'" 0.5 Tc to 0.98 7"c- It may be seen that the points for five gases lie on a common straight line, and therefore one is led to assume that similar points for other on the same straight line. By the term "simple" we Xe, and 2 whose molecules have no dipole moment only a small one) and do not associate in the liquid or the vapor
simple liquids would
mean
lie
liquids such as Kr,
(or at least
lv/T, P{v'"
of so
many
-
v")/T
may be
This relation present form
it is
7.41 6.66 6.05 5.42 4.99
4.20 2.98
3.71
2.55
Din (ed.), ''Thermodynamic Functions of Gases," Butterworth London, 1 956.
that is, the pressure, sort of information found in engineering handbooks entropy, enthalpy, and volume of both saturated liquid and saturated vapor at
4.58 3.65 2.50
.
20
31.1
41.9 6.06 1.33 0.655 0.399 0.262 0.173 0.102
=
(for 0.5
5.4
<
^<
is
therefore of interest to
of this strange proportionality.
We
first
it
ly
=
P(v'"
(12-16)
states.
involves a
In the
knowledge
examine the consequences
write Clapeyron's equation in the
form
dP/P dT/T*
we may write
>
regarded as a law of corresponding
of limited usefulness, however, since
quantities. It
1
Sci-
- v")/T =
(lv/To)Tc
Pf/"
-
v")/T
364
I
Heat and Thermodynamics
12-7
Equation (12-17) is a genuine law of corresponding stales, expressed in terms of reduced temperature and reduced pressure. It was first suggested by E. A. Guggenheim, who plotted the logarithm of the reduced vapor pressure against the reciprocal of the reduced temperature for seven simple liquids, as shown in Fig. 12-16. The points are seen to fall well along one straight line
(0.6), c,"
'a
whose equation
(0.7),
°'#
is
'
- 53
In -(0.8) -rf
r
30 (0.9) o
365
Phase Transitions; Liquid and Solid Helium
-
for ° 55
t)
(
r
< To < *)
,/t
^
\"\
y-
20
o
N, *K
Ar a
~\
CO
N
CH, •
Ne
At
-2
\
0123456789
K
§
v".
P(v'" -v")/T, J/mole-deg Fig. 12-15
and
A
Ar
Kr
states valid at reduced temperature between about 0.5
law of corresponding
-3
1.
x Xe
Notice that the right-hand
member dP
is
equal to 5.4 To-
a
The resulting equation,
o
less
be integrated from than 0.5. Hence,
ln^ = or
,
ln
P
pT
T to
To and from P
5.4Tc {.
\T
V
N2
y o,. D CO
..- dT
-p= 5ATcyi>
may
K
to Pc, provided that
T/Tc is
CH„
not
\^
V
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Tc) Tc/T
5.4 (l
-
ty
(for 0.5
<
£
<
l).
(12-17)
Fig. 12-16
A
law of corresponding
dynamics," Interscience, 1957.)
stales for simple liquids. (E.
A. Guggenheim, "Thermo-
367
Phase Transitions; Liquid and Solid Helium
12-7
Heat and Thermodynamics
366
In Table 12-6, the normal boiling points of 12 simple liquids arc listed critical points, and the reduced normal boiling points,
and the numerical agreement with Eq. (12-17) is quite satisfactory. It is conceivable that liquids whose molecules have large electric dipole moments and exert unusual forces on one another will obey a law of corresponding
along with their
Tb/Tc, are seen corresponding
with a different numerical constant. There is another interesting consequence of the law of corresponding ourselves states as given in Eq. (12-17), which is brought to light if we restrict allow us to from the critical point to to a small region of temperature far enough In this region, normal boiling point. regard l v as constant, say, around the v" is negligible compared with v'", and the vapor pressure is small enough to
to lie
states.
between 0.57 and 0.61, within the range of the law of
we
Therefore, comparing Eqs. (12-17) and (12-18),
get
states
approximate that of an ideal gas, or Clapcyron's equation becomes
v
l
-f Using the data in Table 12-6,
and again a
a
single straight line
= 5ATC,
(12-19)
graph of Ivb/R vs. Tg is shown in Fig. 12-17, is found for fifteen simple liquids. The slope
= RT/P. Under these circumstances, 1
CS 2
.
dp
lv
dT
RT /P
If
we
Tb,
v _
dlnP d{\/T)
d
(P/Pc) d(\/T)
In
2800
a,/ o /
integrate this equation over a very small temperature range around
where
has the constant value Ivb,
lv
i In
P =
t const.
we
-
get
2400
—
HBr
lv »
(12-11
¥c
Heat of Vaporization
12-6
at
the
Normal
Boiling Point, Ivb
Liquid
Tb,
°K
Tc,
°K
TB/Tc
nt/i
\ a:
IVB
°K
RT„
/
/
/
C2H4 C/j
1600
Ivn/R,
/
°s 2000
"
o
Table
//
2
R
l
or
o
3200
/ o Xe
1200
Ne
N
27.1
A
77.4 81.7 87.3
F2
86
2
CO
O
s
CM4 Kr
Xe C 2 H,. C 2 H« HC1 HBr Cl 2
CS 2
90.2 112 120 165 170 185 189 206 239 319
44.7 126 133 151
144 155 191
209 290 282 306 325 363 417 552
0.606 0.614 0.614 0.578 0.597 0.582 0.586 .
574
0.570 0.582 0.604 0.582 0.568 0.574 0.578
210 676 725 786 786 822 988 1090 1530 1640 1770 1920 2120 2460 3220
°Kr
7.7
8.8 8.9
Oy/ 800
°CH„
COo'^Ar
9.1
A,
9.2 9.1
8.8 9.1 9.2 9.6 9.6 10.1 10.3 10.3 10.1
400
/°Ne 100
200
300
Te
,
Fig. 12-17
A
law of corresponding
reduced temperature of about 0.6.
400
600
°K
states for heat of vaporization of simple liquids at
a
I
Heat and Thermodynamics
368
12-8
369
Phase Transitions; Liquid and Solid Helium
is found to be 5.6, in fair agreement with the expected value. (The agreement would be perfect if only the first nine liquids were considered.) Since the reduced boiling point of many liquids is in the neighborhood of 0.6, Eq. (12-19) may be regarded as a law of corresponding states, with the
of this line
statement:
ra
Cat
?
7Vrc ~0.6) = 5ARTC
.
Referring to Table 12-6, note that the ratio given in the last column, conchange due to vaporization at the normal boiling point,
stituting the entropy is
not
YV The increase is small, however, so that provided by taking ly/RTjs to be about 9 a work-
constant but increases with
a rough approximation
known
ing rule
is
as Trouton's
rule,
5 7.7
—
which
is
useful in cases
where Tc
is
not
known.
E to a."
12-8
Critical State 57.6
and vapor phases of a substance may coexist in equilibrium at a constant temperature and pressure over a wide range of volumes from v" where there is practically all liquid, to v'", where there is practically all vapor. At a higher temperature and pressure, v" increases and v'" decreases. At the critical point, the two volumes (and therefore also the two densities) coincide. On a Pv diagram, the critical point is the limiting position as two
The
liquid
points lying on a horizontal line approach each other. Hence, at the critical point, the critical isotherm has a horizontal tangent, or (dP/dvjrc is
—
0.
1.30
As
evident from Fig. 12-18, the critical isotherm must have a point of inflection
at the critical point,
At
all
phases,
including the critical point,
=
,
=
0-
is
finite, so
precise values of Pc,
exactly when the critical state raised in temperature slowly
ume. Consider the tube
is
that i'c,
=
filled
and vapor
properties d>
-(\/v){dv/dP) T
reached. For this purpose, the material
When
the temperature
is
is
is
according to the dashed line 3
so that the initial position of the meniscus
then the meniscus will
fall
to the
bottom as the
is
Schneider,
—
>
4 of Fig. 12-18.
To
observe the
critical
must be filled so that the meniscus will remain in about the middle of the tube, as shown by the dashed line passing through the critical point. As the critical point is approached, observation becomes difficult because of the following factors:
raised, the liquid
expand and the meniscus will rise to the top of the tube as the last of the vapor condenses, according to the dashed line 1 —> 2 in Fig. 12-18. If the is filled
W. Ilabgood and W. G.
point, the tube
will
tube
Isotherms of xenon near the critical point. (II.
1954)
=
in a sealed tube of constant vol-
so that the meniscus separating the phases
near the top of the tube.
gm/cm
0.80
3
are all infinite.
y = cp/'cv is also infinite. and Tc, it is necessary to determine
and uniformly
p,
Fig. 12-18
the three physical
{\/v)(dv/dT)r, and k
the other hand, cv
To measure
initially
(d 2 P/dv 2 )r c
points within the region of coexistence of the liquid
T(dS/dT) P
On
and therefore
0.90
1.00
1.10
1.20
near the bottom,
last of the liquid vaporizes,
1
Since the compressibility
is
infinite,
the gravitational field of the
earth causes large density gradients from top to bottom. These
be offset somewhat by taking observations of the tube
and horizontal
positions.
in
may
both vertical
370
12-8
Heat and Thermodynamics
Since the heat capacity is infinite, thermal equilibrium is difficult to achieve. It is necessary to keep the system at a constant temperature (within 10~ 2 or even 10 -s deg) for a long time and to stir constantly.
2
3
1
35
35
Since the thermal expansivity is infinite, small temperature changes of a local mass element within the system produce large volume
30
30
changes, and therefore there are violent density fluctuations, which
25
25
give rise to a large
amount
is
called
-
20
critical opalescence.
may
be measured at the same time as the critical temperature by noting the pressure at which the meniscus disappears. The critical volume, however, is much more difficult to determine. This measurement is carried out most commonly by measuring the densities of both saturated liquid and saturated vapor as a function of temperature to as close critical pressure
The vapor and
to the critical temperature as possible.
5
15
I
is
usually called the
rectilinear diameter,
o S
/
^ ^V\
20
15
£\T>TC
10
V
A rgon
3$o
Argon
liquid densities arc
then plotted against temperature, and a line representing the arithmetic average of these densities is constructed as shown in Fig. 12-19. This last line linear. Extrapolation to the critical
/
10
----
u
o E
The
^ °v
of light scattering, so that the material
becomes almost opaque. This phenomenon
371
Phase Transitions; Liquid and Solid Helium
and experiment shows it
temperature yields the
|
125
100
175
150
200
0.01
to be virtually
(a)
critical density.
12-20
Fig. log
(Tc
-
(a)
T). (M.
Cy J.
1.0
0.1
(Tc
T, °K
10
-D/Tc (6)
of argon below and above the critical point, (b) Bagatskii, A. V. VoroneV,
Cy
of argon varies
and V. G. Gusak, 1963.)
1.0
Nitrous oxide
0.8
relationship for the saturated phases as shown in believed to show irregularities near the critical 12-19 a time was for Fig. curve is now believed to be smooth, as shown. However, the point. —» 2 in Fig. 12-18 represents a constant- volume 1 line dashed vertical The a volume less than the critical volume. If the heat of temperature at increase
The density-temperature
Saturated liquid -C--0-.,
"--o
°^
measured along this line, it will be found to be quite large, berequired not only to raise the temperature but also to change the relative proportions of liquid and vapor. As the liquid saturation curve is crossed at point 2, c would be expected to take a sudden drop. Of particular capacity cy
e
cause heat
Rectilinear diameter no
d-O-O-o
rt-O-C-O-C
-o Critical point
v
0.4
interest -,-0
0.2
:
-=o-o-"
.-C
Gusak,
the
who showed
This was that the cy
T curve
at the critical volume, very close 1963 by Bagatskii, Voronel', and of argon varied as the logarithm of (Tc — T),
the behavior of the c v vs. first
done
in
1
25
30 t,
Fig. 12-19
is
to the critical point.
as
Saturated vapor
20
is
is
35
40
'C
Measurements of liquid and vapor densities of nitrous mean density through the critical point. (D. Cook, 1953.)
oxide,
and extrapolation of
shown
in Fig. 12-20.
Using a temperature-measuring technique in which a change of 10~* deg Kelvin could be detected, Moldover and Little also measured cv above and 3 below the critical point of both He 4 and He Their results again indicate that cy varies as log (Tc — T), and therefore c v has a logarithmic singularity at .
the critical point. This sort of behavior called a lambda
transition,
which
will
is
peculiar to a type of phase transition in Art. 12-10.
be discussed
372
373
Phase Transitions; Liquid and Solid Helium
12-9
Heat and Thermodynamics
Fusion
12-9
method of measuring the heat of fusion of a solid is to supply energy at a constant rate and to measure the temperature at convenient time intervals. Plotting the temperature against the time, a heating curve is obtained in which the phase transition appears as a straight line at
The
simplest
electrical
constant temperature, of length At as measured along the time
axis.
The
Low
etc., are exactly the same as those required measurement of heat capacity, which were described in Chap. 4. If n
apparatus, shielding, precautions, in the
I
iTemperature bath
moles of solid melt in time At, with electrical energy supplied at the rate £1, then If
Fig. 12-21
= £IAt
The normal melting point of a solid and Ifm is the latent heat of fusion at the normal melting point, the entropy change associated with melting at this temperature is famj&Tu* expressed in units of R. This entropy change is listed in Table 12-7 for 15 nonmetallic solids and 15 metals, and it may be seen that metals show more of a regularity than nonmetals. Very roughly, Ifm/ RTm is about 1 for metals. If 7".u is the
Table 12-7
Nonmctal
Entropy Change Accompanying Fusion Normal Melting Point Tu
°K
Ifm,
hit
kJ/mole
RTm
Metal
Blocked-capillary method of measuring melting pressure.
phases coexist in first
and and one equilibrium determine the melting curve,
and temperatures
various pressures
tasks of the
experimenter
is
to determine the equation
54.4 63.2
N CO 2
68.1
A
83.8 89.9 90.7 104 116
C,H,
CHj
C H., Kr Xe S
161
Clo
ecu
CH ON 4
CaCl,
Gc Si
2
172 250 406 1055 1210 1683
Hg
6.41
0.98 1.37 1.48 1.70 3.83 1.25 3.88 1.70 1.72 4.49
2.47
1.19
4.30 3.28
Ag Au Cu
2.71
Be
2.76
Ni
0.444 0.722 0.838 1.18 2.86 0.944 3.35 1.64 2.30
at the
Ifm,
Ifm
°K
kJ/mole
RTm
14.5
28.8 31.8 46.4
K Na
371
Li
454 594
Cs
Rb
Cd Pb Zn
Mg
601
692 923 1233 1336 1356 1556 1728
2.30 2.14 2.34 2,32 2.61 3.02 5.78 4.86 7.29 7 33 11.3 12.8 13.0 11.7 17.6 .
liquid
of the
of this curve.
In
often
method, illustrated in Fig. 12-21. The the gaseous phase is compressed to a high pressure and is forced capillary, part of which is immersed in a low-temperature bath
measured by the
material in into a steel
blocked-capillary
whose temperature may be adjusted to any desired value by careful choice of the bath liquid and its pressure. Two manometers, A/i before the bath and Mi after, communicate with the capillary. The melting pressure associated with the temperature of the bath is the maximum reading of Mi. Four typical melting curves, those of neon, argon, krypton, and xenon, are shown in Fig.
1m,
234 302 312 336
solid
the region of low temperatures, the melting temperatures and pressures are
12-22.
In 1929 F. E. Simon and G. Glatzel suggested an equation that has been fairly successful in representing the
o.
which the
at
0.98 0.85 0.90 0.83 0.85 0.80 1.17 0.97 1.27 0.96 1.10 1.15 1.15 0.91 1.23
P- P
s
=
data on melting curves, as follows:
a[(T/T3
y-
(1
1]
2-20
P3
are the coordinates of the triple point, and a and c are condepend on the substance. At the high pressures, i°3 is negligible, the equation is usually used in the form
where Tj and
stants that also
so that
P/a= (T/r3 y-\. The
values of a and
c for
the four condensed inert gases
are listed in Table 12-8, and those for S. E.
Babb,
Jr., in
an
many
shown
in Fig. 12-22
other solids have been given by
article in the Reviews of
Modern
Physics (April, 1963).
374
Heat and Thermodynamics
12-9
The
slope of the melting curve
is
negative for those few substances like
which contract upon melting. This behavior is shown by Bi, Ge, Si, and Ga and requires values of T in Simon's equation less than T3 Consequently the values of a arc negative. The various values of a and c for four of the forms of ice are shown in Fig. 1 2-23. In Fig. 1 2-24 arc shown the enormous pressures and temperatures needed to produce gaseous and liquid carbon and also the solid crystalline form of diamond.
18
ice I
/
.
16
-J
14
Theories concerning the exact processes that take place when a solid melts have engaged the attention of physicists for
/
12 1
£
many
years. It
was
first
/
Neoi l
J
/
/
/Argon
1 '
0)
8000
Y
(A Q)
375
Phase Transitions; Liquid and Solid Helium
8
0.
/ Krypton
7000
6
4
/
Xe non 5000
2
/
50
/ 200
150
100
250
300
350
(S.
E. Babb, Rev.
4000
Temperature, °K Melting curves of neon, argon, krypton, and xenon. Phys., April, 1963; G. L. Pollack, Rev. Mod. Phys., July, 7964.) Fig,
1
2-22
Mod. 3000
2000
Melting Parameters of
Table 12-8
Condensed Inert Gases 1000 Solidified
Th
Pi,
fl>
inert gas
°K
atm
atm
Ne Ar Kr
24.6 83.8 116
Xc
161
0.426 0.681 0.723 0.806
1022 2240 3010 3410
1.6 1.5 1,4 1.31
-50
-40
-30
-20
-10
10
Temperature, °C Fig. 12-23 (1
bar=
Phase diagram of water, with melting parameters of 10« dyn/cm 1 .)
ice I, III,
V,
and VI.
376
12-10
Heat and Thermodynamics
Phase Transitions; Liquid and Solid Helium
377
On the basis of such ideas, it is possible to provide a slight theoretical foundation for the Simon equation; as of 1968, however, no complete theory of fusion has been developed. role.
1
10 5
i 10
1
solid
1
in?
Lip
li
Higher-order Transitions
12-10
id
1
1
10 3
*m— So
-i id
—
The Triple point
(diamond)
10* i
processes of sublimation, vaporization,
transitions of the
first
and fusion were
called phase
order because the first-order derivatives of the Gibbs
— (dG/dT)p and V = (dG/dP)r, underwent finite changes .S" = during the transition. In contrast, there are many phase transitions in which function,
—
the entropy and the volume are the same at the end of the transition as they were in the beginning. In such phase changes, T, P, G, S, and V remain unchanged, and therefore H, U, and Falso remain unchanged. If Cp, k, and /3 undergo finite changes in such a transition, then, since
-/-
10 Solid (g raphite)
/ Liquid
/
1
|
/ /
-*r~
'
IT"
.
10
T
Vapor
K,/=
10"'
2000
1000
4000
3000
0V =
and
^-
r\-3
i
=
5000
(*s.\
A. ( - i9\ =
dT\
\dTjp "(c^),.
=
d"-G
dTJ
-^(S)
=
d 2G
_d_fdG
dT
dT\dP)
- dTdP'
Temperature, "K Fig. 12-24
Phase diagram for carbon.
Lindemann
vibrations becomes large
shakes
=
k bar
function.
itself
to pieces."
that a solid melts
enough
With
to
this
would be finite changes of the second-order derivatives of the Gibbs Such a transition is called a second-order phase transition. This transition was first suggested by Ehrenfest, who derived two simple equations that expressed the constancy of S and V. It was first thought that there were many there
10° dyn/cm 1 .)
when the. amplitude of the lattice break down the attractive forces holding together; in more picturesque language, "In melting, a solid
suggested by the lattice
(1
point of view,
Lindemann derived the
formula
=
TM
(12-21)
const..
examples of second-order transitions; but as experimental measurements were made closer and closer to the transition temperature (sometimes to within a millionth of a degree!), neither Cp nor (3 was found to achieve a finite value at the beginning or at the end of the phase transition. It may be that there is only one example of a second-order phase transition, namely, the change from superconductivity to normal conductivity in zero magnetic field.
where
T/l
and
v
are the molecular weight
characteristic temperature,
holds quite well for
many
and
Tm
is
and molar volume,
(-)
is
the
Debye
the melting temperature. This relation
metals and nonmctals, but there are a few of each
Since this transition will be considered in some detail in Chap. 15,
By
far the
lambda
most interesting higher-order phase transition characterized by the following:
transition,
that depart radically from the formula. This suggests that the melting process is
not exclusively a question of lattice vibrations. Dislocations and vacancies
1
law of force
2
liquid, are all considered to play a
3
in the crystal lattice, as well as the quantities specifying the
among
the molecules of both solid
and
we
shall reserve discussion of second-order transitions until then.
and G remain constant. and V (also 6', H, and F) remain constant. Cp, /3, and k are infinite. T, P,
,S"
is
that called the
378
12-10
Heat and Thermodynamics
First
c.
He' constitute the X line shown on the phase diagram of Fig. \2-26a. define a new variable I, where
Lambda
Second order
order
1
*
the slope of the X line at
any
379
Phase Transitions; Liquid and Solid Helium
If
= t - n,
point,
dP/dT (=
be the same at a cor-
P\), will
T
+
t,
I
to
= 7\ responding point on the parallel curve where the temperature = merely the line shifted curve marked I I is X a distance dashed since the the right.
t
<«)
(c)
(6)
2-25
1
Distinguishing characteristic
among
the three types of phase transitions,
(a)
First-order; (b) second-order; (c) lambda.
shall limit ourselves to
very small values of
/.
On the entropy curve of Fig. \2-26b, the slope of the dashed curve marked = is the same as that of the curve = (S\) at a point at the same pressure, I
t
very small. The same is true on the volume diagram of Fig. \2-2dc. since With these facts in mind, let us use the second T dS equation to represent a t
Fig.
We
we
is
A graph of Cp against T serves to distinguish among the three types of transiThe name "lambda transition" is accounted by the fact that the shape of the CP — T curve in the third graph resembles the Greek letter lambda. Among the many examples of X transitions are: tions, as
shown
in Fig. 12-25.
t
for
(1)
=
t
=
t
t
=
t
=
t
"order-disorder" transformations in alloys; (2) the onset of ferroelectricity
in certain crystals such as Rochelle salt; (3) the transition
from ferrornagnetism to paramagnetism at the Curie point; (4) a change of orientation of an ion in a crystal lattice, such as the ammonium chloride transition; and
most interesting of
(5) the transition from ordinary liquid helium (liq. helium (liq. He II) at a temperature and corresponding pressure known as a lambda point. It may be seen in Fig. 12-25a that, as a substance in any one phase approaches the temperature at which a first-order phase transition is to occur, its Cp remains finite up to the transition temperature. It becomes infinite
He
I)
only
all-
T=
7\
T = rk +t
to superfiuid
when
a small
before this takes
amount of
the other phase
is
present,
and
its
(a)
V
behavior
1
place shows no evidence of any premonition of the coming
event. In the case of a X transition, however, as starts to rise before the transition point
is
is
evident in Fig.
1
2-25c,
tfO
=
t
=
t
Cp
reached, as though the substance,
form of only one phase, "anticipated" the coming phase transition. Because the molecules of substances undergoing X transitions interact strongly with one another, even at distances beyond those of their nearest neighbors, in the
a rigorous statistical treatment is very difficult. The one physical process that has yielded to mathematical solution, namely, the transition from ferromagnctism to paramagnetism according to a simple model known as the two-
dimensional Ising model,
shows the typical anticipatory
rise of
Cp before
the
X transition occurs.
The
temperatures and pressures at which the X transition takes place in
T (c) Fig. 12-26
The X line of He* on a phase diagram, (b), (c) Entropy and volume at The slope of a curve at a small temperature difference I above (or below) the same as that at a corresponding point on the X line. (a)
points on the X line.
X
line is the
^
380
Heat and Thermodynamic
small entropy change on
T\
+
t.
Lsing small
the
dashed entropy curve of Fig. 12-266,
letters to indicate
where
381
Phase Transitions; Liquid and Solid Helium
12-10
T
molar quantities, +
+
0.06
ds
=
C
-j,dT
-
+
dP,
vj3
+
-
+
_
0.04
a dP T~dT +V^df
cp
ds
4-
.
-
Helium 4
+•
.***** 0.02
++
1
-
Since ds/dT and dP/d'f are the same, respectively, as the slopes at the corresponding points on the curves where T = T\, we have
•
T
*
(12-22)
vfiK.
^
x
.^"
•
•
-
V »«
...o-°
» ^-* Z**
T> Tk
This equation is true whether T > 7\ or T < 'J\, and it shows that c /T P should vary linearly with vfi as the transition is approached (either from above or below) with a slope equal to P^. Using t as a convenient parameter, experimental measurements of c P and as a function of t, when combined, should
between cp/T and vfi. Because it is not convenient to measure the c p of liquid helium, the molar heat capacity at constant saturation c s is usually measured, and c,- is calculated from a simple relation that
•
^o
0.04
-'
Calculated
asymptotes 0.06
^—
•
10-
10"
yield a linear relation
Fig. 12-28
10"
Volume expansivity of liquid lielium II vs.
log
io-
\T
—
T\\. (Atkins and
Edwards;
Chase and Maxwell; Kerr and Taylor.)
can be found 25
Helium 4
20 E DO
15
10
problems at the end of
h*§
The
^,
even
^
"X.
V
of liquid helium II with
io- 3
t
is
t is
shown
shown
to
I
=
in Fig.
10 1
-6
in Fig. 12-27,
and
seen in Fig. 12-28, where the transitions,
is
displayed
The expected linear relation between cp/T and Some numerical values at three points on the X line
deg.
2-29.
K= X;
Table 12-9
>r
M-.JU.-'
10"
is
down
In the case of liquid
are given in Table 12-9.
"""W,
10"
c.s
this chapter.
quite small.
dependence on the logarithm of t, characteristic of X
T
v@
T>T
variation of
is
the variation of & of liquid helium II with
2s *
10
'
Quantities at Three Points on the X Line of
n,
PX>
atm
°K
g/cm 3
0.05
2.172 2.00 1.762
0.146 0.167 0.180
io-
\T--J\\ Specific heal of liquid helium II vs. log \T — T\\. (Kellers, Fairbank, and Buckingham; Hill. Lounsamaa, and Kojo; and Kramers, Wasscher, and Garter.)
Fig. 12-27
in the
helium, the correction
13
29.7
Pi
=
(dP/dT) x
atm/deg
-130
-
75 60
vi
=
(dv/dTh
cm 3/g
deg
•
13
2.7 1.5
He
si
1
= (ds/di% J/g
•
deg 2
2.3 0.9 0.6
382
Heat and Thermodynamics
Phase Transitions; Liquid and Solid Helium
12-11
-
383
60 aim
in-s
50 aim
40 aim
30 atm
20 atm
2000
mm
1000
mm
•v.
2
1.06
0.04
-0.02
0.02
-0.04
-0.06
-0.08 Fig. 12-30
ft deg-' Fig. 12-29
Straight-line relation between
cp/T and j3
3
Temperature, °K
of liquid helium II.
Phase diagram of
He
1 .
its own vapor pressure; therefore, as the vapor above liquid pumped away, helium remains a liquid down to the lowest temyet achieved (~10 -6 °K). To produce solid helium, it is necessary
be unstable under
helium
12-11
Liquid and Solid Helium
is
peratures
to bring the
helium atoms closer together, to the point where attractive forces cohesion. This requires an increase of pressure above 25 atm,
Any temperature and pressure at which three phases of the same substance may coexist in equilibrium are a triple point. Water and many other sub-
so that there
stances have several triple points, but only one triple point refers to the
in
equilibrium of
solid, liquid,
and vapor phases. Every material that has been
studied has such a triple point, with the exception of helium. As the vapor of liquid He 4 is pumped away, the remaining liquid gets colder; and at the
X point, liquid helium triple point for liquid
The
II forms, so that the
I,
liquid II,
X point
and vapor,
zero-point energy of liquid helium
as
may
shown
may produce
is
an upper
triple point
where two
liquids
and a
solid
may
coexist
equilibrium. As a matter of fact, the phase diagram of Fig. 12-30 shows the
presence of two different crystal structures for solid
He'', hexagonal closepacked (abbreviated hep) and body-centered cubic (abbreviated bcc). About one part in a million of He 4 is the light isotope He 3 shown in Fig. ,
be regarded as a
12-31, along with that of
He Three 4
.
features are quite striking:
in Fig. 12-30.
about 210 J/mole, or about three times as large as the heat of vaporization. If a crystal were to form, it would is
1
The normal 760
mm)
is
boiling point of
3.2°K, so that by
He
3
(at
pumping
which the vapor pressure is He 3 vapor away tempera-
the
384
Heat and Thermodynamics
turcs
may be produced which
10
-7
He
3
The
He would 1
mm, which
is
385
arc lower than those obtainable with
He". Thus, by lowering the vapor pressure of He 3 to 0.024 mm, a temperature of 0.4°K can be reached. To reach the same temperature with
Phase Transitions; Liquid and Solid Helium
12-11
require lowering the vapor pressure to 2.7
5000
—
X
out of the question.
has no triple points at
melting curve, which
2000 all. is
the locus of points representing tem-
peratures and pressures at which solid and liquid
He
3
are in equilib-
rium, shows a minimum at about 0.3°K.
Both liquid He" and liquid He 3 show a negative thermal expansivity at temperatures above about 1°K. In the region of high pressures, from 100 to 5000 atm, other crystal struc1
tures
form
in both
He 4 and He
3 ,
as
shown
in Fig.
1
2-32.
The heat capacity
of
60 atm
50 atm
5
2
1
50
Temperature, °K Fig. 12-32
Phase diagrams for
He and He 1 i
at high pressures.
(A. F. Sekuch
and R. L.
Mills, 1962.)
measured and found to obey a Dcbyc T 3 law, with a Debye depending both on 7" and on the molar volume v. The temperature dependence is such as to enable an extrapolation to be made to zero temperature, so that the volume dependance of (-)o can be studied. This is shown in Fig. 12-33, and the curves have a twofold purpose: these solids has been
1
By using the of a crystal
1
1.5
2
Temperature, °K Fig. 12-31
Phase diagram of He*.
2.5
result of Prob.
is
1
1-20, namely, that the zero-point energy
$/?0o, one can calculate the zero-point energy of solid
with that of liquid helium. At 0o = 30°K, and at 1 000 J/mole. O = 110°K, ffi© Both of these extreme values exceed the zero-point energy of the liquid.
helium and compare |A'0
~ 270 J/mole;
it
~
386
Heat and Thermodynamics
By using
2
Phase Transitions; Liquid and Solid Helium
the result of Prob. 11-23, namely, that the Gruneiscn
T
is
given by
T
= -
d In d In
+ 2.0 O
v +
in conjunction with the curves in Fig. 12-33, the result
that
The
T
~
transition
is
1.5
obtained
2.5.
+ 1.0
between the
particularly interesting.
387
The
experimental values of v"
—
solid
and
liquid phases of
may
latent heat of fusion
and
He 4 and
of
He
3
is
E
be calculated from
+0.5
dP/dT, using Clapeyron's equation. He 4 and by Dugdale on He 3 are shown in Fig. 12-34. Notice that, in both cases, the latent heat becomes practically zero at low temperatures, whereas P(v" — v') does not. As a result, the energy difference u" — u' = If — P{v" — v') becomes
The
results of
v'
of
such calculations by Simon and Swenson on
negative, indicating that the energy of solid helium helium at the
is
greater than that of liquid
same temperature. The melting of solid helium
at such tempera-
2
1
tures
is
in this
a purely mechanical process, since there
temperature region.
An
is
practically
no
latent heat
isothermal reduction in pressure produces
melting; conversely, an isothermal increase of pressure produces solidification.
3
0.4
Temperature, °K
0.8
(a) Fig. 12-34
equation
2.0
°
(b) the
aid of Clapeyron's
He 3
(J. S. Dugdale).
Energy relations for liquid and solid helium, calculated with
and the first law.
1.6
1.2
Temperature,
(a)
He* (F. E. Simon and C. A. Swenson);
(b)
120
Another interesting characteristic of solid helium is
is
that
its
compressibility
greater than that of liquid helium.
100
PROBLEMS 12-1 (a)
occurs (b)
Show if
that in a Joulc-Kclvin expansion,
(dv/dT)r
Show
=
no temperature change
v/T.
that
/MP
= T2
~d(v/T)
dT
In the region of moderate pressures, the equation of state of gas
may be
Pv u,
Fig. 12-33
Debye
mole of a
=
RT + B'P +
C'P2
,
cm-'/mole
@o as a function of volume for solid He* and He 3
1962; Dugdale and Franck, 1964.)
1
written
.
{Heltemes and Swenson,
where the second and only.
third virial coefficients B'
and
C are functions of T
388
I
(c)
Icat
and Thermodynamics
Show
approaches zero,
that, as the pressure
299°K, 300°K;
it
is
Show
that the equation of the inversion curve
435 dyn/cm 2
(b) at
The
Compute
.
absolute zero; and
The heat
12-7 mole. (d)
P= - B' C-
of sublimation of zinc at
is
T(dB'/dT) T(dC'/dT)
and 10
12-2 The Joule-Kelvin coefficient /i is a measure of the temperature change during a throttling process. A similar measure of the temperature change produced by an isentropic change of pressure is provided by the coefficient j»g, where
B
-3
at
mm
cpdT =
Ms
—
fi
12-3
According
to Hill
The vapor
P
(b)
12-4
is
given
What What
600°K
is
4.67
(b)
19.49
-
H
2
is
=
3063/7'.
What What The
12-9 solid
X
the vapor-pressure constant.
,
is
the temperature of the triple point?
are the three latent heats at the triple point?
hydrogen
triple point of
at this point
is
the equation of the heli
In
is
at Ts
0.0810 g/cm. 3 that of liquid ;
P
=
H
14°K. 2 is
The density of g/cm 3 The
0.0710
.
given by
is
w^-y-
(atm)
-21.0
+
5.44
T-
0.3 In T,
0.1327' 2
and the melting temperature
(a)
pressure of zinc at
is
P= where
13,800 .I/molc,
1
-(3D.
and Lounasmaa,
k.I/
of Hg. Assuming zinc vapor to be an ideal gas, calculate (a) the
vapor pressure of the liquid
inversion curve
found to be 130
The vapor pressure in millimeters of Ilg of solid ammonia given by InP = 23.03 — 3754/X and that of liquid ammonia by In P
(b)
v_
1.96.
is
is
12-8
= =
600°K
600°K
such that
is
heat of sublimation at absolute zero, and
(a)
Us
the latent heat of sublimation (a) at at 200°K.
(c)
heat capacity of solid zinc
/
Prove that
389
Phase Transitions; Liquid and Solid Helium
in
is
is
given by
atmospheres.
the
maximum
TM =
inversion temperature?
point on the inversion curve has the
maximum
pressure?
Saturated liquid carbon dioxide at a temperature of 293°K and atm undergoes throttling to a pressure of 1 aim. The tem-
(a)
a pressure of 56.5
perature of the resulting mixture of solid and vapor is 1 95°K. What fraction vaporized? (The enthalpy of saturated liquid at the initial state is 24,200 J/mole, and the enthalpy of saturated solid at the final state is 6750 J/mole. The heat of sublimation at the final state is 25,100 J/mole.)
AU = L
(l
\
-
(a) the
33
the three latent heats at the triple point in terms of R, to
(b)
12-10 in
Compute the slope of the vapor pressure curve of the solid at T 3 The intensity / of the atomic beam emerging from a narrow slit .
an oven containing a solid in equilibrium with shown in Chap. 6 to be
its
vapor at the temperature
T was
the
(b)
(atm)
within 5 percent.
is
12-5 Prove that, during a phase transition of the first order, entropy of the entire system is a linear function of the total volume; energy change is given by
Compute
P 14
/
= V2TrmkT'
-±l\ d
m
is Boltzmamrs constant, and P is the vapor molar latent heat is /, the molar volume of the solid is negligiand the vapor behaves like an ideal gas, show that
where
dlnPj
is
the atomic mass, k
pressure. If the ble,
12-6
Iodine crystals have an atomic weight of 127 g/mole and a specific heat of 0.1 97 J/g deg. Iodine vapor may be assumed to be an ideal diatomic gas with a constant c P At 301°K, the vapor pressure is 515 dyn/cm 2 at •
.
;
1
dl_
I
dT
1 (- --\ T \RT 2/
390
Phase Transitions; Liquid and Solid Helium
Heat and Thermodynamics
12-11 critical
Some
of the properties of saturated water
and steam near the
ST
diagram.
The
heat capacity at constant saturation
defined as
point are given in the accompanying table: 'sat
Properties of Saturated
T,
°K 473 523 573 623 647
P,
10 6
dyn/cm 2
15.550 39.776 85.917 165.37 221.29
and
v".
»'",
h",
h'",
cm 3/g
cm 3/g
J/g
J/g
1.1565 1.2512 1.4036 1.741
127.2 50.06 21 64 8.80
852.39 1085.8 1344.9 1671.2 2085
2793
.
3.1
Plot lv/Tc against P(v'"
3.1
tfl
Show
—
_A= A dT T
2801
2749 2564 2085
£snt
=
— dT " v")/T, determine the slope, and compare
it
— one
12-13
and the other
at
(T
+ dT, P + dP) — are shown
rfi
sat
sat '
T. A -p=I VP
—
AJI ep
,(.")
dP
—
Cp
-
Cf
r<2)
+
(
I
(vfivi
1
WA
-
with
In Fig. P12-1 several examples of two neighboring phase transiat (T,P)
ds (n = ™ T dT
that
w
that of Fig. 12-15.
12-12
dT
Water and Steam
dl
tions
is
391
on an
other at (7"
— one at (T,P) and the
Consider two neighboring vaporizations-
+
the expansivity
dT,
P+
dP).
Assuming
at constant saturation
v"'
» v" and
of the vapor
v'"
= RT/P,
show that
is
S
T\
v'"\dT)^ 12-14
RT)
Using the van der Waals equation of
P== RT —
->
1'
v
state
b
v-
show that Pc
and
=
vc
21b 1
=
36,
Tc =
8a
27bR
calculate RTc/Pci'c-
12-15
Using the Dieterici equation of
v
state
RT —b
show that Pr
and is
Fig.
PI 2-1
calculate
12-16
Solid
=
vc
>
=
2b,
ARb
RTc/Pdc-
When
lead
is
melted at atmospheric pressure, the melting point
600°K, the density decreases from 11.01
heat of fusion
Tc =
is
24.5 J/g.
What
is
to 10.65
g/cm 3 and ,
the latent
the melting point at a pressure of 100 atm?
392
Heat and Thermodynamics
Water
12-17
at
its
freezing point (T^P,-) completely
The temperature
container.
is
393
Phase Transitions; Liquid and Solid Helium
Ts
reduced to
at constant
fills
a strong steel
volume, with the
pressure rising to P/. (a)
Show
that the fraction y of water that freezes v'/
(b)
-
given by
is
v'/
State explicitly the simplifying assumptions that must be may be written
made
in
order that y
yv
=
y"{0"(Tf
-
rj
-
K"(pf
-
p,-)]
"7-"'f (c)
Calculate
X
lO- 6 deg-'; 12-18 (a.)
Prove
;y
k"
for
=
that, for
i
=
0°C,
(b)
=
-
-5°C, 590 atm; 0" = -0.102 cmVg. f
= -67
v'
a single phase,
(dP\ \dTjs
B
=
aim; /
1
12.2 M(atm-'); v"
_£p_ Tvft
Calculate (6P/dT) s for ice at -3°C, where cm 3/g, and /3 = 158 M(deg->).
c,>
=
2.01
J/g
deg,
1.09
(c)
Ice
is
originally at
— 3°C
and
1
atm.
The
pressure
is
increased
adiabatically until the ice reaches the melting point.
and pressure is this melting point? [Hint: is (dP/dT)s cut a line whose slope
slope
At what temperature At what point does a line whose
is
that of the fusion curve,
—133
atm/dcg?] 12-19 Fig. P12-2 shows a thermodynamic surface for water viewed from the high-temperature end. Consider 1 g of ice in the state i (Pi = 1 atm, 1\ = 273°K). If the ice undergoes an isentropic compression to a state /: (a) Why is the state / in the mixture region? In other words, why does
some (b)
Fig.
PI 2-2
of the ice melt?
Show
Calculate
(d)
that the fraction x of ice that
x
=
sf
is
melted
is
given by
—~
1.09 cm»/g,
12-20
*i
4-.
v'
=
a
-,
A
and breadth
steel b is
a:
when 7> = 272°K, P, =
0'
=
bar
158 Mdeg-'), in the
embedded
aid of an external magnetic (c)
State explicitly the simplifying assumptions that may be written
in order that x
must be made
the bar. (a)
bar v
_ _
c'p(Tj
-
T,)
-
Tjv'fJ'jP,
Mi
-
P$ '
(h-)/
133 atm, c'P
=
=
2.01 J/g
•
deg,
331 J/g.
form of a rectangular parallelepiped of height
in a cake of ice as
field,
shown
a constant force
The whole system is at 0°C. Show that the decrease in temperature
is
AT =
JT{v'
-
bclf
v")
^
in Fig. P12-3.
is
exerted
With
the
downward on
of the ice directly under the
394
Heat and Thermodynamics
Phase Transitions; Liquid and Solid Helium
Choose any point where the slope is (dP/dT)^ in (dv/dT)>, in (c). Near this point, where T = 7\
Iron bar frozen
+
into the ice
T
\dTj x
=
(b)
~v
SW +
ds«>
(a)
(T
+ dT,
s
(i)
=
and
_
1 cf - ef Tvpv>-0M' _
j>
(i)
=
and
v if) at (T,P),
v
(i)
+ dv = i{>
P+dP).
Prove that
_ dT ~ dP_
Ice melts (see Prob. 12-19) under the bar, and all the water thus formed is forced to the top of the bar, where it refreezes. This phenomenon (b)
known as regelation. Heat is therefore liberated above the bar, is conducted through the metal and a layer of water under the metal, and is absorbed by the ice under the layer of water. Show that the speed with which the
j (/) at (T,P),
\dTjx
Prove that
W + dv"> at (b)
and
that:
+
+
In a second-order phase transition, V
(b);
{ds/dT)\ in
show
K (i)x " \df) x
dT
PI 2-3
h
In a second-order phase transition, = i !/> + dsM at (T dT, P dP).
dP
Fig.
t,
cv
w
12-22
(a);
395
fr»
- pv>
k«>
-
'
k«>
is
bar sinks through the
ice
dy_
(These are E/irenfest's equations.) 12-23 For a two-phase system in equilibrium,
P
only; therefore,
is
=
U'Tjv'
dr
-
&\-(P JTJs
v")J
dT
plpbc
For such a system show that
where D"
the overall heat-transfer coefficient of the composite heat-
is
conducting path consisting of the metal and the water
U
Km
layer.
U'
is
given by
Kw
regardless of the type of transition between the phases.
where x m and x w are the thicknesses of metal and water layer, respectively, and A'„, and Kw are their respective thermal conductivities. (c) Assuming the water layer to have a thickness of about 10" 3 cm and a thermal conductivity of about 6 X 10 -3 W/cm deg, and the bar •
to be
1
cm
long, with a
the bar descend is
0.6
12-21
W/cm
A
•
when
and
J=
each equal to 1 mm, with what speed will 10 7 dyn? (The thermal conductivity of steel b
dcg.)
substance undergoes a phase transition at the temperatures
and pressures indicated by the points on the curves
in Fig. 12-26a, b,
and
c.
is
a function of
T
13.
face film
is
397
Special Topics
13-2
expanded isothermally until the area A is very much larger than Then, if the surface tension is a function of the temperature only,
the original value.
the heat transferred
is
SPECIAL TOPICS and the work done
From
the
first
is
W = -J(A-0).
law,
U -v.=(j-r%y, is the energy of the liquid with practically no surface and energy of the liquid with the surface of area A. Hence,
where Uo
U
is
the
Stretched Wire
13-1
U-U
two laws of thermodynamics
T
^dj
t
In the case of an infinitesimal reversible process of a stretched wire, the
(13-3)
dT
A
yield the equation
dS
= dU -
J dL.
The
left-hand
member
with the surface only, Therefore, to obtain any desired equation for a stretched wire,
it is
necessary
merely to choose the corresponding equation for a hydrostatic system and replace P by — J^ and V by L. Thus, the second T dS equation becomes
is
interpreted as the energy per unit area associated the surface energy per unit area. It
i.e.,
is
seen that
—
Uo)/A has the same dimensions as J; that is, ergs per square centimeter equals dynes per centimeter. With the aid of the above equation, the surface energy per unit area of a film may be calculated once the surface
(U
tension has been measured as a function of temperature.
TdS = C 7 dT + T The
<&*
(13-1)
At
the critical temperature, the surface tension of all liquids
surface tension of a pure liquid in equilibrium with
application of this equation to isothermal and adiabatic processes
Table 13-1
Surface Tension, Surface
Energy, and Heat of Vaporization of
Water
Surface Film
13-2
obtain any desired equation for a surface film,
replace
V by A and P by — J in
system. Thus, the
first
T dS
T, it is
necessary merely to
°K
J, dyn/cm
the corresponding equation for a hydrostatic
U-Uo /v,
A ergs/cm
2
kJ/g
equation becomes
T dS = CA dT - T Imagine a small amount of liquid
in the
m
dA.
(13-2)
form of a droplet with a very small
surface film. Suppose that, with the aid of a suitable wire framework, thesur-
273 373 473 523 573 623 647
75.5 51.5 29.0 18.9 9.6 1.6
143 138 129 122 111 80
is
zero.
The
vapor can usually be
is
exemplified by some of the problems at the end of the chapter.
To
its
2.50 2.26 1.94 1.72 1.40 0.89
398
Heat and Thermodynamics
13-3
Special Topics
399
represented by a formula of the type
is the critical temperature, and n is a and 2. For example, in the case of water. Jq = 75.5 dyn/cm, c = 374°C, and n = 1.2. Since dJ/clT is the same as dJ/dt, all the quantities necessary to calculate {U — L'o)/A are at hand. The data for and (U — Ua)/A are plotted in water are given in Table 13-1, and both
where Jo
is
the surface tension at 0°C, lc
constant between
1
t
J
one compares the surface-energy curve with the hcat-of-vaporization curve in Fig. 13-1, it will be seen that both quantities vary with the temperature in a similar manner, both becoming zero at the critical Fig. 13-1. If
temperature. 140 -
13-3
Reversible Cell
120
Equations for a reversible cell composed of solids and liquids only may be obtained from corresponding equations for a chemical system by replacing E
100 -
V by Z and P by —£. Thus,
the
first
T
dS equation becomes
TdS=Cz dT-T^ and
for a saturated reversible cell
(13-4)
dZ; z
whose emf depends on the temperature
only, the equation becomes
T dS = CzdT-
T% dZ. dl
In the case of a reversible isothermal transfer of a quantity of
100
200
300
400
Temperature, Fig. 13-1
of water.
°
500
600
electricity
Z;
—
Zi,
700
K
Temperature variation of surface tension, surface energy, and heal of vaporization
where Zf — Zi is negative when positive electricity is transferred externally from the positive to the negative electrode. During this process, the cell delivers an amount of work II'
If
iNg coulombs of positive
= -€{Z, -
Zi).
electricity arc transferred
from positive to
13-4
Heat and Thermodynamics
400
negative externally, where;
is
Zs
the valence
and
A'p
is
Faraday's constant, then
denoted by AH. Therefore, (13-5)
-Zi=
-/A'f;
n v T d£ K —}^vl ^j.>
whence
In the case of a saturated reversible be shown rigorously
(see Prob.
1
cell in
which gases are
liberated,
it
can
3-20) that
W = /%*?•
and
From
401
Special Topics
the
first
law, the change in internal energy
(13-6) is
,d£
V,-lk--jM*-
T T*Y
volume a process takes place at constant pressure with a negligible enthalpy. For of change equal to the energy is change, the change of internal
When
H=
+ PV dH = dU+ P dV
and
I!
V
dP,
accurate values of the heat of reaction. It
AH A
dU.
number
negative
we may
of cells in
AH
Table
-Hi=
-; A'F
dr)
particular case In order to interpret this change of enthalpy, let us take the from the externally electricity positive of the Daniell cell. The transfer of reaction the accompanied by copper to the zinc electrode is
T,
Reaction
°K
+ CuSO., =
Cu + ZnSO., + 2AgCl = 2Ag + ZnCf. + 2AgCl = 2Ag + CdCIs Pb + 2AgI = 2Ag + PbL> Ag + JH gs Cl = Hg + AgCl Pb + Hg,Cl = 2Hg + PbCl, Pb + 2AgCl = 2Ag+PbCl Zn Zn Cd
=
i
3
-
Zn
+
CuSO., -» Cu
+
transferred,
1
mole
Most
of the values of i.e.,
AH are
an exothermic
negative.
reaction.
2
273 273 298 298 298 298 298
Va-
Ernf
lence,
s, V
2 2 2 2 1
2 2
dS dr mV/dcg
-0.453 -0.210 -0.650 -0.173 +0.338 +0.145 0.49001-0.186
1.0934 1.0171 0.6753 0.2135 0.0455 0.5356
AH
AH (calori-
(electric
method), kJ/mole
metric
method), kJ/mole
-235 -207 -168
-232 -206 -165
-
-
+ -
51.1
5.45 96.0
-105
+ -
51.1
3.77 98.0
-104
ZnSO.,.
each of the initial constituents is formed. The change disappears, and 1 of each of the final of mole enthalpy 1 the of enthalpy in this case is equal to initial constituents each of the 1 mole of of constituents minus the enthalpy heat called the This is of reaction and is pressure. at the same temperature and
When jNw coulombs are
compare values of is shown
interesting to
Reversible Cells {Nt = 96,500 C/mole)
therefore write
H,
13-2.
indicates a rejection of heat,
Table 13-2
For the reversible transfer of jNv coulombs of electricity through a reversible atmospheric cell whose volume does not change appreciably at constant pressure,
is
obtained electrically with those measured calorimetrically. This
for a
whence, tinder the conditions mentioned,
dH =
The important feature of this equation is that it provides a method of measuring the heat of reaction of a chemical reaction without resorting to calorimctry. If the reaction can be made to proceed in an electric cell, all that is necessary is to measure the ernf of the cell as a function of the temperature at constant atmospheric pressure. The heat of reaction is therefore measured with a potentiometer and a thermometer. Both measurements can be made with great accuracy; hence this method yields by far the most
of
mole of each of the final constituents
13-4
Fuel Cell
In the reversible cells listed in Tabic 13-2 and also in the well-known automobile battery, the reactants and the products of reaction are stored within
402
13-5
Heat and Thermodynamics
H,0 +
403
Special Topics
which the following quantities apply: The heat of reaction AH at 1 atm and 298°K is —286 kJ/molc. To maintain isothermal conditions at 298°K, heat Q must be transferred between the cell and its surroundings equal to for
Oxygen-lean
H;
air
— 48
kJ/mole.
W
The work output
(
=
jN&&)
of the fuel cell
is
therefore
Potassium hydroxide electrolyte
jN ¥ £ = Q -
AH
= (-48
£ =
whence Porous metal or
=
+ 286) kJ/mole;
238,000 J/molc 2
X
96,500 C/mole
1.23 V.
carbon electrodes with suitable
The
catalyst
1
actual terminal voltage of this cell
.23
V, not only because of the
Ir
when
The
failure to achieve isothermal conditions.
that the electrodes
Hydrogen supply
— Oxygen-rich
\~~-
'
with
air
have pores of just the right
just the right catalyst to initiate the
and that no impurities poison the Fig.
3-2
1
Schematic diagram of a hydrogen-oxygen fuel
cell.
all
these problems
delivering current
drop within the
is
cell
operation of the size,
is
lower than
but also because of the
that they
be.
cell
requires
impregnated
decomposition of perhydroxyl ions,
surfaces
and stop the reaction. To solve much work remains to be done
a very difficult task, and
toward their solution. the cell
itself.
In a modern development, called a fuel
cell,
the reactants are fed
to the cell in a steady
stream and the products are continually withdrawn,
while the
a steady current to an external load.
cell is
cell delivers
depicted schematically in Fig. 13-2.
The
tion of potassium hydroxide chosen to provide ions.
The
electrolyte
is
A
typical fuel
an aqueous
an abundant supply of hydroxyl
electrodes are of metal or carbon specially baked, impregnated,
and treated
so that multitudinous pores are produced, each of
which
is
a few
microns in diameter. The pores are small enough to allow the electrolyte to
Consider a capacitor consisting of two parallel conducting plates of area A, linear dimensions are large in comparison with their separation /,
whose filled
— Z.
from blowing through. electrode, hydrogen gas is adsorbed in Within the pores of the negative the form of hydrogen atoms, which combine with OH ions to form water and electrons. The electrons move through the electrode to the external circuit while the water mixes with the hydrogen stream and is removed. Within the pores of the positive electrode, adsorbed oxygen molecules combine with - By water and electrons to form hydroxyl ions and perhydroxyl ions O2H action of a suitable catalyst, the perhydroxyl ions decompose into an additional hydroxyl ion and an oxygen atom. The oxygen atoms thus formed unite to form oxygen molecules. The net effect of the reactions within the pores of both electrodes is the reaction
done
and the
air at the other electrode
.
H + 2
|O a = H s O,
with an isotropic
solid or liquid dielectric. If a potential difference
established across the plates, one plate
be drawn in by capillary action and to prevent the hydrogen present at one electrode
Dielectric in a Parallel-plate Capacitor
13-5
solu-
If the
charge of the capacitor
is
is
£
is
given a charge -\-Z and the other
changed by an amount dZ, the work
is
&w = -e dz, the minus sign indicating that an increase of charge (positive dZ) requires work done on the capacitor. Work supplied by an outside source accomplishes two objectives'. (1) an increase in the electric field Eva the space between the two plates, and (2) an increase in the polarization of the dielectric. We therefore seek to find the contribution of each of these terms to the total work
-£ dZ. If
E
is
the electric intensity in the dielectric, then the potential difference
£ =
El.
404
13-5
Heat and Thermodynamics
Also,
Z=
C£, where the capacitance
is
amount dE, and the second
given as
dielectric
by
with
eT
equal to the
dielectric coefficient,
or
by the dielectric
relative permittivity.
AW = -EdP' The
close parallelism
emphasized since the displacement /J
=
is
Therefore,
z = r4 s; 4tt/ and
that required to increase the polarization of the
were no matter present in the space between the term would represent the total work; therefore the work done
dP'. If there
plates, the first
Art'
in
Table
(dielectric only).
between the
electric
(13-9)
and the magnetic
is
Comparison of Magnetic and Electric Systems
Table 13-3
Magnetic
and 4ir
Electric
- V EdD
y - j-?fdB
Total work
have, therefore,
Air
Air
dI47
situations
13-3.
t r E,
*-£* We
405
Special Topics
=
-
= -El^-dD;
Work done
to increase the field
Work done
by the material
- ¥-3fd¥ At -% dM
V Aw
EdE
-EdP'
Aw
and
since the
volume V —
Al,
The thermodynamic coordinates of a dielectric are therefore E, P , and 7', and an equation of state is a relation among them. If the temperature is not
AW = - ^-EdD Aw This expression
is
equivalent to
— £ dZ and
represents the total
on or by the capacitor when the displacement But
D =
(13-7)
(total).
is
too low, a typical equation of state
is
?-(•+*)*
work done
changed by an amount dD.
The thermodynamic
equations appropriate to a dielectric
obtained from those of a chemical system by replacing
P' E+Awy
(13-8)
Thus, the second
T
P by
may
be
— E, and
easily-
VbyP'.
dS equation becomes
where P' is the polarization, i.e., the total electric moment of the dielectric produced either by orientation of elementary dipoles or by the production of
[^,j E dE
induced dipoles. Therefore,
This equation
d
W = - -Aw EdE -
may
EdP'
polarization that
where the
first
term represents the work required
to increase the field
be applied to
(1)
a reversible isothermal change of
field
or (2) a reversible adiabatic change of field. The temperature change accompanying the second process is known as the eleclrocaloric effect. Changes of
by
as pyroeleclric
accompany changes
effects.
of temperature are sometimes
known
406
13-6
Heat and Thermodynamics
&,
L-*
The
Piezoelectric
—-E.P'
first
The
is
easily recognized as the heat capacity at
may be
evaluated most readily with the aid of a
expression on the right
constant tension and
407
Special Topics
field; thus,
other two derivatives
Gibbs function, thus:
piezoelectric
T.S Fig. 13-3
Relation between piezoelectric, thermoelastic, and pyroelectric
G'
effects.
Taking the 13-6
differential of this function,
we
1 .
get
Piezoelectric Effect
substance describable with the aid of the coordinates S, 1-, T undergoes adiabatic temperature changes or isothermal entropy changes
An
= U - TS - JL - EP
dG'
= dU
- T dS -
S dT
- J dh
-L&- E
dP'
-
P' dE;
elastic
when
changed. Such
may
be called thermoelastic. An isotropic dielectric whose coordinates are E, P', T undergoes adiabatic temperature changes or isothermal entropy changes when the electric intensity or the polarization is changed. Such effects may be called pyroelectric. If a system in an electric field undergoes isothermal or adiabatic changes of polarization when the tension is varied, or isothermal or adiabatic changes of tension when the electric intensity is varied, the system is said to be the tension or the length
is
effects
These phenomena are called piezoelectric effects. (The first two syllables are pronounced like "pie" and "ease.") It may be seen from Fig. 1 3-3 that piezoelectric effects are really a combination of thermoelastic and pyroelectric effects. The simplest type of piezoelectric material under the simplest type of stress say, a pure tension in only one direction and in a uniform electric field in that direction is described
piezoelectric.
—
with the aid of the five coordinates J such as rochelle
salt,
2
—
and
realizing that
T dS = dU - J dL- E dP\ we
get
dC = -SdT - LdJ Three
P' dE.
results follow:
1
At
const.
J,
dG:7
=-SdTj-P'dEj,
(g^=0^.
2
At
const. E,
dG'E
= -S dTE -
L dJK
(g^ = \^)
3
At cons, T,
dG>T
= -LdJr -
F dEr (f§)^ = (g^-
(2),
the expression for
,
S>B
,
L, E, P', T. In actual cases of crystals
",
and ammonium dihydrogen phosphate, one is stress and of strain, with each compowell as to the several components of the other, so
quartz,
Using
results (1)
and
T dS
becomes
concerned with several components of nent related to the
field as
that the equations are quite involved.
We limit ourselves here to the simplest
possible situation.
For an infinitesimal reversible process, the
TdS = dU -
first
and second laws
T dS = Gwi dT - T (^) dj \ol /y.E
& dL~ E dP'.
which may be used
to calculate isothermal
temperature changes when J or
T,
entropy
J, E; T, J,
is
a function of three variables, any one of the following
P'; T, L, E; or T, L, P'.
l^\
d'J
Choosing the
-(§),/-
first set,
we
T
J (S \"l
dE,
(13-10)
J.7.E
yield
2"
The
I-
sets:
Result (3)
is
£
entropy changes or adiabatic
or both are changed.
a relation between what are
known
get
'GIL*
3L\
dP^
dE/r.^
d-7/r.
as piezoelectric
coefficients,
408
13-7
Heat and Thermodynamics
since
one expresses the
change of electric field on the length, and on the polarization.
effect of a
Special Topics
409
Thermoelectric Phenomena
13-7
the other the effect of a change of tension
When two
If the other sets of coordinates are used, different functions are appropriate,
the
T
dS equations are different, and shown in Table 13-4.
different piezoelectric coefficients are
obtained, as
dissimilar metals or semiconductors are connected
and
the
junctions held at different temperatures, there arc five phenomena that take place simultaneously: the Seebeck effect, the Joule effect, the Fourier effect, the Peltier effect, and the
Thomson
effect.
Let us consider each of these
effects briefly.
Table 13-4
Alternate Expressions for the Piezoelectric Effect
1
Seebeck
.
Appropriate thermodynamic function, (2) T dS equation, and relation between piezoelectric coefficients
We call
C
junction
(1)
Coordinates (3)
Gibbs function:
the
J, E
TdS= ,
dT+T (H)
G>, B
= V - TS - JL -
^
.
-
Piezoelectric coefficients:
d? + T (%)
EP'.
I
1
Thermoelastic Gibbs function:
dE.
-r-75.
(
C" =
1
U — TS —
is
known is
&
s
T dS = P'
Cj, P
Piezoelectric coefficients:
(
-r-r-,
\OP
Pyroelectric Gibbs function:
=
1
/T,J>
C" = U -
-r-=.
1
effect.
T of the
test
The Seebeck
and the other at temperaemf £,\n in the
existence of a thermal
When
kept constant, the thermal
the temperature of the reference
cmf
is
found to be a function of the
junction. This fact enables the thermocouple to be 1.
effect arises
is
a net motion of the charge carriers as though they were driven by a
nonelectrostatic dP>. is
field.
The
line integral of this field
around the thermocouple
the Seebeck emf.
For a given Tr, Sab
—
The
contact with different heat reservoirs. test junction,
diffusion of the charge carriers takes place at the junctions at different rates.
&L.
dj + T (^)
.dT+T (^-)
as the Seebeck
T the
used as a thermometer, as described in Chap.
There T,
at
from the fact that the density of charge carriers (electrons in a metal) differs from one conductor to another and depends on the temperature. When two different conductors are connected to form two junctions and the two junctions are maintained at different temperatures,
di"\ = /
/^A —,
its junctions in
warmer junction
temperature
T,
has
ture Tit the reference junction. circuit
Piezoelectric
In Fig. 13-4 a thermocouple consisting of two different
effect
A and B
conductors
stant value, the relation 1
is
a function of T.
between £ais and
If
7" is
Tr
is
changed
to another con-
the same except for an additive
\0~S/T,P'
constant. It follows therefore that the value of d£Aii /dT is independent of
TS -
and depends only on the nature of A and B and upon T. The derivative dtSAii/dT, at any value of Tr, is known as the thermoelectric power of the
EP'.
thermocouple. T, L,
E Piezoelectric coefficients:
—
(
-—-
=
1
\dLjr.,:
1
-——
I
)
\dEj T.L
Temperature
Mclmholtz function:
T, L, P'
T dS =
C,., P '
dT
.4
= 6'—
+ T (H)
Piezoelectric cocflicicnts:
dL+T ( ||
-r— ) \(IP /r.L I
T„
TS.
=
(
tt
)
I
dP'.
1
\0LJr,F' Fig.
1
3-4
Thermocouple of conductors
A
and
B
with junctions at
T and
Tr.
Tr
1 410
13-8
Heat and Thermodynamics
411
Special Topics
thermal cinf £.\n is not balanced by an external emf, a current I exists whose value may be adjusted by varying the external cmf. As long as a current exists, electrical energy of amount PR is dissipated within the circuit. This is the well-known Joule effect.
The conduction of heat along the wires of a thermo5. Thomson effect couple carrying no current gives rise to a uniform temperature distribution in each wire. If a current exists, the temperature distribution in each wire
Imagine a thermocouple whose junctions arc at tem3. Fourier effect peratures 1\ and 'A, respectively (7\ > T-<), and which has been broken at one point, its two ends being maintained at some intermediate temperature T by means of an insulating reservoir. There is no thermoelectric current, and therefore no Joule effect; but heat is lost by the reservoir at T\, conducted along both wires, and gained by the reservoir at 7'2 with no net gain or loss to the reservoir at T. The wires can be imagined to be suitably lagged so that
tional
2.
Joule
If the
effect
,
is no appreciable lateral transfer phenomenon of heat conduction The
there
of heat across the surfaces of the wires. is
called the Fourier
The
temperature
its initial
altered
extracted laterally distribution
To measure
is
From
The
junction
itself
is
tracting the F-R loss
which heat
and correcting
is
transferred
for the
is
The
rate at
power
first
which
dT
used as a sort of
and
Peltier heat
is
reversed, with the
c
d
The
is
transferred ir/.
The
is
proportional to the
quantity
equal to the heat transferred
same but
reversible.
ir is
when
called the
unit quan-
rate at
which Thomson heat
is
equal to al dT, where a
When
the direction of the current
magnitude remaining the same, the
and
irreversible
B
called the Thomson
coefficient.
Peltier heat
Lord Kelvin was the first to realize that the two phenomena, the Joule effect and the conduction of heat, could
not be eliminated by merely choosing wires of proper dimensions. For,
made very
thin in order to cut
whereas
is
resistance increases;
is
electric resistance, the heat
depends on the temperature and the materials
the accepted convention to regard ttah as positive to
is
application of thermodynamics to the thermocouple has had a long
when an
elec-
causes an absorption of heat by the junction.
if
down
the wires are
if
the
heat conduction, the electric
made
thick to cut
conduction increases. In spite of
down
this,
the
Kelvin
assumed that the irreversible effects could be ignored on the ground that they seemed to be independent of the reversible Peltier and Thomson effects.
junction.
A
transferred into a small region
interesting history.
through a thermocouple
current from
is
Simultaneous Electric and Heat Currents in a Conductor
13-8
By considering
It is
conducted. Since the Joule effect
opposite to the direction of the temperature gradient (low to high
of a junction, being independent of the temperature of the other
tric
is
temperature) causes an absorption of heat by the conductor.
wires are
in the opposite direction.
Peltier coefficient
it is
calculated. After sub-
of electricity traverses the junction.
tity
The the
is
any one wire,
conducted heat, which was deter-
of the current or equal to
Peltier coefficient
b
Peltier heat
at a small region of
The Thomson heat is reversible. The Thomson coefficient depends on the material of the wire and on the mean temperature of the small region under consideration. It is the accepted convention to regard a as positive when a current
The The
Thomson heat
of a wire carrying a current I and supporting a temperature difference
called
mined from previous experiments, the Peltier heat is finally obtained. Extensive measurements have yielded the following results: a
the
can be calculated and the conducted heat is known from previous experiments, the Thomson heat can be obtained. The following conclusions may be drawn from such measurements:
the rate of change of temperature and the heat capacity
of the junction, the rate at
restore the initial temperature
known temperature gradient in the region and to pass a known current either up or down the gradient. The rate at which Thomson heat is transferred is equal to the rate at which electrical energy is
Peltier effect takes place
perature of the junction changes. calorimeter.
al all places along the wires to
called the Thomson heal.
necessary to produce a
whether the current is provided by an outside source or is generated by the thermocouple itself. The Peltier heat is measured by creating a known current in a junction initially at a known temperature and measuring the rate at which the temthe Peltier heat.
is
dissipated minus the rate at which heat
effect.
Imagine a thermocouple with its junctions at the same means of an outside battery, a current is produced in the temperature. If, by the temperatures of the junctions are changed by an amount thermocouple, entirely the Joule effect. This additional temperature change that is not due to Allowing is the Peltier effect. for the Joule effect, the heat that must be either 4. Peltier effect
supplied or extracted to restore a junction to
by an amount that is not entirely due to the Joule effect. This addichange in the temperature distribution is called the Thomson effect. Allowing for the Joule effect, the heat that must be either supplied or is
the purely reversible transfer of unit quantity of electricity circuit,
Kelvin
set the
sum
of all the entropy changes
equal to zero and derived relations which have been amply checked and
which are undoubtedly
correct.
The stubborn
fact remains,
however, that
412
Heat and Thermodynamics
the Seebeck, Peltier,
13-9
and Thomson
effects are inextricably linked
with the
irreversible effects.
Attempts to resolve these difficulties were made by Bridgman, by Tolman Fine, and by Mcixner, but the results were not entirely free of objection. The solution is to be found in the macroscopic treatment of irreversible coupled flows developed by Onsager, which was introduced briefly in Chap.
and
The
following
The
Z. u and £as have simple interpretations in terms of thermal electrical conductivity, respectively. The quantities Ln and conductivity coefficients. They represent the effect of a potential are coupling and Z.21 difference on an entropy current and the effect of a temperature difference on an electric current, respectively. Onsager proved by means of the micro-
coefficients
point of view that
scopic
a simplified version of Onsager's method, based on the work of H. B. Callen. 9.
A
is
AT established
small temperature difference
thermal equilibrium and gives rise to a heat current I Since a cool reservoir Q end of the wire is gaining entropy from the wire at a greater rate than .
that at
which
a
warmer is
reservoir at the other
being produced
dS dr
_ ~
in the
AT _ A> -ji ~
end
I'ii
is
losing
it to
the wire
we
which If
is
AT
now known is
set
A
is
I-I
(13-13)
as Onsager's reciprocal relation.
equal to zero in both Eqs. (13-11) and (13-12), and then the
we
equations are divided,
get
wire at a rate
In
AT *s
~f
I /iJ'=0
'
Also, in the absence of
where Is
—
across a wire disturbs the
at one
say that entropy
413
Special Topics
an
Z.22'
electric current,
Eq. (13-12) provides the relation
the entropy current equal to Iq/T.
small potential difference Ac? established across a wire disturbs the
and gives rise to an electric current /. Since a reservoir at temperature T which maintains the wire at a uniform temperature is gaining entropy and there is no entropy input to the wire, we say that entropy is being produced in the wire at a rate
Acf
electrical equilibrium
A77r=o
Since Onsager's reciprocal relation provides that
_
L12
A£
dS
Lll
T'
dr
£22
We
therefore have
two
Lu =
£sb, let us
write
Lii L*l
different physical interpretations of the quantity
e:
When exist
both a temperature difference AT and a potential difference Ac? across the wire, the rate of entropy production is the sum, or
dS dr
_/A£\ \ATJ I=
departure from equilibrium is not too great, the entropy and electricity flow are coupled in a simple manner, both flows depending linearly on both If
AT/T
and
I /AT=0
AT Ad? = Is— +1 T
Ac?/'/', thus:
e may be regarded as the entropy current per unit temperature, or as the change of potential difference per
called the Seebeck
h
m
j
I
_ -
.
and the nature
Ac?
Because of
coefficient
electric current
at a given
unit change of temperature
this latter physical interpretation, t is
of a substance
and
is
a function of the temperature
of the substance.
(13-11)
13-9
and
\dTJi,
„
Thus,
at zero electric current.
AT
(13-14)
= _(d£\
AT
^21 -jT
+
Seebeck and Peltier Effects
Ac?
Ln-1 Iff'
(13-12)
Consider the thermocouple depicted the conductors
A and B
is
in Fig. 13-5.
The
maintained at the temperature
test
T
junction
e
of
while the two
414
Heat and Thermodynamics
13-9
Temperature
Temperature
T
T "
'
Temperature
To
T
potentiometer
J
415
Special Topics
A thermocouple consisting of two metals is well-suited as a thermometer. At low temperatures a highly sensitive potentiometer must be used, and at high temperatures metals with high melting points are essential. Junctions may be made with very low heat capacity, so that they react quickly to changes of temperature. As a result, thermometry has been the main practical application of thermoelectricity since its discovery by Seebeck in 1 821 Since .
/T Fig.
1
theme
3-5 to
Thermocouple consisting of wires
the binding posts of
a potentiometer.
A
and B connected to copper wires
C and C,
Ts
and
each with copper, are maintained at temperature TR The two copper wires marked C arc connected to the brass binding posts of a potentiometer, forming two more junctions, each of which is at room temperature Ta The potentiometer is supposed to
and
d,
(usually that of
an
c
.
=
and
in
(fA
— «c) dT —
follows that
£ab =
ice bath).
be balanced, so that / The second relation
—
(See Fig. 1-11.) it
junctions
m dT,
0u
£a —
£/, is
the Seebeck
£ab = £ac
or
cmf SabTherefore,
Eq. (13-14), namely,
rr J Tb
emf
if
C
is
—
rr (t B
J 2-s
(13-16)
chosen to be platinum, two tables of values of thermal
,
£a — £ = c
"AIP Handbook"
for
electric current traverses a
junction of two different conductors
uniform temperature, the heat that must be supplied or withdrawn,
at a
over and above the Joule heat, to keep the junction at a constant temperature is
what we have already described above
as the Peltier heat. This
may
be
calculated in terms of Seebeck coefficients as follows. Consider the thermo-
tc dT,
j
in the
different metals.
When an
—
dT,
— one for metal A and platinum and the other for metal B and platinum
many be applied to each of the conductors A, B, and C the Seebeck coefficients being t A , t B and «c Integrating from one end of each wire to the other, we have
ec)
£bc-
— enable one to find £ab- Such tables are given may
—
/T
unction
j
e
of Fig. 13-5, depicted in greater detail in Fig. 13-6.
the junction
is
at a uniform temperature, there
is
Even though
a heat current Iqa into the
junction and a different heat current Iqb out of the junction, both carried
£,-£* = H'tBdT,
from A
-*-£
and
When
along with the electric current
tcdT.
these equations are added, the
the right side, the
first
and
last
left
side
becomes
to B.
The
PRj (where
with
/,
which
difference between Iqa
Ay
is
is
taken arbitrarily in the direction
and Iqb must be
transferred, along
the resistance of the junction), in order to keep the
S a — £ b = £ A n On .
terms cancel, so that
Temperature
r £ab =
If the
Seebeck
[
T
L
{tA
—
tB)
dT.
It
(13-15)
A and B are known as functions of T, the cmf made by joining A and B and holding the junctions at
coefficients of
of the thermocouple
TK and T could be calculated by performing the integration indicated in Eq. (13-15).
Fig.
The
1
3-6
A and B and PRj.
Thermojunction of conductors
Peltier heat is the difference between /q
at
a uniform constant temperature T.
416
Heat and Thermodynamics
temperature constant. Calling
13-10
this heat
current l'Q
,
we have
It supports a
417
Special Topics
temperature difference AT.
supports a potential difference Ac?yi = t A AT. It carries an electric current / = Ac£ A /AR A It
=
1q
By
definition, the Peltier heat
= Iq—
ttabI
Since the Is
+
I*&j
=
first
Iq/T,
it
part of Eq.
{Iqa)at-o
~
(Iqu)*t-o.
.
A A
is
I'Rj
=
(Iqa)at-u
(13-14) defines
e
— to
heat current Iq enters
it.
different heat current Iq
+ A/q
leaves
it.
(Iqii)&t-o.
be
(Ia /l)&T-a
and
sir
follows that
=
Iqa
IT( A
and
IQH
=
JHTtg.
To a balanced
The
Peltier coefficient therefore
T/ifl
Thomson
13-10
Effect
becomes
—
T(e A
potentiometer
—
e B ).
(13-17)
and Kelvin Equations To generator,
Consider the thermocouple of Fig. 13-7 consisting of wires A and the test junction at temperature T' and the reference junction at T,
B {.
term, voltage
with
£A
If the
thermocouple is on open circuit or is connected to a potentiometer that is balanced, there will be no electric current, but there will be a heat current and a temperature distribution throughout the wires. Suppose wire A is now placed in contact point, so that
situation
at each point with a heat reservoir of the
there
is
same temperature as
no heat exchange between A and
that
the reservoirs. This
depicted in Fig. 13-7a, but only one of the reservoirs is shown— in contact with a small portion of wire A, across which there is a temperature difference A 7" and a potential difference is
the one at temperature
=
Ac?,!
tA
T
AT. The heat current
Iq entering this portion
is
Open
equal to the heat
circuit
current leaving the portion.
Now
suppose that the potentiometer circuit is opened and an outside is connected to wire A (with the aid of connecting wires of the same material as A) and that the terminal voltage of the generator is adjusted until T' it has exactly the value £ A =
:
.
.
.
,
wire/1 will be altered both by Joulean dissipation and by the Thomson eflect. Referring to Fig. 13-76, we see that the small portion of wire A in contact with the reservoir at T has the following characteristics:
Fig. 13-7
wire
A
(a) In the
at temperature
the heat transfer between
to
AIQ
=
absence of an
T and its heat
electric current, there is
reservoir at
a portion of wire
PAR- IaA AT.
A
T.
(b)
no heal flow between a portion of
In the presence of an
at temperature
electric current,
T and its heat reservoir at T is equal
418
Heat and Thermodynamics
1
It exchanges heat P ARA (Joule effect) and IaA AT (Thomson with the heat reservoir in contact with it. Therefore,
Q
= IA£A
M
As
Q
before, Iq
=
=
Ie A
-h
AT-
IcA
we differentiate Eq. we get dS A n/dT =
stant,
(13-18)
It follows
is
ITAeA
this
TR
con-
with Eq. (13-17), we see
_ d£ A n dT
(13-21)
''
(13-19)
.
If
Kelvin's first equation.
is
we
d£Ali/dT
substitute the value
o~
AT~
Comparing
given by
from Eqs. (13-18) and (13-19) that -IffA
e,s .
T which
+ ITAeA
—
eA
jus
AT.
JTt A , and therefore a small change
AIQ = h A AT
(13-15) with respect to T, while holding
that
AT,
A
419
Special Topics
(,r,i
PARA - laA AT
=
1
£ AH may be obtained by integration, tA b/ T without further calculation, and —
M or
effect)
3-1
A
for tA
— Oil T
—
«« in
Eq- (13-20),
we
d 2£,
get
(13-22)
dT 2
,
r de A
,
or
which
is
Kelvin's second equation.
The Kelvin equations have been
often verified
by experiment. Similarly, at a point on wire
B
that
Til
=
is
at temperature T,
Thermoelectric Refrigeration
13-11
—T dtn dT .
Metal thermocouples arc not suited to extract heat by the
and
finally,
A
—
ob
= — T -yss (eA —
««)
•
(13-20)
because the difference in the Seebeck coefficients
Let us collect the three equations for the three reversible thermoelectric
effect
(e A
—
en)
—
ec).
Q - 270°K(« A
dT.
(13-15)
esb
X
20
AX
10.
—
«Bi
=
109 /xV/deg,
in
converting a drop of water into
ice.
The
large currents
needed, the large amount of Fourier conduction, and the large Joulean heat A
—
0~B
= — T -pr,
(e A
—
dissipation resulted in a very low coefficient of performance, so that 6fi).
all
expressed in terms of the difference of the
coefficients, so that if
tA
and
es are
known
no com-
(13-20)
mercial Peltier refrigerator appeared on the market for over a century after
The major breakthrough occurred when semiconducting compounds were found to have large Seebeck coefficients, good electrical conductivity, and poor thermal conductivity. Thus, a thermocouple of Leoz's experiment.
Seebeck
tB )
(13-17)
effect
Notice that they are
-
If the thermojunctions were of Cu and Fe, «c u — efB = 13.7 jxV/deg, and Q would be only 0.74 W. In 1838, Lenz tried thermocouples of Sb and Bi, with
and succeeded Thomson
ice
equal to
Peltier effect
tA b — T(e A
Peltier effect
To produce
T would
effects:
SA n = L
so small.
have to be about 270°K. Using a current of about 20 A and about 10 thermocouples in series, we could obtain a Peltier heat current
cubes,
Seebeck
is
two
as functions of T, then
420
13-12
Heat and Thermodynamics Fins to ambient a
Metal connectors
_
temperature, the distribution of frequencies and the energy of each frequency band arc independent of the nature of the walls and depend only on the temperature and the volume. The quantum picture of the radiation in the cavity is that of a system of photons with many different frequencies, but all moving with a constant speed c and completely independent of one another. All photons of the same frequency constitute a set of particles that satisfies the conditions of indislinguishabilily and independence (absence of interaction) better than any system of molecules or electrons. The most striking peculiarity of a system of photons is that the total number photons of of all frequencies is not constant. As equilibrium is approached, photons absorbed by atoms, and other photons perhaps of different frequencies are reemitted. Therefore, during the approach to equilibrium, although the are
•Electrical
/'///>/''>/!'/ ;;;/ //777i
/ /
insulation
///////
(thin mica)
total energy of the
The number
photons
may remain
quantum
of
Fig, 13-*
with the translational kinetic enormously larger than the number of molecules that can occupy these states. Very few of the available states can be occupied at one time, and when a state is occupied, it is most unlikely that it contains more than one molecule. With electrons, however, the num-
chamber
Schematic diagram of a thermoelectric refrigerator.
—
Such a refrigerator
in Peltier refrigeration since
Table 13-5
is
depicted schematically in Fig. 13-8. Progress
1821
is
shown
in
Table
is
quantum states and the number of electrons arc comparable. The operation of the Pauli exclusion principle, which limits each state to at most two electrons, provides therefore the complete filling of all low-lying states ber of
p-type Bi 2 Te 3 joined to n-type Bi 2Te 3 has a value of i «„ equal to 423 /xV/ A deg, so that 10 couples at 270°K with a current of 20 A give rise to a Peltier heat current of 23 W. These values make a Peltier refrigerator economically feasible.
constant, the total number does not.
states associated
energy of the molecules of a gas
Fins to cold
421
Special Topics
13-5.
up
known
to a level
as the Fermi
hold lor photons, so that
Peltier Refrigeration
level.
With photons the
situation
is
still
number of photons is not constant, there is no clearcut relation between the. number of photons of frequency v and the number of quantum states g, available to these photons. The Pauli principle does not different. Since the total
there
may
be any
number of photons
in the
same energy
slate.
The No. of
T,
Thermojunction
(A
— i,
°K
A
couples
Fe-Cu (1821)
270
20
10
Sb-Bi (1838)
270
20
10
109
Bi 2Te 3 (/0-Bi 2Te 3 («)
270
20
10
423
,
,
Peltier tfi,
MV/deg ,
13.7
,
heat rate,
/.
is
5.9 23
where the
are large integers. If the
n's
is
in a
n's
a uniform
+ »*),
»;
are taken to be rectangular coordi-
is
with a radius this
one value
t,
of the
the surface of the sphere
+ nS =
radiation in equilibrium with the interior walls of an
called blackbody radiation. If the walls are at
+
nates, the locus of all points in this space corresponding to
Properties of a System of Photons
evacuated cavity
(nl
8m D
0.74
energy
The electromagnetic
moving
given by the familiar expression in Eq. (10-1),
w
(1963)
13-12
translational kinetic energy of a molecule or an electron
cubical box of side
r
=
2Z.(2me,')*/7'-
space contains one
Since the
quantum
state.
h! n's
are integers, a unit
The number
of
quantum
volume states
g,-
in
da
422
Ilcat
and Thermodynamics
13-13
corresponding to an energy lying between spherical shell of radius r. Thus, g,-
da
=
=
%
2
4irr
•
and
e,
+
da
is
I the
volume of a
dr
AI
|
t,
2
4r -^
~ (&&&* da 71
2mti
is a speck of coal with negligible heat capacity. As the coal partiand reemits photons, any resulting change of energy will be so absorbs cle small that the total energy of the radiation may be regarded as constant. We require an expression for the number of ways in which N, photons may be
cavity there
among g„ quantum states with no restriction on the number of any one state. Let us represent identical indistinguishable photons by crosses, XXX; and quantum states by the spacing between vertical lines, XXX,X'XX, etc. A typical distribution of photons among states would
distributed
photons
=
2vL
423
Special Topics
'&<*>
in
thus be
This result appeared in Prob. 10-2 and also in the discussion of in metals given in Eq. (11-48). This equation must also hold for photons, but form, since photons have zero rest mass. it
in
of
terms of momentum/', where
quantum
states
Then, since L?
=
=
e
corresponding to
To apply
not in an appropriate
it is it
to photons, let us rewrite
p-/2m. hetg p dp represent the number
momenta
lying between p
XXX|X,XX|XXXXX|XX|XXX.
free electrons
and p
+ dp.
This symbolism represents three photons
two
\i
dp 4jtF
p
1
photons and g v states is A> g, — T N, gv — 1 symbols is (A „ g„
+
(2m)'
1m
+
To
large.
2pdp
take care of
h
terms of frequency
much
too
we
o„ K
!,
but this expression
is
all
the frequencies, the
=
n
(13-24)
N,\g,\
v
and
v
+ dv. Then,
is
the probability appropriate to the Bose-Eins/ein
= U=
const.
In
Finally, doubling this result so as to include photons with
and using
= &VF v"
We
Que = S
have In (If,
+ g,) !
2
In A',!
- S In £,!;
=
x In x
—
both kinds of
polarization,
dv.
Stirling's
In Q,
(13-23)
approximation, In
=
2(A',
+ g,)
- 2A\ In After
Bose-Einstein Statistics Applied to Photons
Consider an evacuated cavity with perfectly reflecting walls containing a total of photons, of which A",, have a frequency v. Suppose that w ithin the ;
statistics.
To find the distribution of photons among the frequencies at equilibrium, we render In Slim a maximum subject to only one restriction, namely, that ~.\\hi>
N
1)
hv
Let g, dv be the number of quantum states between
13-13
—
that the photons are indistinguishable,
v
This
g r dv
the. fact
by replacing (g, — 1) by g,', therefore, summing over thermodynamic probability becomes result in
in the second,
must divide by N,l; and to show that the lines representing partitions are indistinguishable, we must divide by (g„ — 1) !. Very little error is introduced
,
Having gotten rid of m, let us now express our by using the dc Broglie equation,
one
number of symbols needed to represent N, 1. The total number of permutations of
sent six states; therefore the total
+
V,
in the first state,
Notice that only five vertical lines arc needed to repre-
in the third, etc.
some In
and
!
(N,
In
.\\
-
a
+ g,) -
+ m, -
x,
+ g,) 2£, In g, + 2g 2(N,
r.
cancellation,
®be =
U=
2(:Y,
+ g.)
ZXJiv.
In
(N„
+g y)
2A', In A',
- Sgr In Sn
424
13-14
Heat and Thermodynamics
Applying the method of Lagrange d or
d
In
In
8BE =
=
QB e
J
+ In
[1
we
multipliers,
+ g,)] 4N, - T
(.\\
*._+*.-, l*H
=
dN.
as always. Substituting into Eq. (13-27) the values of g*
set
+ In
(1
.V,.)
dN,
0,
=
(13-25)
c
If the
by U„ with one
restriction,
energy lying
dv,
in the
y
=
dv
this last
—
equation by
0.
and adding, we
+ jf) -
(l
0kv
=
get
1
v
and
v
+ dv
is
denoted
dv,
SirVh
v 3 dv kT e>"'
3
-
1
get
The 1"
-
N,hv
c
Multiplying
wc
0,
v2
J"> kT
frequency range between
U = Mv dN, =
3
and
then
namely, dll
= 8VK
N,
0,
425
Special Topics
spectral energy density u, is
defined to be U„/V, so that finally
(13-26)
0,
8wh
+ ^ Ml =
1
fBhv
=
N,
and
?.71 (13-2
which
we consider the system of photons
13-14
«**'-
To discover the physical significance of as a
/J,
c
'
N,
=
and from Eq.
k
d
In
=
Sue
we
XNJiv and
dl!
=
=
as the
f)
|f]
= M*.
Optical Pyrometer
volume
is
thermometry is a blackbody at between liquid and solid gold, called the gold point (1336°K), originally determined with the aid of a gas thermometer. The construction of the gold-point blackbody used at the U.S. National Bureau of Standards is shown in Fig. 13-9. The energy needed to melt the gold is provided by an electric current in molybdenum windings. To measure temperatures above the range of thermocouples and resistance thermometers, an optical pyrometer is used. As shown in Fig. 13-10, it consists tool of high-temperature
the temperature of equilibrium
essentially of a telescope IT in the tube of which
kE0hv dN,.
X/iu
dS
Inasmuch
T In M +
get
dS
U=
+
=
(13-28)
'
1
.
(13-26), at equilibrium,
M
But since
k
-
Planck's radiation equation.
The fundamental
=
h '' kT
k In. 0„ F Then, from
Eq. (13-25),
dS
e
1
thermodynamic system with an entropy S
Substituting,
is
3
and
lamp bulb
L.
When
is
mounted
the pyrometer
is
a filter of red glass
directed toward a
furnace, an observer looking through the telescope sees the dark
lamp filament against the bright background of the furnace. The lamp filament is connected to a battery B and a rheostat A'. By turning the rheostat knob, the current in the filament and hence its brightness may be gradually
dN„
—
h$ dU.
constant, dS/dll
a small electric
=
—
increased until the brightness of the filament just matches the brightness of
1/T, and
the background.
From previous calibration of the instrument at known temammeter A in the circuit may be marked to read
peratures, the scale of the
t
=
the
FI-
unknown temperature
directly. Since
no part of the instrument needs
to
426
Heat and Thermodynamics
13-14
Depth of cavity
Quartz cylinders
Radius
of cavity
= 72
blackbody.
opining
When
this
is
Special Topics
427
the case,
«x(7au)
KxCO
2k'
and using the Planck equation, Alumina cone «j/X!TA«
_
1
(13-30) 271
^-Argon
where is
c»
=
hc/k
In v
Quartz wool
I
Graphite
Silocel
Gold
this
7au = 1336°K. Solving Eq. unknown blackbody.
way, a second optical pyrometer
may
(13-30) for
T gives
the tem-
be calibrated so that a value
lamp current at which the filament disappears tells the temperature of a blackbody at which the pyrometer is sighted. If the red filter transmitted only a very narrow band of wavelengths and if all unknown bodies whose temperatures must be measured were blackbodies, everything would be fine and no further physical ideas or instruments would be needed. This is not the case, however. Elaborate and difficult corrections must be made in order to take into account the finite bandwidth needed for a visual match, as well as
of the
powder
Blackbody at melting point of gold; furnace as used at U.S. National Bureau of
Fig. 13-9
cm/°K. The wavelength transmitted by the filter where X = 6.5 X 10 -6 cm (6500 A) and, as
before,
perature of the
///
1.4388
often chosen in the red region,
mentioned
'
=
Standards.
the departure from blackbody conditions. These details, however, are for the
come
into contact with the hot body, the optical
pyrometer
may
be used at
expert.
temperatures above the melting points of many metals. To understand how an optical pyrometer is calibrated at temperatures
above the gold point, let us rewrite the Planck equation, Eq. (13-28), in terms of wavelength instead of frequency. Let «x dX be the energy density of d\. Then, since v = c/X the radiation within the wavelength range X and X
+
and
\dv\
=
(
U\
dX
=
X^dX Sirhc -r-
hc/UT
Suppose that an optical pyrometer noted.
Now
sighted on a blackbody at the gold
is
point and that the current necessary to
(13-29)
[
make
the lamp filament disappear
is
T
is
suppose that another blackbody at a higher temperature
sighted with the
same pyrometer
at the
same current
setting,
equipped with a sector-shaped opening whose angle
8
but through a disk
can be varied at
sectored disk rotates rapidly,
it
n
will.
same fraction 6/2t of the radiation of all wavelengths. As 6 is made smaller and smaller, the radiation gets dimmer and dimmer until it is the same as that from the gold-point
As the
transmits the
Fig. 13-10
Main features
of optical pyrometer.
428
Heat and Thermodynamics 13-15
The Laws
13-15
The Planck
Wien and
of
is
y = l-e-^.
equation, expressed in terms of wavelength, takes the form
d\
plotted in Fig.
1
3-1
—
\-*d\ 8t/w ghclMT
-J
0.6
can be seen
wavelength X„, at which the a maximum. To see how X m depends on '/', we note that when x 6 («* c '** r — 1) is a minimum. Thus,
that, the higher the temperature, the smaller the
energy density «x is a
is
maximum
» ah 5X 4(^c/x t r
-
[x«(«*«/x*r
1)
+
-
1)]
\yciUT
(
=
/x = 4.S
n a
_
for seven different temperatures. It
1
2
0,
_ ill J±. _L = kT\-
)
/
/
0.4
y = x/5
to*
o. 2
1
Fig. 13-12
is
—
1
two curves, y
=
—
1
point of intersection. This
e
whence
Wavelength Fig. 13-11
in
20,000
angstroms
is
*~*
and j> done in Fig.
=
is obtained most simply by and noting the value x m at the where the point of intersection
at/5,
13-12,
xm
r=
he
_
6.63
4.96A
X
10~ 27 erg
4.96 \mT
=
4-96,
X
1.38
0.290
cm
s
X
X
3
Planck equation.
X
=
0.290
X
10 s
'
deg,
which is known as Wien's law (pronounced "Veen"). Angstrom units, where 1 A = 1 -8 cm, \m T
10 20 cm/s
10-" erg/dcg
Variation of spectral energy density of blackbody radiation at different tempera-
tures, according to
X
x/5.
found to 4.96. Hence,
or
15.000
=
1
5UT'
%=
10,000
tf"
5
a transcendental equation whose solution
plotting
is
4
X must therefore satisfy the equation
1
This
3
Graphical solution of the equation
The maximum
5000
429
y 1.0
u\
This
of Stefan-Boltzmann
Special Topics
A -deg.
(13-31)
If X
is
measured
in
13-16
430
T=
At as
Heat and Thermodynamics
=
2000°K, X m
shown
T=
14,500 A; whereas at
6000°K,
X,„
w
5000 A,
v
8irk
in
terms of frequency,
u
total energy density
The quantum
is
/*>
=
,
u„dv
1
due
Jo
c
/" =
/
s
entirely to
its
very similar to
is
we imagine photons of many different fresample N, photons of frequency v. Each photon is
F,
A
c.
speed, so that mc 2
=
photon has no
rest mass. Its
mass
hv, or
v 3 dv
f«
= UtIi z—rh
volume
quencies and pick out as a has an energy hv and a constant speed is
u
picture of the radiation in an enclosure
that of a gas. If the
3
(13-28)
The
Radiation Pressure; Blackbody Radiation as a Thermodynamic System
13-16
in Fig. 13-11.
Let us go back to Planck's equation expressed
431
Special Topics
Jo
e
, hv ' kT
—
m =
1
hv
—„• c-
If
we
let
x
=
hv/kT, then u
Stt^T 4
=
In the kinetic theory of an ideal gas that was treated in Chap.
x *d. X* dx
/•== r«=
1 e
Jo
—
N molecules on a wall of an enclosure of volume
exerted by
1
6,
the pressure
V was shown
to
be [see Eq. (6-8)]
was mentioned
It
11-13, in evaluating the heat capacity of a
in Art.
T approaches
crystal in the limit as
zero, that
/« x 3 dx
„,„,,. J -
«*-l =
6
sw 90
where (w 2 ) was the average of the square of the speed. Since speed
X4 = 15"
frequency
8rrW _,
Hence
"
- 15AV ^
c,
we v
P .!&*:«. 3 V c2 *
NJiv
1
Let us accept without proof that the radiant einittance /8b of the walls of a cavity in which blackbody radiation is in equilibrium is connected with the total energy density u by the following relation:
®* =
4
* 15AP
/eB
is
=
We
=
P
and written
S
/t
4
is
in
u (13-33)
2 15
X
5.67
X
(3.14) »
(6.63)
X
X
3
X
(1.38)
10- 81 (erg
X
4
•
s)
10- 64 (erg/deg) 4 3
X
9
X
10 2l) (cm/s) 2
cm-f deg 4 ,
s
•
the temperature of the radiation, for convenience. Strictly speaking,
it is
not the
radiation to which the temperature applies, but the matter in equilibrium
C g
10- 6
Blackbody radiation is therefore completely specified by the pressure of the radiation, the volume of the radiation, and the temperature of the walls with which the radiation is in equilibrium. This temperature is sometimes called
.
with the radiation.
,
•
Since blackbody radiation
which
=
have 27T
=~
3*'
r*
the famous Stefan-Boltzmann law mentioned in Chap. 4
=
and, for the total pressure,
or
This
V
3
P B
photons have
the expression
(13-32)
•
all
get for the contribution to the pressure P„ of those photons with
good agreement with experimental
values.
may
is
described by the coordinates P, V,
and
T,
it
be treated as a hydrostatic system, and any one of the equations derived
432 in
Heat and Thermodynamics
Chap. 11 may be applied
to
A
it.
particularly important result can be
obtained from the energy equation
U=
Vu and
P=
a/3,
TdS = ibT*dV,
where
u
is
T
a function of
only, the energy
T
du
u
dT
3
.
we
get
In u
or
u
=
=
In
T*
+
In b,
is
contained in a cylinder with perfectly reflecting walls.
is
caused
is
always
radiation
bT",
agreement with Eq. (13-32). u/3 and u = bT*, the equation of takes the interesting form
state of
decreasing, the temperature of the coal also decreases.
V
4
S*
1
T'
FT* =
The above equation shows
•
of existing in
U = VbT\
final
T dS
after integration,
(13-34)
dT
The
in the first
dT
dV =
or
and
dP
=
$bT*dV = -4VbT 3 dT,
(13-37)
const.
volume of blackbody radiation were the radiation would then be capable equilibrium with matter at a temperature which is one-half the that,
if
the
increased adiabatically by a factor of Also, since
the radiation
equilibrium with the coal; but since the energy density of the
in is
blackbody radiation
p=|n /dP\ [dTJr
If
to
temperature of the coal can be found by setting dS equation. Thus,
P=
and
Suppose that blackbody radiation,
Reversible adiabalic change of volume
in
is
Since
Vi).
expand against a piston, the expansion will be adiabatic, since there is no exchange of energy between the walls and the radiation. The work done on the surroundings is accomplished at the expense of the internal energy of both the piece of coal and the radiation. If the grain of coal has an extremely minute mass, its heat capacity may be regarded as negligible in comparison with that of the radiation. During the expansion, the radiation
to
Integrating the above,
-
equilibrium with an extremely small piece of matter such as a grain of
coal,
du dT 4 7" T
which
2.
in
3
q = ib-n(vf
and
equation becomes
and reduces
isothermally, heat will have to be supplied to the walls to keep the tempera-
ture constant. Thus,
(&-*(&-' Since
433
Special Topics
8,
original temperature.
The
first
T
= (vr) = 4VbT3 r
(13-35)
-
dS equation,
PROBLEMS
TdS= C v dT + T therefore
becomes
13-1 The tension in a steel wire of length 1 m, diameter 1 mm, and temperature 300°K is increased revcrsibly and isothermally from zero to 3 1 0" dyn. Assume the following quantities to remain constant: p = 7.86 g/cm 6 12 2 10a = 12.0 X deg '; Y = 2.00 X 10 dyn/cm Cs = 0.482 J/g deg. How much heat in joules is transferred? (a) ;
(13-36)
•
;
which may be used
in
two ways.
(b) (c)
1.
Reversible
isothermal
change of volume
If
blackbody radiation, in is caused to expand
equilibrium with the walls of a cavity at temperature T,
(d)
formed
How much
work
in joules is
What is the change What would be the isen tropically?
done?
in internal energy?
temperature change
if
the process were per-
434
The
13-2
equation of state of an ideal elastic cylinder
CL =
is
and
2 J/deg. Calculate (dT/dL) s for values of L/L equal to
cate
only.
L =
If
is
a constant and Lt, the length at zero tension,
the cylinder
2/-o,
show
(a)
The
is
is
a function of
and isothcrmally from L
stretched reversibly
=
T
Lo to
that:
is
heat transferred
is
X-ray
unstretched,
structure.
When
diffraction experiments indi-
stretched isothermally, a crystalline
it is
found, indicating that the large chainlike molecules are oriented. or negative?
Is (<)S/dJ7)T positive
(£)
Prove that the linear expansivity of stretched rubber
13-6
A
whose cmf depends only on jNg coulombs of electricity. of
reversible cell
isothermal transfer
T
is
negative.
undergoes an
is
Q = -KTL
-
(\
is
Show
AS =
that
/JVjp -=*.
AG = -jNF £. and AH are found to approach the same value as proaches zero, what conclusion may be drawn concerning lim A.S?
%a T),
the linear expansivity at zero tension,
<*o,
rubber
(a)
(a)
where
When
an amorphous
structure
K
1, 1.1, 1.5,
2.
13-5
*$-% where
435
Special Topics
Heat and Thermodynamics
expressed as
(A)
Show
(c)
If
that
AC
T
ap-
T—o 1
13-7
dU
When
zinc sulfate (valence
at a temperature of
the reaction
2.31
is
273°K and
X
=
2) reacts chemically with copper,
at atmospheric pressure, the heat evolved in
10 s J/kmole.
The emf
of a Daniell cell at
273°K
is
X
(b)
The change
of internal energy
AU = 13-3 is
1.0934 V, and the emf decreases with temperature at the rate of 4.533 10~ 4 V/deg. How does the calculated heat of reaction compare with the
is
+%KT*-L a
measured value? 13-8 The emf of the cell Zn, ZnCU, HgaCljs,
.
In the case of the ideal elastic substance whose equation of state
£ =
Hg is given
-
1.0000 4- 0.000094(i
by the equation
15).
given in Prob. 13-2, prove that: («)
(»)
<8V\ = AYaoT. dLji
w)*
=
Write the reaction, and calculate the heat of reaction (a)
Prove
called the elaslocalork
is
effect.
The magnitude
of this effect
(A)
Assuming the equation of
Derive the expression P'
/dT\ = KTY/L {di) s -c: Sju where Cl (b)
T=
is
L?\
- zy ~
,„/L \L a
a° r
state to
be
= X VE,
2Lg\]
+ U) ,
a-
T
is
expressed by the quantity (dT/dL)s(a)
that, for a dielectric,
6U\ dEj T K
An ideal elastic cylinder has the equation of state given in Prob. When the length is changed isentropically, a temperature change
occurs. This
100°C.
Lao 7'.
13-4 13-2.
at
13-9
'
where the susceptibility % IS a function of T only, and the volume V is constant, show that the energy per unit volume of the dielectric is equal to
the heat capacity at constant length of the entire cylinder.
Assume the following values 300°K (practically constant), K ao
=
5
X
for
=
certain sample of rubber:
a
X
1.33
10-< deg-
10 3 dyn/dcg,
where f(T)
1 ,
is
an undetermined function of temperature.
W 436
Prove that the energy per unit volume of an electric
(c)
vacuum
437
Special Topics
Heat and Thermodynamics
field in
a
given by
is
£*
6k A T*
8jt*
Using the relation
(d)
er
=
+
1
show
47rx,
that the
total
energy per
volume
unit
y
This result, called the R, O. Pohl in 1965, If
was observed by G. Lombardo and and Np*/6kA = 1.6 X tQr+ cm* deg*/V2 calctdate AT at the following values of A',: 0.5, 1, 5, and 10 kV/'cm. 13-13 I. Shcpard and G. Feher, in 1965, observed the piezocalorie effect. Derive an expression for this effect in zero electric field.
(dielectric plus field)
=
f(T)
+
M
Sjj£
+
~jff-
Show
that for a dielectric the difference in the heat capacities
is
given bv
C«
-
Gp.
= xVI\ show
-
{d p l/dE)T
-
F If
x = C/T, show
T ( dx X
,
£,
V
deg)t
that
-
(a) {b)
(a)
show
In the case of a dielectric whose equation of state
is
P'
= CVE/T,
"A
—
The
the Peltier coefficient of the
What
The thermal emf of a Ni-Pb thermocouple is given by (1 9. 1 when the reference junction is at 0°C. What is t at 127°C? What is am - o*b at 57°C?
/»V/
(0.030 /xV/deg 2 )* 2
heat transferred in a reversible isothermal change of
field
What would 0-/1
=
be the characteristics of a
thermocouple where
0?
Would
13-17
is
C.V Q= -£(&,-£}).
/T Tk
(b)
junction at 100°C?
test
this thermocouple make a good thermometer? Integrate Eq. (13-20) from a reference temperature temperature T, and derive the equation (b)
that:
(a)
is
13-16
V 13-11
What
refer-
would be transferred at this junction by an electric current of 10 A in 5 min? Would this heat go into or out of the junction? (c) What is the difference between the Thomson coefficients at points at 50°C? 13-15
(c)
-
temperature.
Peltier heat
that
CB — C P
difference between the Seebeck coefficients for Bi and Pb (0.47 ,uV/deg 2)i, where I is the Celsius
-43.7 pV/deg
by
given
(b)
If P'
•
(a) Calculate the thermal emf of a Bi-Pb thermocouple with the ence junction at 0°C and the test junction at 100°C.
0^/372* /
The
13-14 is
(b)
T = 2°K
,
13-10 (
eleclrocaloric effect,
{a a
~
cr,i)
TH
to
any
d'T.
The small temperature change accompanying a reversible adiabatic
change of
field is
(a)
CV ar =
cmf's, (b)
Interpret this equation in terms of a Seebeck emf, two Peltier
and two Thomson emPs. Interpret this equation in terms of the
first
law of thermodynamics,
neglecting irreversible effects.
13-12
A KC1
capacity equal to
doped with Li ions was found to have a heat and a polarization P' = (A'n i/3kT)E, where N is the ions, m the electric dipole moment of each Li ion, and k is crystal
AT
concentration of Li
3
Boltzmamvs constant. If the initial electric field Iu is adiabatically reduced zero, show that the resulting temperature change AT (with A7"
(c) If three different wires A, B, and C arc joined in junctions at the same temperature, prove that
to is
(tab)t
+
(tbc)t
+
(ttca)t
=
0.
series,
with
all
438
Heat and Thermodynamics
Two
(rf)
B
wires A and
other ends of
A and B
Special Topics
form a junction at the temperature T. The C forming two junctions, both
(
Show
that
are joined by a wire
at Tr. Prove that
T
T .
— What
is
I'
1
T
T
AT
.
'nV
'S -j^
M>
1p'
the importance of this result?
13-18
In Fig. PI 3-1 is depicted an idealized apparatus similar to that used to explain a throttling process. There is, however, a big difference. The throttling process is strictly adiabatic, whereas the gas flowing through the
(d)
Express Is and
(e)
Show
— /„
I'M
in Fig. PI 3-1 is
higher than that on the
left.
This process
is
called thermal effusion or the
as linear functions of
_
Us.
Uffljt,
Z.21
WA
and .L»22 i
effect.
(/)
AT/ T and AP/T.
that
maintained at a constant temperature and pressure on the left side and at a constant higher temperature on the right. If the pores in the plug are small enough, the pressure on the right will be found to be
porous plug
Knudsen
439
Show
that
A/A
(Is/In) sr—o
477/„=o Temperature
Temperature
T
r + Ar
The quantity S* is the entropy accompanying the transport of (g) mole of gas through the plug. It is found from kinetic theory to be R/2. Prove that
Q
I',
v. s,
1
where the subscripts
1 Fig. PI 3-1
Thermal
effusion
{Knudsen
effect) of
(/()
a gas through a porous plug.
A
13-19
t.
heat
Q
AQ
at the temperature
Show
T+A
moles of gas be transferred in at temperature T and gains
Q
Apply the
first
= Q *£ - AQ + nAs. T
law and obtain
Prove that, (a)
7".
that the entropy produced in time t
AS'
(b)
let n
In this time, suppose the gas loses heat -
TdS = dU
+
on the right. Under these circumstances,
is
(b)
-
60
id)
if (7
dC -
13-20 n Mi.
cell in
is
by using
the result of Prob. 6-5«.
described with the aid of the five
+ PdV -1/dX.
W
- TS, as usual: = V+ -S dT + VdP + 1/ dX.
<8S\ ,dP)x.T
fdV\ \dTj X j.'
'as\
($f\
,dX)r.r
\dTjp. x
\dXj T P ,
AQ =
2 refer to opposite sides of the plug.
thermodynamic system I', T, 1/, X, where
Suppose that both pistons arc moved to the right simultaneously, thereby P on the left and a constant pressure P AP
maintaining a constant pressure
and
1
Derive the preceding equation
coordinates P,
time
\TA'
Pa
h
_
-
Using the
five
which gases may be
coordinates P, V, T, £. liberated,
show
that:
Z to describe a reversible
440
Heat and Thermodynamics
dG =
(a)
Special Topics
-S dT + VdP + SdZ.
13-24
The volume
of the blackbody radiation contained
with walls at a temperature of 2000°K
»
" (H),, " (£)„• dll = TdS+ V dP +
0)
dZjr.p For a saturated
(e)
13-21
and
film
'
its
W
AH = —
= -SdT + V dP
\dp) T ,A
13-22
\dTJp,z ;.Vf
J
=
„/,
A
to describe a surface
empty boxes
in this tabic:
Ideal gas
Conduction
molecules
electrons
Photons in an evacuated cavity
How
does Nt compare with git
What
the expression
is
for g?.
What
fraction of these
states
I
is filled?
low many particles are in each filled state?
Is
2 A',- =
const.?
What is
kind of statistics appropriate?
13-23 (a)
The
spectral energy curve of sunlight has a
maximum
length of 4840 A. Assuming the sun to be a blackbody,
perature of (b) (c)
what
at a is
wave-
the tem-
emitting surface?
its
What What
How much
(b)
What
(cj
How much
is
(d)
If the
(a)
PV =
heat
a cavity
.
is
transferred?
the change of energy of the radiation?
is
the energy density of the sun's radiation?
is
the radiation pressure exerted by the sun's radiation?
i
work was done?
const., for
(c)
jjj«tm Cp= 00.
(d)
G =
(b)
+ J dA
\dA) T ,p
Fill in the
(a)
cm 3
in
increased reversibly and isothcr-
expansion were performed reversibly and adiabatically, what woidd be the final temperature of the radiation? Prove that, for blackbody radiation, 13-25
'<-*{$),}
Using the five coordinates P, V, T, accompanying liquid, show that:
dG
(a)
cell,
mally from 10 to 1000
is
441
0.
an isentropic process.
14-1
14.
interesting in that,
when
Cryogenics, Negative Temperatures. Third
their valence electrons are
Law
443
removed, the resulting
ion lacks an inert-gas configuration because of incomplete filling of the 4/
which in the crystalline ground state has an /. value
electron shell. In the particular case of gadolinium,
PARAMAGNETISM,
state
CRYOGENICS, NEGATIVE
in the
is
form of a trivalent ion
Gd"'"'"
1
,
the
equal to zero, so that again J = S, When an atom, molecule, or ion possesses a resultant magnetic
TEMPERATURES, AND THE THIRD LAW
in its
lowest energy level and
is
removed from
sufficiently
be only weakly influenced by neighboring magnetic simple way when subjected to an external magnetic into a
number
of separate states, each of which
quantum number m.
The
is
its
fields, it field.
moment
neighbors
behaves
The
to
in a
level splits
characterized by a magnetic
m
allowed, discrete values of
correspond to discrete
magnetic-moment vector with respect to the. external magnetic field. These values of m vary in integral steps from — J to -\-J. Thus, when Fe+++ in its lowest energy level, with J = •§, is subjected to a orientations of the
14-1
Atomic Magnetism
magnetic
An atom
or molecule in
its
lowest energy state and uninfluenced by
its
field may have a net magnetic moment by virtue of two kinds of electronic processes: (1) orbital motion of one or more electrons, and (2) uncompensated spins of one or more electrons. These two effects are specified by the resultant orbital quantum number L and the net-spin quantum number The resultant of L and .Sis denoted by J. It is often the case that the orbital quantum number L is zero, in which case the atomic magnetism is due entirely to electron spin, since ./ = When atoms or molecules are moving freely in the gaseous phase, the magnetic moment of the gas is determined by the /. and S values of each neutral atom or molecule. Thus, oxygen gas is paramagnetic, because the resultant orbital quantum number of each neutral oxygen molecule is zero, and S = 1 for the ground state of the molecule. At lower temperatures, when atomic particles coalesce to form crystals, the atomic structures that occupy the various lattice sites are rarely neutral. In the sodium chloride crystal, the lattice sites are occupied by positive sodium ions and negative chlorine ions. Each of these ions has an inert-gas configuration of closed electron shells, with L = 0, 6* = 0, and J = 0. In a crystal where two or more valence electrons are removed from one atom and taken up by another, the resulting structures may again have an inert-gas configuration with no resultant magnetic moment. There are some atoms, however, such as Cr and Fe, each with three valence electrons which exist as trivalent ions in the crystalline phase and do not have an inert-gas structure because the third electron shell is incomplete. These trivalent ions Cr+++ and Fe +++ have magnetic moments. The effect of orbital motions is said to be quenched by the fields of neighboring ions, but there remains a net electron spin S, so that J = S. The rare-earth atoms are particularly
field
3f, this level
neighbors or by an external magnetic
.S".
.S'.
into six different energy states with
is split
m
equal to .
2j
When '¥ =
0,
1
2
J
+#,
"sf>
these six states all have the
same energy, and
lowest energy level of Fc +++ has a degeneracy of
or 2.1 4-
6,
1
therefore the .
The
effect of
an external magnetic field is to remove this degeneracy. If the energy of the lowest level is arbitrarily set equal to zero, then the magnetic potential energy of the ion in any one of the magnetic states is equal to
* where hh
is
eh
%
is
(14-1)
the Bohr magneton,
Mm
and
= —gmWm,
Amnc
=
0.927
the local magnetic field to
X
10
20
which the ion
practically equal to the external magnetic field is
crg/Oc,
is
subjected and which
when
The quantity g is called the Lande splitting whose internal magnetic field (created by isotropic, g = 2 when /. = 0. The minus sign in
a long cylinder parallel to 9£.
I actor.
In the case of a crystal
other magnetic ions)
is
Eq. (14-1) merely indicates that positive values of tions of the vfi,
and
m
correspond to orienta-
magnetic-moment vector with a component
these orientations involve a decrease of energy.
in the direction
When
negative,
and the
potential energy of the ion
magnetic ion as a small compass needle.
is
of
the orientations
of the ion are such as to have components antiparallel to the is
is
the sample of crystal
increased.
field,
Think
If held perpendicular to a
then
m
of the
magnetic
444
I
Energy
[cat
and The rmodynamics
Cryogenics, Negative Tempcratnres, Third
14-1
Magnetic Ions in Paramagnetic
Table 14-1
e
+ -?(2,,„X)
Magnetic
Paramagnetic
+ 4(2p,,K)
salt
J
L
Cr 2 (SO<) 3 -K 2 SO r 24H 2
8
2
(quenched) Fc +++
l'e.(S0 4 ) 3 -(NH4).,SOr24H 2 (iron
s
f
(chromium poiassium alum)
445
Salts
S
ion
Law
2
ammonium alum)
(quenched)
Gd +++
Gd 2 (S0 4) 3 -8H 2
7
7 2
2
(gadolinium sulphaic)
<:„=
2Ce(iN-0 3 ) 3 -3Mg(NO a ) 2 -24H 2
1
5 2
3
(X)l
GMN)
(cerium magnesium nitrate,
.84
(11)0.02
This apparent "splitting" of a spectral line into a pattern of lines is Zeeman effect. It may be seen in Fig. 14-1 that at a low value of the six magnetic states are close together, so that, if Ae is the energy difference between any two adjacent states, Ae is very small. If is 1 kOe, lines.
called the
%
%
Ae
=
2 MI(
^=
5
~
2
X
0.927
X
"
10
20
crg/Oe
0.2
X
10-' c erg.
=
1.38
X
10- 16 erg/deg
~
1.4
X
10 3
Oc
%, kOe Fig. 14-1
Splitting of the lowest energy level of
Fe+++
ion in a magnetic field into six
separate energy stales.
At a temperature of 1°K, k'F
and then allowed
undergo a quasi-static rotation till its magnetic moment is parallel to the field, it will do work on its surroundings and undergo a decrease in energy. To point a compass needle antiparallcl to a magnetic field, work would have to be done on the needle in twisting it against an opposing torque, and its potential energy would therefore increase. The magnetic splitting, proportional to the external magnetic field W, is shown for Fe +++ in Fig. 14-1, and some of the electronic
Hence, at
energy level of four important trivalent magnetic ions existing inside a
In a large magnetic
Table 14-1. an atom or ion are also split into a number of separate magnetic energy states by an external magnetic field. In the case of a gas supporting an electric discharge in a narrow tube placed between the poles of a strong electromagnet, atoms in one of the magnetic energystates of an upper energy level may undergo a transition to a magnetic state
and
of a lower energy level
equation,
field
X
1
deg
to
X
T = 1°K and 9f = Ae
10-" erg.
10 3 Oc,
< kT
(for
7^ =
1
kOe).
data for the lowest
crystal lattice arc listed in
The upper energy
kT at 1°K
however, say, "X = 20 kOe, Ae 1.4 X lO" 16 erg, so that
field,
is still
=
4
X
10 -16 erg,
levels of
and emit
a spectral line.
of the magnetic field, a single spectral line
What
now becomes
was, in the absence a
number
of spectral
Ae>kT
(for^=
20 kOe).
In the paramagnetic crystals listed in Tabic 14-1, the distribution of magnetic ions among the various magnetic states is given by the Boltzmann
Wi
cc
e
-'
446
1
Icat
and Thermodynamics
Therefore,
when Ae
weak
may
<
kT,
all
the six closely packed states that exist in a
be populated almost equally. Since each state corresponds a different orientation with regard to the external magnetic field, the to field
M of the crystal would then
net magnetic polarization or magnetisation
small.
When
the
field
strong enough to
is
— ./-2/xivV,
lowest state, with energy
make A«
will
>
however, only the
/:'/",
be populated. Since this state
corresponds to the ions lining up parallel to the external magnetic
M would have
magnetization
its
be very
largest, or saturation,
value
M
gat
field,
the
words,
form a subsystem with its own identity, describable with and T, as though the rest of the crystal were a conmagnetic potential energy e, of any ion is equal to — guvP^nii, but
%, M,
The
only energy possessed by the ion. has been mentioned that the lowest energy level of a magnetic ion in l)-fold degeneracy the absence of an external magnetic field has a (2J which is removed by the splitting action of the magnetic field. Even in the this is not the It
+
absence of an externa/ magnetic field, there are internal fields which provide some
.
splitting of the lowest energy level.
There
arc, of course, the
very weak magnetic
provided by the other relatively distant magnetic of these fields is small and sometimes may be neglected fields
14-2
Statistical
Mechanics of a Magnetic-ion Subsystem
the
The
ions occupying lattice sites in a typical crystal such as
sodium chloride
are localized (and therefore distinguishable) and tremendously influenced
by
When
their neighbors.
the vibration characteristics of the crystal as a
found that there are three times as many normal modes as lattice points and that these 3.V normal modes may be treated statistically as 3JV distinguishable but weakly interacting harmonic oscillators. Upon making use of the appropriate expression for the thermodynamic probability (see Art. 11-12) and performing the usual operations, the Boltzmann equation was found to apply. The paramagnetic crystals of greatest use in practical thermodynamics contain paramagnetic ions surrounded by a very large number of non-
whole arc analyzed,
it
is
much
stronger effect of the
of electric states.
We have seen separation '"splitting
ions are very dilute
among them
is
and
far apart, so that
The same is Table 14-1. The
very weak.
magnetic ions in the other three salts listed in the magnetic ions in a dilute paramagnetic salt
is
almost gaslike
We may
therefore apply the statistical
field of, say,
T = tn^¥
2
T«
energy level
by their method appropriate to
X
0.917
Stark-effect splitting in
behavior of
weak-
level to
is split
show
10- 20 erg/Oe X 20 1.38 X 10-' 6 erg/deg
X
is
less.
wc assume
where
9-f
only two states whose energies arc zero and ga
is
so low (in the neighborhood of 1°K) that the vibrational energy
the heal capacity of everything hut the magnetic ions may be ignored. The N magnetic ions constitute a subsystem with which there is associated a temperature T of its own, which may or may not be the same as the temperature
and
of the rest of the crystal.
The
magnetization (or total magnetic moment)
the ions has nothing to do with the rest of the crystal,
magnetic
field
% produces no
effect
on the
M
of
and the external
rest of the crystal.
In other
g\. In zero magnetic field
the ions
10 3
would be
=
0.
is
whose
Oe
Suppose,
in the
and
and the
field,
much
too small
for the
that the effect of the crystalline field
Consider a piece of crystal containing N magnetic ions and about fifty times as many nonmagnetic particles. Let us suppose that the temperature of the crystal
X
not completely removed by the crystalline
into only a few electric states. This splitting
these distinguishable, weakly interacting particles.
and
states
most crystals is of the order of a tenth or a Sometimes the entire degeneracy of the lowest
in Fig. 14-1 at the point
simplicity,
magnetic
3°K.
hundredth of a degree or
in the
rise to
kOe would correspond to a temperature
20
k
The
Ogives
=
2y.n%. It is often convenient to express this energy" in terms of a temperature T, where kT = 2iau9<£. The
true of the
ness of interaction, but the ions are, of course, distinguishable positions.
that a magnetic field
given by Ac
is
magnetic splitting of a
—
chromium
the crystal.
and negative charges of the ions in the lattice, and the interaction between an atom or ion and an electric field, known as the Stark effect, is similar to the Zccman effect in that some or all of the degeneracy of an energy level may be removed by the electric splitting of the level into a number
magnetic particles, as shown in Table Each chromium ion in Cr2(SO.i)s" K2SCV24H2O is surrounded by 1 potassium atom, 2 sulfur atoms, 20 oxygen atoms, and 24 hydrogen atoms a total of 47 nonmagnetic particles. In magnetic interaction
electric field in
but the effect comparison with This field is due to
ions, in
the positive
14-1.
other words, the
447
the magnetic ions
the aid of its coordinates tainer.
Law
Cryogenics, Negative Temperatures, Third
14-2
is
sake of
to provide
and whose degeneracies are
5],
at temperatures far below 81/k, most of
lower energy
state;
whereas at temperatures
much
higher than &i/k, the two states would be about equally populated.
To apply statistical
mechanics rigorously
to a
paramagnetic
into account (1) the magnetic-ion subsystem in a field
%
%,
crystal, taking
(2) its
magnetic
= 0, and (3) its mechanical and electric interaction with the lattice when and thermal interaction with the lattice, is a complicated problem requiring the use of a statistical device known as a canonical ensemble. For details of this calculation, the reader is referred to Chap. 20 of P. M. Morse's "Thermal
448
f
Icat
and Thermodynamics
For a system of M
Physics." Much can be done and understood, however, by applying the simple method of dealing with localized, weakly interacting particles that was used in Chap. 11 with a crystal.
Let
energy of a paramagnetic ion, equal to the sum of the energy 5,- due to crystalline field splitting and the magnetic potential energy ~gpqP&ak due to the presence of a magnetic field 7£. Thus e,
stand for the
e;
=
—
5,
gn n9Arii,
weakly interacting
S = kin where 2A\-
= N=
from A*,-, be the instantaneous, nonequilibrium — J to -\-J. LetATj, Jvs, populations of the various energy states. In the systems we have studied up to this point, namely, an ideal gas, a non metallic crystal, a metallic crystal, and a photon gas, it was found possible to interpret the sum 2;V,-e,- unambiguously as the internal energy U. This is not the case, however, with a paramagnetic ion subsystem in a magnetic field, where .
.
SA',-€i
.
.
=
.
XNiSi
we
in integral steps
A',
is
449
given by
A"!
A',!jV2
•
!
•
•
and taking
= -AS
In
NidNi.
is approached adiabatically, at constant 9£ and at constant usual three equations to satisfy: have the we
equilibrium
.
-#Sln NtdNi =
0,
dNi
=
0,
XdN,-
=
0.
2e,
Using the Lagrange method, we get
-
SfSNigitBttti.
=
Ae-»",
Z=
2
A',-
The sum in the second term on the right M, so that the entire second term, —9&Mt
M
paramagnetic system of magnetization cannot be included within an expression to conclude, therefore, that the first
entropy
constant. Using the Stirling approximation,
dS If
ions, the
Law
get
(14-2)
and 8 h and m, ranges
has only two values,
8;
localized,
total
the differential,
where
Cryogenics. Negative Temperatures, Third
14-2
is
seen to be the magnetization
is
the magnetic potential energy of a
in a field 9-f, a
for internal energy.
sum on
the right
is
where
quantity which
We
are forced
and
the internal energy, or Substituting the equilibrium values of the A"s into the expression for dS,
U= The sum 2A ;e; r
is
WiSt.
therefore
2A'
If
f
e;
= V - WM.
our system of paramagnetic ions approaches equilibrium
(dQ
=
0)
and
(2) at constant field
{£¥ =
0),
(14-3)
(1) adiabatically
and therefore
Ni
Substituting these values of the
= y
.
e-« ltT
.
into the expression for the entropy,
then r
dSNifk
=
dlJ
= SA
r ,:£,-
remains constant.
0,
5 = Nk
-9fdM- M d%
= dQ -
or
it is
a simple matter to show that
M
m
Since
2A
r ,-e,-
=
II
—
9^M, we
In
Z+
2A
,e,-
T
get
- Nk In Z + U - 9fiM T U - TS - 9AM = -AIT In Z. S
we get
450
Heat and Thermodynamics
The faction on
the left is the analog of the Gibbs function and the magnetic Gibbs function,
G* = and we
Cryogenics, Negative Temperatures, Third
14-3
is
only two states: one of energy zero and degeneracy g a , and the other of energy Si and degeneracy gi. Then, rise to
u -TS- KM.
(14-4)
G* = -NkTlnZ.
ZM =
(14-5)
which
is
a function of
In the next section,
dG*
KdM - SdT ~ M d%.
= dU- TdS-
and since
TdS =
M=
Therefore,
-
(g\
The magnetic
.
(14-6)
•-(^'
Also,
T only. we
Z^, which
shall evaluate
will
be found
to
be a
part of the partition function of a magnetic-ion subsystem
given by
Z& =
»<»
=
(14-10)
Magnetic Moment of a Magnetic-ion Subsystem
14-3
is
consequently,
+ git-***,
-KdM,
= -SdT - MdK.
dG*
g*
otK/T.
function
j
dll
451
We have simplified matters by assuming that the crystal field splitting gives
therefore
get finalh
Now,
Law
If
we
a
let
then, since
m,-
Bra**"*/**.
(14-11) (14-12)
--kf-'
— J — J + 1,
can take on the values
.
t
.
.
'
'
,
J—
1,
J,
we
find
and therefore
*-
a
«'(^X
+«>»'•
Z& = (14-7)
Finally,
This
U=
G*
is
X e mi= —J
° m:
=
—
Z?{ -
+ g~* Vr~ + V>
a finite geometric progression with a ratio
+ TS + KM
-MTln Z
~" J
e
g—aJ 1
HO% l^/) + XkTln Z + KM, z» =
(14-8)
The
four boxed equations will enable us to find all the desirable information about our magnetic-ion subsystem, once we have our partition function expressed as a function of T and K. Since = S 6i
Z=
- g^Km<,
t
z# =
or
In i
i
or
aJ .
hence,
g-a/2
_ _
(./
j-
e
°' 2
e
,
we
get
a(J+i)
go/2
£ )a
-}
(14-13)
sinh ia
Therefore,
S^-tf.—«< B #m,)/M'
= 2g e-
e
-
g—aV+l)
sinh
e":
+
gaf.J+1)
Multiplying numerator and denominator by
and
'
Z=
In sinh
(J
+
In
Z
\)a
-
In sinh
-
+
In (g
+ mt-W**).
'l^x e ai' a ^m^kr
z - z na z^.
i
(1
4 _ 9)
Since a
= gnuK/kT,
is
seen to be a function of
K and
T.
(14-14)
452
Heat and Thermodynamics
We arc now in
14-3
a position to evaluate
M with the aid of Eq.
Cryogenics, Negative Temperatures, Third
Law
453
(14-6).
— '"(tb^-Mt^ d In
= NkT
M
Finally,
The
»]-*»(£
\dX)
=
A^jun j- P n sul h (J
=
A'^jUb
=
+
(J
1
7 7
+ $)
[(•/
who
i) fl
cosh (J
l)
quantity in curved brackets
after L. Brillouin,
+
sinh (J -f
AT#Kfr/
is
—
1" sinh
+ *)a -f
*)
-JaJ
sinh
coth (J
5
| cosh la
£)
|)a
-
-Jn
$ coth
(14-15)
jw]
called the Brillouin Junction Bj(a),
extended the
first
T
\
Z<3f fcla
da
named
paramagnetism
classical theory of
(due to T.angevin) to include quantum ideas. Therefore,
M where
Bj(a)
=
-=[(,/
+
•J
Before
we
= Ng^JBj(a), coth (J
-J)
in Fig. 14-2 for a
number
of values of J.
,
=
By
—=
cosh x -.
1, e' >$>
Bj(a)
Since a
e~ x, so that coth x
"j[(J +
= gut^/kT,
i)
e*
—.
sinh x
»
-
\
coth
-
e
x
=
1
which
M, is
let
us
plotted
definition,
+ —
e~ z e~ x
>
= 1. Therefore,
-J]
(14-17)
>o].
features of the Brillouin function,
coth x
for x
i)a
consider the consequences of Brillouin's equation for
examine the mathematical
and
+
(14-16)
(for a
for large values of a,
»
large values of a are achieved
Fig. 14-2
The Brillouin
junction.
1 )
•Since
M
=
.\fi HgJBj(a),
it
follows that at values
M
of 7
when or
or
W-»l>
gJ
— » — F
^gPB y-
1.38 >;>
2
oI'M/T
greatly in excess
M
kOe/deg, the magnetization has its saturation value e&i where Mat = NgunJ, and the magnetization per magnetic ion expressed in terms of Bohr niagniions, A/ s „t/A'/iji, becomes
X
X
0.927
10-»erg/deg X 10-" erg/Oe
Mm* = J g
(for
%l T » 7 kOe/deg)
» 7 kOe/deg. This conclusion
was
tested
by W.
E.
Henrv with the
first
three salts listed
454
Ileal
and Thermodynamics
14-3
Therefore, for small values of
2+
l i
(J
\
455
a,
7.00
BM = j
Law
Cryogenics, Negative Temperatures, Third
+
2
7 6
a
i)a
-JW + ^-H =
—
(
p
4-
r 4-
±
-
'-
—
±)
2>I
and
5.00
Bj(a)
finally,
Small values of a are achieved 4.00
j— a
(for a
«
1).
when
Y « 7 kOc/deg. When
I
this
is
the case,
WW
+Aa = v r^ Mw = ##«aJ —3—
3.00
+ l)^ M = iV^47(7 T 3k
or
This
is
Curie's equation,
+
lfff
2T7
!
(fory«7kOc/deg\
and the Curie constant Cc
is
(14-18)
seen to be
1.00
\7(J Cc = Ntfig
+ 1)
(14-19)
3A
A
30
20
10
9f/T, kOe/deg Plot of magnetization per magnetic dipole, expressed in Bohr magnetons, against §); (//) iron ammonium alum (J f); anil for (I) chromium potassium alum (J Henry (1952), experimental results W. E. points are The of (HI) gadolinium sulfate (J -J). and the solid curves are graphs of the Brillouin equation. Fig. 14-3
•
=
=
%/T
mass of crystal containing exactly A',., (Avogadro's number) magnetic ions is known as 1 gram-ion. The gram-ionic masses of the four crystals listed ion is given in Table 14-1 are given in Table 14-2. The Curie constant per g
=
Table 14-2
Curie Constants and Heat-capacity Constants m, gram-ionic:
in
Tabic
14-1
The
.
experimental results, plotted in Fig. 14-3, agree very well
with the Brillouin equation and
When
x
<3C
1
,
it is
with the limiting value of
easy to show that
coth x
—
Paramagnelii:
sail
x
,
-f-
*
v 3
J(J+l)
Cc (measured) cm' cleg
Cc
(calculated)
AIR,
cm 3 deg •
deg'
g -ion
g -ion
crystal
M/NpaCrs(SQi)»-KfiO», 24H«0 I-'ci(SOi)j-(NHjhSOi'24HiO Oil-(SOi)iSH;0
-1
mass of
2Ce(NOj)j-3Mg(NO!)i-24HjO
499 482 373 765
3.75 8.75 15.75
1.84 4.39 7.80
-0
1.88
4.38 7.87 (!|)
0.317 (±)
0.018 0.013 0.35 6.1
X
10-'
456
Meat and Thermodynamics
by
= Cc _
6.02
X
3
Upon
X (0.927) X 10~* X 1.38 X I0-" erg/deg
10 23 1/g
2
ion
•
9
erg 2 /Oe 2
-g*J(J
At constant M, the left-hand member becomes Cm, and since of T only, dU/dT is also a function of T only. Therefore,
+ l).
Oc has the same dimensions as erg/cm gram-ionic Curie constant becomes
=
o.i25
sinl^sijw + D g ion
and provides the calculated values measured values is very good.
in
8 ,
Thermal Properties
The
and from the expression
for
U in
=
If A'
N^k =
jVa, then, since
R
given by Eq. (14-8),
"« f-B^ + **
we have
R,
the heat capacity per gram-ion
given by
of a Magnetic-ion Subsystem
This
known
is
(go/gtV>* T
Si
k*T*
as Schottky's equation
[1
and
+ is
plotted in Fig.
To
=
JSy(a)
_
= 0.1°K and go/gi = 1. From the middle curve of Fig. 14-4 we see that, at T = S,/k = 0.1°K, c M/R = 0.2. Since R = 8.3 J/'g ion deg, cm = 1700 mj/g ion deg. If the rest of the
gjj.y.%
U=
+
go
-Ngv. a 7fJBj{a)
to
Eq. (14-16), the
first
[In feo
-Jf
JB.
+
•
^1*-''/*')]
+
1.0
0.8
+ gie~ N8i
+ 1
According
Zint
dT
-
(
kT* Therefore,
1
Q In
term on the right
is
+ 7fM.
(14-21)
0.6
—3PM, which cancels
0.4
(go/gi)*'"
^^"N^..
the third term, so that
///
0.?
Nh
U= 1
Since
+
(go/giW'" T
(14-22)
*
dT"
**
^oT^-
L 0.2
0.4
0.6
0.8
—^1.0
1.2
1.4
1.6
1.8
2.0
TAS,/k) Fig. 14-4
dT
&l
n
TdS= dU-PPdM, dT
4-4 for three different
appreciate the numerical aspects of Schottky's equation, consider a
•
gg /da\ da \dTfa
In
1
magnetic-ion subsystem in which &\/k
have
/g In Z\ _ d V ar )k
(14-25)
/W* T !
(*
value: of the ratio gi/go-
,
We
(14-24)
+ (go/ay^y
[i
c,u
is
U is a function
Eq. (14-22)
(14-20)
Table 14-2. The agreement with the
internal energy of a magnetic-ion subsystem
namely,
457
the
«•_ 14-4
Law
dU dT
Cm =
2
recalling that the unit
Cc
Cryogenics, Negative Temperatures, Third
14-4
(14-23)
absence of
Schottky heat capacity of a magnetic-ion subsystem whose lowest energy
an
external magnetic field, is split into tivo states with degeneracies go
several values of Si/go.
level, in the
and
gi, for
458
Heat and Thermodynamics
crystal
14-4
had a Dcbyc of about 250°K and, say, 1 00 times would be, from Eq. (11-47),
as
Cryogenics, Negative Temperatures, Third
Law
459
many particles, 0.008
the heat capacity
3
100
The
/125V
T*
«
100
X
i
X
3
(0.1)
w
0.013
heat capacity of the magnetic-ion subsystem
mJ/molc
Copper polass urn
deg.
therefore over 100,000
is
times as large as the rest of the crystal
Many and
above
than
less
°K but
0.1
,
When
this value.
7
»
are of interest
Sj/it,
Eq. (14-25)
reduces to
R
go/gi
{h/kY
const.
(l+go/gi)*
T*
J2
"tail" of the Schottky curve
an interesting and important
0.003
is
0.002
therefore an inverse square curve.
fact that this inverse-square behavior
is
0.001 It
is
true at
when Stark-effect splitting gives rise to two degenerate levels but also when there are any number of closely spaced levels. = 0, such as magnetic interactions Also, other causes of splitting when among the ions, are found to give rise to a \/T"- curve. It is therefore the
2
not only
higher temperatures
%
custom
^ 0.004
K <$
cm _
The
0.005
o
subsystems have values of Si/k
of use at temperatures
to represent the heat capacity
at temperatures above the Schottky
per gram-ion of a paramagnetic
maximum by
salt
the equation
R
T
Fig. 14-.5
netic salts,
listed in the last
and the 1/T 2 law
is
6
5
10
7
12
11
c/R « 1/7"2 {Ashmead.) .
The most important thermal property of a magnetic-ion subsystem entropy. According to Eq. (14-7),
S = XtT
is
the
(il^ +Sth ,Z;
(14-26)
2
column
4
Test of the relation
and from Eq. Values of A/R are
3
1/T ?
A/R
C_M_
sulfate
of
Table 14-2
verified in Fig. 14-5 for
for four
paramagIn
copper potassium
(14-14),
Z =
In sinh (J
+
i)a
-
In sinh |
+
+
In (g
gie-^ kT).
sulfate.
The
cerium magnesium nitrate (abbreviated CMX), plays an important role in low-temperature physics. It is a complicated crystal with a structure that is highly anisotropic. Parallel to the axis of symmetry (the last salt,
so-called trigonal axis), the
constant
Lande
splitting factor
almost zero. Perpendicular to
is
g
an external field, rotation through 90 deg is all that over, the heat capacity constant
is
in the
this splitting
is
is
very small.
A
field. It is
believed that
to ionic interaction rather than the Stark effect.
Z
is
introduced into the expression for
S, the resulting
is
when
equation
the temperature
divided into two parts:
CO
T»
8i/k
and
(2)
T < h/k
and
(14-27j
S = S,{T,K).
simple
the lowest energy level of the
absence of an external magnetic
due
however, the Curie placed with its trigonal
needed to magnetize the crystal. Moreis remarkably small, indicating a very split
In
so small that the Curie
A/R
small value for the various factors that
cerium ion
is
magnetization
its
range
this axis,
constant has an appreciable value. If a single crystal axis parallel to
is
When
looks rather formidable but has simple properties
If
S is
plotted as a function of T, at various values of 7f the resulting curves
resemble those of Fig. 14-6.
The ST curve
for
%=
%
is
particularly signifi-
Within the range, indicated by vertical dashed lines, ,$Va0 s constant because the temperature is too low to provide an appreciable heat capacity of the nonmagnetic particles, but too high to allow crystalline field splitting cant.
'
460
Heat and Thermodynamics
Cryogenics, Negative Temperatures, Third
14-5
S
and
field,
9^i.
At a temperature
in the
heat capacity of the nonmagnetic particles
salt
magnetized isothermally
second
T dS equation
in the process
derivative
i:
—
negligibly small, the
is
For
> j.
this process the
gives
T
T dS = The
461
neighborhood of 1°K, where the
lattice is
Law
(dM/'dT)^
is
a
m^-
measure of the change of alignment of a
system of magnets accompanying a rise of temperature (disordering effect)
when
the external field (ordering effect)
is
kept constant.
The
derivative
is
therefore negative for all substances of the type dealt with in this chapter.
(In the next chapter
we
shall learn
increase of magnetic field
The second
~
1 °
K
about a superconductor for which an
a disordering effect!) Since
(dM/dT)^ is
nega-
heat goes out during an isothermal magnetization.
tive,
~&;/k
is
step,
»—*/,
is
a reversible adiabatic reduction of 3^ in which
= CW d'f
T
r
(5&"
Fig. 14-6
Entropy of a magnetic-ion subsystem as a function of temperature for various values of the magnetic field. In the region between the two dashed lines, S%=q is constant,
Since
U=
This change
constant,
and
Cm =
0.
or ionic magnetic interaction to show an appreciable effect. Since the only
term involving the dependence of (14-22)] field
is
internal
energy on temperature [Eq.
provided by the interaction of the ions with cither the crystalline
or with
one another, the internal energy
ture region. Finally, since
is
also constant in this
Cm = T dS/dT = dU/dT,
it
(SM/dT)^ and d7f are in
temperature
is
both negative,
it
follows that
dT is
negative.
Experiments of America and were then taken
called the magnetocaloric
effect.
this sort were first performed by Giauque in up by Kurd and Simon in England and by de Haas and Wicrsma in Holland. In these experiments a paramagnetic salt is cooled to as low a temperature
as possible with the aid of liquid helium.
A
strong magnetic field
is
then
tempera-
follows that
Cm =
K=
in this range.
14-5
Production of Millidegree Temperatures Ionic Demagnetization
by
Almost every property of matter shows
interesting changes or variations in
down to about by causing He 4 to evaporate rapidly. With He 3 it is possible to reach about 0.3°K. In 1926 it was suggested independently by Giauquc and by Debyc that much lower temperatures could be achieved with the aid of paramagnetic salts. The principle of the method is presented graphically in Fig. 14-7, where the entropy of a magnetic-ion subsystem is plotted against the temperature for two values of the external magnetic the temperature range below about 20°K. Temperatures
1°K are
easily obtained
l'K Fig.
14-7
demagnetization
In the isothermal magnetization i
—» /
k—ri,i i, the j
T
entropy decreases. In the adiabatic
{to zero field), the temperature decreases
462
Heat and Thermodynamics
1
applied, producing a rise of temperature in the substance
flow of heat to the surrounding helium, some of which After a while, the substance possible.
At
this
The magnetic
moment,
field
is
Cryogenics, Negative Temperatures, Third
Law
463
and a consequent
thereby evaporated.
both strongly magnetized and as cold as
the space surrounding the substance
now reduced
is
is
4-5
to zero,
evacuated.
is
and the temperature
of the
paramagnetic salt drops to a low value. The paramagnetic salt is either a single crystal, a pressed powder, or a mixture of small crystals in the form of either a sphere, a cylinder, or a spheroid. It is placed in a space that may be connected at one time to a pump or, at another time, to a gas supply. This space is surrounded by liquid helium whose pressure (and therefore temperature) may be controlled. Surrounding the liquid helium is liquid nitrogen, and the intervening space is
evacuated. Helium gas
is
admitted into the space containing the para-
magnet is switched on. The rise of temperature produced by switching on the magnet causes a flow of heat through this magnetic
salt before the
helium gas into the liquid helium. In other words, the helium gas as a conductor of heat to enable the paramagnetic
ture equilibrium rapidly. It
temperature equilibrium
is
is
salt to
come
the paramagnetic salt thermally insulated. In
many
is
used
As soon
therefore called the exchange gas.
attained, the exchange gas
is
tempera-
to
pumped
as
out, leaving
past experiments, the
adiabatic demagnetization was accomplished by swinging the entire cryostat out of the magnetic field produced by a huge electromagnet such as that
made
depicted in Fig. 14-8. In recent years, solenoids have been
of
many
thousands of turns of fine superconducting wire, through which from 10 to 100 A may be sent. There is no power dissipation in the coil so long as it is maintained at a temperature below which the wire is superconducting. The physical principles involved in the construction
and operation
of supercon-
The conventional more modern one in
ducting magnets will be explained in the next chapter.
is compared roughly with the With a superconducting magnet, (1) the sample space may be
experimental apparatus Fig. 14-9.
made
larger; (2) the magnetic field
may be made
stronger; (3) the cost of
magnet may be one-quarter that of a conventional magnet; and (4) the power requirement is negligible. The next step is to estimate the temperature. For this purpose, separate the
coils of
wire surrounding the paramagnetic
susceptibility
M/%,
by means of a
special bridge circuit.
which
is
with the aid of Curie's equation. magnetic temperature,
is
salt are used.
The paramagnetic
a function of the temperature,
A new
temperature scale
The new temperature
defined as
Curie T* =
constant
susceptibility
Cc%
M
is
is
measured
now
7"*,
defined
called
the
Fig. 14-8 The famous electromagnet of the Laboratoire Aim'e Cotton at Bellevue, near Paris, used by Simon and Kurti in some of their pioneer experiments. (Courtesy of A'. Kurti.)
464
14-5
Heat and Thermodynamics
Vacuum.
Values of
3
M/%
Cryogenics, Negative Temperatures, Third
Law
%/T are always small enough that Curie's law
is
465
obeyed,
=
Cc/T. The external magnetic field is reduced to a low value at which the magnetization is Mj. Both ?// and f may be made equal to zero.
4
M
Under
these conditions
we may
use the
first
T dS equation
as follows:
T dS = Cm dT - T (jp\ dM. i
Vacuum
or
exchange gas
From
Curie's equation,
we
Sample space
get
b9f\
_
dT) M and we
also
M Cc
'
have
r
Table 14-3
- A
Pioneer Results in the Magnetic Production of Low Temperatures Final
Superconducting
Experimenters
Date
Paramagnetic
salt
magnet
With conventional magnet Fig. 14-9
With superconducting magnet
Oiauque and MacDougall
1933
Dc Haas, Wiersma,
1933
and Kramers is
seen that, in the region where Curie's law holds,
T*
temperature, whereas in the region around absolute zero differ
is
T*
is
expected to
Tabic 14-3
There
is
a set of conditions, easily achieved
in
the laboratory, under
temp.,
r*
1.5
0.25
27,600 19,500
1.35 1.35
0.13 0.12
27,600
1
35
0.085
8,000
The
initial
is
low enough
to
make
the contributions of
The temperature never gets where C« = A/T-.
lower than the
of the Schottky curve
.
1.16
0.031
1935
Iron
ammonium alum
24,075 24,075 24,075
1.20 1.29 1.31
0.018 0.0044 0.0055
5,400
1.15
0.35
8.000
1.23 1.23 1.23 1.23
0.09 0.038 0.072 0.114
Cesium titanium alum 1935
Cadolinium sulfate Manganese ammo-
nium Iron
tail
ethyl sulfate
Alum mixture
Kurti and Simon
temperature
fluoride
24,600
which
the nonmagnetic particles negligible. 2
°K
Chromium potassium alum
possible to calculate the final temperature achieved after an adiabatic
demagnetization:
1
temp.,
Oe
1934
Wiersma
(compiled by Burton, Grayson-Smith, and Wilhelm).
it is
field,
magnetic
sulfate
Cerium
Dc Haas and
early results of the great pioneers in this field are listed in
Cerium
Dysprosium ethyl
the real Kelvin
somewhat from the Kelvin temperature.
The
sulfate
Initial
Adiabatic demagnetization experiments are carried out more readily with a super-
conducting magnet.
It
Gadolinium
Initial
Iron
Iron
sulfate
ammonium alum ammonium alum ammonium alum
14,100 8,300 4,950
466
Heat and Thermodynamics
Substituting these values into the
T
first
dS equation and setting dS
=
for
= 4-,dT- T — dM l -
[T,dT
.
1
467
450
Ln
CM,
Law
500
an adiabatic demagnetization, we get
or
Cryogenics, Negative Temperatures, Third
14-5
„ 400
where the substance
is
demagnetized
to magnetization
M
f.
Upon performing
the integrations, 350
\Tj
Tf)
2C<
(Ml
-
A/?) />>
/° son
and using Curie's equation, 1
J_
T)
T\ 7> _
When K, =
Cc (Ofl
_
A\Tf
250
9ff\ T))'
(CC/A)K}
+
1
(14-28)
200
(14-29)
150
0,
®
Cc
A
CMN
which shows that tlie lowest temperatures are reached by salts in which A is small and Cc is large. From the values otA/R and Cc in Table 14-2, it may be seen that cerium magnesium nitrate is the most favorable salt by a factor of over 100. The graph in Fig. 14-10 shows that the conditions under which Eq. (14-29) was derived hold well for CMX down to 0.01°K. The remainder of the graph (down to 0.002°K) is the result of experiments by Daniels and Robinson; Hudson, Kaescr, and Radford; and dc Klerk, combined with experiments and theoretical calculations of Frankcl, Shirley, and Stone. The entire graph enables CMN to be used as a thermometer down to
100
50
o' 2
4
6
0.002°K.
Most experiments on adiabatic demagnetization to achieve the lowest possible temperature, in
proceeds until
?fs =
0. If,
arc performed in order
^
therefore, during
Fig.
14-10
Adiabatic demagnetization
Fremiti, D. A. Shirley,
and N. J.
which case demagnetization
instead of demagnetizing to zero field,
we
stay
within the region of field and temperature designated by the space between the vertical dashed lines of Fig. 14-6 (e.g., or 7^ -> , then 5 2 curious results are obtained. First of all, the entropy is a function of 9f/T only,
10
8
HCi/tti,
->^
an isentropic reduction of 9f
,
;(
^J
{not to zero), since
S
is
12
14
16
18
kOe/deg
to zero field
of cerium magnesium nitrate. (R. B.
Stone, 1965.)
T must remain constant. That this is true under the condidemonstrated by the experimental results of Hill and Milner, 14-11. Second, \{9£/T is constant, since is a function of
constant, the ratio 7/7 tions specified
shown
in Fig.
%/ T only,
M
is
M
is constant! If,
therefore, the magnetization does not change, the
term adiabatic '•demagnetization"
is
a misnomer. As a
result,
the
word
Heat and Thermodynamics
468
14-5
and
if
Mi
-MdPA
ZNidn =
Now 2A
r;€,-
Law
469
the population of the w,th state,
is
or
U
Cryogenics, Negative Temperatures, Third
is
the
total
energy of the
ions, that
and the magnetic potential energy
2«,
M+
2.V,-
is,
— WM.
the
sum
du
of the internal energy
Thus,
= U - 9fM, = dU - % dM -
SJViQ
and
(14-30)
M &,
fe
Since 2A'j
dt<
= —M
reduces to
d9
2«,-
= dU -?f dM,
dXi
It
Tr,°K Fig. 14-11
9£s
,
Constancy of the ratio
Ti (always
the same) to
9fs
,
%j/Tj
during adiabatic "degaussing" of
7>. (J. 5. Hill and J.
H. Milner,
CM N from
= T dS.
dNi
Iti
follows that an isothermal decrease of entropy,
in
M
capacity at constant magnetization Cat is equal to (dU/dT).\i, then zero. If the first T dS equation is applied to this process,
Cm
In Fig. 14-12A, the magnetic-ion subsystem is in zero field, and therefore the magnetic energy states arc extremely close together, with some of the
N,
N,
JV.
is
7f=
we have
dS,
Cm, and
dM all
that these results apply only
in the
—
9f=?f
XV Cm d'f - T r 9
which takes place
first step of any experiment for producing low temperatures, must involve changes in A',-, that is, population changes. During the second step, however,
which an adiabatic demagnetization takes place, dS is zero, and therefore that is, the populations remain unchanged. the dNi's are zero
1957.)
does not change (and this has been "degaussing" has been suggested. If Wiersma, and Casimir, 1936 and 1940), Haas, experimentally verified by de — adiabatic process, the work done is But in an then the work ffi dM is zero. that M! is zero. Since the heat internal energy, so equal to the change of
T dS =
(14-31)
t
K=
/»
equal to zero. Of" course,
when temperatures
it
must be understood
are not too low
and
the field
not reduced to zero. These conditions will be found to be important in certain experiments on nuclear magnetic subsystems. Let n be the energy of an ion with total quantum number m, in a field vf. is
When
the
field
changes a small amount d%, the energy change de,
=
—gutfflU d9-f\
dt {
V)
Cft)
is
Isothermal magnetization Fig. 14-12
k —* i and
Changes
and
in energy states
adiabatic demagnetization
i
—
»
/.
-(f) Adiabatic demagnetization
ionic populations during isothermal magnetization
470
Heat and Thermodynamics
14-6
degeneracy being removed by internal
electric
and magnetic
Since
effects.
—
the states are so close together, they arc equally populated. From k > i the populations change so that the low-energy states become highly populated, as
shown
in Fig. 14-12;'. In the last step,
i
—
»
/,
the populations remain con-
Cryogenics, Negative. Temperatures, Third
pcratures as low as 0.7°K have been reached, but this
471
Temperatures
rare.
is
Law
lower than 0.7°K cannot be reached by pumping liquid helium because a film of liquid helium II creeps up the walls of the pumping tube, vaporizes, and then recondenses. This phenomenon will be. discussed in the next chapter.
In studying the properties of matter at low temperatures, the experimental
stant in order to keep the entropy constant.
apparatus
is
usually surrounded
by a bath of liquid
He and 4
,
a measurement
of the pressure of the vapor in equilibrium either with the liquid bath itself or
Low-temperature Thermometry
14-6
When
of some liquid helium in a separate bulb, in conjunction with a vapor-
normal boiling point (or, in the case, of carbon dioxide, the normal sublimation point) and the triple-point temperatures and pressures must be determined with great accuracy. Temperatures arc usuallymeasured with a helium-gas thermometer according to the principles developed in Chap. 1 The bulb of the gas thermometer is often incorporated within the liquefaction apparatus. Pressures are usually measured with a a gas
is
liquefied, the
.
mercury manometer, sighted through the telescope of a cathetometer. Once these temperatures and pressures are measured, they may be used as fixed points for the calibration of gas thermometers of simpler design. Some of the most important fixed points of low-temperature physics are listed in Table
pressure-temperaturc table, serves to determine the temperature of the apparatus and of any secondary thermometer mounted on the apparatus.
how accurately the relation between helium vapor and temperature is known. Up to 1955, various approximate thermodynamic PT relations, similar to the sublimation equation given in Art. 12-5, were used and provided values Everything depends on pressure
many of the research projects. In the early 1950s, however, the development of secondary thermometers such as carbon resistors and paramagnetic salts gave rise to a host of temperature measurements that of temperature adequate for
PT tables. Small anomalies began to show up, and these could be explained only as discontinuities and
could be compared with the results obtained from
inconsistencies in the vapor-pressure-temperature relationship. In the spring
14-4.
of 1955,
Table 14-4
Clement and
his
developed an empirical
Useful Fixed Points in Low-temperature Physics
coworkers at the U.S. Naval Research Laboratory
PT relation
that resolved practically all the existing
discrepancies within a few millidegrecs.
Gas
Critical
point
Carbon dioxide
Nitrogen
Helium IV
126.26°K
atm
33°K 12.8 atm
III
216.6°K 5.11
atm
77.35°K 760
63.14°K 94
20.26°K 760
13.96°K
mm
mm
54.1
Not long
H. van Dijk and P and T, vapor was taken into account as after,
)urieux derived the following thermodynamic equation between
which departure from ideality of the He well as the pressure dependence of the enthalpy of the in
194.7°K 760
mm
In
1
P= -
37.86
3.32°K 890
3.195°K 760
? >
P °" dP
5
RT + 2 » T -FrL**" dT + WrL l
,
,
Pv'"
'Rf where
P
in
is
millimeters of mercury,
/
=
59.50 J/mole, and
ia
=
.
Zo '
12.2440.
the latest values for the various properties of liquid
and gaseous helium, van Dijk published a tabic of P and T values which was just as successful as Clement's in resolving the existing difficulties and which had the further advantage of a thermodynamic background. In 1958, the table of van Dijk and Durieux was accepted by the Advisory Committee on Thermometry of the International Committee on Weights and Measures and L'sing
2.172°K
mm
lowest temperature that can be reached easily with liquid helium
liquid:
IB
mm
mm mm
sublimation point
1
mm
4.215°K 760
mm
The
Triple point
5.20°K 1718
mm
Helium
point
304°K 73 atm
33.54
Hydrogen
boiling
M.
Normal
Normal
is
about 1°K. This is achieved by pumping the vapor away as fast as possible through as wide a tube as possible. With special high-speed pumps, tem-
designated as the
Measurement isotope
He
3
"He
4
Scale of 1958," abbreviated
of the vapor pressure of the
7'
5S
.
much more
expensive light
enables one to measure temperature from the critical temperature
472
Heat and Thermodynamics
The
3.32°K. clown to 0.2°K.
14-6
very tedious measurements and calculations
He vapor 3
leading to a table of temperatures and
Roberts, Sherman, Sydoriak, and Brickwedde. scale,"
The
was accepted by the Advisory Committee
T i9
as equivalent to the
in
pressures table,
were made by
known
as the
"r62
1962 and was recognized
scale in the region of overlap.
gas thermometer and the vapor-pressure thermometer are elaborate, exacting, and sluggish devices. To measure heat capacities, thermal conduc-
The
tivities,
and
several other physical quantities of interest at low temperatures,
of small temperature changes must be made quickly and accurately. For these purposes, secondary thermometers must be used. One of the first to be employed was a resistance thermometer made of carbon. Pieces of paper with carbon deposited on them or strips of carbon prepared by painting with colloidal suspensions have two advantages. They have extremely small heat capacities and can therefore follow temperature changes quickly, and their electric resistance, which increases rapidly as the
many measurements
temperature
is
reduced,
main disadvantage
They must be
is
insensitive to the presence of a
of such
calibrated
thermometers
is
magnetic
field.
The
their lack of reproducibility.
anew each time they
are used.
In 1951, Clement and Quinell discovered that carbon composition radio resistors, made by Allen-Bradley and rated from ^ to 1 VV, had all the proper-
a low-temperature secondary thermometer- -namely, and insensitivity to magnetic fields. The reasons for these desirable properties are not understood, but to such thermometers are attributed the accuracy of much of the work done in lowten iperaturc physics since 1 951 In using small radio resistors as thermometers,
ties
high
most desired sensitivity,
in
reproducibility,
.
the plastic coating
The thermometer
is first is
removed and replaced by a
thin coat of lacquer.
then attached to the experimental apparatus or placed
The resistance is measured at a number of known from measurements of the helium vapor presnumber of such measurements has been made by J. R.
in a hole drilled for that purpose.
temperatures that are sure.
The
Clement,
greatest
who
tried
Laboratories.
The germanium
Law
473
"doped" with excess arsenic atoms and then an analysis of several of these semiconduc-
is
sealed in helium-filled capsules. In
thermometers along with radio resistor thermometers, Lindenfeld gives upper limit to the variability of the temperature corresponding to a given resistance the value +0.4 mdeg. This is better than that possible with carbon resistors. The main advantage of germanium thermometers is their tor
as an
reproducibility.
They
retain their calibration after
any number
of cycles
between helium and room temperatures. Two germanium thermometers A and B arc compared with an encapsulated carbon thermometer C (0.1 W, 68 fi at room temperature) in Fig. 14-13. Some paramagnetic salts obey Curie's law very closely down to 1 °K and even lower. If the Curie constant is determined by measurements at known temperatures, then the magnetic salt may be used as a thermometer. In the
helium range, the change of magnetization of all paramagnetic salts is To measure temperature to mdeg, the magnetization must be measured with tremendous accuracy with the aid of a specially constructed ac mutual-inductance bridge. The magnetic thermometer is of most importance in the temperature range below 1 °K, where its magnetization changes by a larger amount for a small temperature change. At temperatures below the region in which Curie's law holds and which have been represented by the so-called magnetic temperature T* = Cc%/M, the corresponding Kelvin temperatures must be determined. To understand liquid
rather small.
how
this is
1
done, consider the
isothermal magnetizations
(1
ST curves of Fig. 14-14 showing a number — 2, — 4, or — 6, etc.) and a number »
adiabatic demagnetizations (2
1
—
>
3,
4
>
1
—
or 6
> 5,
>
-
>
7, etc.).
of of
By measuring the 5, 7, etc., and by
magnetic susceptibility at each of the points numbered 3, using the known value of the Curie constant, the T* values may be obtained at these points. Taking the process 1 —* 2 as a typical isothermal magnetization, we may calculate the entropy change Si — S\ by applying the second
T dS
equation,
several empiric equations to represent the relation
between the resistance R' and T.
One
Vlog R'/T =
of the most satisfactory equations
a
+
b log R',
TdS=
is
CyfdT
+ T (W)
&f. •
(14-32)
Integrating from
1 to 2,
the experimental results within a few millidegrees. If a carbon thermometer is calibrated in the liquid helium range and then kept below 20°K, it will retain its calibration indefinitely. After wanning to room temperature, however, it will have to be recalibrated when brought back to
which
Cryogenics, Negative Temperatures, Third
we
get
fits
*-*~jr-(s&"
resistance
helium temperatures.
Since the paramagnetic
salt in
Higher reproducibility is attainable with the aid of germanium resistance thermometers developed by Kunzlcr, Gcballe, and Hull at the Bell Telephone
the right-hand integral
may
calculations
we may
(14-33)
question obeys Brillouin's equation accurately,
be evaluated, and therefore after a series of such
obtain Si
—
Sq,
.S"e
—
St,
Sg
—
St, etc.
But these entropy
474
14-6
Heat and Thermodynamics
Cryogenics, Negative Temperatures, Third
4k^
*/--
A"
T5«
Tf Above
I
known at as shown
—
>
T„iar
T,*
.S' 5
the points 3,
let
7'
max are Kelvin temperatures; below
changes are equal to
2
/•
1/
T
Isothermal magnetizations and adiabatic demagnetizations on an entropy tempera-
Fig. 14-1 ture diagram.
Now
T7*
475
Law
—
S\,
— S Ss —
S3
a graph of.?
—
Si
5, 7, etc.,
s,
,
max
etc. .S3
are magnetic temperatures.
Since the 7"* values are
can be plotted against
7'*,
in Fig. 14-1 5a.
wc have performed the adiabatic demagnetization minimum temperature, characterized by the magnetic
us imagine that
3 and arc at the
s-s
3
(c) 2
3
4
5
Temperature, °K Fig. 14-13
Comparison of carbon and germanium thermometers. (P. Lindenfeld, I96d.]
Fig. 14-15
Kelvin temperatures at the points 3, 5, 7,
slope 0} the curve in (c) at the designated points.
etc.,
are obtained by measuring the
476
Heat and Thermodynamics
temperature
Ta
.
With the
14-7
some
aid of
Cryogenics. Negative Temperatures, Third
Law
477
suitable heating device (such as a
gamma
gold heating coil used by Giauquc, the absorption of
Kurti and Simon, or magnetic hysteresis used by dc Klerk),
rays used by
let
300
us measure
the heat absorbed at zero magnetic field in going from 3 to
5, from 3 to 7, and from 3 to 1. Since these values of heat arc measured in zero magnetic field, no work is done, and hence the heat is equal to the internalenergy change V — V%, These internal-energy changes are plotted against
from
3 to 9,
200
7"* in Fig. 14-156.
Combining graphs
U—
U$
is
(a)
and
plotted against
S
(b),
—
S3.
we
get the graph of Fig. 14-1 5c, in
Since the point 3
which
always the same, Us of the curve at any point is
and 5a are constants, and therefore the slope dM and %? (dU/dS)v=o- Since T dS = dU -
%
(dU/dS)v=o =
=
is
100
0,
T, 100
200
procedure has been followed and a table of values of 7'* and T, such as Table 14-5, has been obtained, other thermometers such as radio
Once
300
400
500
this
1/T, deg-' Fig. 14-16
Relation between
T* and T for
a sphere of
CMN.
(Frankel, Shirley,
and Stone,
1965.)
Calibration Data for a Sphere of
Table 14-5
CMN
(Compiled by R. B. Frankel, D. A. Shirley, and N. J. Stone, 1965.)
14-7 %i/Ti, kOe/deg 1.0 1.9 2.9 3.8 4.6 5.4 6.2 6.9
S/R (calculated)
1/T*
0.691
20
0.686 0.678 0.667 0.654 640 0.625 0.610
40
.
60 80 100 120 140 160
1/7; deg-' 20 40 60 80 100 120 140 160
Ki/Tt, kOe/deg 7.8 8.75 10.2 12.1
14.2 15.5 18.0
S/R (calculated)
0.590 0.567 0.529 0.477 0.420 0.384 0.321
i/r*
1/7",
deg-"
Magnetic Refrigerator
1
1
At the conclusion
180
181
200 230 260 290 300 310
210 249 305 383
430 500
of an adiabatic demagnetization,
the temperature because of the unavoidable heat leak that is present in every cryostat, no matter how well designed. If the experiment under consideration can be performed quickly and does not itself involve too large a dissipation of energy into the system, adequate results may be obtained. When, however, the experiment itself requires a dissipation of about
immediately
starts to rise
50 ergs/s, a continuous refrigeration
is needed. For this purpose, Daunt and Hccr designed a magnetic refrigerator, operating in a cycle, in which iron ammonium alum could be magnetized while in contact with one helium bath (hot reservoir) and demagnetized when in contact with another helium
bath (cold reservoir). resisters,
germanium
crystals, or other
paramagnetic
salts,
can be calibrated.
The most widely used thermometric salt is a single crystal of CMN cut into the shape of a sphere. The calibration curve between T* and T, shown in Fig. 14-16, is the combined effort of many physicists, but the final corrections were made by Frankel, Shirley, and Stone, who used radioactive, oriented cerium ions whose temperature was measured by noting the asymmetry of 7-ray emission.
A
schematic diagram of the apparatus is shown in Fig. 14-17. Notice that two thin metal rods marked "thermal valve." These rods arc made of lead and connect the paramagnetic salt contained in the "working cell" with the upper helium bath and with the lower "reservoir cell," respectively. there are
At any temperature below 7.22°K, lead becomes superconducting and, at the same time, becomes a very poor heat conductor. If, however, the superconductivity is removed with the aid of a magnetic field of only a few hundred
478
Heat and Thermodynamics
Cryogenics, Negative Temperatures, Third
14-8
— Vacuum
Heer have been incorporated into a magnetic
line
Law
479
refrigerator (constructed
A. D. Little Co. of Cambridge, Mass.) composed of
all
by
necessary parts that
perform automatically. Helium bath
Thermal valve
Polarization of Magnetic Nuclei
14-8
V alve magnet
The magnetic moments
— Working .
cell
— Main magnet
e
magnet
of the chromium, iron, and gadolinium ions are uncompensated spins of the electrons that surround the respective nuclei. When a paramagnetic salt containing, for example, the chromium ion with J = | is placed in a magnetic field of the order of lO 4 Oc and the temperature is lowered to about 1°K (so that the ratio 9f/T is of the order of 10 4 Oe/deg), the ionic magnets become partially oriented in the direction of the field. A convenient measure of this partial orientation is provided by the ratio of the magnetic moment to the maximum value or saturation value
due
to
M
sat.
mal valve
Reservoir
M
When
polarization.
this ratio
From Eq.
is
expressed in percent,
M
cell
and under the conditions
specified,
m^¥
X
2
kT In Fig.
14-2,
/}j(1.3)
often called the magnetic
R (g^K = Dj kT
M,
— Experimental space
it is
(14-16),
=
X
0.9
1.4
X
0.76; that
X x
10- 20
10-is
is,
10< 1.3.
j
the ionic magnets are 76 percent
polarized.
The
particles inside the nucleus of an
nuclear magnetism.
number Helium bath
The
/ that plays the
nuclear magnetic
same
moment
is
role as the
much smaller
of almost 2000. In the expression for a Fig. 14-17
Magnetic
oersteds (provided
refrigerator of
Daunt and Heer.
substitute the
by the two "valve magnets"), the lead immediately
becomes a good heat conductor. During magnetization of the salt in the working cell, the upper thermal valve is magnetized and the lower is not. During demagnetization, the reverse is the case. Each cycle takes about 2 min, and at the end of about 30 min the lower reservoir reached a temperature of about 0.20°K, with a heat extraction rate of about 70 ergs/s. The design features of Daunt and
mass of a proton,
M.V
X
have spins that give
is
rise to
quantum atomic quantum number J. A characterized by a
than that of an atom, by a factor
Bohr magneton
j*b
=
eh/Amnc,
if
we
get a nuclear magneton where 0.9
m,,/m B 5
also is
X
IP" 20 1840
10- 24 crg/Oe.
nuclear magnets are subjected to an external magnetic
field,
the lowest
a number of separate states, each of characterized by a nuclear magnetic quantum number that may
energy level of the nucleus
which
we Mb
=
« When
atom
resultant nuclear spin
splits into
480
Ilcat
and Thermodynamics
14-8
take on discrete values corresponding to discrete orientations of the nuclear
magnetic moment with respect to the external magnetic field. These values range from —I to -\-I in integral steps an exact replica of the atomic situation, where m took on values from — ./ to -\-J in integral steps. The calculation of the nuclear magnetic moment M.\ parallels perfectly that for the atomic moment, so that
—
Mx
= Xnxgxllh
(
(14-34)
where git is the nuclear splitting factor, which wc shall take as equal to 2. Suppose that we have a nucleus with / = | and want to produce a nuclear polarization of 76 percent. Then,
Cryogenics, Negative Temperatures, Third
Law
481
The method
of nuclear polarization that has been most fruitful was sugby Gortcr and Rose and involves the behavior of a nucleus in the very strong local magnetic field (over 100 kOej produced by its own electronic structure. To understand this, let us consider the implications of the theo-
gested
first
and experimentally verified result that the magnetization of an isolated spin system (whether atomic or nuclear, provided that neither nor T is close to zero) is a function oi%jT only. If, for example, 50 percent magnetic saturation is achieved with a particular paramagnetic salt at a temperature of 1°K with a field of 10,000 Oe, then at a temperature of 0.01°K a field of only 100 Oe would be needed to produce the same magnetic retically derived
%
polarization. In other words,
1°K and demagnetizing field
if
the proper material were used, starting at
from a field of 10,000 Oe to a would drop (p^/T = const.) to about
(or "degaussing")
of 100 Oe, the temperature
0.01 °K, but the magnetic polarization would
Bi{
fxwWfkT) =
and
kT 7f_
or
1.3
T
2
X X
1.4_X 10-' 5 X 10- 24
still be the same. One could also demagnetize to zero field, thereby reaching a somewhat lower temperature, and then raise the field to 100 Oe and still have available about 50 percent polarization. Goiter and Rose realized that these polarized ionic magnets would give rise to a unidirectional local field at the nucleus of each ion which
0.76
= 2
1.3
X
would be much larger than any
10 7 Oc/dcg.
field
then achievable in the laboratory
(before the advent of superconducting magnets)
— of the order of 100,000 Oe
or more. therefore,
If,
we
used a
field of 2
X
10* Oe, the nuclear-spin system
-3 °K have to be at a temperature of 1
in
order
to
would
provide a nuclear magnetic
we could settle for a smaller polarization, we 10~ 2 °K. Lining up nuclei in the same direction is a very valuable procedure for the physicist. If the nucleus is radioactive and emits alpha particles or beta particles or gamma rays, it is important to know whether these emanations arc emitted as abundantly in one direction relative to the magnetic moment as in another. If they are emitted equally in all directions, one refers to their symmetry or isolropy. A preferred direction indicating asymmetry or anisotropy enables the nuclear physicist to accomplish the polarization of 76 percent. If
could start
at, say,
following:
1
Test various
3
This experiment was carried out several times by Roberts and his coworkers Oak Ridge Laboratories. They polarized Mn and Sm nuclei and detected
their polarization
trons from the
When
by measuring
Oak Ridge
their ability to scatter polarized slow
radioactive cobalt nuclei
with different intensities
neu-
reactor.
Co 60
are polarized, 7 rays are emitted
By comparing the
in different directions.
intensity
one direction with that at right angles to this direction, a quantity called the anisotropy is defined. Experiment shows that the anisotropy is a sensitive in
function of the temperature, so that,
from previous experiments,
a
if
this function
measurement
has been determined
of y-ray anisotropy enables
one
to obtain the temperature. conservation laws that are
supposed to hold during nuclear
disintegrations.
2
at
Obtain information about the shape of the nucleus. Obtain numerical values of certain constants needed
One of the most spectacular experiments in nuclear cryogenics was performed by Ambler, Hudson, and Wu at the National Bureau of Standards in Washington, IXC, in 1957. Lee and Yang, who were later awarded the Nobel prize, suggested that the nuclei of Co 60 in undergoing radioactive decay, might emit beta particles (electrons) more abundantly toward one magnetic pole than toward the other. To test this hypothesis, polarization of ,
in the
theory of
nuclear processes.
Nuclear polarization has been achieved by using a very low temperature and a very large field; but this procedure, known as the brute-force method, has not been available in many laboratories.
the cobalt nuclei was necessary. Beta-ray counters were then used to show whether there was a different reading in the direction of the north poles than in the reverse direction. The apparatus is shown in Fig. 14-18. The Co 60 obtained by neutron bombardment of nonradioactive Co 59 was introduced ,
,
482
Heat and Thermodynamics
Cryogenics, Negative Temperatures, Third
14-8
483
Law
CMN
-—
crystals was horizontal, so that The strong magnetic axis of the isothermal magnetization and adiabatic demagnetization to zero field could be accomplished with a horizontal magnetic field. By this means, the
Pumping tube
CMN
—
A
was then slipped over the outer dewar, and a vertical field of about 100 Oe was used 60 ions without appreciable warming of the CMN. The to polarize the Co housing and
Lucite rod
Co 60
layer were cooled below 0.01 °K.
solenoid
60 strong local fields at the nuclei of the ions polarized the Co nuclei, the direction of polarization depending upon the direction of the current in the sole-
noid. The north poles could therefore be made to face either toward the anthracene crystal or away from it. The beta-particle ejection from Co 60 is accompanied by gainma-ray emission. Although they are not shown in Fig. 14-18, there were two gainma-ray
-Vacuum space
counters to detect and to measure any Lucite light pipe not in
the
gamma-ray
anisotropy. Since
gamma-
ray anisotropy had already been measured as a function of temperature, it served as a convenient thermometer. The curve in Fig. 14-19 shows that there
vacuum space
are
many more
beta particles emitted in a direction opposite to the solenoid field, that
the south poles of the
Co 60
nuclei emit beta particles
the north poles. This result
known
in direct
is
is,
more abundantly than
contradiction to a nuclear principle,
which had assumed that cerbehave the same way for one configuration of
as the principle of the conservation of parity,
tain nuclear processes should
its mirror image. The experimental proof that parity is not conserved in beta decay has had a profound effect on theoretical and experi-
the nucleus as for Scintillating
anthracene
crystal in the
vacuum space
Cerium magnesium
nitrate
mental physics.
(CMN)
crystals (strong magnetic axis
1.3
horizontal) 1-2
00
Fig. 14-18
polarized
Co eo
c
Apparatus of Ambler, Hudson, and Wu to measure (i-particle emission from nuclei. (Liquid He and Mt dewars and fi-ray counters not shown.)
into the crystal lattice of cerium
magnesium
nitrate in the
particles emitted
by Co
110
(in
small crystal of anthracene.
The
decaying to Ni 60) produced scintillations
X
10
OO
•
P »°n
'
-
9
0.8
•/^
r
a
•
i
7f
\
beta in
0.7
a
4
6
The light flashes traveled up a light pipe, consisting
of a 4-ft lucite rod whose upper end communicated with a photomultiplicr
tube and counter.
X
c 3
form of a thin
layer lying in the bottom of a cup-shaped housing of this material.
1.1
\
Time Fig.
14-19
Asymmetry
in
denotes increasing temperature.
10
8 in
12
14
16
18
20
minutes
beta-particle emission
Maximum polarization
from polarized
nuclei.
Increasing time
is at the lowest temperature.
484
Heat and Thcrmodv
14-9
14-9
485
Cryogenics, Negative Temperatures, Third T.aw
Production of Microdegree Temperatures
by Nuclear Demagnetization
'/
Electronic
3
stage
10- Z °K
Kr 2o K
10- ? °K
K=
9f< 200 0c
K=
Since nuclear magnets are only about one one-thousandth as strong as
around 0.01 °K and from 50 to 100 kOe. We have seen how local fields of this magnitude may be provided by the uncompensated spins of the electrons circulating outside each nucleus itself. If these polarized nuclei could then be made to undergo a reversible adiabatic demagnetization, they would cool off to a temperature in the neighborhood of 10 -5 °K. Since these nuclei occupy a thin layer on a large crystal at 0.01 °K, however, any loss of polarization would be
ionic magnets, their polarization requires temperatures fields
much more
isothermal than adiabatic.
You cannot
rO
-5 cm
£
Q i
expect a few nuclei to
cool off a big crystal.
The only method 10
-3
°K
that has been used so far to achieve temperatures below
involves a double process, consisting of an ionic demagnetization
followed by a nuclear demagnetization.
Two
separate magnetic fields sup-
two separate magnets are used, as shown schematically in Fig. 1 4-20, a diagram prepared by Kurti of Oxford, in whose laboratory such experiments have been carried out. In each of the four parts of this figure, the electronic stage represents a mass of chromium potassium alum in which are embedded 1500 enameled copper wires, each with a diameter of 0.0003 in. The copper wires continue for a distance of about 8 in. and are then bent over and bound together to form the nuclear stage itself. The first part of the cooling is done with the aid of chromium ions, and the second part by copper nuclei. The fine, plied by
insulated copper wires serve three purposes:
1
2 3
Provide a heat-conducting
medium between
the nuclear
and
the
1"K
Nuclear
^~0
stage
(a) Fig. 14-20
10 _2o K
~10- 2 °K
~10" 5o K
K=
^
^=0
(c)
(d)
(£>)
The four
steps in nuclear cooling. (N. Kurti.)
that even a minute amount of heating such as results from a small pin dropping through a height of one-eighth of an inch would warm a bulky specimen of several ounces from one-millionth of a degree to the starting temperature of one one-hundredth of a degree and thereby spoil the experiment." Even the eddy currents induced in the copper wires by virtue of slight
electronic stages.
variations of current (ripples) in the
magnet
Minimize eddy currents induced by demagnetization. Produce a low temperature by nuclear demagnetization.
aid of the metal ripple shield
shown
in Fig.
supplied
by solenoids
in
must be prevented with the The magnetic fields were which currents of thousands of amperes were coils
14-21.
maintained.
The
four steps in Fig. 14-20 are as follows:
The
lowest temperature ever obtained anywhere (1968) is 1.2
One (a)
Isothermal magnetization of the electronic stage.
demagnetization process
(b)
Adiabatic demagnetization of the electronic stage and cooling of nuclear stage to 10" 2 °K.
tronic stages.
to
Isothermal magnetization of the nuclear stage.
(d)
Adiabatic demagnetization of the nuclear stage, with an accompanying temperature drop to about 10~ 5 °K.
The experiment
is
not as simple as
it
sounds.
To
quote Kurti, "The
10"~ G
°K!
is the heat transfer between the nuclear and elecDuring the isothermal magnetization of the nuclear stage, this transfer must be good. During the following demagnetization, it must be poor. In the experiments of Kurd's group, the fine copper wires represent a compromise that served both purposes only moderately well. Another diffi-
culty
from the nuclear stage by a distance great each magnetic field to its own paramagnetic particles. Both of these problems can be partially solved by a clever method conceived by Blaissc. Suppose the nuclear stage constitutes a core completely surrounded is
enough stringency of the conditions to be satisfied can be illustrated by remarking
X
of the biggest experimental difficulties to overcome in a double
to separate the electronic to confine
486
14-10
Heat and Thermodynamics
—
,
Liquid
in
N,
at
-3 °K.
Then demagnetization from
-7 °K. This should give about 10
would be no
Vacuum space He"
cooling to about 10
Cryogenics, Negative Temperatures, Third
method would have
1CH
Law
Oe
487 to zero
the advantage that there
heat-transfer problems, since all the operations are
made on
the
nuclear magnetic subsystem.
0.9° K
He 3 ato.35°K KCr alum
— Ripple
at
Negative Kelvin Temperatures
14-10
1CT 2 °K
Let us recall the original definition of the Kelvin scale of temperature: Two Kelvin temperatures are to each other as the heats transferred during isothermal processes at these temperatures, provided that these isothermal
shield
processes terminate on the same adiabatic surfaces. If
Q and Q s are the absoand T„ respectively,
T
lute values of the heats transferred at temperatures
the original Kelvin definition provides the relation
Copper
nuclei at
10
6
°K
r-r.£ an arbitrary standard, the choice of a number for Ts is also arbitrary. If it is chosen to be negative, then all temperatures would be expressed by negative numbers. Whether Ts is chosen positive or negative, as Q is made smaller and smaller in any unordered way, the limiting value of Q is zero (i.e., the least amount of heat that can be transferred is no heat If T, refers to
at
all),
and
temperature Fig. 14-21
Cryostat for nuclear cooling (symbolic).
1
Magnetize isothcrmally at 1°K Insulate thermally.
3
Rotate the field
to the
in this direction,
4
it
in the
x
own
absolute zero, and
when
A
mean temperatures
the Kelvin scale
is
CMN
is practically nonmagnetic y direction. Since therefore undergoes an adiabatic demagnetization
adiabatic demagnetization.
Another method, suggested by Kittel, is to polarize the nuclear magnets by microwaves at a temperature of 1°K and in a steady magnetic field of, say, 10 4 Oe. Removal of the high-frequency field adiabatically should result
if
T
In other words, the lowest
is zero.
negative temperatures have any meaning
colder than absolute zero!
defined in the usual
way with
But what
is
meant
T = +273.16°K? s
clue as to the meaning of negative Kelvin temperatures
the expression for temperature used
in statistical
is provided by thermodynamics,
r=
-
direction.
and its temperature drops, even though the. field is still there. Wait until the cold CMN has cooled the nuclear core, and then reduce the magnetic field to zero, thereby cooling the core by its
is
at all, they cannot
by a crystal (or a group of crystals identically oriented) of cerium magnesium nitrate with its strong magnetic axis pointing toward, let us say, the x axis. Suppose that we perform the following operations:
2
therefore the lowest value of
The most mole of
familiai-
thermodynamic
mole of ideal gas or a As the temperature is higher levels. This requires more
systems, such as a
crystal, have an infinite number of energy levels.
more and more atoms are raised to and more, energy, and results in greater and greater disorder as the atoms arc distributed over more and more states. As the energy goes up (positive dU), the entropy also goes up (positive dS); hence the ratio dU/dS is positive. For raised,
T
to be negative, an increase of energy would have
of entropy! This obviously cannot take place
number
of energy levels.
to be
when
accompanied by a decrease
a system has an infinite
488
14-10
Heat and Thermodynamics
Cryogenics, Negative Temperatures, Third
Law
489
energy level, which is a state of minimum disorder, or zero entropy. When the two energy levels are equally populated, the internal energy of the system is Me/2 and there is maximum disorder and, hence, maximum entropy. If and r atoms are in the upper energy level, U = A e, and again we have when all curve has a positive left half of the entropy. The minimum disorder, or zero
N
slope,
and therefore T(dU/dS)
is
positive.
The
right half, with negative slope,
is the region of negative temperatures.
As we
start at the origin in Fig. 14-22
and go
to the right,
we proceed in the
direction of increasing energy, increasing hotness, and therefore increasing temperature. At the position of maximum entropy, where both energy levels
are equally populated, the temperature is infinite. Beyond the maximum the temperature must be hotter than infinity. Hence, negative temperatures are hotter
than infinity/ If we object to this conclusion,
we
are really objecting to the
original definition of the Kelvin scale. Everything would be much neater if we defined a new quantity, the negciptemp, equal to the negative reciprocal
of the Kelvin temperature; thus, Fig.
1
4-22
two energy
Relation between entropy and energy of a system of particles that can exist in only levels.
71= Another way of looking
at the
matter
is
with the aid of the Boltzmann
equation,
When S
is
Fig. 14-23. is
til A'i
T=
at
= f— c«*-«i)/W
-
X
is shown in Here absolute zero is at an infinite distance to the left; the origin =o, and negative temperatures are to the right of the origin, with
plotted against the negciptemp, the resulting curve
If the system has an infinite number of energy levels, an increase of temperature produces increased populations of higher and higher energy levels, but
no energy level ever gets populated more than the one below it, so that the ratio Ni/Ni is always less than 1 and T is positive. At V = =c A 2 would be equal to ivj, but this would require an infinite amount of energy because of the infinite number of energy levels! Evidently, for T to be negative, N2 would have to be larger than iVjj that is, the upper energy levels would have to be populated more than the lower ones. This would require even more than infinite energy which is even more than nonsensical. We conclude, therefore, r
,
that in the case of an ordinary system which has an
infinite
number of energy
levels,
negative temperatures are an absurdity.
But what about a system which has only a finite number of energy levels? Suppose, for the sake of argument, that a system were capable of existing in only two energy levels. Let the system consist of particles and the levels
N
have energies external
field.
and
e,
where
an atomic constant, independent of any the relation between entropy 5 and internal 14-22. At zero energy, all A' atoms are in the lower
in Fig.
-200
100
Negciptemp
e is
The curve showing
energy £7 is shown
-300
Fig. ture).
1
4-23
Graph
71
=— \/T
of entropy vs. 71, the negciptemp {negative reciprocal of Kelvin tempera-
490
Heat and Thermodynamics
"minus zero" an infinite distance to the right. It is psychologically more appealing to have the coldest possible temperature represented by - =o and the hottest conceivable temperature by tively,
with both
+
and
oo
— °o
+ eo, instead of by
finite
number
producing a papulation
and —0, respec-
between.
Therefore, to achieve negative Kelvin temperatures,
tem with a
of energy levels and,
— that
somehow
we must
find a sys-
or other, succeed in
an equilibrium (or near-equilibrium) state in which there are more particles in upper states than in lower ones. To reduce the temperature of a substance far below 1°K, the magnetic and thermal properties of a magnetic subsystem (ionic or nuclear) were used. The purpose of such experiments was to cool the entire substance, not just the subsystem. To accomplish this, it was necessary to satisfy the followinginversion
is,
The magnetic
ions must interact among themselves with sufficient strength and speed that (like the molecules of a gas) statistical equilibrium can be assumed and a definite temperature can be attributed to the ionic subsystem.
2
3
The nonmagnetic
particles (called, for simplicity, the lattice) must have practically no heat capacity in the low-temperature region under consideration. Equilibrium between the magnetic-ion subsystem and the lattice must be attained fairly rapidly.
achieve negative temperatures, we must make use of the magnetic and thermal properties of a nuclear magnetic subsystem under the following conditions (one of which
is
the same
as,
in emitting a photon, and the nucleus absorbing this photon would go from a lower to an upper state. These interactions play the same role in the achievement and maintenance of equilibrium as the collisions
lower state
between gas molecules. Experiments on the nuclear magnetic subsystem of LiF take place in a region of temperature and field similar to that enclosed within the vertical dashed lines of Fig. 14-6. In this region both S and arc functions aiC¥/T only; therefore, in an adiabatic demagnetization, since S is constant, both
M
M
much lower than
which the experiments were made. During an and dS are both zero, and therefore the. expression for the temperature dll/dS becomes indeterminate. The previous analysis of negative temperatures presupposed that the energy-level spacing was an atomic constant. With a nuclear (or ionic) magnetic subsystem, however, the level spacing At depends upon %, and the magnetic energy gp.% is not internal but external potential energy. To obtain a useful and appropriate expression for T, we use the magnetic enthalpy H*, where those at
adiabatic change of
field,
dll
wc
The
nuclear magnetic subsystem comes to equilibrium with lattice
is
room temperature, with
a large heat capacity.
Equilibrium between the nuclear magnetic subsystem and the attained slowly
enough
(from, say, 2
min
lattice
to several hours, that
experiments can be performed on the nuclear subsystem in this time interval, as though the subsystem were isolated.
The system found by Pound, Purcell, and Ramsey in 1951 to satisfy the conditions for the production of negative temperatures is the subsystem consisting of the nuclei of the lithium ions in a LiF crystal. These were found to come to equilibrium among themselves in 10~"B sec, to require about 2 min more
to
come
to equilibrium with the lattice,
and each
to
m
m
T=
get
Choosing at
= du - ?fdM = T dS - M d9f,
(14-35)
(14-36)
itself
very rapidly.
The
dH*
Since
while two are entirely different from, those
just listed):
or
491
energy level that is split into only four nuclear magnetic states (7 = #) by an external magnetic field. The weak interactions among these magnetic nuclei involved emission and absorption of photons produced by transitions between some of the four states. That is, one nucleus would go from an upper to a
H* = U - %M. To
is
Law
are constant. The internal energy is a function of V only and TfifT and has an appreciable value and variation with T only at temperatures very
conditions:
1
Cryogenics, Negative Temperatures, Third
14-10
have a lowest
U
to
S and %. In
Fig.
1
4-24,
five different values of
H* =
M
—?
have the value zero,
functions o(9f/T, then
M
is
%,
.S"
a function of .S",
with the lowest curve (at 9£\) referring to the
(dH*/dS)&, the slope of any constant-field curve, and that the vertical line a —> b, at constant S, repreT sents an adiabatic demagnetization from a large field %± to a small one 9^=, during which and 9&/T remain constant. In the upper half of Fig. 14-24, -\-7fM is plotted against S for reversed fields, that is, for negative values of 7£. The slope of every upper curve at every point is negative, and the process c — d represents an adiabatic magnetization during which the nuclear subsystem cools from —10 to — 4()0°K. largest field. Notice that is
the temperature
M
•
492
Heat and Thermodynamics
14-11
^r^^^
Cryogenics, Negative Temperatures, Third Lai
The Experiment
14-11
of Pound, Purcell,
493
and Ramsey
d(7 = -400 °K) in
In the experiment of Pound, Purcell, and Ramsey, the crystal was placed a magnetic field of about 6300 Oe and allowed to come to thermal equilib-
rium at room temperature, 300°K. Under these circumstances,
~^^^^^^^ -a?
\
\\
—^^ Region
of
^^\^^
\
~~~~~~^-^
a,
Bt
\
\
and the
\\\
,T
fractional polarization
5
(small)
/
___
^-"/y/
'
^^**^'
6(r=5°K)
S
U Mni
= -.
T= +
.
///
^4 j*'
Region of positive
^^^
temperatures
/ 1 / /
5
1.4 1.5
X IP" X 6300 X 10- 16 X 300
X
24
10-".
=
(fl)
—^
a,
is
«
-
1
3
=
^
X
2
the Brillouin function reduces to
\
\
^~\
temperatures
-^ c(r=-io°K)
At such values of
>>
v.
negative
_ gvHJ&P _ kT =
1.3
X
X
1.5
10~ 6
X
10-»
.
Although this is a very small value, the methods of nuclear magnetic resonance (nmr) are still effective in showing the difference between the number of nuclei whose magnetic moments are in the direction of the field and those whose moments are opposite. The crystal is placed inside a small coil that is connected
in series
with a variable capacitor.
The
coil
and capacitor form which
the resonant circuit in a radio-frequency oscillator, the frequency of
may be varied by adjusting the variable capacitor. The output of the oscillator is
Jf±——
is
^^*"
JF and
Xj^^-
AM
observed with an ordinary receiver. If the frequency of the oscillator adjusted to a value v%, where h>$f
=
Ae
vipf
=
6300 Oe)
=
gxnx?^,
2X5 X 6.6
1Q- 2 '
X
63 00
10-"
X
then some of the Li nuclei with their spins parallel to the so that their spins
#,
Fig.
1
4-24
become
antiparallel,
with an
10'
Hz.
field will
absorption of energy,
be flipped
and some
of the Li nuclei with their spins antiparallel to the field will be flipped to the
(large)^
a(T=
3C
with an emission of energy. But since these two processes occur with equal probability, and since there are slightly more nuclei with parallel position,
0°K)
Magnetic enthalpy
H* vs.
entropy
S of nuclear magnetic subsystem at five different
—
—
magnetic fields and for the same fields reversed, a > b, adiabatk demagnetization; b magnetic field reversal; c—>d, adiabatk magnetization; d e * a, cooling through
— —
constant large magnetic field.
>
c,
rapid
infinity at
their spins parallel than antiparallel, there
is
a net absorption of energy, which
observed as a drop in the amplitude of the oscillator output, and hence as a drop in the output of the receiver. This drop in output corresponded to a positive temperature for the nuclear subsystem at 300°K. is
AM
494
Heat and Thermodynamics
14-11
Cryogenics, Negative Temperatures, Third
taining the LiF crystal, the magnetic field was reversed in
+ 6300 Oe
a time of 0.2
(Such a
y.s,
during which time the
Law
—
a value of about nuclear magnets could not follow
field reversal is highly irreversible.)
to
In this process,
the slight polarization parallel to the field at
/;
495
/;
—
>
1
00
Oc
the field.
c in Fig.
14-24,
(due to more nuclear magnets
lower states than in upper ones) became a polarization opposite the field (more nuclei in upper states than in lower ones), with a temperature about — 10°K. The next process, c—*d, represents the adiabatic magnetization
in
accomplished by putting the crystal back into the reversed field of 6300 Oe, during which the temperature cooled from —10 to — 400°K. The last step,
Nuclear magnetic polarization
—> e —* a, was the inevitable cooling due to interaction with the lattice, in which the temperature decreased from — 400°K to — to (which is the same as +«>) and then went back to + 300 o K. The success of the experiment depended on performing the field reversal in a time less than the Larmor precession period and bringing the crystal d
+ 100 Oe
Fast passage 6 (0,2 X 10" sec)
ow passage 2 sec
-lOOOe
Time
back to its place in a time less than the relaxation time for equilibrium between nuclear subsystem and lattice. A symbolic ficld-vs.-time graph is shown in Fig. 14-25, with a comparison between the fast-passage (negative temperature) result and a slow-passage one. The small arrows indicate the magnetic polarization of the Li nuclear subsystem.
-10°K
In the 2-min period
which the nuclear subsystem was at negative nmr apparatus showed an increase at v^, indicating a net emission of energy from the Li nuclei and
temperatures, the in signal
in
AM
receiver of the
thereby proving the existence of negative temperatures. Figure 14-26 shows the changes in the spacing and the populations of the energy levels during the experiment of Pound, Purccll, and Ramsey, which
^down
N,
Nt 7ta = 6300 Oe
Ni
%h=
100 Oe
K
c
= -100
0e
^- = -6300 Oe rf
-6300 Oe
Fig. 14-25
Relation between nuclear magnetic polarization
and
the magnetic field for slow
passage compared with that for fast passage.
The next step, a —> b in Fig. 14-24, was to remove the crystal from the magnetic field of 6300 Oe to a coil in a field of about 100 Oe, reversibly (slowly) and adiabatically, during which the polarization (parallel to the field) remained constant and the temperature presumably fell to about 5°K, although no attempt was made to measure this nuclear-spin temperature. In a field of 100 Oe, a lithium nuclear "magetic top" undergoes precession with a period of about 1 /*s. By discharging a capacitor through the coil con-
- (&)•
(a) Ad. demac
field
Fig. 14-26
(d)
(c) Very fast
Ad. mag.
reversal
Steps in the experiment of Pound, Purcell, and Ramsey.
496
Heat and Thermodynamics
in the opinion of this
of
modern
Cryogenics, Negative Temperatures, Third
14-13
author ranks
among
the most significant experiments
times.
behaves, but at negative temperatures. Since, by definition of the Kelvin
Te
Qc
Thermodynamics
at
when thermodynamics takes some peculiar twists at negative temperatures, but much remains the same as at positive temperatures. Take, for example, the entropy principle, which states that the sum of all the entropy changes accompanying a natural irreversible process is positive. Suppose
Q
units of heat leave a hot reservoir at a temperature, say, of
and enter a colder
reservoir at, say,
— 100°K,
(Recall that the hottest negative temperature
is
shown and the
as
—0
whereas the entropy change of the colder reservoir entropy change is
-Q
which
is
positive, just as it
In Fig. 14-276
is
is
=
-100
i
V
coldest
= + J2_ -100
is
is
+ Q/ — 100.
done
by the engine,
on the engine in
to flow naturally
— ».)
— Q/ — 50; The
total
To
100
a heat engine
get
from the hot
W units of work
at negative temperatures,
to the cold reservoir.
a heat engine operating between reservoirs you would have to make use of the device shown out of
where Qc. units of heat arc taken from the cold reservoir were a refrigerator). Then a smaller quantity Qn would go into the hotter reservoir, and the rest would be available for work. But the hot reservoir could be dispensed with, for the Qn units of heat would naturally flow back to the colder reservoir. The net result would be that Qc — Qh units of heat were extracted from the colder reservoir and converted comin Fig. 14-27c, (as
way
to be
\4-27a. is
_Q_
to imitate the
work would have
much heat must
work IV being done
— 40°K
with positive temperatures.
shown an attempt
Q/i units of heat leave the hotter reservoir, twice as
enter the colder one. Therefore, instead of
order not to violate the principle of the conservation of energy. But the device depicted in Fig. 14-27 b is an expensive gadget for doing a job that requires no device at all. If all you want is to push heat into a cold reservoir, it is sufficient merely to allow Qn
in Fig.
Since heat leaves the hotter reservoir, the entropy change
-50
~ 5Q
Negative Temperatures
Classical
that
497
scale,
9dL='IJi= 14-12
Law
though
it
pletely into work, in violation of the Kelvin-Planck statement of the second
law. This
is
the only principle of classical physics that
at negative temperatures
Up
—but
it is
an important and
is
violated
by systems
interesting one.
to the present time, the only real use for systems at negative
temperabeen in the rapidly expanding field of mascrs and lasers. Perhaps, the future, experiments on heat engines and refrigerators will be per-
tures has in
formed at negative temperatures. Then
14-13
it
will truly
be fun
to
be an engineer.
Third Law of Thermodynamics
We have seen how the Joule-Kelvin effect is employed to produce liquid helium at a temperature below 5°K. The rapid adiabatic vaporization of liquid helium then results in a further lowering of the temperature to about 1°K with He 4 and to about 0.3°K with He 3 The magnctocaloric effect is .
then used to lower the temperature of a paramagnetic compound (magneticion subsystem plus lattice) to about 0.001 °K. In principle, it is possible to
achieve
still
lower temperatures of matter by repeated applications of the effect. Thus, after the original isothermal magnetization,
magnctocaloric Fig. 14-27
(a)
Spontaneous flow of heat in the direction of decreasing temperature, (b) something that requires no device, (c) A heat engine thai could be used
costly device for doing
convert heat
Qc
— Qn from
the cold reservoir completely into work.
A to
the
first
adiabatic demagnetization might be used to provide a large
of material at temperature
Tn
amount
a heat reservoir for the next isothermal magnetization of a smaller amount of material. A second adiabatic to serve as
498
demagnetization might then give
The
rise to
a lower temperature 7W,
and
so on.
question which naturally arises at this point
caloric effect
is whether the magnetoused to cool a substance to absolute zero.
may be
Experiment shows that the fundamental feature of all cooling processes is that the lower the temperature achieved, the more difficult it is to go lower. For example, the colder a liquid is, the lower the vapor pressure, and the harder it is to produce further cooling by pumping away the vapor. The same is true for the magnctocaloric effect: if one demagnetization produces a temperature 7/t, say, one-tenth the original Ti, then a second demagnetization from the same original field will produce a temperature 7/2 which is also approximately one-tenth of Tf%, Under these circumstances, an infinite number of adiabatic demagnetizations would be required to attain absolute zero. Generalizing from experience, we may state the following: By no finite series of processes is the absolute zero attainable. This is known as either the principle of the. unattainability of absolute zero or law of thermodynamics. Just as in the case of the second law of thermodynamics, the third law has a number of alternathe unattainability statement of the third
Another statement of the third law is the experiments leading to calculations of the way the entropy change
tive or equivalent statements. result of
of a condensed system during a reversible, isothermal process
T
ASt behaves
approaches zero. For example, the entropy change of a
may
reversible isothermal compression
or be calculated from the second
AST = Since
/3
S(T,;px)
p,
fW
was agreed upon. Nernst
and (dAIfdT)^ decreases with support the view that, as
a
is
therefore accepted:
7".
is,
= =
r(W) \drjv
approaches zero. He did not think in terms of entropy and, moreover, was of the opinion that this statement and also the unattainability statement could be derived from the second law with the additional assumption that the heat capacities of all materials approached zero as the temperature
approached zero. Nernst also maintained that both statements were true both reversible and irreversible. It was mainly the experiments and arguments of Simon in the period from 1927 to 1937 that
for all kinds of processes,
made
precise the region of validity of the third law.
In order to ability
show
that the Nernst-Simon statement
statement are equivalent,
it
is
reversible process. Let us return to a paramagnetic salt isentropic demagnetization,
i—*f
Fig. 14-28
Eq. (14-37).
of Fig. 14-28.
as
during a 7"'s
as '/'decreases.
d?f,
is
a condensed system.
The
very strong to that the system
following principle
associated with any isothermal reversible process of a condensed
system approaches zero as the temperature approaches zero.
Let us call this theorem the Xernsl-Simon statement of the third law of thermoBoth this statement and the unattainability statement have had a
dynamics.
long and checkered career since the original paper by Nernst
in
1907. It
and the unattain-
necessary to derive an equation for
the limiting value of the entropy change accompanying an isothermal
Jo
ASt decreases provided
Helmholtz
function during an isothermal process approaches zero as the temperature
because
Experimental evidence
7 decreases,
— that
is
.9(7,0)
499
originally stated, as
the third law, that the temperature derivative of the change of
dP.
dT
example of A.SV decreases
T decreases,
-
AST = S(T,7^)
The entropy change
resolved and the statement
leading to
f*
Law
took 30 years of experimental and theoretical research, during which time there were periods of great confusion, before all differences of opinion were
of a paramagnetic salt during a reversible isothermal
magnetization also decreases as
solid or a liquid
this
solid
be measured at different
cither
T dS equation,
- s(T,o) = -
decreases as y'decreases,
The entropy change
Cryogenics, Negative Temperatures, Third
14-13
Meat and Thermodynamics
Diagram
to derive
The
and consider any
entropy change between
500
Heat and Thermodynamics
the point
T=
(
0,
7f =
where Cyf values of
and /
is
and the
9^i)
$-
14-13
state
=
5(0,^,)
i is
the heat capacity at constant
Jfi.
The change
x=
field,
^=
^ = ^,
a positive quantity for
entropy between the point
in
S o
dT,
r
fo
501
Cryogenics, Negative Temperatures, Third I-aw
(
T=
0,
9f =
all
0)
is
Sf Since
=
Si
,S>
and S(0,Wi)
lim im [S(T,!¥i)
wo T-
=
5(0,0)
f
-
=
5(0,0)
- S(TM =
dT.
T
Jo
-
lim [S(T,9Ai)
'^±dTI
S(T,0j],
we have
^=2fi d T.
f Jo
5(0,1*0
(&)
Ti TiC& =i
T \ Jo
8(0,0}
(14-37)
J
Fig. 14-29
5
—
*
6 and 6
—
(a) *
law were, not true, the processes Diagram to show the equivalence of
// the Nernst-Sirnon statement of the third
1 could be used
to
attain absolute zero, (b)
the three statements of the third law.
To
prove the equivalence of the unattainability and Nernst-Sirnon
ments of the third law, we proceed in the same manner as Kelvin-Planck and Clausius statements of the second law.
U=
Let
—U =
N= —N = As
falsity
of the unattainability statement;
falsity of the
netization
exactly the
— U Z)
same way
would be gained by repeat-
suppose that
possible to find a value
systems are capable of undergoing an isothermal reversible decrease of entropy followed by a reversible adiabatic decrease of temperature. Furthermore, the Nernst-Sirnon statement also applies to materials in frozen meta-
negative, thereby violating the Nernst statement.
stable equilibrium, provided that the isothermal process in question does not
To
prove that
— N Z) —U,
Sirnon statement.
it is
suppose that the left-hand
Then
it
Eq. (14-37) that would
would be
make
member
of
would be
possible to find a value of Ti in
the second integral equal to this
negative number. As a result, the
first
integral
would vanish and
Tf
zero, thereby violating the unattainability statement.
all
disturb this frozen equilibrium.
Referring to Fig. 14-29/;,
fact that
the point (0,^7)
—
>
7 could
—N 2) — U may
also
be readily seen from
Fig. 14-29a. If
below the point (0,0), then the adiabatic demagnetizabe used to lower the system to absolute zero.
lies
we
see that, in the isothermal process
1
—
> 2,
a decrease of entropy and that in 3 —* 4 there is another decrease, and so on. If the entropy of the system at absolute zero is called the zeropoint entropy, we see that a third equivalent statement of the third law is as there
is
follows:
By no finite
tion 6
Since the proof proceeds in
as before, however, nothing
which makes 7'/ = 0, thereby violating the unattainability statement. Then, from Eq. (14-37), the left-hand member would be
Eq. (14-37) had any negative value, thereby violating the Nernst-
The
of temperature during an adiabatic magnetization,
in the intermediate stale.
N
of Ti
2
and a decrease
A paramagnetic substance was invoked in the proof of the equivalence of £/and only for convenience and concrctcness. By means of a slight change in symbols, the same proof may be applied to any system whatever, since
-A'D -U.
and
~~ -V,
and the
ing the details.
U= N
prove that
the proof of the equivalence of the unattainability
such as a superconductor
Xernst-Simon statement.
-UD -N
To complete
Nernst-Sirnon statements of the third law, one ought to consider a type of system that undergoes a decrease of entropy during an isothermal demag-
truth of the Nernst-Sirnon statement;
when
To
in the case of the
truth of the unattainability statement;
before,
1
state-
series of processes
can the entropy of a system be reduced
to its zero-
point value.
The
equivalence of all three statements of the third law
in Fig. 14-296.
is
clearly displayed
502
Heat and Thermodynamics
14-13
There are many physical and chemical
facts
which substantiate the
third
Table 14-6
Cryogenics, Negative Temperatures, Third
Law
503
The 100-year Journey toward Absolute Zero
law. For example, using Clapeyron's equation,
Dale
Investigator
Temp.,
Development
Country
°K
dP_
dT
1860
Kirk
Scotland
First
toward deep refrigerreached tempera aires
step
ation:
conjunction with a phase change that takes place at low temperature, the statement that
below freezing point of Hg.
in
lim
(s<-f
-
*<«)
=
1877
i
,(/)
stantiated
—
v <-' )
is
used throtpressure vessel, obtaining line mist only.
First liquefied oxygen:
1898
Dewar
England
First
1908
Kamerlingh-
Netherlands
First liquefied helium:
First property measurements at low
temperatures: used small tities
=
A 0,
not zero for a first-order phase transition. This
by the melting curve of solid helium shown
in Figs.
As a matter of fact dP/dT of solid He approaches zero very by the experimental result of Simon and Swcnson, that 1
12-30
is
sub-
to 12-32.
rapidly, as
dP_
=
0.425
T7
1927
Simon
Germany &
933
Giauque & MacDougall
U.S.
Kapitza
England
First adiabalic demagnetization:
10
first
4.2
&
U.S.S.R.
Developed
helium
expansion engine:
Made
1946
'
room temperature, and the temperature in the interior of the hottest star, about 3 X 10 9 °K, is 10 million times room temperature. Cryogenics is still ahead by a factor of 10. A chronological account of the progress toward lower temperatures is given in Table 14-6 (reproduced with the kind permission of International
using
4.2
possible
liquefaction of helium without liquid Hj precooling.
Collins
U.S.
-.
Cryogenics has therefore enabled us to get to one one-hundred-millionth of room temperature. The temperature of the sun, 6,000°K, is only 200 times
in 1926.
liquefier
Developed commercial helium lique-
used
expansion and COUnterllow heat exchangers.
ffl
0.25
proposed by
Giauque and Debye 1934
1956
Simon &
England
First
England
Reached
Kuni
and Technology)
liquefier: used adtabatic expansion from pressure vessel with liquid Hs
Principle
(300°K) equal to
Science
Developed helium
fier:
X
-1.2
as Dewar; shortly
precooling.
.
There are many other applications of the third law in the fields of physical chemistry and statistical mechanics. For further study, the writings of Simon and of Guggenheim arc recommended. The fact that absolute zero cannot be attained is no cause for misgiving. A temperature of 3 X 10 -6 °K represents a fraction of room temperature
3
used same
20.4
thereafter, lowered pressure over liquid to get 1°K.
shown
1
1*J£ =
0;>.
hydrogen:
liquefied
method
England
dT
N\ and
of liquid
77.3
quan-
used Joule-Kelvin effect and counterflow heat exchanger.
Onnes
90.2
from
process
Poland
implies that
since
France
Wroblewski & Olzewsld
1884
dP V lim -7=
Cailletet
tling
0,
T->0
234.0
nuclear experiments; used adiabalic demagnetization of nuclear stage of a paramag-
netic
1960
Kurti
2.0
engines
lO^ 5
salt. lowest, temperature so far:
Nuclear cooling methods.
10
5
504
Heat and Thermodynamics
Cryogenics. Negative Temperatures, Third
PROBLEMS
(b)
In a magnetic
14-1
field
9Y,
where the
total
energy of an ion
As 7"—>0 at constant 9£,
dH *\
is
M)T =
( \d r
(c)
«i
the
number
of ions
A";
=
+
—gmaP&mt
14-6
with magnetic quantum number m, total magnetization is
M
Boltzmann equation. The
M Show
Si,
=
is
given by the
Prove that,
=
Bj(a)
In
Z
Prove is
that,
when
when J =
U=
2A';5,-
Prove
that, for
dM.
tanh|-
splitting of the lowest
If the gas
energy level by the
la:
gi «
_
- •*&<«)
,- t .
Show
|=
ideal
and obeys Curie's law, show tha at
In (2.7
+
1)
-
=
In
(2J
+
1)
-
Sketch a reversible adiabatic surface on a Tl'M diagram, assuming
A
paramagnetic
salt
obeys Curie's law and also
Cm = A/T*.
that:
Sv-
^^ +
-| ~R
-~M dM.
to be constant.
„, T
~
1
A
2
T*
M
+ ~ 2CC*J-
C% — Cm —
(b)
C ° nSt
-
2
Cc 3£_ y2
'
2
(c)
The
equation of an adiabat
is
M=
9f
const.
Vl +
{Cc/A)?f*
the Curie constant.
Convert the second energy equation (Chap. 11) into a form = f(9t^/T), M, and 7". Prove that, since U of a magnetic-ion subsystem must be a function of T only. 14-5 For a paramagnetic salt obeying Brillouin's equation, prove that: 14-4
appropriate to the coordinates JK,
w
is
T dS = C V .M dT + „RT~^~
Cv,m
Whena«1,
is
gas,
-J,
14-8
where Cc
a paramagnetic
and
(c)
(6)
a function of temperature only and is a function of temperature only
is
neglected,
g=
to
Tc, an antiferromagnetic roughly proportional to the tem-
is
14-7
(b)
14-3
M.
\9T/i
Between absolute zero and the Curie point
that (a) the magnetic enthalpy
[)?/-
Derive Eq. (14-22) directly from the relation
crystalline field
s
that (b) the heat capacity at constant field
(a)
(b)
Cm.
ZNigixnnii.
M = NkT
dM
505
perature: thus,
Show
(a)
T
oxide has a magnetic susceptibility that
that
14.2
C^—>
Law
Cj»r
— Cm — — yfi BM dT w
M
14-9 function of
Make
T
a graph of the heat capacity of gadolinium sulfate as a 1.5 to 5.0°K and for values of equal to 0, 2, 5, and
W
from
lOkOe. 14-10 Take a paramagnetic Carnot cycle, and verify that
Qc
solid
Tc
obeying Curie's law through a
506
Cryogenics, Negative Temperatures, Third
Heat and Thermodynamics
From
14-11
the microscopic point of view, the disorder of a para-
may
two ways: by increasing the temperattire and by decreasing the field. From this point of view, explain why a reversible adiabatic demagnetization should be accompanied by a decrease magnetic
in
solid
be increased
in
temperature.
At zero magnetic field, the lowest energy level of the magnetic paramagnetic solid is split by the crystalline electric field into two states with degeneracies go and gi and with energies and Si. (a) Calculate the temperature at which Cm is a maximum when 14-12
ions of a
=
gi/g"
1-
Calculate the
(b)
value of
Cm when
=
gi/go
gram-ions of gadolinium sulfate obeying Curie's law are a magnetic field of 20 kOe and at a temperature of 1 5°K. The field is
reduced reversibly and isothermally to zero.
How much How much
(a) (b)
(a)
is
work
is
(d)
reversibly
14-14
order that an adiabatic demagnetization
CM.X
(a)
What
(b)
After
is
performed
ammonium alum be from an initial field of 10 kOc iron
K
is
demagnetized adiabatically
to
a
field
of
1
kOc, what
is
the final temperature?
What
14-16
initial
field
chromium potassium alum
If,
deg.
What
is
field
is
is
at
1°K and
at
the final temperature of the combination?
at the conclusion of the demagnetization in part (a), the
immersed
in 36
He
moles of liquid
equals 0.1047" 6,2 J/g
•
deg,
4
what
at
2°K, whose
specific heat
alum
capac-
the final temperature of the
is
Sketch a Carnot engine cycle on a
1
to
diagram:
(b) (c)
Why is it impossible to operate a Carnot engine between
at a negative temperature
14-21
and one
=
a reservoir
at a positive temperature?
Suppose the magnetization of a paramagnetic
M
M
solid
is
given by
sat/(fl ),
when a = 0, and J(a) —> 1 when a —> co where a = gii%/kT and f(a) = The change of entropy during an isothermal reversible increase of field from to JV,- is given by
required in order that an adiabatic de-
should lower the temperature from
9fM
At positive temperatures. At negative temperatures.
magnetized isothermally from zero
the heat of magnetization per gram-ion?
magnetization to zero
of
temperature of 0.2°K?
originally at 1.5
it is
is
(a)
kOe.
field to 5
(b)
14-20
done?
and adiabatically? At what initial temperature should
14-15
One gram-ion
combination?
transferred?
to zero final field yield a final
field ?fC
10 kOe. After undergoing an adiabatic demagnetization to zero field, it is immersed in 36 moles of liquid He 4 at 4.2°K whose specific heat capacity
ity
What is the internal energy change? What would be the final temperature if the process were
(c)
in
heat
placed in an external magnetic
14-19
equals 2.5 J/g
1.
is
507
Each proton has I = k ai1 d a magnetic moment p.. An applied radio-frequency field can induce transitions between the two energy states if its frequency v is equal to 2y.9£. The power absorbed from this radiation field is proportional to the difference in the number of nuclei in the two energy levels. Assuming that k'f 2> 2jj.%, how docs the absorbed power depend on T?
Two
14-13 in
maximum
A sample of mineral oil
14-18
Law
A.SV
=
I0~* °K,
using the following: (a)
Chromium
(It)
CMN? A
14-17 field
If
and g
["'
solid at
temperature
T
is
=
the solid contains weakly interacting magnetic ions with
1,
to
what temperature must one
J =
Determine whether the Xernst-Simon statement of the third law •£
cool the solid so that 75 percent
of the atoms are polarized with their magnetic
moments
parallel to the
external magnetic field? (b)
(but to
If,
X
were paraffin with many protons = l,gx = 2, and a magnetic moment equal what temperature must the paraffin be
on the other hand, the
no magnetic
1.41
aj'ia) da.
g^
placed in an external magnetic
kOe.
of 30 (a)
potassium, alum?
ions), with / 10~ 23 crg/Oc, to
the following cases:
=
(a)
Curie's equation: /(a)
(b)
Langevin's equation: f(a)
(c)
Brillouin's equation,
with
J=
(d)
Brillouin's equation, with
any
a.
=
coth a
solid
lowered to achieve a nuclear polarization of 75 percent?
-
^: f(a)
=
tanh
f(a)
=
Bj(a).
./:
is
true in
508
Heat and Thermodynamics
14-22
Using the Ncrnst-Simon statement of the third law, prove For a reversible cell or a thermocouple,
(a)
that:
SUPERFLUIDITY AND SUPERCONDUCTIVITY
Sf" For a surface
(b)
film of liquid
S
He 4
15.
or
He
3 ,
o (rr) -
14-23 (a)
At
T=
0,
=
(dS/dV) T _ a
0,
and
also
Superfluidity of Liquid
15-1 _a_
=
3F/ r
Helium
II
0.
Helium gas
at
a pressure of
1
atm
liquefies at
4.215°K. If the temperature
of a system consisting of liquid helium in equilibrium with
From
this fact,
prove that '3(1
lim
A)
=
(b)
k
is
solid
Pv
where G(v)
is
whose equation of
+
G(v)
=
is
is
negative.
The
violent bubbling that takes
state
is
is
a constant, prove that
when helium
its
X point. This provides a simple visual method of observing the attainment of the X point and indicates an anomalously high thermal conductivity for liquid helium II. One of the most significant properties of liquid helium II is its abnormally boiling point ceases
Fa,
a function of volume only and T T approaches zero.
cy approaches zero as
own vapor
place by virtue of the unavoidable heat leak into a very cold liquid at
the isothermal compressibility.
In the case of a
volume expansivity
that the
T-.0
where
its
lowered to the X point (2.172°K), liquid helium II is formed. As the temperature is further reduced, the density of liquid helium II decreases, indicating
passes through
its
low viscosity. According to Poiseuille's law, the rate of flow of a liquid through a tube or an annular space is inversely proportional to the viscosity, other factors remaining constant. If the rate of flow of liquid helium through a fine annulus is measured as a function of temperature, the solid curve of Fig. 15-1 is obtained. At the X point, the rate of flow increases abruptly and provides experimental evidence of a very low viscosity. The result of a similar experiment on the light isotope of helium of mass number 3 is shown as a dashed curve. No such phenomenon has been observed for liquid helium III. A theory of liquid helium II was advanced by Landau in 1941 to account for its peculiar behavior. Landau assumed that the energy levels of liquid helium II (not of separate helium atoms) consisted of two sets of overlapping continuous energy states: one representing the levels for sound quanta, or phonons,
and
the other for
quanta of vortex motion, or
phonon
level
was assumed
to be below the lowest roton level
gap.
On
the basis of other assumptions,
quantum hydrodynamics of liquid helium II.
Landau was
that accounted quite well for
rotons.
The
lowest
by an energy
able to construct a
many of the properties
510
Heat and Thermodynamics
Superfluidity
15-1
511
and Superconductivity
has the virtue of simplicity and has proved very helpful to experimentalists.
tLZt
According
20
to this picture, liquid
helium
II is to
be imagined as a mixture of
composed of (1) normal atoms with normal viscosity and (2) superfluid atoms with zero-point energy and entropy, capable of moving through the normal atoms without friction. If p is the density of liquid helium II, p„ the density of the normal part, and two
'
liquids
p, the density of the superfluid part,
o (ft
P
E
=
P«
(15-1)
Pa-
QO
-
15
o
At the X zero
5 o
If
He 4 o TO
/
/
/
/ 1 !
5
/
1 /
12 ^.~*'
Fig. 1 5-1 annulus. (D.
=
the atoms arc normal and p„/p
and p n /p
set of concentric disks
is
=
1,
whereas
at absolute
0.
set in oscillation in liquid
helium
the
moment
^f-X poin 1 for
normal atoms
depend on the density of
temperature chosen. By combining the results of these two experiments and by applying plausible equations, Andronikashvilli was able to measure the ratio p n /p as a function of the temperature. At
/
at the particular
lower temperatures, de Klerk, Hudson, and Pellam were able to infer values
He 4
non
phenome-
helium II to be described later. The combined results of all these measurements are shown in Fig. 15-2, where it may be seen that p„/p
1
3
II,
through the superfluid atoms; as a
disks
of inertia of the disk system will
of p„/pfrom their measurements of the speed of "second sound," a 4
Temperature, °K
Rale offlow of liquid helium
W.
a disk or a
result,
/
/ (
\
all
normal atoms are dragged by the
/ i
point,
the atoms are superfluid
the amplitude of oscillation will decrease with the time because of the viscosity of the liquid. If the disks are extremely close together, many of the
/
te 3
all
HI and of liquid helium IV through a M. Abraham, 7949.)
7
X
1
in liquid
0~ s-cm
Osborne, B. Weinstock, B.
1.0
An
equally plausible point of view was developed from
quantum
statistical
mechanics by F. London, who envisaged the formation of liquid helium II from liquid helium I at the X point as a peculiar type of quantum condensation, known as Bose-Einstein condensation. This is a phenomenon which takes place not in ordinary space like the familiar condensation but in momentum space, where the condensed particles have zero-point energy and momentum. Many of the statements of both Landau's and London's theories rest on plausibility arguments rather than on derivation from quantum theory. This situation has been considerably improved by the researches of Feynman, who has succeeded in deriving, on the basis of quantum statistics, the energy-level and energy-gap picture of Landau while at the same time retaining the BoseFinstein condensation idea of London. Both Landau's and London's points of view were redescribed by Tisza in picturesque language by a phenomenological theory of liquid helium II
known
as the two-fluid model
—which, although not meant
to be
taken
literally,
10
1^
"'
/ /
10~ 2
10- 3
SI
)pe
=
5.3
/
io-"
/-<
10~ 5
IO" 6 J
10"'
s*
—
SI ape
^ s*
\
i
i
X point-^J
-•
10 _a
1
0.15
0.2
1
0.3
1
1
1
1
t
0.4 0.5 0.6
l-i-
Temperature, Fig. 15-2
1
°
-i
i
1
0.8 1.0
1.5
^
2.0
K
Variation of normal fluid concentration with the temperature. {Measurements of Andronikashvilli, de Klerk, Hudson, and Pellam.)
512
15-2
Heat and Thermodynamics
Superfluidity
and Superconductivity
power of T up to about 0.6°K, and from there on the more complicated. It is an interesting fact that the heat capacity
varies as the fourth
variation
is
Fountain of liquid
helium
and the thermal conductivity of liquid helium II also abruptly change their temperature dependence at 0.6°K.. These facts lead us to believe that, at temperatures below 0.6 K, only phonons play a significant role. Since p n /p increases so markedly as the temperature rises, a temporary atoms may be produced locally i.e., in a small region
—
scarcity of superfluid
of a vessel containing liquid helium II
—
if
the temperature of that region
513
II
..Vessel containing liquid helium II
is
by heat leaking in from the outside or by heat supplied by an As a result, superfluid atoms diffuse into the region. This motion of superfluid atoms relative to normal atoms accounts for the abnormally high heat conductivity of liquid helium II. It is more a transport of mass than a transport of heat. In a mixture of liquid helium 4 and liquid helium 3, a temperature increase at one place causes a flow of superfluid He 4 atoms to that place but not of atoms of He' since liquid He 3 shows no superfluidity. This has been used to increase the concentration of He 3 in a mixture to a value far above raised, either
electric heater.
Liquid
^Heater
helium II
Emery powder
Capillary
normal.
^~—f— Fountain Effect
15-2
One effect,
Surrounding bath of liquid helium II
of the most peculiar
phenomena
discovered by Allen and Jones in
liquid in a vessel
1
of liquid helium II
938.
They observed
communicating through a narrow capillary
is
the fountain
that the level of to a
surrounding
when heat was supshown in Fig. 15-3o. A more
bath of liquid rose above that of the surrounding bath plied electrically to the liquid in the vessel, as
can be demonstrated with the apparatus shown in Fig. \5-3b. The helium between the grains of finely ground emery was warmed by the radiation from a flashlight, and the consequent increase of pressure gave rise to a fountain of liquid helium, which has been observed to be as high as 30 cm. striking manifestation of this effect
The
may
two chambers or two parts of a vessel communicate with each other through a very narrow capillary, a thin slit, a small hole, or the spaces between a closely packed powder, etc., and if the liquid helium II in one chamber is maintained at temperature 7", and pressure P h then, if the temperature of the helium in the other chamber is maintained at T 2 the pressure will automatically become fountain effect
be epitomized as follows:
If
,
P2, a positive
temperature gradient giving
The numerical measure
rise to a positive pressure gradient.
of the fountain effect
1
T2 -
Ti
is
taken
to
be
Fig. 15-3 served.
(ft) Apparatus of Allen and Jones with which the fountain eject was first obApparatus of Allen and Misener for showing the fountain effect.
(Ii)
or, as it will
where g
is
When
be shown
later,
the Gibbs function.
liquid helium II
is transferred through a very small hole, slit, or generally accepted that only superfluid atoms are transported, carrying with them their zero-point energy and entropy. If we choose as a
capillary,
it is
standard state for the calculation of entropy of liquid helium the state at absolute zero and arbitrarily assign zero entropy to this state, then, since liquid helium at absolute zero consists exclusively of superfluid atoms, we
may ascribe zero entropy to superfluid atoms. Therefore, escapes through a narrow space, no entropy is lost.
when
liquid
helium
Suppose two vessels, each containing liquid helium II, are connected by a narrow capillary, as shown in Fig. 15-4. If the vessels are maintained at
514
Heat and Thermodynamics
15-2
again that there
is
Superfluidity
and Superconductivity
515
no entropy change accompanying the flow of matter, we
have
-^
So
am =
D,
' o
clQo
or
Fig.
1
5-4
dm
Reversible transfer of mass
T
different temperatures
and
=
ToSo dm.
Similarly, for the right-hand portion, which gains dm grams of superfluid atoms, thereby gaining no entropy from the incoming atoms but gaining it instead from the reservoir, we have
of superfluid atoms.
dQ = Tsdm.
T, respectively, with the aid of suitable
then for a given pressure on one mass of helium there will exist a on the other. Suppose both pistons are moved to the right very slowly so that P and 7~ are maintained constant on the left
reservoirs,
Considering the system as a whole,
definite equilibrium pressure
P and T constant on the right, these being equilibrium values. Let m, the mass of liquid helium II on the right, increase by dm. Assuming that (1) only superfluid atoms, carrying no entropy, move through the capillary and
and
Net heat transferred
=
Ts dm
Net energy change
=
(u
Net work done
=
(Pv
is no friction in and no heat conduction through the capillary, then each portion of helium undergoes an isothermal, isobaric reversible
that (2) there
change of mass. It
where
should be remarked that the behavior of a quantity of liquid helium II
which
atoms only, and thus loses no entropy by virtue of this flow, is entirely different from the behavior of the usual "open system" of classical thermodynamics, where entropy is transferred by virtue of the flow loses superfluid
stant v,
u
and
T and
and the temperature tends
therefore distributed over a smaller mass
This
is
to
T$ dm
—
is
a flow of heat must take place to the reservoir. That is, when liquid helium II in the left-hand chamber loses superfluid atoms (loses mass), it loses no entropy through the capillary but loses entropy to the reservoir of an
fore,
—
dm)
=
somo
s
is
.
dm
=
(u
go
=
g-
—
ita)
dm
+
(Pv
—
Povo) dm,
(15-2)
left to
remain constant and varying
+ vdP
=
P and T
0,
—so dm, (15-3)
finally,
the specific entropy of liquid helium II at the constant tempera-
ture To and constant pressure P Since remains constant during the process. .
The entropy change transferred
ToSo
-s dT and
where
and volume, respectively, at conT and P Obviously, u, uo,
refer to constant
Considering PQ and To on the on the right, we have
amount
—
v
P&o) dm,
rise.
prevented, however, by the reservoir at constant temperature. There-
So{ma
and
—
dm,
remain constant during the process. Applying the first law of thermodynamics, we get
the pistons move, only superfluid atoms flow through the capillary.
Since these atoms carry no entropy, the entropy in the left-hand vessel
represent the specific energy
«o)
7Vo dm,
Vo all
of matter.
When
v
P; while uo
—
—
from the
reversible, the
sum
of
of the reservoir
is
is
a function of To and Po,
it
which
dQ
/7'
,
where dQ
is
the heat
helium to the reservoir. Since the process is the entropy changes is zero. Remembering once
liquid all
so
meter,
is .r
pressure
P is expressed in dynes per square centigram-degree, and v in cubic centimeters per gram. If the measured in centimeters of helium, then letting y be the height
the fountain-effect equation. in ergs per is
of the helium column, a„ the acceleration of gravity,
and p the density
of
516
Heat and Thermodynamics
15-3
Superfluidity and Superconductivity
517
rapid, local temperature fluctuation gives rise to a rapid, local variation in the ratio pn /p» without altering the sum p„ p s = p. An ordinary longitudi-
+
wave (first sound) is the propagation of fluctuations in p throughout a medium. Fluctuations in the ratio pn/ps, without change of p, propagated through liquid helium II were predicted independently by Tisza and Landau; these are called second sound. Second sound may be compared with
nal
sound in another way: in first sound, vibrations of normal and superfluid atoms together are propagated through a medium; whereas in second sound, vibrations of normal and superfluid atoms relative to each other arc propagated. First sound may be produced in liquid helium by a vibrating piston and, when so produced, is propagated with a speed w\ in agreement with the laws of classical physics, except that 7 = 1 and there are no temperature differences between compressions and rarefactions. The temperature variation of w\ is shown in Fig. 15-6, where it may be seen that a striking disfirst
continuity of slope occurs at the X point.
may
Second sound
be produced by a temperature pulse or by periodic
temperature fluctuations produced in a 1.6
1.4
2.0
1.8
appropriate electric current.
2.2
Temperature, °K Fig.
15-5
Fountain
affect.
(Experimental points from
Cryophysics" p. 324, 1961.) Solid curve represents
s/a,,.
D. V. Osborne, "Experimental
(Kramers, Wasscher, Bots, andGorter,
This method
flat, is
disk-shaped heater by an
very simple but has the
1960.)
helium,
P=
ypa„,
and the fountain-effect equation
may
be written as 210
(2L\ = JL. . JL \vT/g)
vpa„
(15-4)
may be calculated from data on
TcdT T
1
^-237 m/sec
a
230
Values of the entropy
Liq
Hell\
the specific heat. Thus,
8
220
in
'
Jo
E Liq
The measurement for the increased
(which
of the fountain-effect pressure
vapor pressure
may amount
to as
in the
much
warmer
vessel.
as 10 percent)
agreement between the experimental values, as shown in Fig. 15-5.
results
must be corrected to allow is
When
this correction
made, there
is
excellent
S
H el\
210
Q. in
200
and the calculated entropy 190
15-3
Second Sound
According
number
of
to the two-fluid
normal
180
theory of liquid helium
to superfluid
II,
the ratio of the
atoms depends strongly on temperature.
A
dis-
advantage of providing a continual supply of heat, which boils away the liquid helium. To avoid this, it has been found possible to generate second sound by variations in temperature of a paramagnetic salt produced by
12
\ 1
point 1
3
1
Temperature, °K
Fig. 15-6
Speed of first sound in liquid helium. (Van Itterbeck and Forrez,
195-1.)
518
15-3
Heat and Thermodvna
superposing an alternating magnetic field on a steady field. Under these conditions, the temperature fluctuations arc alternately above
and below no net input of heat. The speed of second sound was first measured by Peshkov, and the most recent measurements are those of de Klerk, Hudson, and Pcllam. In some of the measurements, stationary second-sound waves were set up in a cylindrical column of liquid helium II between two flat disks wound with resistance wire. An alternating electric current was used to produce the necessary temperature fluctuations in one disk, and the fluctuations in resistance produced by the temperature changes at the other disk served to detect the nodes or antinodes. In other measurements, a pulse of current in one disk caused a temperature pulse to be propagated along a cylindrical column, and the velocity of this pulse was measured with electronic techniques. The mathematical expression for the speed of second sound may be derived in a simple manner from the laws of thermodynamics, by making use of the two-fluid model. Consider a column of liquid helium II indefinitclv lon°- of cross-sectional area A, and at temperature T. Suppose that heat Q is supplied at one end by a source at temperature T AT for a time r. Both AT and r may be chosen arbitrarily small. The temperature rise AT will be propagated along the column with a speed w%, the speed of second sound. It is assumed the ambient temperature, and there
If
H
H= where
heal current,
current.
turned
If,
after the time interval r,
the region of elevated temperature
off,
and
relative
but this should not be confused with a convection or conduction heat produces a slight increase in the number of normal
H
The
atoms and a consequent slight decrease in the number of supcrfluid atoms. The expression might therefore with more reason be called an excitation-energy current.
The
contribution to the entropy change of the system due to the increase
of temperature
AS —
=
is
t+at
Since
d'T
pcw-vrA
''
T
pcwirA In
T\
in Fig. 15-7.
atoms have a velocity vH in atoms have a velocity vs in the opposite direction. The total momentum per unit volume is pn v„ + p r v s where p„ and p„ are the densities of normal and
is
(15-5)
motion of normal and superfluid atoms appears successively at neighboring points in the fluid, and thus is propagated with a speed vo% The energy // disappears at one place and appears at another. We may therefore speak of a
and without attenuation. In time t, a column of length w»t and of volume w 2tA will be affected. This volume of fluid, which wc shall call the system, is shown as the shaded region
and supcrfluid atoms. The normal the direction of propagation, and the super-fluid
pcw%rA AT,
the specific heat capacity of the system.
c is
the heat source
that the propagation takes place without dispersion
system differs from the rest of the liquid helium not only in tempera-
by
A T, then
+
The
519
and Superconductivity
the heat necessary to increase the temperature of the system
is
is
ture but also in the motions of the normal
Superfluidity
(
AT = =
H
1
pew -IT A
AT _ 1 /A7"Y T 2\ T
IT)
(15-6)
amount of entropy crosses area A in time r, the entropy current H/TtA, correct to the first order in AT/T. According to the two-
this
density is
fluid model, this entropy is carried by the normal atoms only, the superfluid atoms being assumed to possess zero entropy, referred to a standard state at
,
superfluid atoms, respectively, the total density p being equal to p„ Ps In the propagation of second sound, the total momentum per unit volume is
+
zero, in contrast with that of
first
absolute zero, Since the normal atoms, of density possess entropy per unit PnSnVn,
sound.
But
equal to
ps
=
mass
From
p n s„, since
$,
=
0,
and II.
is
vn
and
given by
and hence the entropy current density
is
psv„, or
Eqs. (15-5)
and
(15-7),
it
=
P su„tA T.
(1
5-7)
follows that
AT T
Propagation of a second-sound pulse in liquid helium
move with speed
the entropy current density
sn ,
//
Fig. 15-7
p„,
.
_ H AT T
sv„
CWn
psV„rA (15-8)
&oi
520
Heat and Theiiuodvnamics
The
kinetic energy
K
15-3
of the system
may
K =
+
iP"l
JJ
But
+
P„Vn
psVs
=
be calculated as
.Substituting for
folic
Q
its
value given by the
H
iPsvl
1
we
2
T
get
'
HAT _ T
2
pn vn
law,
K
0,
=
v*
first
521
and Superconductivity
T + AT r T\
T + AT This gives
or
Superfluidity
T AT =
2
;
'
H^
P.<
2K.
(15-10)
3- l4i+rK = ^-^
whence
WoT.4
Substituting the results of Eqs. (15-8)
K=
and
*P
- v°-w
2tA.
(15-9)
psh'lrAT
and
=
P* «t>2
If the
source
is
turned off after time
r, this
kinetic energy travels
through
the liquid without dissipation. Thus, at
any moment, there exists a quantity of liquid of volume wnA with excitation energy per unit volume H/wvrA = pc AT, and with kinetic energy per unit volume K/witA = ipp n v 2n /p s Neither
Wt
or
p
(15-9) into Eq. (15-10),
we
get
— V'WotA, pn
„
Pi
=
(15-11)
-»/
\P»
c
.
energy
transformed into the other; both coexist and remain constant in time as they
is
which
The temperature variation of the speed of second sound is shown in Fig. Above 1°K, the speed remains fairly constant at a value of about
proceed through the liquid.
To derive an expression for the speed to find
a relation between
of second sound,
H AT/T and
K. Let
Q
it is
merely necessary
be the heat leaving the
T -\- AT; //be the heat necessary to raise the system in temperature by A T; and K be the kinetic energy of the system. Then, from the first law of thermodynamics,
the equation for the speed of second sound.
is
15-8.
18
m/s
until
almost 2°K.
It
then decreases to zero at the A point. As the tern-
source at
250
1
The
entropy change of the source
200 is °
Q
been shown
to
of the system
due to the temperature
rise
has already
^ i
be Liq uiti hel
If
T\ Assuming, as usual, that the process is
1
100
Kf. _ 1AT'
the universe
%
1
150
T + AT The entropy change
1
— First sound velocity w
_.
Q = H+K.
zero;
is
um II.
i
50
reversible, the total entropy
change of
"N
n
whence
0.2
0.4
0.6
0.8
1.0
Temperature,
AT
k-m
TV
2
T
Fig.
1
5-8
1.4
1.2 °
1.6
1.8
2.0
2.2
K
Temperature dependence of lite speed of second sound in liquid helium II. (De Klerk,
'
Hudson, and Pellam, 1954.)
522
I
feat
and Thermodynamics
15-5
pcrature goes below 0.6°K, it is seen that the speed of second sound, like other properties of liquid helium II, undergoes a change of slope. Using the equation for the speed of second sound in conjunction with experimental measurements of w«, c, and s, de Klerk, Hudson, and Pellam calculated the temperature variation of p„/>, plotted in Fig. 15-2.
and
it is
these values
Superfluidity and Superconductivity
523
240
220
which are 200
Fourth Sound
15-4
180
In first sound, normal and superfluid atoms vibrate together, and the propagation of these vibrations involves pressure and density variations but no temperature variations. In second sound, the normal and superfluid atoms vibrate with respect to each other, and in the propagation of such
160
140
vibrations there occur variations of temperature but not of pressure or of density. In 1948, J. R. Pellam suggested that liquid helium II might support a wave in which only the superfluid atoms undergo vibrations, the normal
atoms being immobilized Atkins
came
to the
in the
narrow spaces of a porous material. In 1959, called this type of wave motion
same conclusion and
100
fourth sound. (Third
sound will be discussed in the next section.) When the normal atoms are free to move, both first and second sound exist. When the normal atoms are damped, first sound degenerates into fourth sound, whereas second sound fails to propagate. In the experiments of
I. Rudnick and K. A. Shapiro, a resonator was with densely packed porous material having channels of less than 0.03 in diameter for frequencies less than 10 kHz. For this purpose, Polypore filter paper and Cr 2 3 ("green rouge") were used. The results are
filled
mm
shown
in Fig.
5-9, where the speed of fourth sound is seen to and second sound. The speed of fourth sound 1
limits of
first
sufficient
accuracy by the equation
\—w\ + ,P
P=
lie is
within the given with
(15-12)
1.5
P
1.7
Temperature, Fig.
15-5 It
Creeping Film; Third Sound was observed by Kamerlingh-Onnes, in 1922, that the levels of liquid II in two concentric vessels placed one within the other soon became
helium
equalized without there being any opening or communicating tube between the two vessels. Rollin, in 1936, showed that this phenomenon was due to a thin film which formed on any solid surface placed in liquid helium II.
15-9
Speed of first, second, and fourth sound
in liquid
helium
II.
(I.
Rudnick and
K. A. Shapiro, 1964.)
The
film creeps with a definite critical velocity, without friction,
higher to the lower
from the
shown in Fig. 15-10a to c. Starting with the pioneer experiments of Daunt and Mendelssohn in England and of Kikoin and Lasarew in Russia, a tremendous amount of ingenuity, skill, and energy has been displayed by many physicists in devising and carrying out experilevel, as
524
Heat and Therniodyiiar
15-5
n
nr
Filling
10-6 and 15
and Superconductivity
525
X 10~ 5 cm 3/s cm. Everyone is also in agreement measurements of L. C. Jackson, establishing that the 6 thickness of a helium film is of the order of 10~ cm, or about 100 atoms thick, film is from the liquid level. decreases the farther the and about 6
X
•
with the excellent
In 1959, Atkins predicted the existence of a surface wave on a liquid helium film and called such a wave third sound. He assumed that the superfluid
<»)
(<") Fig. 15-10
n
Superfluidity
and
C°)
—
emptying of a vessel by a creeping film of liquid helium II.
Helium bath
ments on helium films. Unlike the fountain effect and second sound, where a combination of the two-fluid picture of liquid helium II and thermodynamics enables one to calculate simple equations that are found to agree with experiment, there liquid
is
no
theoretical derivation of
by film transport,
and
an equation
for the. rate of flow of
no two experiments agree concerning the
ence of flow rate on, for example,
source level from the top of the beaker over which the film
temperature, (4) material and of possible temperature differences,
depend-
(1) difference of level, (2) depth of the is
creeping, (3)
smoothness of the creep surface, (5) existence (6)
method
of filling or
emptying one of
the vessels, and (7) effect of radiation. The most recent series of experiments on rate of film creep, conducted
under carefully controlled conditions in which purity and uniformity of temperature were emphasized, was carried out in 1965 by R. W. Selden, J. H. Werntz, P. J. Fleming, and J. R. Dillinger in Wisconsin. They used five slightly different
experimental systems,
all
variations of the system
,-—'-"p'
Helium
shown
symbolically in Fig. 15-11. All glass surfaces over which the film could creep constituted one side of a thin wall, the other side of which
was
in contact
with liquid helium in which temperature differences could be kept
about 10-6 deg. Measurements were made of the rate of flow
R
per unit length of periphery, as a function of the level difference different temperatures
emptying.
and
different values of
H, both when
filling
down
of
to
/
volume
AH
for
and while
A few of the results are shown in
Fig. 15-12, where it may be seen from emptying rates and that both arc dependa simple linear manner. The only agreement among
that filling rates arc different
ent on
AH, but not
in
different experimenters
point and increases as
is
that the rate of film transfer
T
decreases, assuming values
is
zero at the lambda
somewhere between
Fig. 15-11 to
Symbolic sketch of apparatus of Selden, Werntz, Fleming, and Dillinger (1965)
study film transfer rates.
526
Heat and Thermodynamics
15-6
=
H
r
12
cm
heat. Since a small
"""
22.4 cm
"
1
.
»—••
H
2
_•-
r 10
2.0
_ -•—
>-*""" 2
^
Suppose that a thin-walled glass container, packed 3 in size, is with a fine powder whose particles arc of the order of 10~ temperature and is then raised in above the filled with liquid helium II angular temperature the container is given an lambda point. If at this (1) momentum and (2) is then cooled below the lambda point say, to about
£j5>-»
1°K, where the ratio of superfluid density p3 to the total density p is almost unity and if, finally, (3) the rotation is stopped, experiment shows that superfluid flow persists. Increasing the temperature to within a millidegrcc
i
25
15
A//,
/«» o/
respectively
Dashed
.
component wall.
The
and 3
Werniz, Fleming, and Dillinger on
refer to filling at temperatures
1.479, 1.441, and
\A8l"K,
curves 1, 2, «nrf 3 ie/«r /o emptying.)
while the normal component remains locked
oscillates,
result
—
30
cm
the experimental results of Seidell,
1. 2,
is
an
Persistent flow
—
10
^1
.
mm
,•*'
Fig. 15-12
film creep, however,
the other hydrodynamic effects should at least be mentioned for perhaps two reasons: (1) they provide examples of quantum effects on a macroscopic level, and (2) they have exact analogs in some of the properties of supercon-
1
L.
**
film flow. {Solid curves
had been devoted to
of space
527
ductors to be suidied in the remainder of this chapter.
•
-
amount
and Superconductivity
Superfluidity
oscillation in the thickness of the film
to the
with secondary
and temperature. The phenomenon is somewhat Third sound was produced and detected by C. W. F. Everitt, K. R. Atkins, and A. Denenstein in 1962 and was studied again in 1964. It was excited in a horizontal film of liquid helium II by pulses of infrared radiation and was detected by measuring the variations in film thickness by the same optical method that had been employed by Jackson to measure the thickness of static films. The speed of third sound was found to be in the range of 20 to 500 cm/s, depending mostly on the variations in pressure
lambda point causes a corresponding decrease in angular momentum, which comes back rcversibly to its former value, however, when the original ratio of p s /p is restored. Experiments of this sort, or similar in principle, go back as far as 1950 and are associated with some of the great pioneers in lowtemperature physics: D. V. Osborne, C. T. Lane, and E. L. Andronikashvilli. The results shown in Fig. 15-13 were obtained in 1965 by J. B. Mchl and of the
similar to classical shallow-water waves.
height ji of the film above the bulk liquid. to
be in
fair
The
0.2
i_
w
o- Cooling
*= Average
where a„
3c
ft
CD
3
& 0.1
(15-13)
3(Pb /p)«^
i
9 o
the acceleration of gravity.
is
•
e
experimental results seemed
E
=
i
•
agreement with a simplified theoretical equation
w\
;»
1.24 K
at
o
• •
15-6
Other
Effects of Superfluidity
i
o
•
The
fountain effect and the propagation of second sound are outstanding
thermal effects of superfluidity. Film creep,
on the other hand,
is
more with the hydrodynamics of the superfluid component in liquid helium and is therefore not of primary interest in a text devoted to the subject of
.6
.4
concerned
1.0
P,/l> Fig. 15-13 density.
(./.
Proportionality between angular
B.
Mehl and
II".
Zimmerman,
1965.)
momentum
of persisting flow
and
superfluid
528
Heat and Thermodynamics
W. Zimmerman, who found
15-6
that the helium current persisted for 5 hr
without any measurable reduction
in
momentum. An undiminished
angular
circulation of supcrfluid helium in a container circulating electrons.
A
is
like a large
better parallel, however,
is
atom with
the persistent electric
It
was
suggested
first
jfp, ds
say, a.
K
is
A
inner radius
a,
whose
particles
shown
move
rnv
minimum
=—
in Fig. 15-1 4a, of outer radius in circles
(for a
R
<
v
constitutes a vortex
line.
=
(for
r
by Fcyniuati,
Bohr-Sommer-
=
(15-15)
n/i,
<
If the vortex line is
nh, nil
1
2-wrn
r
(15-16)
with speeds such that
<
r
=
and with
K=
nh/lvm. Jin
=
1, 2tt A"
has
its
smallest value:
R) (15-14)
and
later substantiated
size,
satisfying Eq. (15-14),
V
in
}
constant. Since v cannot be infinite, r must have a
cylinder, such as that
shown
where p s is the tangential momentum of a helium atom of magnitude rnv, ds is an element of its circular path, and n is a quantum number that may take on integer values. This becomes
r
where
as a vortex ring, as
quantum condition
Quantized vortex
K ——
known
by Onsagcr, and
axis of
v
is
that vortices should exist in supcrfluid helium, satisfying the feld
lines and rings When a solid cylinder rotates about symmetry, the linear speed of a point v is directly proportional to the distance r from the point to the axis, according to the elementary relation v = oir. When water is draining out of a sink through a hole in the center, the water rotates in a manner which is quite different and which is known as irrotalional, where 2.
doughnut), the system
529
and Superconductivity
Fig. 15-146.
its
current in a superconductor.
its
circle (a torus or
Superfluidity
h 6.63 X 10-* erg-s r **»-!)--6,67X10-4 9
a),
f
t,
g
=
bent so as to form a complete
The
0.997
X
10
'
*em*/sec
(15-17)
attempt to look for such vortices in superfluid helium was made F. Vinen, who used a fine wire stretched along the axis of a cylindrical vessel containing liquid helium II. He reasoned that, when the system is set in rotation, a vortex line should be set up around the wire which in 1961
first
by W.
should affect
(a)
its
frequency of transverse vibration. Vincn's experiment was
reasonably successful and was followed by a
number
of researches involving
methods of measurement. The beautiful experiments of G. W. Rayfield and F. Rcif in California and of G. Careri, S. Cunsolo, P. Mazzoldi, and M. Santini in Rome involved the motion of positive and negative helium different
ions through superfluid helium.
is
Rayfield and Reif were able to show that a positive or negative helium ion quickly surrounded by many helium atoms in the form of a vortex ring
and that such rings move through superfluid helium with almost frictionlcss motion at a temperature of 0.28°K. The quantity R in Fig. 15-14 was taken to be the radius of the ring
itself, in
which case the value of v
velocity of the vortex ring perpendicular to
Fig. 15-14
(a) Vortex line, (b) Vortex ring.
its
at /
= R
is
the
plane. Classical hydrodynamics
provides the following equations for the energy E and velocity v of a vortex ring, whose radius R is much greater than a, moving through a medium of
530
I
lout
Superfluidity
15-7
and Thcrmodynai
Motion of He 3 through Superfluid
15-7
531
and Superconductivity
He
4
140
phase diagram of He 3 (Fig. 12-31) shows that there is only one liquid phase. Many attempts have been made to discover a new phase at very low
The
1
O
120
100
r
this or any other liquid phase. have met with failure at all 'emperatures down however, may be dissolved in liquid He and
temperatures and to detect superfluidity in
+ charges A — Cnargeb
o
Up
to 1968, all such attempts
to about
1
He
mdeg. Liquid
3
1
,
,
the study of such solutions has led to 80
some very
When He
as seen in the following examples: (1)
i1 s.
60
(2) If
\\ o
the
\ 40
when
absorbed; and heat
is
interesting 3
and
dissolved in
is
He 1
If a solution of
\
S
i
^a-^ "~~— a-^—
1
4 ,
heat
He
4
less
is
mm,
than 0.001
the dissolved
is
is
0.6
He will mm. (3) 3
lowered in temperature below 0.87°K, the solution will separate
into
two phases with a distinct boundary- -one phase a concentrated solution
(of
He
3
in
He
4 )
and a
denser phase, a dilute solution (G.
Walters and
K..
o
n
20
10
40
30
Diffusion
E, electron-volts Fig.
1
Relation between velocity v and energy
5-1 5
mental data for positive and negative charge by Eqs. (15-18) and (15-19). (G.
carriers.
W. Rayfield and
E of a vortex ring.
The
The points
are experi-
curve is the theoretical relation predicted
He 3 gas (0.02
j
P. Reif, 1963.)
He 3
fias l
(30
torr)
i
and
V
lP
(27rKy-R(ln~-ll
= K
pump
torr
density p:
E=
Pump
(15-18)
--
(
1
-h to-r)
He 4 bath
at
1-3°K
(Hot reservoir)
/
2RV
n
8R V. --4 V
(15-19) Capillary <0"^tr cdon
v as a function of B for both positive and negative and obtained the experimental points shown in Fig. 15-15. The curve is the theoretical relation obtained by combining Eqs. (15-18) and (15-19), and the agreement is remarkable. The experiments gave the values
Rayficld and Reif measured ions
i. HeaT|
Evaporator at
•
6
°
K
"^containing diluted solution
^1
of
He 3
in
Dilution
He 4)
Concentrated. He 3 liquid Diluted.
corresponding to n
=
1,
1.00
is
10" 3
cm
2
/s,
-i
He 3
Concentrated 3 liquid
-He
Diluted
r.-d|j
Liquid
and
— He 3
liquid
Heat exchanger
a
which
X
chamber
<0.1°K
at
c -
2tK =
=
1.28 A,
of the order of atomic dimensions.
is
applied to the solution at a temperature (say, 0.6°K) at which
vapor pressure of
vaporize out of the solution, because the Fie vapor pressure
20
He
the process takes place adiabatically, the system cools.
3
\
useful effects,
Fig. 15-16
"Harwell" He*-He'
dilution refrigerator.
532
I
W. M,
Icat
and Thermodynamics
Fairbank, 1956).
(4)
He 3 atoms
diffuse through supcrfluid
He 4 from
one place to another where the concentration of He is smaller. An ingenious refrigeration cycle was developed by Hall, Ford, and Thompson and independently by Xeganov, Borisov, and Liburg in 1965; this is now used in an actual refrigerator, built by the Oxford Instrument Company in England. The refrigerator is capable of maintaining the temperature of an experimental space at about 0.1°K, with a refrigerating capacity of about 20 ergs/s, by circulating 2 X 10-8 moles of He 3 per second. A symbolic diagram of the apparatus is shown in Fig. 15-16. Let us assume that the apparatus has been running for sufficient time to establish a steady state, with the He 4 bath at 1.3°K, the evaporator at 0,6°K, and the dilution chamber at about 0.1 °K. The pump delivers pure He 3 (point a), which on its way 3
Simple 3
4
metals
Be
Li
533
Superfluidity and Superconductivity
15-8
5
6
7
B
C
N
13
14
15
Al
Si
P
9
10
F
Ne
16
17
18
s
CI
A
s
Noble metals
12
11
Transition metals
Mg
Na
s*
\
1.18
^
*"
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
0.39
5.30
0.88
1.09
53
3/
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
Rb
Sr
Y
Zr
Nb Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
0.47
0.56
3.40
3.72
0.49
9.20
0.95
8.22
'
54
Xe
to point b is cooled to almost 0.1 °K. It enters the concentrated solution at
55
56
3?
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
and then some of it crosses the phase boundary into the dilute solution Both the solution b —» c and the phase separation c —> d involve the
Cs
Ba
La
Hi
Ta
W
Re
Os
Ir
Pt
Au
Hg
TI
Pb
Bi
Po
Al
Rn
4.87 6.00
0.17
4.48 0.012 1.70
0.67
0.14
94
95
102
point at
c,
d.
absorption of heat from the materials
He
chamber. The
3
He
plied to vaporize the
brings the
He
3
in
then moves through
back
to
3
the space surrounding the dilution
He from d 4
at a pressure of
its
to
about 0.02
original pressure at
e,
where heat
mm.
A pump
is
sup-
87
88
89
90
91
92
Fr
Ra
Ac
Th
Pa
u
1.37
1.4
1.8
finally
97
98
99
100
101
Np Pu Am Cm Bk
Cf
Es
Fm
Md No
93
96
7.19
a.
Fig.
Superconducting Transition Temperature
15-8
4.15 2.39 3.95
1
5-1 7
Positions of superconducting elements in periodic table; transition temperatures
is confined to the simple metals and the There are no superconductors with only one valence electron, and there is also none with six valence electrons. When alloys of superconducting elements are formed, the values of Tc often lie between those of the separate components, as with Mo and Nb on the left of Fig. 15-18. The reverse may also occur, however, as shown on the right side of Fig. 15-18, where an alloy of 70 percent Mo and 30 percent Re has a Te of 11°K, although neither constituent has a Tc above 2°K. The largest
in Fig.
15-17 that superconductivity
transition metals.
As
the temperature
is
reduced, the electrical resistance of
decreases until a residual value
is
many
metals
reached. Further reduction of temperature,
even to the millidcgree range, often shows no further change of resistance.
About known
half the metals, however, exhibit a most extraordinary behavior as superconductivity: that
has been reached
and
is,
after the residual
the temperature
is
value of the resistance
further reduced, the resistance sud-
denly drops to zero. Superconductivity has been found in 26 metallic
atmospheric pressure or lower) and
ments
(in bulk, at
alloys
and compounds. The
transition
in well
ele-
over a thousand
from normal resistance
to zero resist-
ance
in the case of pure, strain-free
in a
very narrow temperature interval, about 10 mdeg or less; hence the Tc can be measured with accuracy. If the rcsistance-vs.-
metals (often single crystals) takes place
Tc have been obtained by forming compounds of Nb with other compound of Nb, Al, and Ge; 18.07°K for Nb 3 Sn; 18.0°K for Nb 3Al; 17.91°K for NbSn. Some of the most peculiar superconductors are compounds in which none of the elements is a supercon-
values of
metals: for example, 20.1°K for a
ductor, including Bi 3 Sr (5.62°K),
many
Many
Ge 2 Sc
(1.30°K), PtSb (2.1°K),
RhTe 2
of the superconducting alloys and com-
transition temperature
(1.51°K), and
temperature curve shows a more gradual decrease from normal to zero resistance, the transition temperature is taken to be the temperature of the mid-
pounds have been planned, caused to be made, and studied by B. Matthias at the Bell Telephone Laboratories. It is his contention that, someday, every metal will be found to be superconducting, provided that it is sufficiently pure and the temperature is low enough. One of the first attempts to apply modern quantum theory to the problem of superconductivity was made in 1950 by II. Frohlich, who for the first time took into account the interaction between the free electrons and the lattice,
point of the curve.
Values of 7c are printed below the chemical symbol
each superconan interesting the present time, superconductivity has never been observed
ducting clement fact that, until
in the periodic table
shown
for
in Fig. 15-17. It is
in the alkali metals, the alkaline earth metals, or the noble metals. It
is
seen
others.
534
15-8
Heat and 1'hermodvnamics
f
-J-,
Alloys with
Re /
and other results are listed in Table uranium isotopes is +2.
15-19,
\
n
Isotope
Element
/^deg/atm
a
V, 0. 2
0. 4
0.6
31
0.8
Concentration of
0.8
1.
0.6
40 Zr 42 Mo 44 Ru 48 Cd 49 In
Mo
5-1 8
Superconducting transition temperatures of alloys of (B. T. Matthias, 1963.)
Nb and Mo
and of Re and
-
1
81 Tl
-
3
82
Pb
-20 -43 -46
90 92
Th
thus,
Tc log
cc
(15-20)
ffl«,
Mixtures of
tin
isotopes
reverse
is
is
true, as
is
shown
where a
is
much
less
either small or zero. It
The superconducting
x 0.565
0.550 2.05
2.06
2.07
2.08
for
Mo
isotopes.
-
-0.21
-0.5 -0.5 -0.5
U
+
+2
1.8 37 23
-
38
-
17
+ 130
Table
is
15-1. In this table
when
it
may also be seen
the isotope exponent
is •£,
than ^ (Mo, Ru, Re, and Os) the pressure not certain whether this
is
that
but in four effect
a real correlation or not.
transition temperature of a thin film
shown
in Fig. 15-20/;,
is different from where the Tc of a Pb film is
of film thickness and of pressure that some non-superconductors become superconducting when formed into a film (Be, 7°K; Ge, 8.4°K; Bi, 6°K), and others when the pressure is raised to thousands of atmospheres (Sc, 6.8°K; Te, 3.3°K; Bi, 6°K), 2.09
1.96
1.97
mass
Log-log plots showing (a) that
1.98
log isotopic
(a) Fig. 15-19
2.6
seen to decrease as the film thickness decreases. So profound are the effects
&4
log isotopic
in
a large pressure effect
that of the bulk metal, as
¥, ^
-0.38
transition temperature depends also on pressure, as superconducting compounds in Fig. 15-20a. In most cases, an increase of pressure causes a decrease in Tc; but with Tl and U the for three
is
0.575
judeg/atm
The superconducting
cases
\-
0.580
-0.5
shown
there
Tc
-0.40
50 Sn
instead of limiting himself to the electrons only. Although a complete theory
could not be formulated at that time, one result that was obtained created great interest: namely, that the transition temperatures of the various isotopes of an element should vary inversely as the square root of the isotopic mass T/l\
-0.34
dTc/dP
cfFcct
a 73 Ta 75 Re 76 Os 80 Hg(a)
-0.5
Zn Ga
Element
-29 -16 -18 +15
13 Al
30
Isotope
dTc/dP,
effect
>
Mo.
was shown
it
Isotope Effect and Pressure Effect of Superconducting Elements
Table 15-1
V Hoys with Nb
1
15-1. Recently,
that a for the
«-
Fig.
535
and Superconductivity
with a = —J. The measurements made in the 1950s on Sn, Hg, Tl, and Pb agreed quite well with Frohlich's equation, but recent measurements on Mo, Re, and Os give a values much less than — and Zr and Ru show no isotope effect at all. Experimental results for Sn and Mo are shown in Fig.
8
6
Superfluidity
1.99
2.00
mass
(b)
a
= —^ for
(T. H. Geballe and B. T. Matthias, 1963.)
Sn
isotopes; (b) that
a =
—
Besides strains and impurities, there is one more condition that affects the superconducting transition temperature, namely, the electrostatic charge on the metal. This effect has not been studied extensively, but it has been found to
be
reversible. If the addition of a
A7"e, then the
removal of
this
charge has raised the temperature by
charge will bring Tc back to
its
original value.
536
Heat and Thermodynamics
dTc
dP 0.25
•
« a
£
0.2
15-9
is finite;
deg/atm
and Superconductivity
537
T = 0, by extrapolation, is zero. The relation between and temperature for the soft superconductors is almost parabolic. to a first approximation,
(3) the slope at
critical field
-°Nb 3 Sn -1.4-KT'/ a V3 Ga -2.4-10 -5 / _ \-a V Si -2.M0 *4 3
That
is,
Kc = K*\\-
0)
Q.
/
E
-
Superfluidity
(15-21)
/ 0.15
where 0.1
is
it
d%
o S 0.05
/
^— c
/
parabolic relation. Values of Tc , 7P&, and the slope c/dT at T = Tc are given in Table 15-2, along with some of the thermal properties of soft superconductors.
~-*^z ""'
A
1000
2000
500
kgm/cm 2
Pressure,
9^ and Tc was discovered by H. W. Lewis in superconductors but also for some compounds that, like the soft superconductors, do not have a partly filled d or f shell. The line plotted in Fig. 15-22 shows that, for these superconductors,
1000
Thickness, angstroms
and E. J. Saur. 1964.) It.
(b)
soft
y
(a) Decrease of transition temperature
films. (P. Hiisch,
simple relation between
1956 not only for the
(a) Fig. 15-20
with hydrostatic pressure. (C. B. Midler
Transition temperatures of lead films superimposed on thick copper
Hilsch,
and G.
v.
Minnigerode. 1963.)
Magnetic Properties of Type
15-9
absolute zero. Only when great accuracy is necessary to take into account departures from this simple
7f<, is the critical field at
desired
Superconductors
I
Of the 26 elements shown in the periodic tabic of Fig. 15-17, which become superconducting without being subjected to a high pressure or deposited as a thin film, the 9 simple metals on the right side of the table arc of particular Because these have low melting points and are physically soft, they soft superconductors. Suppose that a soft superconductor is made in the form of a cylinder about ten times as long as its diameter, is placed in a magnetic field longitudinally, and is kept at a constant temperature less than interest.
are labeled
Tc
If the field starts at
zero intensity and
remain superconducting
until a critical field
become normal. The
is
then increased, the metal will
7fc
is
reached, at which point
it
from the superconducting state to the normal state covers a very narrow range of 9£ and may be said to occur at the value 7£C Furthermore, if 9^ is now reduced, the transition to superconductivity takes place at the same value 9^cThe magnitude of the critical field required to destroy superconductivity depends only on the temperature. Thus 9£c is a function of T only and is will
transition
-
shown the
in Fig. 15-21. It should be noticed that
each
W
c -vs.-
12
T curve divides
%T plane into two regions, superconducting and normal,
like
a phase-
on a PT diagram. All curves show the following properties: always negative; (2) the slope at the point T = Tc (9fc = 0)
transition curve
Fig. 15-21
(1) the slope
superconducting transition metals.
is
Threshold field
vs.
4 5 6 Temperature, °K 3
temperature for eight soft superconductors
and for
three
538
15-9
Heat and Thermodynamics
Table 15-2
The remaining
Properties of the Soft Superconductors
pure and
and preferably
Tc,
\ dT )t= tc,
Oe
°K
0.
7',
°K
mJ/mole deg 2
cm 3 /mole
•
Oe/deg
of 7(^0
and Tc do not
satisfy
Zn Ga 48 Cd 30
31
49 In 50 Sn
80
Hg
81
Tl
82
Pb
-158 -121
105 53
93 86
428 310 320 209
-156 -147
108 199
-
51
-
30 283 306
1.35
9.93 9.20 11.8 13.0 15.4
0.65 0.60 0.69 1.66 1.78
16.1
411
-194 -126 -226
339 171
803
72 79 105
1.81 1.5
?/"„
3
cc
There are and
.
'/'J,-
other properties that will be described later, such as coherence length
so that
it is
now
sets of
the practice to refer to
all
superconductors have such metals as
type
Magnetic and thermal properties of the superconducting are given in Table 15-3.
Table 15-3
13.8 17.3 18.3
3.0
made very
if
form of single crystals, also show superconductors, although their values in the
the simple relation that
penetration depth, which these two 1.175 0.88 1.09 0.56 3.41 3.72 /4.15(a) 13.95(3) 2.39 7.19
13 Al
539
and Superconductivity
seventeen superconducting transition metals,
free of strain
properties similar to those of the soft
Metal
Superfluidity
Metal
common,
in
I superconductors.
transition metals
Properties of the Superconducting Transition Metals
Ta,
Oe
°K
\ dT )r.r c
7',
e, ,
°K mJ/mole
V, •
deg 2
cm 3/'mole
Oe/deg 22 Ti
/
1000
_
500
Hg(«)/
f
/
Ah X/3)
-
/•l\
/
Al
100
^ .
40 Zr 41 42
Nb Mo
43
Tc
44
Ru
57
La
72 73
Hf Ta
74
W
75 76
Re Os
77 90
Ir
Th
91 Pa
Zn./ Ga
92
50
U(7 )
0.39 5.30 0.49 9.20 0.95 8.22 0.47 f(a)4.87 1(0)6.00 0.165 4.48 0.012 1.70 0.67 0.14 1.37 1.4 1.8
56
1310 47 1985 97
66
-300 -482 -170 -453 -350 -196
789 1095
—
-
-
830 1.07
-334
201
-235 -183
80 19
162
~2000
-
420 380 291
275 450
3.5 9.8 2.80 7.80 2.0 -
600 152 140 252
240 400 430 500 420 170
-
-
-
182
3.3 9.4 11.5 2.16 5.9 1.21
2.3 2.4 3.1
4.7
_ 10.3
10.6 8.39 14.1
10.8 9.40 8.6 8.40 22.4 13.5 10.9 9.54 9.07 8.45 8.54 19,8 15.0 12.5
/
"
,'Cd
By forming
20
alloys of type I superconductors,
by forming compounds, by
depositing as thin films, by introducing strains through cold-working, and
/
1
by introducing impurities, superconductors with entirely different properties may be fabricated. Known as type II superconductors, these will be found to have
till
I
0.5
1
9f a
(H. W. Lewis, 1956.)
I
1
I
great practical importance.
2
1
Tc Fig. 15-22
V
In^Sn
200
-
23
,
The
°K
I superconductors was discovered by Applying Maxwell's electromagnetic equations a material of infinite conductivity, it was concluded that a superconducting
zero electrical resistance of type
Kamerlingh-Onncs 3
7"c
for soft superconductors that do not have a partly filled
d or f shell.
to
in 1911.
540
Heat and Thermodynamics
/s it
541
Superfluidity and Superconductivity
15-9
follows that
M= Beyond
- W-f
(15-22)
the critical field, however,
Phase transition
/S
The /$
=
8=
K
M=
Of
M
and
— V%/4t
=
means
0.
that a superconductor
is
a
perfect diamagnelic.
K
^c 5-23
fact that
=
The
Behavior of magnetic induction 13 and magnetization
Fig.
1
when
the field is increased isothermally
M
of a superconducto.
if placed in a weak magnetic field at room temperature and then down below To while still in the field, would retain its magnetic flux as as it was maintained at its low temperature. The argument based on
transition
temperature
T
from superconductivity
and
normal conductivity
to
at constant
at constant critical field 9^c, performed
under the ideal circumstances just described, is & phase transition. Since experiment shows a latent heat, the transition is of the first order. If the magnetic Gibbs function is defined in the usual way, that is,
metal,
long
electromagnetic theory was so convincing that no one checked the conclusion experimentally for twenty years. In 1931, Meissner cooled a superconductor in a weak magnetic field to a temperature below its Tq and classical
discovered that, at the netic lines of force result
is
moment
the metal
were expelled and
now known
as the Meissner
became superconducting,
the magnetic induction effect,
all
mag-
/3 became
zero.
This
=
is
con-
and the property /S
sidered to be just as fundamental as the property R'
=
0.
we
T dS =
Since
At constant
Suppose that a type I superconductor is made in the shape of a long thin and placed longitudinally in a magnetic field. The behavior of the magnetic induction /3 and the magnetization M, as the field is increased isothermally,
is
shown
in Fig. 15-23.
the particular temperature chosen. It
superconducting, /3
=
0;
and
9^c
the critical field appropriate to
is
may be
since
= 9f +
seen that, while the metal
this
T
and
9£,
dG*
=
M £>f.
reduces to
-SdT -
M d%.
and
C*m + dG* M = ,/G'*w
Therefore,
-S M dT -
=
M M d^c =
and
MM
-
MM
G*<">
-f-
dG* w
,
dG*<-"K
SM dT -
P
dT
Since
=
+
-
M<»> (Df0i
A/<«>
4r
'
is
we 4;
7f dM,
dT -
,
transition of the sample.
cylinder
—
S
where the superscripts (s) and (n) stand for superconducting and normal phases, respectively. At a temperature T + dT and a field %c + d%c
create an "unbalance" in a bridge circuit,
which then causes a signal to appear at the input to a detector. The detector amplifies this signal, and its output is plotted as a function of temperature to show the superconducting
dll
TS
dG* =
In fact, the most
commonly used method of determining whether a material is superconducting and, if so, the value of Tq employs the change in self-inductance of a solenoid when the sample under investigation is located inside the coil. The solenoid is made part of a parallel resonant circuit that controls the frequency of an oscillator, and changes in frequency corresponding to changes in inductance
dG*
get
U-
- WM, = dll - T dS - % dM -
0* =
cooled
get
dKc - dT =
, 4
*
(15-23)
M The
third
law of thermodynamics requires that the entropy difference
542
Ifcat
SM
—5 7=0.
at
(l)
Since S
15-10
and Thermodynamics
vanish at absolute zero: therefore the slope
M—
=
S U)
d%c/dT must be zero
Superfluidity
10 Tin
L/T, where
/.
the latent heat,
is
we
543
and Superconductivity
>"
get
/-
TV9fB £¥g 4tt dT
0>
efi
(15-24)
Superconducting a
s*
JjjP
D
m-€
.-**
6 £>-'
This derivation
is
in all respects
equivalent to the derivation of Clapcyron's
T=
extremes of temperature: namely, at
T=
Tc, where 9£g
=
0,
where dP^c/dT
=
0,
and
field takes
therefore a higher-order phase transition
place with no latent heat,
— one
it is
measurements, to be treated in detail later, show of the second order and may possibly be the only example
for
8
12
1
14
16
18
tVkv Fig. 1 5-24 circles,
tin by Keesom and van Laer. 1938 {open and by Corak, Goodman, Satterthwaite, and Wexler, 1956
Measurements of heat capacity of
super; open squares, normal),
(solid circles, super; solid squares, normal).
Heat Capacities of Type
The apparatus
J,
of three types depicted in
of a transition of this type.
15-10
Normal
at
Fig. 12-25. Heat-capacity is
fti
0.
Since the phase transition in zero
that the transition
''a
-&
The value of dTfofdT may be obtained from the ^c^H curve, and hence L may be calculated. Values of L calculated in this way agree very well with those obtained calorimetrically. It is interesting that L vanishes at two equation.
measurement
I
Superconductors
of the heat capacity of a superconducting
substance differs from that for an ordinary substance only in the provision
a magnetic field to destroy superconductivity at temperatures below Tc. As a rule, the heat capacity is measured in the absence of a field at temperatures above and below Tc in order to show the transition at Tc, as well as the temperature dependence below To- Then a field larger than Pf is applied, and the complete temperature dependence of the normal heat capacity is measured. Temperature is measured usually with a carbon or germanium resistance thermometer that is inserted into the sample, and heat is supplied by a momentary current in a heating coil wound around the sample. Resistance-vs.-timc curves arc plotted automatically on a recording instrument, and the data read from the records are fed into a computer. A typical set of experimental results is shown in Fig. 15-24, where the measurements done for tin by Keesom and van Laer in 1938 are compared with those of Corak, Goodman, Satterthwaite, and Wcxler in 1955. As is usual at low temperatures, cp = cv = c, and c/T is plotted against 'T* since, in the normal state, c in mJ/molc deg 2 is given by Eq. (11-53), for
125
e
T»
+ y'T.
Experimental curves of the same character arc. obtained for transition metals of great purity and freedom from strain, as shown by the results of H. A.
Leupold and H. A. Boorse (1964) on a single crystal of Nb in Fig. 15-25. Values of the electronic constant y' and the Debye (-) are listed in Tables 1 5-2 and 15-3. The behavior of the heat capacity in the superconducting phase is entirely different from that in the normal phase as shown in Figs. 15-24 and 15-25.
From
these experimental graphs,
it is
possible to read values of c
M —
c
that
may be
used in conjunction with simple thermodynamic equations, as follows. It has been shown with the aid of the magnetic Gibbs function that, for a reversible transition from superconductivity to normal conductivity at temperature
T and
holds. Thus,
field
9f =
?fc, an equation similar to Clapcyron's equation
from Eq. (15-23),
SM
- Sw =
4i
Differentiating both sides with respect to
T
dS (n)
TT-
dSM T dT _ "
T
c
dT
and multiplying by T, we get
VT d_( 4a-
dT\
c
d?fc\
dT/
544
Heat and Thermodynamic
15-10
545
Superfluidity and Superconductivity
Equation (15-25) may be used with more confidence to infer the magnetic from calorimetric data. Thus, performing the
properties of superconductors differentiation indicated,
we
get
dVf
_ ,M =
,£«)
At \ d'f / At
T=
—
Tc, 5Kc
and f
this equation,
i
as the Rutgers equation, the slope of the y^c-vs.-
may
c u)
_
c co)
T=
dr
to
,
member may be
j-
Td {*•%)
\
integrated by parts, with the result that
"Tc
Normal and
5-25
Lcupotd and
II.
superconducting heat capacities of a single crystal of A. Boorse, 1964.)
Nb.
(II.
the left-hand side are heat capacities, so that for
1
mole of
dl
4w
ITe Jo
-
aV-f i:
4Kc dl AtJtsc
term on the right vanishes at both the upper and lower the second term reduces to a simple integral; thus,
The The terms on
dWc
A.
get
/()
v
=
7"
be found from calorimetric
T = Tc we
Jo
1
zero. Thus,
(15-26)
known
Tc
Fig.
is
\''Ic
Integrating Eq. (15-25) from
light-hand
the right
£¥o
curve at the transition temperature To measurements.
The
c %C -j-fqdT-
member on
the second
dl It=Tc
and from
4jt
first
dT.
limits,
and
material
M-M.'JL±(3xr 0£o This
is
a very valuable equation, but
it
must be used with care.
and
is
difficult,
9^ =
and If
calorimetric
measurements are particularly difficult, as they are with mercury, it is necessary to use Eq. (15-25) in conjunction with magnetic measurements (9fc as a function of T) in order to find y'. If the material is very pure and quite free from strain, the phase transition from superconductivity to normal conductivity may take place reversibly, and the equation may be used with confidence. It
r *"—•>«-£•
(15-25)
however, to get some of the hard superconductors pure
strain-free. When this is the case and the measured values of 9fe are introduced into Eq. (15-25), the calorimetric data so derived may be in error by a large factor.
J«*r^-c^)dT. Jo X
(15-27)
v
and vanadium the values of •Vo, and indeed the entire %c-v%.-T curve calculated from calorimetric data, agree very well with purely mag-
With
tin
netic measurements.
metal before and after the superconducting transition indicate no change in any property of the crystal lattice. It is therefore assumed that a superconducting metal has the
X-ray diffraction studies of the
same Dcbyc (125/©)
3
3
J"
as in the
crystal lattice of a
normal phase.
If we subtract the lattice heat capacity
from the superconducting heat capacity, therefore,
we ought
to
546
Heat and Thermodynamics
15-11
0.1
equation
N Eh
0.01
~
0.001
1\
Tin
,(s>
-vr,iT
approximately. This behavior has been observed for many type I superconductors and has an important theoretical significance in terms of an energy
X
0.0001
gap, as
15-26
wc
15-11
123456789
shall see in the next section.
Energy Gap
10
Tc /T Fig.
(15-29)
f
y'Tc
\
0.00001
properties of superfluid helium were adequately described by which He 4 below the X point was imagined to consist
Many of the
Superconducting electronic heat capacity of
(H. R. O'Neal and N. E-
tin.
Phillips, 1965.)
the two-fluid model, in
normal atoms and superfluid atoms with densities p n and p., the temperature was reduced, normal atoms underwent a type of "condensation" and became superfluid. Conversely, the absorption of heat provided the energy to raise superfluid atoms to the normal state, as
of a mixture of respectively.
get a heat capacity associated with the superconducting electrons, c[f. Thus,
#
= cM ~
(jf)*
T3
(15-28)
-
As
well as to raise the temperature. These picturesque ideas are legitimate interpretations of the rigorous statements and equations of quantum statistical
mechanics.
A
similar situation exists with regard to superconductivity.
free electrons in a superconducting metal below
In Figs.
Tc/T.
1
5-26 and
Many
1
5-27, the logarithm of the ratio c'i'/V
of the experimental points
lie
on a
Tc is plotted
against
straight line, indicating that
the electronic contribution to the superconducting heat capacity obeys the
in 1934, to
state as well as to raise the temperature.
\\ -
Nio Dium
electric
and Casimir enabled the description of existing phenomena of superconductivity and suggested further experiments leading to other
phenomena.
\
Valuable as they have been and are, two-fluid models of superfluidity and of superconductivity cannot, however, be regarded as physical theories. No adequate microscopic theory of superconductivity based on quantum ideas was
o *
\
-
-3
o V)
r-
0123456789 1
10
Tc/T Boorse, 1964.)
account for persisting
of Gorter >
8 -2
-4
To
currents in superconductors, F. London and H. London in 1935 made addimodel tions to Maxwell's equations, which in conjunction with the two-fluid
\
-l
The
Tc were imagined by Gorter
be a mixture of normal electrons and superconducttemperature was reduced, normal electrons underwent As the ing electrons. superconducting. Absorption of heat was became condensation and a type of electrons to the normal superconducting the energy of needed to raise the
and Casimir,
1
Fig. 15-27
547
Superfluidity and Superconductivity
Superconducting electronic heat capacity of niobium. (H. A. Leupold and
H. A.
developed until 1957, forty-six years after the discovery of superconductivity, when J. Bardeen, L. X. Cooper, and J. R. Schrieffer succeeded in applying quantum mechanics to the behavior of free electrons in a crystal lattice. If the free electrons arc regarded as an ideal gas obeying the Fermi-Dirac behavior and statistics, only repulsive forces are taken into account, and the it would be circumstances, Under these are ignored. properties of the lattice quite impossible to account for the isotope effect or the heat-capacity results.
548 It
Heat and Thermodynamics
1
was pointed out by Cooper, however, that the presence
of
an electron in a which may
lattice of positive ions gives rise to a slight distortion of the lattice,
be described
as a slight
crowding of positive
1-0
This clump of positive ions
ions.
5-1
Superfluidity and Superconductivity
1
u^o— CqI a'-o±
+,
then serves to attract another electron, so that attractive forces among electrons may be deduced. The attractive force produced by lattice interaction
-rr,
Hf
can take place over a distance of as much as 10~ cm, in agreement with an idea previously postulated by A. B. Pippard, namely, that the phenomenon
fc
of superconductivity pointed to the existence of a long-range influence which must exist over a coherence length of the order of 10~ 4 cm. Cooper was able to
0-6
show theoretically that the effect of electron attraction in a crystal lattice is to produce pairs of electrons (now known as Cooper pairs) which move as a unit without exchanging energy with the vibrating lattice ions. According to the theory of Bardcen, Cooper, and Schrieffcr (hereafter abbreviated BCS), when energy is supplied to the superconductor, some of the Cooper pairs may be broken apart into separate electrons. There is therefore an energy gap of 2e between the Cooper pair and the two separated electrons. It was found, in the last section, that the superconducting heat capacitycould be regarded as the sum of a lattice portion (125/0) 3 7" 3 and an electronic portion <$', which roughly obeyed the equation
y'Tc This equation has the form needed
gap
e,
to describe
BCS
curve
a.
Tin
o
Tantalum
+
Lead
D_Niobium
\
—
V 0-2
0-2
0'6
0-4
0-8
T/Tc 1
5-28
(J. Sutton
.
—
0-4
Fig.
ae- bT- IT
^
0-8
4
=
Variation of energy gap with and P. Townsend, 1963.)
temperature. Solid curve is derivedfrom
BCS theory
a process involving an energy
where €
bTc
kT
T
Table 15-4
Energy Gap
at
T =
0;
and therefore
To
2e
Element (15-30) 13 Al
If the various portions of the curves in Figs. 15-26
and 15-27 are used
vide values of
e— is found
b,
this
quantity
—and therefore
also
to pro-
have a tem-
to
perature variation. Heat-capacity measurements, as a rule, provide this temperature variation at the lower temperatures, but fortunately there are other methods that provide the complete temperature dependence of t. Points obtained by several different methods are shown in Fig. 15-28, where seen that the experimental results arc in good agreement with the curve derived from the BCS theory. The quantity e„ is the value of e at absolute it is
zero.
Values of 2e
23
V
30 31
Zn Ga
41
Nb
for
1
2 superconductors are given in Table
1
5-4
and arc
kTc
&
3.3 3.5
0.0027 0.014 0.0028 0.0034 0.033 0.0027 032 0.019 0.019 0.058 0.030 0.069
3.1
3.4 3.7
48 Cd 49 In
3.1
50 Sn 73 Ta
3. 5
80 Ilg(a) 81 TJ 82 Pb
/kTc
Values of 2e /kTc
'
&-*
549
3.6
3.6 4.6 3.6 4.2
.
9
Tc (
~K
dKA dTJT,
\ 3.5 3.9 4.0 4.0 4.0 3.2 3.6 3.5 3.6 4.1
3.6 4.2
550
Heat and Thermodynamics
15-12
was first measured by D. Shoenberg in mercury, by subinto minute droplets. As was to be expected, X was mercury dividing the 6 of the order of 5 X 10" cm. It was also found to vary small, very found to be
5.0
penetration depth,
oHg(«)
R
1
-as
551
Superfluidity and Superconduct:
.
with
T
according to the equation
X
Pb
Xo
=
Vl -
4.0
where X
Nb o
Ta 3.5
Tl,
50
It In
Cd
the penetration depth at absolute zero. in the previous section that the electron-electron attrac-
was mentioned
produced among the free electrons in a superconductor by and which is responsible for the formation of of about Cooper pairs is active over a distance the coherence length -4 is that superconductor of a type I features characteristic One of the cm. 10 a type I Another feature of in Fig. 15-30a. schematically shown as X tion
Ga o
is
(15-32)
T*/T<
Sn
which
is
interaction with the lattice
Zn
«
3.0
—
—
t,
?,
in is provided by the behavior already studied, depicted which the isothermal transition from the superconducting the normal one takes place at constant 9^c (when the sample is a
superconductor Fig. 15-306, in 0.002
0.005
0.01 log
Fig. 15-29
0.02
0.0b
0.10
T/
Relation between energy gap 2«o and
To/
phase to
long cylinder parallel to the
& for type I superconductors.
plane
into only Iwo regions,
The coherence
Tc/& in Fig. 15-29. The rough relation shown on the graph has some theoretical significance in terms of the BCS theory. It was pointed out by A. M. Toxen in 1 965 that there was a good correlation between values of 2i a /kTc and the quantity in the last column of Table 15-4. plotted logarithmically against
Type
II
Superconductors
%
The
inclusion
abruptly according to the equation
y
f
,v
is
for 9fi to
curve divides the
TfT
of a superconductor
is
a measure of the
mean
free
the average distance between two consecutive scattering of the electron. Electrons can collide the result in processes which defects, or boundaries, and such crystal impurities, scattered by with or be is,
of impediments can be deliberately created in a crystal. The greater the frequency the same shows that Experiment length. coherence the scattering, the smaller
magnetic field penetration, creating crystal defects by so that it is the substance as a very depositing or by alloying, through cold-working, by length coherence of the £ and an increase reduction thin film to effect both a also increase the
which increase scattering
to
1
X.
superconductor in which X
depicted schematically
»
£ is called a type II superconductor
in Fig- 15-3 la. It has
and
(15-31)
%
outside.
The
and X
is
quantity
the distance X,
called the
is
most unusual properties and
— important practical applications. Instead of starting the isothermal s > n sudden phase transition at c it starts at a lower field 9fC \\ then, instead of a transition is smeared isothermal phase the critical field, transition at constant than 7fc that is larger value to a of from ?f, over a range of values %c\
%
as
= 7ff^\\
fe of the value
A
,
%n
the distance perpendicular to the boundary
be reduced
of penetration depth
-
by the Londons of an additional term with the classical electromagnetic equations enabled them to derive the result that the field %i inside a superconductor drops from its outside value 7£ to a value of zero very
where
9feT
that the
possible by injecting impurities,
When a superconductor is placed in a magnetic field of intensity %, no magnetic lines of force penetrate so long as 9f < C The transition from the finite outside value 9f to the inside zero value cannot, however, be discontinuous.
£
path of the electrons, that
factors
15-12
length
field), so
superconducting and normal.
shown
in Fig. 15-316.
As a
result, the
7fT
plane
is
,
subdivided into
three
superconducting: (b) above 9fic% normal; and (c) where the material is said to be in the mixed phase, in between 9£c\ and Wet, as well as normal regions and can theresuperconducting which it contains regions: (a) below
fore carry
an
9fC
\,
electric current
without resistance. This
is
usually a large
552
15-12
Heat and Thermodynamics
•*i
attraction
process on a macroscopic scale, like the motions of the extranuclear electrons of a giant atom. If the ring of current is made smaller, it encloses a smaller magnetic flux and, when very small, may be considered a vortex ring
\ ..
-
f
553
quantum
Type I superconductor
-
and Superconductivity
magtaincd at the appropriate temperature. Careful measurements of the over 1 00,000 would last the current netic field of this persistent current show that example of a years. It is generally believed that such a persistent current is an
Electron
K
Superfluidity
3^=*'
of current. F.
London
first
realized that the magnetic flux
produced by a
current vortex should exist in discrete quanta, and he calculated a quantum of flux to be hc/e. A large current vortex would contain so many flux quanta
xV(a)
Electron attraction
.0
/
Type
II
superconductor
Normal
M''
Fig. 15-30 tion depth transition
Fundamental properties of a type I superconductor,
(a)
Magnetic field penetra-
X and coherence length £ are such that X
isothermal phase
and %ci may be as much as 1 times as large as 9^c As a result, at the proper temperature, a type II superconductor can remain superconducting in a large field (up to about 200,000 Oe !) or can carry a huge current region,
.
without becoming normal. Because of this property, superconducting solenoids may be wound and, once a current of about 20 to 30 A is set up, will provide a field of, say, 60,000 to 100,000 Oe without any power consumption.
The most striking aspect of type II superconductors is the mechanism of magnetic flux penetration in the mixed state, and its analog with quantized vortices in liquid
helium
ring, the current persists
a current is induced in a superconducting undiminished for as long a time as the ring is mainII.
If
Fig. 15-31
Fundamental properties of a type II superconductor,
»
£. (b) The X and coherence length £ are such that X 9/- c 2 starts at a field %c.\ smaller than 9fc and ends at a field phases exist: superconducting, mixed, and normal.
tion depth
(a)
Magnetic field penetra-
isothermal phase transition larger than
9^c, and
three
554 that
any change
current or flux would appear continuous. If the vortices are made extremely small, however, such as in a vanishingly thin superconducting cylinder, and if such a cylinder is placed in a weak field, as the field is increased the flux should increase in discrete steps. Difficult as this experi-
Germany
(b)
in
ment sounds, it was performed by two experimental groups independently— B. S. Deaver and W. M. Fairbank in America and R. Doll and M. Nabauer in
in 1961.
The
555
Superfluidity and Superconductivity
Heat and Thermodynamics
results of these notable
experiments confirmed
London's prediction, but the size of the flux quantum (or fluxoid) was found to be only one-half as large as expected. This is regarded as a direct experimental verification of the existence of Cooper pairs. Since the current is carried by pairs of electrons, each with a charge of
2e,
the flux
quantum
Calculate the speed of third sound at 1.5°K and 10
level of the
A
15-5
R =
radius
1.28
g/cm 3
X
10~
3
has a quantum number n
4
cm, and a hollow core of radius a
Using the curves
15-6
(b)
the energy,
if
cm
mercury, and (c) lead. 15-7 Show that,
G*< 8 >
in Fig. 15-21, calculate the
=
a
1,
1.28 A. (a)
He
is
work done and the
of material during the transition
3
if
(at
G*
from the superis
(a)
tin,
(b)
denotes the magnetic Gibbs function,
=
T,9fc)
G*<*> (at T,0)
In the transition of a type
15-8
=
the density of
conducting to the normal phase at 3°K when the material
Little in 1964.
the
.
heat absorbed by 100
should be hc/2e. Flux quantization was also shown to exist very close to the transition temperature Tc of a superconductor by R. D. Parks and W. A.
The penetration of the magnetic field through a type II superconductor is considered to be accomplished through minute cylindrical current vortices, with axes parallel to the field and each enclosing one quantum of magnetic
He
vortex ring in liquid
Calculate the velocity of the ring and 0.146
cm above
bulk liquid.
+
VTfl
superconductor from the supercon-
I
ducting to the normal phase at a constant temperature T, show that:
flux.
V_d_ 5 (.>_SW=_JL
(«)
PROBLEMS A
15-1
dT (^).
8tt
C« -
(b)
thermally insulated vessel contains liquid helium at 4.2°K and
atmospheric pressure. Helium vapor is pumped away iscn tropically until the remaining liquid and vapor achieve a temperature of 1°K. What fraction of
Assuming
15-9
C<»>
^ £, (&$.
=
37 dt-
that, for a type
superconductor,
I
amount
the initial
of liquid helium evaporates during the process? (In the temperature range from 1 to 2.2°K, the specific heat of liquid helium equals 0.104 r 6 2 J/g -deg, and from 2.2 to 4.2°K it remains approximately con-
M=
-
stant at about 2.5 J/g
15-2
•
deg.
Liquid helium
tain effect.
is
The
heat of vaporization at 1°K
contained in a vessel designed
to
is
T*
20.1 J/g.)
show the foun-
found to be 30 cm when the temperature increase is 5 mdeg, what is the entropy in J/g deg? 15-3 A column of liquid helium II is at a temperature of 1.5°K. (a) Find p„/p, p,/p, and p K/p„. If the fountain
'125\ 3 ,.(»>
- -
and
Vr
+
is
(f)"
r>
•
(b)
If c
=
1.13 J/g
show
that,
when Tis near
absolute zero, the entropy change from the super-
conducting to the normal phase
deg, calculate the speed of second sound. (c) small region at one end of the column undergoes an increase of temperature of 10 mdeg. Calculate v n and vs
is
expressed as:
•
A
yfe)
_
,(.)
=
y'T.
Tfil
=
7f\
.
(d)
The
density at 1.5°K
is
0.146
g/cm 3 Calculate .
the excitation
(b)
- 4*y
.,, 7" 2 .
energy per unit volume and the kinetic energy per unit volume. (e)
Verify that
HAT/T =
2/C.
15-4 (a)
ic)
Calculate the speed of fourth sound at
1
.5°K.
Explain
how
it is
possible to
measure
noncalorimetrically and magnetically.
7' of a
superconducting metal
556
Heat and Thermodynamics
15-10
16.
Prove that:
M
dT.
T
Jo
Jo
J
CHEMICAL EQUILIBRIUM
(*)
w\ (c)
-/TcXTg)
Zty'Tl
2
Figure PI 5-1 a and b displays the difference between a paramag-
15-11 netic crystal
and a type
I
superconductor.
Dalton's
16-1 Type
I
superconductor
Law
Imagine a homogeneous mixture of inert ideal gases at a temperature T, a and a volume V. Suppose there arc n\ moles of gas A\, n 2 moles of gas /1 2 etc., up to n c moles of gas A c Since there is no chemical reaction, the pressure P,
.
,
mixture
in a state of
is
PV = P = It
is
equilibrium with the equation of state ( Bl
+ n )RT,
+m +
c
^RT + n1?RT+
+ ^RT.
clear that the expression
fXT Tc T (a)
(b)
represents the pressure that the kth gas would exert
V
Fig. PI 5-1
alone. This
is
if it
called the partial pressure of the kth gas
occupied the volume
and
is
denoted by pk
.
Thus, (a)
Discuss the isentropic process
i'
—*f in
point of order and disorder. (b)
Assuming the
lattice
Pi
each figure from the stand-
contribution to the heat capacity to be
negligible, so that
The above equation e
W =
y'Tc ae- bT
>
p*
is
P*
+
,
= "fRT, (i6-i
+Pc-
expresses the fact that the total pressure of a mixture of
equal to the
sum
of the partial pressures. This
IT
Now,
V=
derive an expression for the temperature Tj achieved by an adiabatic magnetization.
= yRT,
P =pl+p2
and
ideal gases
and
= fRT,
=
(«1
+ H8 + RT
2 «*
+n
RT
,
c)
-y-
is
Dalton's law.
Chemical Equilibrium
16-3
558
559
Heat and Thermodynamics
the partial pressure of the kth gas
and
is
h-$XT. Substituting the value for V,
The
ratio «*/S«*
is
we
get
and
called the mofc fraction of the fcth gas
is
denoted by
.v*..
Bunsen
Thus, Xl
=
«i
Xi
= ~—
burner
Tie >
To
2nk
and
pi
The mole
=
XiP,
pi
=
x 2P,
pc
=
xj>.
*i
Xl
is
which
membrane equilibrium
said to exist.
is
+
same on both •
2S«*
sides of the
Whether
this is actually so is
1.
is
determined, the
last
can be calculated
Semipermeable Membrane
If
a narrow tube of palladium
sealed into a glass tube, as
shown
is
shall
suppose that there
not important.
membrane
as
which we have
exists
a
to deal.
We shall make use of the princian
ideal device for theoretical
Gibbs'
Theorem
the aid of a device equipped with two semipermeable membranes, possible to conceive of separating in a reversible manner a mixture of two
With
closed at one end
in Fig. 16-1, the
We
to each gas with
purposes.
16-3 16-2
Membrane equilibrium is achieved when membrane is permeable is the
membrane.
membrane permeable
special
ple of the semipermeable
Hence, if all but one mole fraction from the above equation.
hydrogen.
the partial pressure of the gas to which the
sn*
=
to
to
clear that
"1
+
Palladium tube permeable
Fig. 16-1
fractions are convenient dimcnsionless quantities with
express the composition of a mixture. It
pump
and the open end
system
may
be pumped
is
to
a very high vacuum. If the palladium remains at room temperature, the vacuum can be maintained indefinitely. If, however, an ordinary bunscn burner is placed so that the blue cone surrounds part of the tube, with the rest of the flame causing the palladium tube to become red-hot, hydrogen present in the blue cone will pass through the tube, but other gases will not. Red-hot palladium is said to be a semipermeable membrane, permeable to hydrogen only. This is the simplest laboratory method of obtaining pure dry hydrogen.
Experiment shows that hydrogen continues to flow through the red-hot palladium until the pressure of hydrogen in the vessel reaches a value equal to the partial pressure of the hydrogen in the flame. When the flow stops,
it is
inert ideal gases.
The
compartments by a gas A\.
vessel depicted in Fig. 16-2
rigid wall,
Two pistons coupled so that
is
divided into two equal
membrane permeable only to the they move together at a constant distance
which
is
a
all gases, apart are constructed of materials such that one is impermeable to state is depicted initial The to the gas permeable only At. and the other is chamber, and the in Fig. 16-2/. A mixture of Ay and A 2 is in the left-hand
right-hand chamber
Now
is
evacuated.
imagine pushing the coupled pistons to the right
in
such a manner
that the following conditions are satisfied:
1
The motion at all times.
is
infinitely slow, so that
membrane
equilibrium exists
560 2 3
There is no friction. The whole system is kept
U, at constant temperature.
to the left
=
(pi
of the forces to the right
=
(Pi
Force
+ pi) X + Pi) X
-
T(Sf and
since
T
is
-
We
Si)
=
0,
-
S,.
not zero, Si
Now
coupled pistons are:
Sum
both reversible and isothermal, the U- Finallv, since the process was have, therefore, the result that Si). is equal to T(S, transferred
Q
heat
These conditions define a reversible isothermal process. Consider the system at any intermediate state such as that depicted in Fig. 16-2£. If pi Zn6.pi ar e the partial pressures, respectively, of A\ and A t in the mixture, Pi is the pressure of A\ alone, and Pi the pressure of A* alone, then the forces acting on the
and
=
561
Chemical Equilibrium
16-4
Heat and Thermodynamics
Si is the
and the volume entropy of the mixture at the temperature T at the same temeach gases, the two of sum of the entropies
V while S, is the alone. If we define the partial perature and each occupying the volume V that the gas would have entropy as the entropy of one of the gases of a mixture temperature, then we same the alone at occupied the whole volume
area, area.
if it
Since
membrane
equilibrium
resultant force acting
moved
all
the
way
exists,
pi
—
Pi and pi
on the coupled pistons
to the right, the gases are
is
=
P-y,
whence
zero. After the pistons
the
have
completely separated as shown
obtain the result that the entropy of a mixture of entropies.
This
of gases
is
is
known
as Gibbs' theorem.
ideal gases is the
sum
generalization for
The
of the partial
any number
obvious.
in Fig. 16-2/.
Since the resultant force was infinitesimal in the beginning and zero = 0. Also, since the process was throughout the remainder of the process,
W
isothermal and the internal energy of an ideal gas
is
a function of
j
and
A
?
only,
Imagine
Vacuum
of inert ideal gases separated
from one another by
suit-
of the separate
perature Initial
number
temperature T and pressure P. able partitions, all the gases being at the same of moles Ai, etc., up to n c moles of A Suppose there arc Bl moles of gas A h of the whole system St is the Before the partitions are removed, the entropy of the fcth gas at temmole entropies. The entropy of 1
sum
(t)
a
m
r
A
T
Entropy of a Mixture of Inert Ideal Gases
16-4
T
and pressure
P
is
equilibrium state
therefore,
sk
= j
S,
=
c Pt
-J?
£ m (j
+
c Pk -jr
(k) Intermediate equilibrium state It is
convenient
- R
5oa-
to represent the first
+
in P;
sm
-M
In
Pj
two terms within the parentheses by
«r,
thus:
"*
Then (/") Final
Fig. 16-2
equilibrium state
Reversible isothermal separation of two inert ideal gases.
5,
~2/
dT k
~T
= RSntfa -
,
s ok
"*"
(16-2)
~R In P).
(16-3)
temperature and pressure remain the After the partitions are removed, the reaction, but the gases diffuse and, by same because there is no chemical
Chemical Equilibrium
16-5
562
Gibbs' theorem, the entropy of the mixture
the
is
sum
of the partial entropies.
The if it it
partial entropy of the kth gas is the entropy that the kih gas would have occupied the whole volume alone at the same temperature, in which case,
would exert a pressure equal
entropy of the mixture
Since
/>*
=
to the partial pressure
p k Therefore, the .
total
is
= R2n k (o-k —
Sj
In
The change
of entropy
due
-
In
P-
In x k ).
(16-4)
to the diffusion of
any number
-
**,
Each
of the
rithm.
mole
fractions
The whole
we mav
Si
= -R2n k In
a number
is
expression
is
application of mathematics to the macroscopic processes of nature that, as the usually gives rise to continuous results. Our experience suggests
The
change due to two diffusing gases become more and more alike, the entropy approaching zero as the gases smaller, smaller and get should diffusion identical. The fact that this is not the case is known as Gibbs' paradox. has been resolved by Bridgman in the following way: To paradox The
of inert gases
is
therefore equal to
S,
the diffusion of two identical gases
same degree of disorder, whereas introduces no element of disorder. the
become
pk)
xk P,
Sf = R2n k (rrk
(16-5)
recognize that two gases are dissimilar requires a set of experimental operdifficult as the gases become ations. These operations become more and more principle, the operations arc possible. in at least but, more alike; and more
when the gases become identical, there is a discontinuity in the instrumental operations inasmuch as no instrumental operation exists by which the gases may be distinguished. Hence a discontinuity in a function, In the limit,
such as that of an entropy change,
be expected.
to
is
than unity with a negative loga-
less
therefore positive, as
it
nk
nn,RT
nkPv
nkv
2n
VnRT
PV
V
should be. Since
'
16-5
Gibbs Function of a Mixture of Inert Ideal Gases
The enthalpy and and pressure
P
the entropy of
mole
1
of
an ideal gas at temperature
are, respectively,
write
S,
—
- S = mR In +nR n\V 4
The above result shows number of ideal gases is
2
In
—
h
=
+ f
h
n-zv
change due to the diffusion of any the same as that which would take place if each gas were caused to undergo a free expansion from the volume that it occupies alone at T and P to the volume of the mixture at the same T and P. The validity of this result was assumed in Chap. 9 in order to calculate the entropy change of the universe when two ideal gases diffuse. The assumption is therefore seen to be justified. As an example, consider the diffusion of 1 mole of helium and 1 mole of that the entropy
and
c
cP
™+
therefore the molar Gibbs function g
=
kt
cpdT- r
+
Applying the formula
dT j„-i?lnP;
=
h
Ts
dT dI
f
for integration
—
is
Ts
equal to
+ RT In P.
by
parts,
=
-T
/
-
+ RT In P
neon. Then, Sg
- Si =
-R(l
= 2R In
563
Heat and Thermodynamics
this expression there arc
In
$
+
1 In
ce
any two
dT - T
d'l I
cP
-=
f
eP
dT
T
In 2.
no quantities such
as heat capacities or entropy
The result is the same for the no matter how similar or dissimilar
constants that distinguish one gas from another. diffusion of
,
|)
inert ideal gases,
they are. If the two gases arc identical, however, the concept of diffusion has
no meaning, and there is no entropy change. From the microscopic point of view, this means that the diffusion of any two dissimilar gases brings about
we
get
g
=
4,
-T
f
*-
dT ,-n— T
1
= RT
RT
n
R
r
(
Ts
c diri
cpdT
T
,
R
dT,
T
564 It is
Chemical Equilibrium
16-6
Heat and Thcrmoclvna
convenient to denote the
three terms within the parentheses by
first
16-6
f cr
dT (16-6)
R
Chemical Equilibrium
thus:
-1 A RT R
565
Consider a homogeneous mixture of 1 mole of hydrogen and 1 mole of oxygen at room temperature and at atmospheric pressure. It is a well-known temperature, that this mixture will remain indefinitely at the same fact
and composition. The most careful measurements over a long state. One period of time will disclose no appreciable spontaneous change of represents a system might be inclined to deduce from this that such a mixture If a not the case. in a state of thermodynamic equilibrium. However, this is spark is electric small piece of platinized asbestos is introduced or if an sudden involving a created across two electrodes, an explosion takes place change in the temperature, the pressure, and the composition. If at the end and of the explosion the system is brought back to the same temperature oxygen, no pressure, it will be found that the composition is now -i mole of measurable amount of hydrogen, and 1 mole of water vapor. The piece of material such as platinized asbestos by whose agency a chemipressure,
The molar Gibbs
=
g
where
is
may
function of an ideal gas
a function of
T
RT{
+
therefore be written as
In P),
(16-7)
only.
Consider a number of inert ideal gases separated from one another, all at the same T and P. Suppose there are moles of gas At, «2 moles of A%, etc.,
m
up
A
to nc moles of
system
G
is
the
c.
sum
Before the gases are mixed, the Gibbs function of the of the separate Gibbs functions, or
cal reaction
is
started
is
1 mole amounts and different kinds of catalysts and if the final composition of the mixture is measured in each case, it is found that: (1) the final composition does not depend on the amount of catalyst used; (2) the final composition the same does not depend on the kind of catalyst used; (3) the catalyst itself is to the at the end of the reaction as at the beginning. These results lead us
started in a mixture of
d
= 2n kgk
= RTSnk (4>k + where the summation extends from k function after mixing
by the
Gf
,
—
\
In P),
to k
=
c.
To
express the Gibbs
necessary merely to replace the total pressure
it is
P
following conclusions:
bk (Why?) Thus,
partial pressure
,
1
Gj
G/
Therefore,
—
+ = RTSntfo* + In P + In =
RTStibifa
In/;,..)
2 a-,).
3 Gi
=
RT~Ln k
In xk
,
The
initial state of the
where the expression on the
right
is
a negative quantity. It
that the Gibbs function after diffusion
holds for
This
will
be shown
all irreversible
We have shown temperature
T
is
is
seen therefore
than the Gibbs function before
be an expression of a general law that which take place at constant T and P.
is
a state of mechanical
and thermal
equilibrium but not of chemical equilibrium. final state is a state of thermodynamic equilibrium. transition from the initial nonequilibrium state to the final equilibrium state is accompanied by a chemical reaction that is too
agency of the
when
it
Through the take place more
takes place spontaneously.
catalyst, the reaction
is
caused to
rapidly.
later to
processes
that the Gibbs function of a mixture of inert ideal gases at
and pressure P
G= where
is less
mixture
The The
slow to be measured
diffusion.
as a catalyst. If chemical combination is of hydrogen and 1 mole of oxygen with different
known
is
RT~2n k { k
given by Eq. (16-6).
+ In P +
1
n xk ),
(16-8)
two compartments by a removable partition that one compartment contains a dilute Suppose as solution of sodium chloride and water maintained at a pressure of 1 atm and 0.01. at a temperature of 20°C— the mole fraction of the salt being, say, Under these conditions the solution is in thermodynamic equilibrium. SupImagine a
shown
vessel divided into
in Fig.
16-3a.
a pose that the other compartment contains solid salt in equilibrium also at partition pressure of 1 atm and a temperature of 20°C. Now imagine that the and temperature of the whole is removed (Fig. 16-3/;) and that the pressure
566
16-7
Heat and Thermodynamics . t
= 20°C
V
of constant mass.
and
= 20°C
aim
1
The salt solution is a liquid phase of two chemical is variable. constituents whose mass, when in contact with the solid-salt phase,
Partition in t
567
Chemical Equilibrium
1
Although the
atm
initial states of
both these phases arc nonequilibrium
states,
it
in terms of thermodynamic coordinates. Since is possible to describe them thermal equilibrium, a definite P and Tmay and mechanical cacli phase is in
each has a definite be ascribed to each; since each has a definite boundary, composition of each phase may the homogeneous, each is volume; and since constituent. In of each of moles number the specifying be described by and mechanical in constituents chemical of c consisting general, a phase coordinates the aid of with the described be may thermal equilibrium P, V,
nh nh
'/\
Under 1
atm
1
atm
system are kept constant at the original values. Experiment shows that some solid salt dissolves;
i.e.,
the mole fraction of the salt in the solution increases
spontaneously at constant pressure and temperature. After a while, the change ceases and the mole fraction is found to be about 0.1. Focusing our attention on the solution from the moment it was put in contact with the solid
1
The
initial state
with the
we
salt,
are led to the following conclusions:
moment it was put in contact one of mechanical and thermal equilibrium but
of the solution (at the
solid salt)
is
not of chemical equilibrium.
2
The
final state of the solution
is
a state of thermodynamic equi-
librium. 3
The transition from librium state
is
the initial noncquilibrium state to the final equi-
accompanied by a transport of a chemical constituent
.
,
nc
.
a given set of conditions, a phase
may undergo
Thermodynamic Description
and
all
the
n's.
Since entropy
is
number
is
a measure of the molecular disorder of the
We
two of P, I", and T and all the n's. During a change of state the n's, which determine the composition of the of a transphase, change either by virtue of a chemical reaction or by virtue general, both. In phases, or between port of matter across the boundaries phase is n's which the for values of the under given conditions, there is a set of chemical and therefore in thermodynamic equilibrium. The functions equilibrium that express the properties of a phase when it is not in chemical when the equilibrium thermodynamic must obviously reduce to those for in
equilibrium values of the
n's
are substituted.
We are
therefore led to
of
Nonequilibrium
of chemical constituents satisfying the requirements (1) that
it
is
has a definite boundary. The hydrogen-oxygen Art. 16-6 is a gaseous phase of two chemical constituents
(2) that
mixture described
in
assume
Consider, for example, a phase consisting of a mixture of ideal gases.
States
is
defined as a system or a portion of a system composed of any
homogeneous and
m
have system, the entropy of a phase that is not in chemical equilibrium must the therefore phase, and of a the entropy shall assume that a meaning. any functions of expressed as be also, can Helmholtz and Gibbs functions
the gases are inert, the equation of state
A phase
a change of state
While this is going on, the which some equilibrium but of thermodynamic not of states phase passes through states are connected by only. These equilibrium thermal mechanical and the n's. Whether V, T, and among P, relation is a an equation of state that internal energy definite it has a or not, equilibrium a phase is in chemical functions of P, V, T, and as regarded H may be and enthalpy. Both U and the equation means of coordinates by the one of the n's; and upon eliminating two of P, V, and T function of any a expressed as of state, U and // may be
represented that any property of a phase in mechanical and thermal equilibrium can be used to denote as that the same form by a function of any two of P, V, T, and the n's of equilibrium. the same property when the phase is in thermodynamic
into the solution.
16-7
.
or all of these coordinates change.
Transport of matter across the boundary between two phases.
Fig. 16-3
.
it
the entropy
PV =
Xn,..RT;
S =
H2Wfc(
is
—
In
P—
In xk);
When
and the Gibbs function
perature
is
T
and undergoes an infinitesimal
irreversible process involving
+ In P +
In xk).
an
dQ with the reservoir. The process may or both. Let dS denote the reaction or a transport of matter between phases change of the reservoir. entropy change of the system and d.S„ the entropy dS', and since the The total entropy change of the universe is therefore dS an increase in the an irreversible process is attended by involve a chemical
exchange of heat
G = RTZn^fa
569
Chemical Equilibrium
16-8
Heat and Thermodynamics
568
+
assumption just made, these same equations may be used in connection with an ideal-gas phase in mechanical and thermal equilibrium when the gases arc chemically active, when the phase is in contact with other
According
to the
performance
of
entropy of the universe,
we may
phases, or under both conditions, whether chemical equilibrium exists or not.
Under
these conditions the
rc's
and
a's
arc variables.
Whether they arc
mass of the phase is variable, the number of n's that arc independent depends on the number of other phases in contact with the original phase and on the chemical constituents of these other phases. A system composed of two or more phases is called a heterogeneous system. Any extensive property such as V, U, S, H, F, or G of any one of the phases may be expressed as a function of, say, 7", P, and the n's of that phase. Thus, if the
Gibbs function of the
first
G<'>
dS dS B
Since
we have
dQ
or
>
0.
= - dQ T
>
0,
- TdS <
0.
T
+
dS
'
the During the infinitesimal irreversible process, the internal energy of performed. dV is work P amount of and an system changes by an amount dU, The first law can therefore be written in its usual form,
dQ = dU + P dV,
phase, id the inequality
=
+
all
independent variables or not is a question that cannot be answered until the conditions under which a change of state takes place are specified. It is clear that, if the mass of the phase remains constant and the gases are inert, the rc's and x's are constants. If the mass of the phase remains constant and the gases are chemically active, then it will be shown that each n (and therefore each x) is a function of only one independent variable, the degree of reaction,
for the
dSa
write
„d) function of (T,P,n \\n
becomes
(i) (
•);
dU +
PdV- TdS <
0.
(16-9)
for the second phase,
G< 2 >
etc.
The Gibbs
=
function of (T,P,n l ?,n l l\
.
.
.);
function of the whole heterogeneous system
is
therefore
therefore, during This inequality holds during any infinitesimal portion and, process. According to the assumpall infinitesimal portions of the irreversible made in the preceding article, U, V, and 5 may all be regarded as func-
tion
thermodynamic coordinates. During the irreversible process for which the above inequality
tions of
G=
C<"
+
G'
2'
+
•
•
.
may change. If we restrict the irreversible that two of the thermodynamic coordinates remain condition imposing the Suppose, for constant, then the inequality can be reduced to a simpler form. Then the example, that the internal energy and the volume remain constant. system at inequality reduces to dS > 0, which means that the entropy of a and V increases during an irreversible process, approaching a constant maximum at the final state of equilibrium. This result, however, is obvious
or
This
16-8
result holds for
any extensive property of a heterogeneous system.
Conditions for Chemical Equilibrium
some process by
holds,
all
of the coordinates
U
Consider any hydrostatic system of constant mass, either homogeneous or heterogeneous, in mechanical and thermal equilibrium but not in chemical equilibrium. Suppose that the system
is
in contact with a reservoir at tem-
L'and V is isolated and own universe. The two most important sets of
from the entropy principle, since a system is
therefore, so to speak,
its
at constant
570
Icat
1
16-9
and Thermodynamics Const. T,
conditions arc the following:
//
1
T
V are
and
the inequality reduces to
constant,
d(U
-
<
0,
dF<
0,
TS)
or
1-72
Ftt
VII
172
T
T
expressing the result that the Helmholtz function of a system al constant
T
and becomes a minimum
at
decreases during an irreversible process
T
and
P
are constant, the inequality
d(U
+ PV-
or
reduces
final
equilibrium
The student will mechanical system
<
0,
dG
<
0,
irreversible process
dynamic potential
Fig.
1
6-4
Taylor
volume
2
r
r
!=/,.+ fit
•
Transition from a state of thermodynamic equilibrium to a stale characterized by
function ft of the left half may now be expanded in a its equilibrium value F m „J2 as follows:
about
scries
at constant
T and
ft.
1
f0\
(&0
!
-
which may be terminated after the. squared term if Sv is small enough. But there is no difference between the behavior of half the system and that of the whole system, or
minimum. The
Gibbs called the function
and
(df\
r
and becomes a minimum at the
that the potential energy shall be a
at constant
=j + Su
I'll
(16-11)
state.
this reason,
V
Sv
Nonequilibrium
Gibbs functions are therefore seen to play a similar role in
Helmholtz and thermodynamics. For
fit -
=F
The Helmholtz
TS)
recall that the condition for equilibrium of a conservative is
_ V
Equilibrium
Fmin an
"L
to
expressing the result that the Gibbs function of a system
P decreases during
V
a lack of mechanical equilibrium.
the final equilibrium state. //
571
(16-10) ./••
and V
Chemical Equilibrium
the function
G
F
=
the thermo-
\Jv) r
[dVjr
the thermodynamic poten-
tial at constant pressure.
Similarly, for the right half,
Condition for Mechanical Stability
16-9
In discussing the important equation
cP - c r = - r in Art. 11-7, the
dvy (dp
Adding
BT h> \dV
remark was made that (dP/dV) T
is
the
total
Helmholtz function of the system
always negative. This
be proved for any system of constant mass. Consider the system depicted symbolically in Fig. 16-4. At first, each half of the system is in equilibrium as well as the entire system. Suppose, now, that the left half is comwill
these two equations,
in the nonequilibrium state is
now
pressed by an
amount
8v
and
with each half remaining at
remaining constant.
or
ft
+ fn. -
expanded by the same amount, constant temperature and the total volume
the right half
=
F,
SFr.v
is
The
relation
between
F and
the
volume
=
(«»)
2 -
(j]ty r v of either half of the system at
572
Heat and Thermodynamic
16-10
573
Chemical Equilibrium
If the constituents are inert, the phase is in chemical and therefore in thermodynamic equilibrium. Imagine the performance of an infinitesimal reversible process in which the temperature and pressure are changed by dT and dP, respectively, and the numbers of moles of the various constituents dn Since we have assumed that arc altered by the amounts dti\, dni, .
.
.
,
l:
.
changes in the n's arc to be regarded as accomreversible addition or withdrawal of the constituents with the plished by the membranes. The resulting change in the Gibbs semipermeable aid of suitable the constituents are inert, the
function of the phase
dG
where
it is
=
^dT
given by
is
dG
dG dP dP
understood that
dG_ dni
dn\
G is
+
dm
dm
a function of T, P,
and the
partial derivative implies that all variables other than the Fig. 16-5
when
Dependence of Helmholtz function of the entire system on the volume of half of
the total volume
and
it,
the temperature are constant.
T and V \s shown
As a
in Fig. 16-5. Since SKt.v
>
\dV*/T
dF
But
is
positive, it follows that
special case, consider
This
is the
Under
it
9£
condition for mechanical stability.
Thermodynamic Equations
dT dP
for a'Phase
Consider a phase composed of c chemical constituents, of which there arc moles of Ai, etc., n c moles of A c ; the phase, is in Hi moles of substance A\, thermal equilibrium at temperature T and in mechanical equilibrium at pressure P. The Gibbs function of the phase can be written as follows:
m
G=
composition and mass).
(for constant
-SdT+
VdP.
It follows, therefore, that
0.
5? = 16-10
which all and the mass of
infinitesimal reversible process in
has already been shown that
dG =
finally,
<
an
these conditions, the composition
«.*«+«*
0.
= -p.
\dVJT
n's and that each onc indicated are
the phase remain constant, and the equation becomes
But, for this case,
and hence,
on c
to be kept constant.
the dn's are zero.
constant
-r- dn c
function of {T,P,n\,m
,n c ).
and
dG =
/dG\ \dTJ,
(& \dP
M
/r,n,,iis
_
(16-12)
= v
(16-13)
,
»,
dG
-SdT + VdP + |dn, + |^ dm + an\ an
,
i
an c
Now consider the effect upon the Gibbs function when a small amount of one of the constituents (say, the kth constituent, Ak) is introduced into the phase, T, P, and the other n's remaining constant. If dn moles of Ak are introl:
574
Meat and
Th ermouynaimcs
duced, the effect on the Gibbs function
is
where dX
nk
the
n's.
is
called the chemical potential of the kth constituent of the phase in
a substance
a function of
is
not present in a phase,
7',
P,
and
in
it,
be
Mi
2
-
n 2 dX
+
•
+ Mc dn
•
c,
G
dX
= mnt
dX
+
ix 2
+ nn
•
c
c
d\,
it
G —
tnni
+ n%n2 +
+
Hcfic
(16-16)
or
which case the Gibbs function would be altered and the value of p. would
finite.
We may now
write
Equation (16-16) shows that
dG = for
+ M2 dn +
dm
we have
all
docs not follow that its chemical potential is zero. The chemical potential is a measure of the effect on the Gibbs function when a substance is introduced. Even though the substance is not present in the phase, there is always the possibility of introducing is
the proportionality factor. Since
dGr ,i' =
A chemical potential of one constituent
If
is
(16-14)
dn k
question.
575
expressed by the partial derivative
dG where
Chemical Equilibrium
16 11
an
—S dT + V dP +
infinitesimal
change
in
dm
+ m dm
-4-
the chemical potentials are intensive quantities, for if
the n's are increased in the same proportion at constant T and P, the n's must remain constant in order that G increase in the same proportion.
Mc dn c
all
the Gibbs function of any phase consisting of
in
inert constituents.
n's is
Suppose, now, that the constituents are chemically active. Changes in the may then take place because of a chemical reaction. Although the phase always considered to be in thermal and mechanical equilibrium, an
infinitesimal process involving a
change
in
7',
P,
and the
n's will, in
general,
be irreversible, since chemical equilibrium may not exist. In accordance with our previous assumption as to the form of the expressions denoting properties of a phase in thermal and mechanical equilibrium but not in chemical equilibrium,
we
= -SdT + VdP +
correctly expresses the
the
as
= dG
Mi-
dnk
may
phase or by
dm
+
ix 2
dn 2
+
4-
n c dn c
change in the Gibbs function for
(16-15)
any
is
a function of T, P, and
infinitesimal process in
Xk
from
T and P in which all constituents are increased same proportion. Since the Gibbs function is an extensive quantity, it also will be increased in the same proportion. Infinitesimal changes in the mole numbers in the same proportion are represented by
in the
ni d\,
=
— Zn 7lh
'
of a chemical reaction, or both.
Imagine a phase at constant
dm =
all
clear that the n's
it is
be caused to change either by the transfer of constituents to or
the agency
the n's. In order that /j t may be an intensive must be combined in such a way that, when quantity, same factor, the value of n k remains the multiplied by the them are all of constituent, fraction of the Am mole same. The
and M!
= -SdT+ VdP+2n k dn k
the n's
The chemical potentials play a fundamental role in chemical thermoThe chemical potential of the A'th constituent of a phase is defined
dynamics.
shall assume that the equation
which
Chemical Potentials
16-11
dn 2
=
n% d\,
dn c
=
it is
to
be expected that p&
is
a function of
actual form of the function depends, of course, on the
The T, P, and x nature of the phase. Consider the following phases: l:
1
n c dX,
requirement; hence
satisfies that
.
Phase consisting of only one
constituent.
In this simple but not trivial
case,
and the corresponding change
in
G
G=
is
dG = G
dX,
and
im
G
(16-17)
57G
i.e.,
the chemical potential
tion of
2
16-12
Heat and Thermodynamics
T and P
the molar Gibbs function
is
G = RTZn k (4>k when compared with
G =
+
In
P
In
which may be written A
and
n
«
In xk),
in
a mixture of
+ lnP + ln.v),
shall assume that
=
RT(
=
+
An
solution.
gk
=
is
defined as one is
+ RT In x 1
go
of the form
k,
(16-20)
mole
of the kth constituent in the
T and
P.
,v 2 ,
.
.
.
,
it
can be shown
+ RT In xo,
w-
is
=
T
got
(16-21)
and P. For any one of the
it
+ RT In Xk,
(16-22)
^H
2
+
JOj, those on the
symbols and
and
H
2
are
proportional
being any number whatever), then n moles of hydrogen and /2 moles of oxygen are formed. Similarly, if the reaction proceeds to the left to the extent that «„ moles of hydrogen combine with n' /2 moles of oxygen, then n' moles of water vapor arc formed. In general, suppose that we have a mixture of four substances whose chemical symbols arc A h /1 2 , A 3 , and A,. Let Ai and A t be the initial condissociate (n n
and A 3 and A A
the final constituents, with the reaction being repre-
sented by ViAt
+
V%At r^ "3^3
+
ViAi.
We
have chosen four substances only for convenience. The equations to be developed are of such a character that they can easily be applied to reactions
T and P only but depends upon the nature upon the solute. By defining functions known as fugacity and activity
in
possible to express the chemical potentials of con-
If
a function of
is
2
to the number of moles of the constituents that change during the reaction. Thus, if 1 mole of water vapor dissociates, then 1 mole of hydrogen and % mole of oxygen are formed; or if 0.1 mole of water vapor dissociates, then 0.1 mole of hydrogen and 0.05 mole of oxygen are formed; or if n moles of water vapor
solutes,
of the solvent as well as
Other phases.
indicated by the notation
where the quantity on the left is called an initial constituent and right final constituents. The numbers which precede the chemical which "balance" the equation (it is understood that both H 2 preceded by unity) are called the stoichiometric coefficients and are
stituents
expressed as a function of
is
number of moles and oxygen, the chemical reaction capable of
H
the molar Gibbs function of the solvent in the pure state,
is
introduce into a vessel a mixture of any arbitrary
(16-19)
ideal solution
expressed as a function of
Mo
we
In p)
+ RT In x.
?
=
If
taking place
two alternate forms
that, for the solvent,
coefficients,
it
of water vapor, hydrogen,
each of the mole fractions of the solutes x h
0/!
also of concentrated solutions.
Degree of Reaction
16-12
(16-18)
Phase consisting of a dilute solution. In the case of a dilute solution in which the mole fraction of the solvent x is very much larger than
where g
and
book.
is always possible to express the chemical potential of any a function of T, P, and the x of that constituent. It will be a surprise to the student to discover how much valuable information can be obtained with the aid of this assumption, without knowing the exact expressions for the chemical potentials of the constituents of a phase.
the general equation
the Gibbs function of
is
state,
where go
this
constituent of any phase as
which the chemical potential of each constituent
pure
beyond the scope of
have, from
We
+
is
SStjU&j
in the
Phase consisting of an ideal
where g k
stitucnts of a mixture of real gases
a func-
577
is
H = RT(4>
in
we
case
this
shows that the chemical potential of one ideal gas ideal gases
is
This
Phase consisting of a mixture of ideal gases. Art. 16-5,
which,
and
only.
Chemical Equilibrium
which any number of substances participate. The v's are the stoichiometric which arc always positive integers or fractions. Let us start with arbitrary amounts of both initial and final constituents.
coefficients,
we imagine
the reaction to proceed completely to the right, at least one of
578
Heat and Thermodynamics
the initial constituents (say, Ai) will completely disappear. to find a positive
number
of the initial constituents
and where N-2
is
combine. If
Then
it is
possible
reaction consists in the dissociation of one initial constituent,
n
such that the original number of moles of each
degree oj dissociation;
is
expressed in the form
« is
=
n
i>i
m
=
n
»-2
+N
called the degree of ionization. Expressing
_
one of the final constituents (say, A 3 ) will completely disappear. In another positive number n' may be found such that the original number of moles of each final constituent is expressed in the form
and, solving for m, «i
The number
this event,
following expressions:
=
=
If the reaction
is
maximum amount amount
to
+
,\
=
«2(max)
=
(n
+n
(n
+
proceed completely to the
imagined
is
minimum amount
to
m(tairi)
)"i>
n<,)i>2
+
r
A 2,
n.i(min)
there
left,
is
the
and the minimum
= =
0,
A' 4
.
n 2 (min)
= A2
0,
(max)
=
(n
n 4 (max)
=
(n
72 3
r
,
called the
in terms of
+
(«o
—
n' )vi
+ n' )vi(l —
e),
=
(tic
+ n'o)vi(\ —
e)
get
ni
«n)"i
I
+ «o)"i0 —
*)
of moles of each of the constituents
(n a
we
+ Ms,
n3
=
»i
=
is
therefore given
+ («o + n )nt + ^4(no
by the
n'o)"« e »
(16-24)
+ n )v3, + n' + )v.,
When
a chemical reaction takes place,
ently.
The
all the n'$ change, but not independimposed on the n's are given by the above relations. These equations therefore are examples of equations of constraint. The equations of constraint are equally valid whether the system is heterogeneous or homogeneous. If each constituent is present in, say,
restrictions
is
the
ni
=
'
1'
ni
n,
+ n? =
2)
)
+ ni)v,(l
(n
-6),
and the maximum
of each final constituent:
=
m(max) and m(min)
1
proceed completely to the right, there
possible of' each initial constituent
ni(min)
e is
initial constituent,
.
possible of each initial constituent
ni(max)
one
r
of each final constituent:
If the reaction
amount
imagined
n' Vi
=
n = n' u 3 n->
n t (original)
(no
(no
least
and
consists in the ionization of
2,
a constant representing the number of moles of A* that cannot we imagine the reaction to proceed completely to the left, at
n 3 (original)
it
the constants that express the original amounts of the constituents,
n\ (original) (original)
when
579
Chemical Equilibrium
16-12
the other constituents. For the present, however,
selves to
homogeneous systems, reserving heterogeneous systems
Since
system Nt,
,
t
all
all
the n's are functions of
e
only,
it
these expressions arc
when
shall limit our-
for Chap. 18. homogeneous An example will show
follows that in a
the mole fractions are functions of
how simple
Suppose that the reaction proceeds partially cither to the right or to the left to such an extent that there are n\ moles of A\ n> moles of A 2 n 3 moles of A 3 and n 4 moles of Aa present at a given moment. We define the degree of reaction e in terms of any one of the initial constituents (say, Aj) as the fraction
we
etc., for
e
only.
the starting conditions are simple.
Consider a vessel containing no moles of water vapor only, with no hydrogen oxygen present. If dissociation occurs until the degree of dissociation is e, then the n's and .v's arc shown as functions of e in Table 16-1 Since the chemior
.
,
cal potential of
each gas
that every chemical potential If the reaction
Hi (max)
«i(max)
It follows
the
left
from
and
e
this definition that
=
1
when
e
—
=
the reaction
—
is
is
«i
(16-23)
the reaction
is
When
the
is
is
a function of T, P, and
a function of
7",
P,
and
x,
it
follows
e.
+
ing amounts:
drii
= — (n
+n
)vi de,
dn 3
=
(no
dn-i
=
+ »o)"2 *&i
dn\
=
(n
completely to
completely to the right.
mixture
imagined to proceed to an infinitesimal extent, the degree from t to e de, the various n's will change by the follow-
of reaction changing
ni(min)
when
in the
—(no
+n + n„)iM
)i<3
de, de.
580
H
Table 16-1
^±
2
H + i0 2
The
2
m's are functions of T, P,
and
At
V
- H,0
*,=
1
"i
=
no(l
A.-.
=
H
-
1
—
«)
•
V '
A*
Let us imagine that the reaction
-V
>:
l
+
= Os
v,
=
=
«3
1
= me
Y3
me
*4
i Xn =
+
» (1
= =
a function
therefore follows that
is
allowed to take place at constant T i.e., during an
is
function decreases; and P. Under these conditions, the Gibbs de, e infinitesimal change in e from e to
+
e
e/2
<
dGr.p Vz
2
G
581
It
e.
3
A
Chemical Equilibrium
16-13
Heat and Thermodynamics
r+172
0.
We have shown that, for any infinitesimal change to which a phase in thermal
e/2
and mechanical equilibrium
T-W2
-S dT + V dP +
dG =
6/2)
subjected,
is
mi
dm
+ w dm +
•
.
Therefore, for this mixture of four constituents,
These equations show that the changes v's,
in the n's are proportional to the
with the factor of proportionality being, for the
— (no +
n
+ (no + n
de and, for the final constituents,
)
of writing these
is
D ) de.
dn\
dn-i
— "2
which shows perhaps more
dn-i
dti\
=
-
-
.
=
+
(no
V\
v-i
J
(16-25)
de,
+ M2 dm + Ms dm + Mi dm,
clearly that the dn's are proportional to the
)"i de,
dm =
("0
n'o)"2 de,
dm =
(n
+n
dm = -(n
,.
B
d„ 2
= - (no
we
change of obtain a general expression for an infinitesimal
+
Gibbs function at constant
T
and
Equation of Reaction Equilibrium
16-13
=
dGr.p
Consider a homogeneous phase consisting of arbitrary amounts of the four and A. capable of undergoing the reaction
constituents A\, Ao, A3,
t
,
viAi + v%A% ^ v$A$
Ml«l
m
+ M2«2 + M3«3 + Ml"
I-
=
(no
+ n )vi(l —
«2
=
(no
+
n'a )v-i{\
—
e) 9 e)
+
iV2 ,
+
Ko)(-"
equation that
this
KiMi
Conversely,
if
is
-
when
"2M2
+
"aMs
+
(16-26)
"iM-0 de.
the reaction proceeds spontaneously
positive, then, in order that dG-r.p
+ »m > Wi +
function
is
a
"«***
<
0,
(reaction to right).
the reaction proceeds spontaneously to the
ciMi
The mixture
arc given by the equations of constraint: «i
from
to the right, so that de
,
G =
It follows
(«o
P. Thus,
-\- v.\A\.
Suppose that the phase is at a uniform temperature T and pressure P. If «i, «2, n 3 and m denote the numbers of moles of each constituent that are present at any moment and m, M2, /j 3 and are the respective chemical potentials, then the Gibbs function of the mixture is ,
+ n§n de, + tfav* de.
v's.
the
n's
dm
form: with the equations of constraint in differential
Substituting,
The
Mi
Another way
as follows:
— v\
dGT ,p =
initial constituents,
+
"2M2
<
"sM3
+
"Hi*
(reaction to
at
which an
infinitesimal
change in the Gibbs function. Therefore, for
«a
=
(no
+ n'o)"3e,
n.,
=
(n
+ «4)"4« + N
left).
and P when the Gibbs change in e will produce no
will be in equilibrium at the given
minimum
left,
T
dCT ,r =
at equilibrium,
we
have I'lMl 4.
+
"2M2
=
"3M3
+
VtflA
(at equilibrium)
(1
6-27)
582
IT
Heat and Thermodynamics
which
is
B have been excited, diffusion change of the universe. After the nuclei of gas B have been excited, a time much larger (b) than the lifetime of the excited state is allowed to elapse, and then diffusion takes place. Calculate the entropy change of the universe.
done
Show
(c)
in the case of ideal gases in the next chapter.
problem
16-1
is
that the answer to part
due
to
M.
J.
(/>)
is
larger than that to (a). (This
Klein.)
16-7 Consider the system depicted in Fig. PI 6-1, where the whole system and also each half are in equilibrium. Consider a process to take place
PROBLEMS Prove Gibbs' theorem by using the apparatus depicted
16-2 in such a
after the nuclei of gas
takes place. Calculate the entropy
necessary merely to substitute the appropriate expressions for the chemi-
cal potentials into the equation of reaction equilibrium. This will be
Immediately
(a)
called the equation of reaction equilibrium. It should be noted that this
equation contains only intensive variables. Evidently, to determine the composition of a homogeneous mixture after the reaction has come to equilibrium, it is
583
Chemical Equilibrium
way
that the gases are separated rcversibly
and
in Fig.
Const. U,
V
adiabatically.
16-2 What is the minimum amount of work required to separate 1 mole of air at 27°C and 1 atm pressure (assumed to be composed of gO> and 5N2) into each at 27°C and 1 atm pressure? 2 and N 2 16-3 Calculate the entropy change of the universe due to the diffusion of two ideal gases (1 mole of each) at the same temperature and pressure, by
S/2
S/2
U/2
U/2
V/2
V/2
,
calculating
J
dQ/T
over a series of reversible processes involving the use of
16-4 sure
P
n\
arc in
is
moles of an ideal monatomic gas at temperature T\ and presone compartment of an insulated container. In an adjoining
removed:
(a)
Show
Calculate the entropy change
the gases are identical.
(c)
Calculate the
the gases are different.
that the final pressure of the mixture
when entropy change when
is
P.
16-5 «i moles of an ideal gas at pressure Pi and temperature T arc in one compartment of an insulated container. In an adjoining compartment, separated by a partition, are n% moles of an ideal gas at pressure P> and tem-
When
Fig.
sL
+
s„
Transition from a state of thermodynamic equilibrium to a state characterized
PI 6-1
by a lack of thermal equilibrium.
which each half of the system remains at the constant volume V/2, but amount of heat 8k (at constant volume dQ = dU) is extracted from the left half and transferred to the right half, as shown in Fig. PI 6-1. Realizin
a small
(b)
perature T.
S =
O = o n \ax
compartment separated by an insulating partition arc n-> moles of another ideal monatomic gas at temperature T-< and pressure P. When the partition
Nonequilibrium
Equilibrium
the apparatus depicted in Fig. 16-2.
the partition
is
removed:
(a)
Calculate the final pressure of the mixture.
(b)
Calculate the entropy change
the gases are identical.
(c)
Calculate the
the gases arc different.
when entropy change when
Prove that the entropy change in part (r) is the same as that which would be produced by two independent free expansions. 16-6 Suppose that we have 1 mole of a monatomic gas A whose nuclei are in their lowest energy state and 1 mole of a monatomic gas B consisting of exactly the same atoms as A, except that the nuclei are in an excited state whose energy e is much larger than kT and whose lifetime is much larger than the time for diffusion to take place. Both gases are at the same pressure and are maintained at the same temperature T by a heat reservoir.
ing that
\dU)v
\du), (a)
Expand the entropy
sr.
of the
left
half
by means of a Taylor
about
the equilibrium value Smax /2, terminating the series after the
term.
Do
(b)
the same for
Show
—
SSa.v
(c)
Show (a)
s/{ .
that
(d)
16-8
series
squared
^i.
+
fit
—
Snout
—
(*)*.
> 0, which is the condition for thermal stability. that the molar Hclmholtz function of an ideal gas
that C\-
Show /
=
«o
T
I
Uv d'T - .,,—
df — Ts — Rl
In
v.
is
584
Heat and Thermodynamics
(b)
Show
that the Helmholtz function of a mixture of inert ideal gases
Show
16-14
Show
(c)
Sn*(/*
+ RT In
fa =
**).
that the change in the Helmholtz function
due
to diffusion
Fi
= RT2m-
In xk
at
.
Consider a hydrostatic system of constant mass maintained in thermal and mechanical equilibrium. It is not in chemical equilibrium, however, because of chemical reactions and transport of matter between phases. these circumstances, the system undergoes
an
irreversible cycle.
infinitesimal reversible process in
Prove
dq
and the
total heat transfer
By means of
the equations
— SdT + V dP
dG =
-\-
2pk
dun
and where (b)
-SdT + V dP = — s dT + v dP =
(a) (b)
Show
16-11 function of (a)
S,
V,
that,
m,
«2
,
if •
•
,
2xk
where ph
V = TS - PV + S »s -SdT+ VdP=Zn k d>i
(*)
tfs
(c)
is
Show
that, if the
as a function of T, V, n\, n 2 ,
.
=
(dU/dn k ) s
.
,
dn.
For any
7
mole
dQ =
n dq
du
+ P dv,
= nT ds.
dl!
+ P dV -
to the entire system,
h dn,
but h
is
the molar enthalpy.
Prob. 16-1 la, show that
How
(d)
Show
v „th« ,
,
does
g
dn.
T dS differ from dQ? T and P remain constant, dQ =
that, if
Starting with no moles of
and
T dS does not.
but
no moles of
H
2
capable of
.,-»•
CO + H
k
CO
undergoing the reaction
.
2
^ C0 + H 2
2
.
Helmholtz function of a phase .
From
(c)
expressed as a
in the
16-12
changes by
T dS = dU+ P dV -
16-16
3fe dn k
n
dfik-
n e , then:
dU = T dS - P dV +
U and V refer
S« t dp*
the internal energy of a phase •
= T ds =
dQ =
prove that:
which
Prove that
if<* S/i/jS*,
i: .
of substance,
(a)
16-10
+RTd In x
dP
Consider a uniform substance of variable mass. Suppose that of substance and that the system undergoes
that
G=
-f vk
any moment there are n moles
an 16-9
Under
*~ hh dT
l
mixture of ideal gases,
in a
is
16-15
Ff -
an ideal gas
that, for
is
F=
585
Chemical Equilibrium
is
expressed
gaseous phase, set up a table of values of A,
v, n,
and
.v,
similar to that
of Table 16-1.
16-17
nr , then:
Starting with no moles of
H»S and
2«
moles of
H
2
.v,
similar to that
0, capable of
undergoing the reaction (a)
dF =
—SdT — PdV+Xiik dn
F = -PV +
(b)
where fik
k
(dF/dn k ) T
,
r^ttor
„„
H S^jfc.
-SdT + VdP =
(c)
=
Zh*
in the
/**•
of
16-13
Prove
that:
-j,dT-PdV +
(a)
(b)
= 5 dT + P dF +
Z»k
2s* fa.
dn k
.
gaseous phase,
Table
16-1.
set
2
S
+
2H 2
?±
3H 2
+
SO,
up a table of values of A,
v, n,
and
Ideal-gas Reactions
17-2
17.
The right-hand member perature. Denoting
IDEAL-GAS REACTIONS we
by
it
111
is
587
a quantity whose value depends only on the temwhere is known as the equilibrium constant,
K
In K,
K = — ("303 +
"404 -* "101
—
"=
K,
(17-1)
"202),
get finally
Xi Xi •1 **2
which
is
to the
(i>3 -f-
pr,+ r,~
\
called the law of mass action. »4
—
"i
—
(17-2)
/ t=t
"2)th power.
K
has the dimensions of pressure raised
The
fraction involving the equilibrium
—
—
a function of the equilibrium value of e that is, of e„ and hence the law of mass action is seen to be a relation among ee P, and T.
values of the x's
is
,
Law
17-1
It is
of Mass Action
obvious that,
if
final constituents, the
It has
there are
more than two
been shown that when four substances which are capable of tinderprt+r,+
going the reaction viAi
and which
+
=
"2jU2
"3^3
+
= KTlfa 4-
where
at constant temperature
we
A'
is
given by
111
K=
and
+ mm
—("303
+
"404
+
In
P
+
In *j)
+
vafoi
to
P+
\nxk ),
+ In P +
A- 3
+
v.\
In
+ or
111
*.•
("3
—
T
wi In x\ "•!
—
+ inP+
— Pi—
„P"+"-"-" =
~-;' 4
vl
-
x l~
v 2 In "2)
—
"101
—
"202
—•'•)•
and equilibrium takes place quickly. If the mixture is then cooled very suddenly so as not to disturb the equilibrium, an analysis of the composition of the mixture yields the values of the mole fractions corresponding to equi-
In Xj)
+
k 4 (* 4
+ In P +
In
.v 4 ).
-("303
= — ("303
+
K4*<
-
+
"404
"1*1
-
—
"101
—
used.
reacting gases are mixed in the mixture is caused to flow
The
proportions at a low temperature, and slowly through a long reacting tube at a desired temperature. The gases remain at this temperature a sufficient time for equilibrium to take place.
known
"202.),
the
methods of chemical
It
"20 2 ).
is
equilibrium has, so to speak, been
The mixture is then allowed to flow through a capillary, where it is suddenly cooled. The equilibrium values of the mole fractions are then measured by
A'2
In -P
The
"frozen." Sometimes a flow method
In * 2 )
Rearranging terms, "3 In
'
has been pointed out that a mixture of hydrogen and oxygen will remain indefinitely without reacting at atmospheric pressure and at room temperature. If the temperature is raised considerably, however, water vapor forms
get
= nfo
•
It
librium at the high temperature.
mOi +
K,
Experimental Determination of Equilibrium Constants
17-2
temperature only. Substituting in the equa-
0's are functions of the
tion of reaction equilibrium,
*-• =
-
f-iA.t,
by expressions of the type
m
and two
arc ideal gases, then the chemical
If the constituents
satisfied.
potentials are given
where the
»«4a
to equilibrium, the equation of reaction equilibrium
"lMl
must be
A 2 ^-
v2
homogeneous phase
constitute a
come
pressure,
+
initial constituents
law of mass action becomes
is
analysis.
a consequence of the law of mass action that the equilibrium constant
corresponding
to a
given temperature
is
independent of the amounts
ol
588
17-2
Heat and Thermodynamics
products which are originally mixed. For example,
in the case of the
If
"water-
we
initial
gas" reaction,
C0 + H 2
2
?S
CO + H
2
X
with n u moles of
start
volume
is
2
Oi
Ideal-gas Rcactic
at temperature
T
and pressure
589 P, the
expressed as
0, I"o
RT
=
no
the law of mass action requires that, at equilibrium, If
*CO*H t O pi+i-i-l #C0 s#H,
= K
=
denotes the volume at equilibrium, with the temperature and pressure
I'",,
remaining the same, then
const.
—
Vc
—
[«o(l
e„)
XCGj#H,
at a constant temperature, independent of the starting conditions. For condi-
where
with arbitrary amounts of CO-> and H> and maintaining the temperature constant at 1259°K, the equilibrium values of the mole fractions are given in Table 17-1, where it may be seen that K is quite
be written as
constant.
or
tions that involve starting
tc
is
the value of
the.
RT
+
2rc t e ]
—p-
>
degree of dissociation at equilibrium. This can
V,= te
=
+c
(1
V„ t?
e
)V\
1
r
Table 17-1
Equilibrium data
C0 2 + H
2
for the
Water-gas Reaction
Since the density p
^ CO + HtO at 1259°K
inversely proportional to the volume,
,.
A = vco.
All.
•vco.
0.101 0.310 0.491 0.609 0.703
0.899 0.699 0.509 0.391 0.297
0.0069 0.0715 0.2122 3443 0.4750
*CO
~
*H s O
.
1.60 1.58 1.60 1.60 1.60
an example of a reaction that does not involve a change in the total number of moles. If such a reaction were to take place at constant temperature and pressure, there would be no change in volume. There are, however, many reactions in which the total number of moles varies. In such cases, it is possible to measure the value of the degree of reaction at equilibrium by merely measuring the volume (or the density) of the mixture at equilibrium. If t c is known, the equilibrium constant can then be calculated. As an example of this procedure, consider the dissociation water-gas reaction
get finally
P«
Now,
All,
0.805 0.4693 0.2295 0.1267 0.0685
0.094 0.2296 0.2790 0.2645 0.2282
ATcoVlljO
at equilibrium,
xx s o t
=
"o(l
—
(e)
«o(l
+
««)
and
therefore the law of mass action
(1
-
No0 ^±2N0 4
2.
=
2«oe«
+«.)'
becomes
+
P=
te)]-
e,)/(l
is
of nitrogen tctroxide according to the equation
*no.
m>(l
[2«./(l
The
we
Equilibrium mixture
Original mixture
•
is
+
K,
«,)
A'-* The
pressure
is
always measured
tion are given in
ture of
323°K
Table
17-2,
in
atmospheres. Numerical data for this reac-
where
it is
seen that at the constant tempera-
the equilibrium constant remains fairly constant for three
different values of the pressure.
There are many other methods of measuring equilibrium constants. For a complete account of this important branch of physical chemistry, the student is referred to an advanced treatise.
590
I
[eat
and Thermodynamics
Table 17-2
N.O.,
17-3
^ 2NO» d
and Temp.,
P.
°K
atm
323
0.124 0.241 0.655
Pa
Pc (air
=
(air
1)
=
,
4g
K=
1)
(17-5)
P,
atm
The preceding 3.179 3.179 3.179
0.777 0.678 0.483
1.788 1.894 2.144
0.752 0.818 0.797
equation, called the van' I Hojf
Heat of Reaction
K
In
In
Differentiating In
d_ -pr.
dl
K = — (C303 +
K with respect to K= —
In
dr
_Ao_
Since
R
-j^
Vi
dT
—
dfo "2
curve obtained by plotting log K against 1/7', multiplied by 2.30/?, is the heat of reaction at the temperature corresponding to the point chosen. As a
dT
S°
rule, log
~R'
K can be measured
case the curve
is
where
The
all
dl
this
procedure, consider the dissociation of water vapor
according to the reaction cp
J
dT}
H
,
(17-3)
2
K = -f^=r„ Rl
(vzfis
-
e±
2
H + $O a 2
.
Starting with no moles of water vapor
the h"& refer to the
+
Vih.\
—
V\la
—
in
v-Jli),
Table
16-1.
any value
and no hydrogen or oxygen, the mole shown
of the degree of dissociation are
At equilibrium,
T
and the same pressure P. m moles of Ai and 1/2 constant temperature and pressure into 1*3 moles same temperature
t
xl
/«=«.
right-hand term has a simple interpretation. If
moles of /I2 arc converted at of
only within a small temperature range, in which
usually a straight line.
As an example of
'
RT 1
-jrrAn
(17-6)
enables one to calculate the heat of reaction at any
fractions corresponding to
Therefore,
d log K d(l/T)'
h
<£
~dl
—
-2.30/?
desired temperature or within any desired temperature range, once the temperature variation of the equilibrium constant is known. The slope of the
RT*
Ycr<
= - xfi (/'o + and
The van't Hoff isobar
get
ut
'
AH =
get
V'ifc)-
_ JcpdT
ho
it
dt
we
T,
wc
—
''ltf'l
f fcpJT dT JT T-
l
RT
we have
—
»*4>l
c 3 -ppr -f- Vi -jt^
J
~R'
d{\/T)
defined by the equation
is
A and 3
im
enthalpy rgAj
one of the most impor-
AH R AH
dlnK
or
equilibrium constant
is
Rewriting the equation as d
The
isobar,
tant equations in chemical thermodynamics.
dT/T17-3
591
AH
„
,
Ideal-gas Reactions
moles of At, the heat transferred will be equal to the final V2I10. Calling this heat vt/u minus the initial enthalpy pjii
+
+
the heal of reaction and denoting
AH =
yj/tj
it
1
+
e./2
+
rJu
—
vihi
—
vdh,
K,
Pi = + «.)*(! - O
K.
1
by AH, we have
m\,
=
or
F*
+ «*
,/2
(17-4)
.
(2
592
17-3
Heat and Thermodynamics
When
e,
is
much
very
-2
H
^ H, + K>
2
°K
X 10"'
6x10"
7
x 10
_*
%.
ee
are given at a
(P
=
1
ee
number
Nr
-3
of tempera-
Dissociation of
atm)
Slope
= -13100
_4
r
-
(2 4- e ,)*(l
(measured)
5
water vapor 2
y '
_
IP
Oil
Temp.,
4 X 10
\
In Table 17-3 experimental value of
Table 17-3
593
l/r
smaller than unity, this equation reduces to
K=
Ideal-gas Reactions
1
K
log
«,)'
T
(atm) 5
-5 1.97 3.4 1.2 1.2 1.77 2.6
1500 1561
1705
2155 2257 2300
X X X
10-'
1.95 4.48 2.95 9.0 1.67 2.95
10-' 10-*
X10- 2
X X
10~s 10- 2
X X X X X X
10- 6
-5.71
10~ s 10- 6
-5
.
6.67
36
6.41 5.87
-4.53 -3.05 -2.78 -2.53
10-'
10-1 10~ 3
4.64 4.43 4.35
X X X X X
\0~*
X
10-*
10-
1
10-' 10-"
-6 2300° 2100"
turcs
and
at constant atmospheric pressure, along with the corresponding
Fig. 17-1,
where
and \/T. The graph
it is
equal to —13,100.
seen that the points
lie
AH = = It is rather difficult to
calorimetric method.
1
900°K
is
-2.30 R
against
\/T
is
shown
in
means
of a reversible
tures a reaction cither docs not proceed or, is
cell.
In a few cases
the temperature dependence of the heat capacities of
villas
+
all
the reacting
—
A//
=
vjis
k
=
h
+ fc P dT,
—
Vikoi
-\- v.\In
—
villi
Villi
vihoi
—
vihoi
+
j(v3Cp 3
+
viCp.\
—
v\Cp\
—
v&pz) dT.
Denoting the constant part by A//n thus:
if it
=
vzhtn
+
Vihot
—
viAoi
—
"ihui
(17-7)
ACp =
v-ffp-i
+
v.\cpi
—
viepi
—
viCi'i,
(17-8)
d/Y
in
and defining
we
get
A//
does, the equilibrium value
=
A//
+
/
A6> dT.
(17-9)
too small to measure. Thus, in the case of the
dissociation of water vapor, the degree of dissociation at room temperature and atmospheric pressure is about 10 _2V which means that under these ,
conditions not even one molecule of
To
dissociation of water vapor.
heats of reaction are obtained cither with the
possible to
of the degree of reaction
—
A/2
measure the equilibrium constant over a very wide temperature range, the graph of log K against 1/7" is found to have a variable slope, indicating that the heat of reaction depends on the temperature. Such a graph is impossible to obtain in most cases, however, since at low temperais
1
then
-13,100
250 kJ/mole.
aid of the van't Hoff isobar or by it
K against /T for
gases. Since
and
measure the heat of reaction accurately by a direct
Most
log
heat of dissociation at the
equal to
X
know
to
Graph of
on a straight line with a slope
It follows, therefore, that the
average temperature of about
which
K
of log
1500°
Temperature, °K Fig. 17-1
values of K, log K,
1700°
1900 =
10-"
HoO
dissociates.
obtain the heat of reaction at any desired temperature,
it is
necessary
The
can be determined by substituting for the cp's the empiric equaA// is known at one temperature, therefore, Ai/ can be calculated and the equation may be used to provide All at any temperature. integral
tions expressing their temperature dependence. If
594
Nernst's Equation
17-4
Ideal-gas Reactions
17-4
Meat and Thermodynamics
and electrons may be regarded
ions,
as a mixture of three ideal
595
monatomic
gases undergoing the reaction
The
equilibrium constant
is
defined by the equation
+
A ?£ 4+
K= —
In
=
.
where
Wc
+
("3>3
—
"404
"101
—
fc P dT ____j -—. dT --. 1
/(o
Starting with n
so
/"
shown
in
moles of atoms alone, the
Table 17-4. For
have, therefore,
K = —
1
,
~ "Ai — "a/'iw) j("3CP3 + PiGm — viCp\ —
+
jp=, (vjio-i
f
vtho4
RI
Defining
A.S'o
the equation
dT
=
+
"3*03
"4*04
—
-j\
("3*03
"1*01
—
+
"4*04
—
"1*01
—
1 '
/"
P
+
e,)][e./(l
+
(1
-
e.)/(l
+
£j
ln
-e
An Saha
is
known
r>
P.
AH
is
equal to %R. Therefore
(17-10) fl
and
$£,
Ifl^Sa-iUT.
A.S"„ ,
&Cp =
Let us put
A5o
which
6.)]
AHo is the amount of energy necessary to ionize 1 mole of atoms. If wc a = NyE, denote the ionization potential of the atom in volts by E, then where AV is Faraday's constant. Since the three gases are monatomic, each
"2*02,
jACP dT dT r2
[6./(l
"2*02)-
cp
K = - AHn AT
is
=*£- /»»+»«-*
1
becomes
la
K=
or 1
equilibrium
dT
J-2
+
In
_ " ln ViCn)
state of affairs at
this reaction,
K=
In
In
e.
"><£•>),
R
as NernsCs equation.
was made by Mcgh Nad monatomic gas. If a monatomic gas is a high enough temperature, some ionization occurs and the atoms,
=
In B.
interesting application of Nernst's equation
to the thermal ionization of a
heated to
Introducing these results into Nernst's equation,
In
--^-_ 1
A =t A +
Table 17-4
+
X
«
V
",
x
=
Bi
1
=
h (1
-
er)
"
1
-«,
A4
= A+ =
"3=
c
"3
+
«3
1
".,
=
1
"4
—
"I
«4
=
|lnT+ P=-^ RI +
In B.
I
atmospheres, changing to
B
from
!
+
1
33a
=
no««
P(atm)
= -
= mu = ne(i
"*
+
e.)
statistical
common
logarithms,
«,
1
+
1
+7,
and introduc-
mechanics, Saha finally obtained the
formula
log A*
P in
ing the value of
A = A
t;.
get
t
Expressing
A
—
we
—^-5
(5050 deg/V)
% + | log
T
+
log
te
—-
6.491,
(17-11)
= where to,, «,., and w a arc constants that tron, and the atom.
refer, respectively, to
the ion, the elec-
596
Heat and Thermodynamics
In order to apply Sana's equation to a specific problem,
know
and the
quantities
co's.
beyond the scope of this listed in Table 17-5. The constant
is
it is
necessary to
A
complete discussion of these book. Values of these constants for a
the ionization potential
few elements arc
co
for
an electron
is
2.
where the
Values of E and
AG is defined
as the free-energy change.
when mixed
co
is
AC = RT(v
The
It should be Gibbs functions of the
in terms of the separate
gases, not in terms of the mixture.
gases
Table 17-5
completely separated at T, P.
g's refer to the gases
emphasized that
Among American
AG
connection between
chemists,
+
-
v t 4>\
-
vi4>i
of the
+ RT In
P#a)
AG
is
known
and the behavior
shown by introducing the values
z -i
597
Ideal-gas Reac
17-5
.g's.
of the
Thus,
P'**'t-«t
1
£,V
Element
But
Ola
In
AC =
therefore,
Na
5.12 3.87 6.09 8.96 9.36 6.07
Cs
Ca Cd Zn Tl
2
1
2
1
1
2
1
2
1
2
2
1
The
K =
—(vj4>:i
—
"it
"2)th
power.
K also contains
stellar
equation to the determination of the temperature of a atmosphere. The spectrum of a star contains lines which originate from
atoms
(arc lines)
also those
which originate from ions
(spark lines).
A
com-
parison of the intensity of a spark line with that of an arc line, both referring to the
same element,
gives rise to a value of the degree of ionization
ing a star as a sphere of ideal gas,
it is
pressure of a stellar atmosphere. Since
e„.
Treat-
an estimate of the the other quantities are known, the
possible to obtain
all
It
T
P
pressure
g
=
is
molar Gibbs function of an ideal gas at
equal to
RT(4>
+
+
In P).
(17-14)
.
G =
in Art. 16-5 that the
and
:i
Let us imagine that ci moles of A\ and v« moles of At are mixed at uniform temperature Tand pressure P and that chemical reaction takes place, thereby forming constituents A3 and A t At any moment when the degree of reaction is e, the Gibbs function of the mixture is
+ ftgKg +
/urn
p«8g
+ uttii,
Hi
=
ci(l
—
e),
«:i
=
1-36,
n-i
=
v-i(\
—
e),
m —
vte,
Affinity
was shown
raised to the (v
_
where
temperature
P
AG = -RT\nK.
temperature can be calculated.
17-5
the factor
(17-13)
It follows, therefore, that the
his
and
V'ifyi);
above equation will be satisfied when both P's arc measured in the same units, whatever the units are. If we express P in atmospheres as usual and calculate AG when each gas is at a pressure of 1 atm, the second term on the right drops out. Under these conditions, AG is known as the standard Gibbs function change and is denoted by v.\
AG". Therefore,
Saha applied
"i0i ""
-RT In K + RT In pn*-f-H-n.
student will recall that
— vi —
+
and each chemical potential is a function of T, P, and e. It follows that G is a function of T, P, and e; therefore, at constant T and P, G is a function of e only. The graph of G against e has somewhat the form shown in Fig. 17-2. At the equilibrium point where e = t c the curve has a minimum at which ,
If
we have
four gases that can engage in the reaction
"t.Ai
+
vtA*
f)
^ v%A% + vtdt, The
we
define the Gibbs function change of the reaction by the expression
AG =
vjgj
+
figi
—
uigi
—
v 2g2,
(17-12)
-0
slopes of the curve at the points
e
(at
=
e
and
= e
«,)•
=
1
may
be calculated from
the equation derived in Art. 16-13, namely,
dGr.p
=
(no
+n
)(i'3/X3
+
"4Mi
—
"iPi
~
"2M2) de,
598
17-5
Heat and Thermodynamics
G
no
and therefore both x 3 and * 4 arc
final constituents,
m
=
^) de Ir.r
On
the other hand,
fore xi
and
when
e
=
-«
(at
e
=
zero.
Hence,
0).
and there-
there are no initial constituents
1,
599
Ideal-gas Reactions
Hence,
X2 are zero.
dG\
= +00
(at
€
=
1).
de Jt.i
The graph
in Fig. 17-2 has these properties.
of A2, v-i/2 moles of
arc present in Mixture of i/,
v2
J'"ig.
1
7-2
which in
Mixture of
A moles of A moles
Graph of
this case
G
of
,
2
Mixture
A
">/2 moles "2/2 moles "3/2 moles
of
of
A3
"4/2 moles
of
A4
against e at constant
of
A2
v,,
moles
e
/I3;
and
of
k 3 moles of
{
=
£ at which there are v\f1 moles of A%; vz/2 moles «m/2 moles of At. At this point the constituents proportion to their stoichiometric coefficients, and the mole
Consider the point
fractions are
A3
of .As
Xl
=
Vl
V2
T and P.
becomes
The VifXi
—
Since
m m RTfa + In P +
and
g„
the chemical potential
=
RT{4>,
:
may be
+ In
(17-15)
P2M2-
Vifll
"4 = s~ Zv
slope of the curve at this point indicates whether such a mixture
equilibrium or not. If this slope
\de Jt.p
Xi
Zv
left; i.e.,
when both
the initial
is
and
final constituents are
In xk )
the equilibrium point is to the right; or when both initial and final constituents are mixed, there will be a tendency for the reaction to proceed tive,
/>),
written in convenient form
=
gk
+ RT In
to the right, causing final constituents to be formed. Finally,
a mixture of both initial and final constituents is there is no tendency for the reaction to proceed at all.
xk
.
Therefore,
It is seen, therefore,
'dCi
—
"3g'l
+
fig 4
—
"lgl
~
stituents
"2g2
+ RT(vi In *3 + or
^
c 4 In
= AG + RT
0* /r,p
It
should be borne in mind that the
x.|
—
more, i>!
In Xi
"2 In #2),
X In
(17-16)
*?'*?
x's in this
rium values but correspond to any value of
e.
equation are not equilib-
Now, when
e
=
0,
there arc
in
this slope
composed of
initial
and
e
=
-g is
final
is
and an
con-
proportion to their stoichiometric coefficients. Further-
obvious from the curve that the magnitude of
this slope is a
measure
from or nearness to equilibrium of such a mixture. We shall the slope of the Ge curve at the point e = i the affinity of the reaction,
of the departure call
-^±-
mixed
it is
if
in equilibrium,
that the sign of the slope of the Ge curve at
indication of the behavior of a system
\de)r.P
mixed, there will
be a tendency for the reaction to proceed to the left, causing initial constituents to be formed. Conversely, if the slope of the curve at e = £ is nega-
zero, y-k
in
is
positive, the equilibrium point is to the
which
is
equal to
,-]
fir)
(«-*)- AG + RT In
(v 3 /Zv)"(lHfZi>)
n (17-17)
600
I
leat
This equation
T=
298
3
K
is
true at
P = When all
and
each reaction.
much
1
temperatures and pressures.
all
The
atm. the
T.et
us choose
term on the right is a constant for are unity, this term is zero. In general its
e's
last
than AG° 9S Consequently, AG°9S has much more effect on the quantity dG/dt than the other term. We may therefore write:
value
is
less
.
where
Aj>
=
v 3 -\-
—
y«
J»i
—
niay be proved rigorously that the expres-
»*• It
and x's. The proof, however, complete generality but choose as on is starting conditions no"i moles of Ai and n v-i moles of A 2 , with no amount of the following simple form A-i and A t then the preceding equation reduces to sion in brackets
is
positive for all values of the v's
we do not
rather lengthy. If
insist
,
(this calculation constitutes
Prob. 17-14c):
no
(17-18)
Therefore,
the standard
and amount
Gibbs junction change
298°A'
at
which a reaction will proceed at
an indication of the
is
For example, water-vapor reaction is a large positive number, meaning that the equilibrium point is far to the left of t = jr, and therefore ee is very small.
AG|98
to
this temperature.
for the
Again, AG?98
is
a large negative
number
therefore the equilibrium point
is
NO ^± JN2 + £Oj;
for the reaction
far to the right of
«
=
£,
and
e„ is
almost
unity.
The
mum
a minimum. In order to verify that G is a minimum, it is necessary to show that (d'2 G/de 2 )r,p is positive at the equilibrium point. For a mixture of arbitrary amounts of four chemically active ideal gases, wc have as well as
a oe
for all values of
e.
=
)
The
right-hand t
found
be
to
+ RT In (1
Therefore, since
de 2 ) T
value of
»d)(.AG
("°
Jt.p
member
=
AC
(no
.
a function of
is
+ n' )RT dt
and
"W
dt
xl'x?
--
_
«
Q
28*
9
X\
*>
v +l Xl
•>
_j.
P
(17-19)
and
P,
X-i
xt
always positive for
all v's
K=
+
In
where In Kisa function of of
t„
know that among 7",
at ec ,
which may be written in the form
only.
T only,
("3
+
and the
v.\
first
—
vi
—
vz)
In P,
term on the right
is
a function
Now, d*e
\
(dlnK\
dltiKjp\ ST )p
Using the van't Hoff isobar to evaluate the numerator and the law of mass action to evaluate the denominator, wc get
$1
is
Since
>
v +l
is
only,
1
"1
(17-21)
(dlnK)/dT
'
"l
"4)
We
of this equation can be easily evaluated for any
+ n'
+
e)
(3 In K/dt.) P
(the details of this calculation constitute Prob. 17-1 4a),
-^ In
-
reaction of a change of temperature at constant pressure. equilibrium the law of mass action provides us with a relation
x2 /
T
«(1
of this equation
dT/p dG\ -
)(v*
the temperature or the pressure is changed. Let us consider first the effect on the equilibrium value of the degree of
In
equilibrium value of the degree of reaction was obtained by setting is the condition that G be a maxi-
S«*
i>2
when
Displacement of Equilibrium
(dG/de) T ,p equal to zero. This, however,
member
+
("1
2 2 and all values of e, it follows that (dG /de )r,p is always positive; hence, maximum. It will be seen that this = not a when t e„, G is a minimum and displacement of equilibrium determining the plays an important role in
and 17-6
xl'x?
dt
Since the right-hand
direction
601
Ideal-gas Reactions
17-6
and Therniodviia
_ - (Ac) 2
(17-20)
1
d[\n
(17-22)
(*&$/}&&)]/&,
already mentioned that the denominator on the right is posifollows that the sign of (de c/dT) P is determined by the sign of AH.
we have
tive, it
AH
RT
602
17-7
Heat and Thermodynamics
Therefore, an
increase of temperature at constant pressure causes
rium value of the degree of reaction
a
shift in the equilib-
which the heat of reaction
in the direction in
is
change of temperature takes place
that equilibrium
and the enthalpy
absorbed.
To determine we write
at constant pressure in such a
Then e c will change change by the amount maintained.
is
will
603
to the value
te
+ de
e,
the effect of a change of pressure at constant temperature,
dHp = d In
Bt,
dPJr
{0 In
Using the law of mass action
2nt- dlh
+ 2/;* dn k
.
A'
Kfp \ OP
d In (d In
we
infinitesimal
way
Ideal-gas Reactions
Since
)„
dlik
K/dP) K/de e ) P
=
cpt
dT and
dnk
= +
+ n )vk
(no
dee
a
,
c,
to evaluate
dllp
both numerator and denominator,
=
2n k crk
dT
and the heat capacity
+
(no
+n
B
+
)(v 3 hz
—
i>Ju
of the reacting gas mixture
—
vjii
vzhi) de„,
is
get
de«A
_
v*
dPjr ~
+
P 41n
"4
—
vi
—
v-i
(17-23)
Q&ft'fflxpd/de^
From
The numerator on
the right
proportional to the change in the
is
moles of the constituents as the reaction proceeds to the this
means
that the
volume increases
at constant
increase of pressure at constant temperature causes
a
T
right. If
and
number
it is
the preceding article,
of
positive,
RT
2
d[\n (xl'x^/xl'x^j/de;
shift in the equilibrium value of the
degree of reaction in the direction in which a decrease of volume takes place.
consequently,
Heat Capacity of Reacting Gases in Equilibrium
17-7
AH
de,
dTJp
P. Therefore, an
r Cp =
Vn >
Li
k c.pk
+
+
J. I(n
-L n >\) „,„, l
— Will>
„
, d\\n (x 3 'x A "/x x 2,.m -)\/dtc
Rl
,,,
/
,,
/
,
.
'
(17-24)
1
Let us consider, as usual, a mixture of arbitrary amounts of four ideal gases capable of undergoing the reaction
PlAl
+
V-lA t
^
V-iA-j
+
As an example, consider the equilibrium mixture of HsO vapor, Ha, and 0» caused by the dissociation of 1 mole of H 2 at 1 atm and 1900°K. We have 8 = = = » 250 kJ/mole, R 8.31 T/mole deg, e c = 3.2 X 10" , 0, AH 1, m, = = — — = and 2nt«o(l t> vt 0, j>s 1, va ««/2), n
V.l/1.1.
?,'„
At equilibrium, the enthalpy
of the mixture
H
=
•
+
is
-J-,
2n,A-
Hence
and
«i
=
("o
n*
=
(no
e e is
+ n )u\(\ — + n' )vi(\ —
e„),
ee)
+ Nt,
x
de e
where n%
=
(no
n*
=
{tin
+ tkdv&e, + n' )vit + a
e
JV4 ,
the equilibrium value of the degree of reaction. Suppose that an
r
V
t
'x,2
-
2n,
;
_
ij
+
(A/f)'(l til '{vi
=
X
t,:
)
«./2)6.(l
-f-
(250,000) 2 8.31
=
-
e„(l
^2>
X
(.
ej
^3 -r ^4)
3.2
(1900)
4.32 J/deg.
-
2
X 10~ X|
3
604
Heat and Thermodynamics
ldcal-gas Reactions
PROBLEMS
17-7
Calculate the degree of ionization of cesium vapor at 10 _e
605
atm
at the two temperatures 2260 and 2520°K.
Show
17-1
that the law of mass action
_
pvp*
may be
17-8
written
K
17-9
where the
equilibrium values of the partial pressures. Starting with n u moles of 3 , which dissociates according to the |-H 2 , show that at equilibrium 8 S
17-3
Starting with n
CO + 3H A'
27(1
A mixture of n pressure
P
ll
vl
^
2
and 3n moles of H 2 which react CH 4 + H 2 0, show that at equi,
where the
x's
17-10
Calculate
+
V2A2
I'e
~
^
j<
3 /1 3
T and V\
^3
moles of A 2 at temperature at
the reaction
7=
At 35°C and
17-5 is
1
mole of
+
(a)
P, the volume
is
Vc Show .
K=
675°K,
17-12
-f v u\\
+
"1
0.132 and
Starting with
At any value
G= ~
"1
atm, the degree of dissociation of
X
2
Calculate K.
(t>)
Calculate
te
at the
e(i-3M3
is
100
+
^->2 + i0
Equilibrium constant
800
900
1000
1105
0.0319
0.153
0.540
1.59
(d)
At
2
e
=
0)
—
v\ixi
—
moles of As, show that:
viia)
= vm u +
-
< (
+
van
+
viy.%.
"2^2,-,
+ In
In
x[%*
-
In x",Xl.
0,
— G(min) = RT
AteG\
Determine the average heat of dissociation graphically.
v*
denotes an equilibrium value.
g-gw = Go
Kelvin temperature
e
{c)
N
SO3 ^± SO s
van
G(min)
mm.
(c) The equilibrium constant for the dissociation of 2 4 has the values 0.664 and 0.141 at the temperatures 318 and 298°K, respectively. Calculate the average heat of reaction within this temperature range.
17-6 The equilibrium constant of the reaction has the following values:
(de e /dT) r at
At equilibrium,
where the subscript
same temperature when the pressure
2950 J/mole. Calculate
t,
at equi-
4
AH =
|l2
moles of A\ and
v\
of
0.27.
(a)
dissociates according to the reaction
that
V«
Vi
HI
temperature.
(b)
librium
1
HI?±iH 2 +
n^v-i
.
Vo
I'D
heat capacity of the equilibrium mixture of
the
- ie yp*
V When
has come to equilibrium at the same
-firing
are equilibrium values.
When
17-11
this Vj/it
that
Prob. 17-7 at the temperature of 2260°K.
- tey
moles ofA% and
occupies a volume
Show
AG =
4e»(2
=
(b)
P.
CO
moles of
according to the equation librium
17-4
m
that
AG = AH + T
NH ^|M +
K = V27
and
Show
(a)
p's are the
NH
17-2 equation
'/'
Calculate the degree of ionization of calcium vapor in the sun's
chromosphere. The temperature and pressure of the sun's chromosphere are approximately 6000°K and 10~ 10 atm, respectively.
"1
In vi
+
In xJjKjJ.
"z
'I
+
"2/
v,
"i
+
f.,
1,
—
G(min) ~
RT
=
Vj lr
+
—
In Xj;.v^.
606
Heat and Thermodynamics
17-13
f„\
In the case of the ionization of a
G— At
=
£
-
—+
t
=
.
r
1
— In — —
:•
;
+
1
e„
1
lp
gy
'
ef
,
4-
ln
(TToi-
—
G(mia)]/2.30i?r against e for the ionization of cesium vapor at 2260°K and 10 -6 atm, using the result of Prob. 17-7. 17-14 Prove that, for a mixture of reacting ideal gases, (a) (a?)
HETEROGENEOUS
1,
d - G(min) Plot [G
.
-ell
.1-6. = - ln iT7e
G(min)
*T
«
+ In
In
-
18.
that:
SYSTEMS
—
At
show
0,
Go (e)
gas,
G(min) 1
(6)
monatomic
18-1 It
Thermodynamic Equations was shown
—
5i!£i!
1
=
"° j;
at temperature
I
T and
and
pressure P,
is
G = *1
and
Aj»
If
we
start
=
with n
J>3
ei
#3
+
C4
*3
—
*4
—
Pi
moles of Ai, n
•0 ("1
Prove
that, for a
aF BTJr
(*)
+
S/i*nt,
v2
moles of
A-<,
and no
/1 3
or A, h
'-' ,
T
each of the
-
+
dS=V n
k
that, for a
accompanying change
dG in equilibrium,
in the
Gibbs function
is
equal to
+ n' )RT(Av)*
(no + n'a ) Ap AH PT(d/du) In to**?/*?*?)
all
the phases;
G =
1
+
«'n
Sjujj
Rfc
over
!Sj4 n|r
+ S^j."'"*
equilibrium,
4/AHAp dp ~~RT P'
2/**
dnk
.
i.e.,
+ mixture of reacting ideal gases
-SdT +VdP +
Suppose we have a heterogeneous system of
(ft Prove
the
V4)
PAW (c)
n's,
•)
Vi){Vi
(no
potential
of the respective constituent,
mixture of reacting ideal gases
V_
dV\ ~ dPJr
17-16
thermal and mechanical equilibrium
is a function of T, P, and the mole fraction and the summation is taken over all the constituents. Furthermore, if the phase undergoes an infinitesimal process involving a change of temperature dT, a change of pressure dP, and changes in
where each chemical
|»J,
show that
17-15
in
equal to
d + j£
where
(A)
Heterogeneous System
16-10 that the Gibbs function of any homogeneous
in Art.
phase, consisting of c constituents "°
for a
'
all
over
the constituents of the 1st phase all
the constituents of the 2d phase
hq
11
over au tne constituents of the wth phase.
an infinitesimal process takes place in which all the phases undergo a change in temperature dT and a change in pressure dP, then the change If
608 in the
Heal and Thermodynamics
Gibbs function
dG =
-
-.S < 1) -.?<«
18-2
is
dT + K<" dP + 2/4 dn^ dT + V» dP + Jfyp dni
plicated. In the
first
variable besides
T
place there
and
is,
as a rule,
Heterogeneous Systet
609
more than one independent
P. In the second place, the equations of constraint
1
'
(for 1st
2)
-aw *r + p»
a^f»
(for
it is either impossible or exceedingly cumberattempt by direct substitution to express G in terms of the independent variables only and dG in terms of differentials of these independent variables. Finally, instead of only one equation of equilibrium, there may be
arc usually of such a nature that
phase)
2d phase)
some
(for ipth phase).
to
several,
This equation evidently reduces to
depending on the type of heterogeneous system.
We are therefore confronted with a type of problem that requires the use of Lagrange's method of undetermined multipliers.
dG -
-sir
KrfP
+ 2^i
1}
fcf *f +
A|»
***,
J5«f°
(18-2)
Phase Rule without Chemical Reaction
18-2
where S and V are the entropy and volume,
respectively, of the
whole hetero-
geneous system.
The problem
of the equilibrium of a heterogeneous system
is
to obtain
an
equation or a set of equations among the p's that hold when all the phases are in chemical equilibrium. If the system is assumed to approach equilib-
rium at constant T and P, then G is a minimum at equilibrium, and the problem can be stated thus: To render G a minimum at constant T and P, subject to
whatever conditions are imposed on the n's by virtue of the constraints of the system.
The mathematical is
condition that
G
be a
minimum
at constant
T
and
P
Consider a heterogeneous system of c chemical constituents that do not combine chemically with one another. Suppose that there are
geneous system
is
that
dGr.r
=
0.
G = 2*fvf +2* (2W2> Hence, the equation that must be
dG TJ = .
where the
2,4" dni
1 '
dn's are not all
satisfied at
+ S/*f
•
•
equilibrium
•
+
Jttf*
*f» =
where 0,
(18-3)
independent but are connected by equations of
To return for a moment to the system treated in Chap. 16— namely, one phase consisting of a mixture of chemically active substances- -we found that the equations of constraint were of such a simple form that, by direct sub-
all
the summations extend from k
We
+w
)(»<3(U3
+
VitH
—
villi
—
G
is
=
1
to k
=
c,
since all the con-
a function of T, P,
have as our equations of constraint, therefore,
G could be expressed as a function of T. P, and only one other independent variable e and that dGr ,r could be expressed in terms of only one differential de, thus: (»o
)
and the n's, of which there are op in number. Not all these n's, however, are independent. Since there are no chemical reactions, the only way in which the n's may change is by the transport of constituents from one phase to another, in which case the total number of moles of each constituent remains constant. stituents are present in all the phases.
stitution,
—
V-h
k
is
constraint.
dGr.p
+ 2n
,
M
,o>
J u1
)
f „m + 4.
=
const.
+ «$** =
const.
P>
«i
n..
V2H2) dt. „< 2 >
At equilibrium, when dGr.p equilibrium, was obtained.
=
0,
only one equation, the equation of reaction
In the case of a heterogeneous system, however, the situation
In order to find the equations of chemical equilibrium, is
more com-
render
G
a
minimum
at constant
T
and
it is
necessary to
P, subject to these equations of
610
Heat and Thermodynamics
constraint.
dG =
tf» Xi
18-2
Applying Lagrange's method, we have
*j* +
•
•
+ M <" Af +
•
•
•
+
•
& dn^ + +
•
•
dGr,p must be negative. Therefore, while
Xi
•
n[
»
Af
=
>
l)
+ fif»
-
p\
2)
may
phases
The chemical
equal.
611
the flow is taking place,
(flow of matter from phase
Obviously, the transfer of matter ceases
become
Heterogeneous Systems
when
to
1
phase
2).
the two chemical potentials
potentials of a constituent in
two neighboring
be compared with the temperatures and pressures of these
phases, thus:
+ X *«
Xf <&!«
c
=
where there are c lagrangian multipliers, one for each equation of constraint. Adding and equating each coefficient of each dn to zero, we get
pP
= -x, »
rf»- ~x 2 -
rf"
=
--x,
rf»
=
-
Ml
-X.2
ii
W
— — Aj -
temperature of phase 1 is greater than that of phase 2, there a flow of heat that ceases when the temperatures are equal, i.e., when thermal equilibrium is established. If the pressure of phase 1 is greater than that of phase 2, there is a "flow" of work that ceases when the pressures are equal, i.e., when mechanical equilibrium is established. If the chemical potential of a constituent of phase 1 is greater than If the
1
0,
is
2
3
—\
that of phase 2, there
is
a flow of that constituent which ceases when i.e., when chemical equilibrium
the chemical potentials are equal, is
**«
2'
-x,,
=
Mi
=
tf>4" =
--x.
Mi
,<<")
=
2'
-
• •
The
rip
-
2'
Mi
=
.
1
tions
- Mr.
. •
which express the important fact that at equilibrium the chemical potential of a constituent in one phase must be equal to the chemical potential of the same constituent in every other phase.
As a simple example, suppose that we have only one constituent present two phases. Then,
and
since
ohf =
— dn[
=
rf»
make G
for
c
in all the
tp
phases are obviously
constituents, there are. altogether c(
—
1)
—1
in
equa-
n's.
c
equations of constraint for the
minimum. These
a
T
ipe
values of the n's
values should, of course, be functions of the
We
and P. find, however, that the equations do not contain the n's in such fashion as to determine their values, because of the fact that parameters
the equations of equilibrium are equations
which are
intensive quantities
among
and depend on the
the chemical potentials,
x's,
which contain the
n's
form
in
nk
Xk
" 5'
Af + *»f (UP', is another way of saying that the chemical potential for a constituent phase depends on the composition of that phase but not on its total mass. There are many different sets of n's that satisfy the equations of phase equi-
This ,
in a
=
(/4
U
~
21
Mi
)
*P.
librium and give
This
Now,
before equilibrium
phase
1 to
2.
the
in the special
l)
dGr, P
phase
among
equilibrium and the
equations of phase equilibrium,
dGr.P
any one constituent
Following out the procedure of Lagrange's method, we should complete our solution of the problem by solving the c(tp — 1) equations of phase that
These are the
equations of phase equilibrium expressing the equality of the chemical
number. Therefore, (18-4)
ro fc
established.
potentials of
= Mr
**P
or
_ -Ar =
Then
is
reached, suppose there h is
dn\
is
may be
rise to the
same minimum value
of the Gibbs function.
seen from the fact that
a flow of matter from
negative; and since this flow
is
irreversible,
g =
4M
IJ
+
/4V"
+ rfV + 9
V-Tnf,
612
Heat and Thermodvna
18-3
but at equilibrium the chemical potentials of the same constituent are the in all phases and hence may be written without any superscripts.
same
Factoring out the
=
G(min) Therefore,
n's,
„,(»<»
we
The
+ «n
+ M»
fi)
value of the Gibbs Junction is the
mass among
Since
the phases.
of the »*s at equilibrium, therefore,
wc may
same for many
we cannot
it, Gibbs considered not only but also those produced by gravity, capillarity, and nonhomogeneous strains. It stands today as one of the most profound contribu-
human
different,
we can
obtain any precise information about a heterogeneous system in equilibrium. As we have seen, the state of the system at equilibrium is determined by
simple matter to remove the restriction that every constituent must is absent from phase 1.
It is a
be present in every phase. Suppose that constituent A\
Then
the equation of equilibrium that exists
=
c(p
+ 2.
these variables there are
two types of equations: (1) equations of 1) in number; and (2) equations of the each phase, and therefore
phase equilibrium, which arc
=
type S*
for
1,
e(tp
—
of equations
=
c(
—
1)
now
many equations as there are variables,
If there are as
however,
-j-
is
of variables exceeds the
variance of
1
.
is
In general,
number
of equations
we
its
it is
worth while
the problem, de novo. Before this
to consider a
few simple applications of
is
done,
the.
phase
present form.
is
the excess
called monovariant
and
of variables over equations
is
is
said to
is
arbi-
have
a
As simple examples of
1.
=
(number
/ =
«*
of variables)
f= c known
Willard Gibbs, University.
The
(number
lp
+
2
Pure substance
[e(
.
-
of equations), 1)
we
shall consider a
pure
In the case of a pure substance such as water, the
phase rule merely confirms what
is
already known. If there arc two phases in
equilibrium (say, solid and vapor), the variance
-
{<=¥>+ 2)
whence
—
the use of the phase rule,
substance, a simple eutectic, and a freezing mixture.
called the variance f.
Thus, Variance
Simple Applications of the Phase Rule
18-3
by one, then the equi-
not determined until one of the variables
Such a system
trarily chosen.
is
to describe the composition of the first phase,
then the temperature
of the whole system at equilibrium arc determined. called nonvariant and is said to have zero variance. If the
librium of the system
This
However,
and composition
Such a system
number
lacking.
To remove the second restriction more difficult and requires solving
rule in
pressure,
present,
—that no chemical reaction take place
is
number
is
need one mole fraction fewer than before. Therefore, since both the number of equations and the number of variables have been reduced by one, the difference is the same and the phase rule remains unchanged.
Hence, Total
the constituent
$ = -x„ is
Among
when
namely,
the temperature, the pressure, and op mole fractions. Hence,
Total number of variables
thought and, along with Gibbs' researches in
vector analysis and statistical mechanics, places him with the greatest of the world's geniuses.
find the values
inquire as to whether
the Equilibrium of Heterogeneous
effects
tions to the world of
+ »<«).
On
original paper, entitled
Substances," was almost 300 pages long. In
chemical
+
minimum
the
distributions of the total
get
emy.
613
Heterogeneous Systems
+
is 1.
There
is
one equation
of equilibrium, namely,
?];
(18-5;
rule, which was first derived in 1 875 by Josiah then professor of mathematical physics at Yale phase rule arose from a general theory of the equilibrium
as the phase
who was
of heterogeneous systems that Gibbs developed during the years 1 875 to 1 878 and published in an obscure journal, The Transactions of the Connecticut Acad-
/('/;/>)
-
n'"(T,P),
where one prime stands for solid and three primes for vapor. We have already that, when a phase consists of only one constituent, the chemical potential is equal to the molar Gibbs function. Hence,
shown
614 is
Heat and Thermodynamics
the equation of equilibrium
18-3
among
T
the two coordinates
and
P
is
1200
nonvariant, and the two equations of
g"
and
=
g'"
T and P. The phase rule shows that the maximum of phases of a one-constituent system which can exist in equilibrium three. The various triple points of water confirm this result.
o
serve to determine both
number is
615
which
will be recognized as the equation of the sublimation curve. If three phases
arc in equilibrium, the system equilibrium = g'"
Heterogeneous Systems
2*
a Let us consider a system of two constituents which neither combine to produce a compound nor form a solid solution but which, in the liquid phase, are misciblc in all proportions. A mixture of Simple
2.
E
eutectic
j2
gold and thallium has these properties. Suppose that we have a liquid alloyconsisting of 40 percent thallium and 60 percent gold in an evacuated chamber originally at about 1000°C. A mixture of thallium and gold vapors will constitute the
vapor phase, and we
that the variance
is
shall
have c
=
2
and
=
2. It
follows
2; hence, the
chosen, the vapor pressure
is
composition and temperature having been determined. If the temperature is now progres-
phase of pure gold will separate from the liquid at a temperature of about 600°C, and the percentage of thallium in the solution
20
60
73 80
100
Percent thallium
sively lowered, a solid
is
40
Phase diagram for
Fig. 18-1
eutectic system of
gold and thallium.
At any given concentration, there will be one and only one which the three phases—vapor mixture, liquid solution, and
thus increased.
temperature at solid gold
—
will
be in equilibrium, because
now
c
=
2 and
=
3; therefore
/=!•
is
By
covering the metals with a piston on which any desired pressure may be exerted, we may exclude the vapor phase and study the variance of the system when only solid and liquid phases are present. In this way, the tem-
and compositions at which equilibrium exists among various be measured and the results plotted on a phase diagram such shown in Fig. 18-1.
peratures
phases
may
as that
Point
and
B
thallium.
A is
the melting point (strictly speaking, the triple point) of pure gold, that of pure thallium. When two phases, solution and vapor, are
is divariant; and equilibrium may exist at any temperature and composition represented by a point in the region above AEH. When the three phases (solution, vapor, and solid gold) are present, the system is
present, the system
monovariant; and equilibrium may exist only at those temperatures and compositions represented by points on the curve AE. Similarly, curve BE represents temperatures and compositions at which the monovariant system
phases— solution, vapor, and solid thallium- -is in equilibrium. The complete curve AEB is known as the liquidus. At E, there are four phases present: solution, vapor, solid gold, and solid consisting of three
known
Hence
c
=
2,
=
and / = 0, or the system is nonvariant. This and the composition at this point as the eutectic
4,
as the eutectic point,
composition.
and solid gold may coexist at all temperatures and compositions represented by points in region ACE. Below line CE, however, no liquid can plus free exist, and the system consists of a solid with the eutectic mixture ED we below BED, and gold. Solution and solid thallium coexist in region Solution
have eutectic plus free thallium. Line CED is known as the solidus. There are many different types of eutectic system, each with phase diagrams of different character. All of them, however, may be understood completely in terms of the phase rule. 3.
Freezing mixture
A
number
of years ago, before the commercial use
cream of solid carbon dioxide ("dry ice") as a cooling agent, foods such as ice salt. common ice and were packed in a container surrounded by a mixture of mixture was thermally insulated and covered, it maintained a constant melt temperature of about — 21°C. Another practice that is still current is to phenomena on it. These the ice which forms on the sidewalk by sprinkling salt may be clearly understood on the basis of the phase rule. If the
616
18-4
Heat and Thermodynamics
some.
salt.
But
librium with
this saturated solution will
ice.
Heterogeneous Systems
617
be too concentrated to be in equi-
Ice will therefore melt, lowering the concentration of the
which will then dissolve more salt. While this is going on, the temperature of the whole system automatically and spontaneously decreases 21 °C is reached. Such a system is known as a until the temperature of
solution,
—
freezing
mix lure.
At the transition point B, where NaCI forms, there are three constituents; hence one might expect a maximum of five phases to coexist. This is not the case, however, because a chemical reaction
takes place.
We shall see in the next article that the presence of this reaction
causes the system to behave as
C
Ice
+
-25
+ solution
^± NaCI
NaCl-2H 2
if
there were only two constituents, so that
only four phases coexist at the point B: solid NaC12H->0, solid NaCI, solution, NaCI -2H
I
\E
?
20
10
and vapor. There are a number
29,'!
I'
30
40
of freezing mixtures that are often used for preserving
materials at low temperatures. These are listed in Table 18-1.
50
Percent NaCI Fig. 18-2
Phase diagram for mixture of
NaCI and
H 0. 2
Freezing Mixtures
Table 18-1 Consider the phase diagram of NaCI and water shown in Fig. 18-2. A is the triple point of pure water, and B is the transition point where the dihydrate NaCl-2H 2 changes into NaCI. Except for the upper right-hand portion of the figure, the
diagram
is
Second constituent
NH
in all respects similar to the simple eutectic
4
C1
AE, the system is monovariant and and ice. Similarly, on EB wc have
Alcohol
Ice
CaCl 2 -6H 2
Ice
and dihydrate. Point is the eutectic point, which, since in the system contains water, is called the cry ohydric point. The mixture of dihydrate and water that forms at the cryohydric point is called a
Alcohol Ether
Solid
Fig. 18-1.
At
all
points on
E
solution, vapor,
this case
°C
-15.4 -21
Ice
consists of three phases: solution, vapor,
perature,
Ice
NaCI
diagram of
Lowest tem-
First
constituent
Solid
CO C0
-30 -55 -72 -77
a 2
cryohydrate.
Only
at points below the solidus
CED
and NaCl-2H 2 exist as solids together. Consequently, when they are mixed at a temperature above — 21°C (as on the sidewalk), they are not in equilibrium, and as a result the ice melts and the salt dissolves. It should be mentioned at this point that, if the system
is
can
ice
open
to the air at atmospheric pressure, then there is one more which would ordinarily increase the variance by 1. Since constant, however, this extra variance is used up, and the
constituent (air),
the pressure
18-4
is
c
Phase Rule with Chemical Reaction.
Let us consider a heterogeneous system composed of arbitrary amounts of constituents, assuming for the sake of simplicity that four of the constituents
are chemically active, capable of undergoing the reaction
system behaves as before. If ice, salt, water,
and vapor
at
0°C are together in a thermally insulated and some ice will melt and dissolve
container, they arc not in equilibrium,
V\A\
+
VtA-l
,
— V-iA
3
*M
At
.
618
Heat and Thermodvnai
Suppose that there are all the
ip phases and, as only a temporary assumption, that constituents are present in all the phases. As before, the Gibbs function
of the system
Adding, and equating coefficients of the — lj equations of phase equilibrium,
o-l H*
_ —
T Z4n4*>
"t
I
M
a»
_ —
.
.
= g* =
.
.
.
=
•
•
•
= Mf.
.,(2)
Ml
The
equations of constraint for those constituents which do not react are same type as before; i.e., they express the fact that the total number of moles of each inert constituent is constant. In the case
*« =
of the
=
ff»«
of the chemically-
active constituents, however, the total
number
of moles of any one
is
not
a function of the degree of reaction. Hence, the equations of
is
Equating the coefficient of dt librium, namely,
ni"
+ *i +
,(')
,m
„(')
„< 2 >
2
which, since Xi
>
*P = v) = »i
n?
nT =
+ *f +
=
ni
(«o
+ «o)"2(l
-e)
+
=
—m
of
m
any phase,
\j
•
•
•
+ «<*> =
"2M2
=
.XV)
Ml
get
—
an extra equation of equi-
=
X4V4
= — n% "3M3
(,»
M
+
0,
of any phase,
etc.,
becomes
*MM-i-
This equation will be recognized as the equation of reaction equilibrium,
46
which
—
was found to lead to the law of mass action. same lines as before. There are
in the case of ideal gases
The
const.
rest of the 1)
argument
follows the
equations of phase equilibrium,
and f equations of the type
S.v
=
1
.
1
equation of reaction equilibrium,
Hence, the
total
number
of equations
is
const..
efa
where n n' A\ and Ar4 have their usual meaning. Applying Lagrange's method, we get the equations below (reading across pages 618 and ,
get the usual
(n n
c(
+ »?' +
+
_
.
we
X3C3
VlMl
+ n' )^e (n„ + « > + if
to zero,
+ X2V2 —
Xiei
constraint are:
»f
we
is
„(1>
«P
dn'& to zero,
c(ip
Mi
constant but
619
Heterogeneous Systems
18-4
—
1)
+1+
v>.
,
619).
4 X,
l)
dn™
+
•
•
•
+ „f> *p +
H
-
dn?
+ X 2 dni
1]
X3
*F X4
A? X5
Since the variables are the same as before, namely, T, P, and the
M dn w +
•
#iT
dnT
+
+
+ +
+ +
X 2 dniv)
X3
dnf X4
to
drCi
X,-
rffij"
+
+
Xi(n
+ n' )vide =
+
X2(n
+
—
X 3 («o
+ «o)"8«fc =
—
X 4 («
+
n'o)
"2 de
n'a)vidt
= = -
dnT
Xr
(«>)
«<'
of
=
+ M^ *?>
X, (fef
A£D
Xc
+
x's,
°
620
Heat and Thermodynamics 18-4
which there arc c^
+ /
2 in number, the varia anance
=
«*
/-
or
The phase
+ 2 - Hr -
(c-
l)
-
v
+
1)
H.
2.
(18-6)
-
rule in this case
is seen to be different, in that c 1 now stands formerly stood. For the reason given before, this form of the phase rule remains unchanged when every constituent is not present in cverv phase Vvc sec therefore that when there are c constituents, present in arbitrary amounts, and only one chemical reaction, there is one extra equation of equilibrium and the variance is reduced by 1. It is obvious that, if there were two independent chemical reactions, there would be two extra equations ol equilibrium; whence the phase rule would become/ = (c ~ 2) 2 v
where
For
c
+
independent reactions, we would have
r
/
The argument up of equations
=
(e
now has been have existed among the
Fig.
1
(18-7)
based on the fact that onlv three kinds variables T, P, and the *'s: equations
(b)
Dissociations of a salt that give rise to additional restricting equations
mole jr actions,
— r) - + 2.
to
(a) 8-3
which expresses the
fact that the solution
dissociation takes place according to the
NH HS
of solid
stituents
form according
into
4
an evacuated chamber and two new con-
NH, and H 2 S
exists that
- NH, + H
#8
=
*s
+ *6 +
Xi
=
Xg
+
and
x1
=
x$.
Adding
=
.s
in Fig. 18-36,
the
x's,
then
namely,
x-\,
Xf),
we
get the dependent equation
+
Xf,
+X = 1
3*8
+
2.X 6
+ X.h
expressing the fact of electrical neutrality.
X Hl s-
Let us call equations of the preceding type restricting equations, and let us suppose that z of them are independent. Then we may list four types of equalisted
tions
Another example
Suppose that the
these equations,
*3
of additional restricting equations among the *'S is provided by the phenomenon of dissociation in solution. Suppose that we have a heterogeneous system one of whose phases is a solution of salt 2 in
cipitate
among
2 S.
This constitutes a fourth type of equation, to be added to the three at the beginning of this paragraph.
*ig. 18-3a
electrically neutral. If multiple
are in the same phase, the restriction always ;
*NH,
1
the
to the reaction
NH..HS Since gaseous
is
scheme shown
there are three independent restricting equations
m
amount
among
(a) Single dissociation; (b) multiple dissociation.
of phase equilibrium, equations of reaction equilibrium, and equations of the type 1x - 1. It often happens, however, that a chemical reaction takes place such a manner that additional equations expressing further restrict.ons upon the *'s are at hand. Suppose, for example, that we put an arbitrary
solvent
621
is
+ + 1
Heterogeneous Systems
salt dissociates
according to the scheme shown in
All the ions, of course,
remain in the liquid phase, and no preformed. Consequently, we have the equation
among
=
x's,
thus:
3
Equations of phase equilibrium [c(
4
Restricting equations (z in number).
1
2
Hence, the x3
T, P, and the
total
number
of equations
x. h
«(j»
-
1)
is
-Kr
+ <(>+ z;
622
Meat and Thermodynamics
and, as usual, the total
or
number
f=
Cip
/ =
(c
of variables
is
op
- [c(
+
Heterogeneous Systems
18-5
2
r
r
+ 2. ip
Therefore
623
NaH,PO,
NaH,PO J
+ z],
H,PO, H 2 POT
Na* If
minus
we the
define the number of components
number of independent
equations,
c'
as the total number of constituents
minus
the
—r—
z.
reactions
number of independent
restricting
i.e.,
c
=
c
H.,0
(18-8)
(a)
we may always
write the phase rule in the same form, thus:
f=c'-
+ 2.
(18-9)
(6)
Fig. 18-4
Multiple dissociation of S'aHzPOi. (a) Without dissociation of
dissociation of
HiO.
2
Assuming
H
2
0.
H
,
2
Number
The problem of determining the number of components in a heterogeneous may be somewhat difficult for the beginner. As a result of experience
with the behavior of typical heterogeneous systems, the physical chemist is able to determine the number of components by counting the smallest number of constituents whose specification is sufficient to determine the composition of every phase. The validity of this working rule rests upon a few
Example First let
vapor.
It is
demonstrate rigorously in
*Na+
us consider a heterogeneous system consisting of a liquid phase composed to
NaH 2 P0 4 show
in water
and a vapor phase composed of water
that, so long as no precipitate is formed by virtue of a reaction
between the salt and the water, no matter what else
number of components
1
Hence, 3
,.
=
is
we assume
to
take place in the solution, the
There are two constituents, no chemical reactions, and no restricting equations. Hence,
=
2.
dissociation of the salt.
There are seven constituents and shown in Fig. 18-4a, and two inde-
pendent restricting equations,
Xgt.*
—
*poj
xa*
=
2xpo,
h *hpo4
h ^H.rOi-
h XstO, —
(By adding the two equations, the dependent equation expressing electrical obtained.) Hence,
neutrality of the solution
is
Assuming
water
dissociation of the
also.
independent chemical reactions, as
xss.-*
=
*po4
xr*
=
2*po,
and Hence,
c'
=
c'
=
8
-
4
-
2
=
2.
r
a'hpo,
\-
7
—
3
—
2
=
2.
There are eight constituents and four shown in Fig. 18-44, and two inde-
restricting equations,
Neglecting all dissociation.
2-0-0
*H,POi-
three independent chemical reactions, as
pendent
two.
/ =
=
4-1-1=2.
Assuming multiple
and
important
2
this article.
1
of a solution of the salt
+ H P07;
restricting equation,
system
shall
Na+
of Components and one
fundamental facts whose truth we
(b) with
POi"; one chemical reaction,
NalljPO., ?±
Determination of the
18-5
t
There are four constituents, NaHjPO*,
single dissociation of the salt.
Na +
H O;
xU¥0l
h
.*H.ro.,-
1- .t
0H -.
624
Heat and Thermodynamics the water. A ninth constituent, (H 2 0) 2 independent reaction,
Assuming association of
5
a result of a
fifth
H "lherc are c'
=
0;=i|(HoO) 2
same two independent
the
still
9-5-2
2
=
,
is
formed as
and add
the
first
two and subtract the sum of the
restricting equations,
and hence
which
is
is
clear, therefore, that
take place in the solution.
it is
a matter of indifference as to what chemical changes of components is always two, provided that
The number
no precipitate forms.
which
is
JUlWHi,
+
get
3,«iici>
+
A1C1 3
^ Al(OH)
3
+
3HC1,
therefore seen to be a dependent reaction.
Since some of the Al(OII)j has precipitated out, there tion,
we
last two,
the equation of reaction equilibrium corresponding to the reaction
2.
3H,0 It
+ fAici, =
3^ir.o .
625
Heterogeneous Systems
18-5
is
only one restricting equa-
namely, that expressing the electrical neutrality of the solution,
Example 2 3*ai+-h-
+
*ii+
=
*oa-
+A
'ci--
To
investigate the effect of precipitate, let us consider a mixture of A1C1 3 and water. In this case, the A1C1 3 combines with the water to form Al(OH) some of which 3,
precipitates out of the solution, according to the reaction
are eight constituents and only four independent reactions. imagine that there are five independent reactions, but if
shown in Fig. 18-5. There At first thought, one might
we
=
and
As a
=
M*i +t *
+
3Mci-,
/JA1(HO)i
=
/*JU++*
+
3/JOH-,
=
3/Jh*
-f-
last
example,
let us
consider a system consisting of water vapor and a solution
NaCl and
containing arbitrary amounts of
Maici,
3/*llfl
Example 3
+ 3/ioH-,
3/tu+
ponents.
write the equations of
reaction equilibrium corresponding to the four dissociations:
3jB8*o
Consequently, c' = 8 — 4 — 1 = 3, and we have an interesting situation in which a heterogeneous system, formed originally by mixing two substances, has three com-
1
3yUci-,
There are
Neglecting all dissociations.
XaX0
KC1; one
3,
KNOj
in water.
five constituents,
HjO, NaCl.
KN0
3,
reaction,
+ KNOj ^t NaNOs +
NaCl
KC1;
0Hand one
restricting equation,
*NaNO,
Hence,
3H,0
c'
=
5
-
1
-
1
+ 3HCI 2
Considering all reactions.
=
=
-XKC1-
3.
There are eleven constituents, which undergo
reactions:
+ OHNaCl P± Na + + 0t" KNOj ^ K T + NO? NaNOj ;=5 Na+ + NO? H
Fig.
18-5
Dissociation, reaction,
and
precipitation that occur
when AlC'h
is dissolved
2
?± H+
water.
KC1 ?± K+
+ car.
these
626
Heat and Thermodynamics
It
should be noticed that the reaction
Any
is
+ KN0
3
?±
NaNO + KC1
the
sum
may
five but that its equation of reaction be obtained by adding the second and third and subtracting
of the fourth and
There are three
is
equal to
a
not independent of the preceding
equilibrium
change in temperature, pressure, and accompanied by a change in the Gibbs function
infinitesimal process involving a
composition of the phases
NaCl
dG = -S dT
+ VdP +
pi?
+ /4
dn\+
+
+
•
fifth.
restricting equations.
The
11
dn? tf» dnf
T *N«NOj = *Cl
h
+
•
•
+ Mf>
dn™.
first,
In general, during such an infinitesimal change, there *Na+
627
Heterogeneous Systems
18-6
among
*KC1,
is
neither equilibrium
the phases nor equilibrium with regard to the chemical reaction.
Complete chemical equilibrium would require both phase equilibrium and amount of sodium lost by the NaCl (to form amount of chlorine lost by the NaCl (to
expresses the fact that the
Na + and XaN0 3)
is
reaction equilibrium. Suppose we assume phase equilibrium only. Then,
equal to the
HP = pf =
form CI- and KC1). The second,
XK*
+
AfKCl
=
Ano,
+ *NaNO„
expresses the corresponding fact concerning the loss of potassium from KNO3. The third is
xn<
=
and
nitrate
^
(The dependent equation of
we
In the event that (H..O, NaCl,
and
five
c'
electrical neutrality
=
start
KN0 3 NaN0 ,
3,
11
—
5
—
3
=
is
dG =
with arbitrary amounts of all five substances and KC1), there are still eleven constituents
which expresses
18-6
c'
=
11
Act
electrical neutrality,
*11+
Hence,
=
-
5
-
2
=
=
—h
Ano,
—
•
in the
= #T
•
= #»,
• •
Gibbs function becomes
rf»i
)
dnT)
"
•
*P +
h *on-,
*f + + *P +
and
•
•
•
•
dn?
Aon-.
A? +
+ rf«i« = — (n + *|" = — (no + itf = + (n + dn = + (n
a
•
•
•
+ d„r
+ n'
)
vi de
+ «o) "2 de + n )p de + n' )vi de 3
=
4.
Consider a heterogeneous system of
tp
phases and
c constituents,
four of
+
v.\A>„
0.
Therefore, the change in the Gibbs function during an infinitesimal process in
v$Ai ^2 V3A3
+ dnf =
dn O-i
which undergo the reaction
+
•
•
•
Displacement of Equilibrium
v\A\
«f =
'
~SdT+VdP + m(dn? + + w + dnT) + Ho(dn'» + + + w (*$» + + + *«>
But
+ #K* t *h+
=
= wf
obtained by adding these
3.
independent reactions but only two restricting equations, namely, •*«»*
itf
•
*oh-.
and the change three equations.) Hence,
=
a4"
•
which there
dG =
is
phase equilibrium but not reaction equilibrium
—SdT+
VdP -f-
(no
+
"'oX^Mn
+
"tin
—
"1M1
—
is
given by
v&a)
de.
628 Since,
Heat and Thermodynamics
under these circumstances
Heterogeneous Systems
G is a function
of T, P.
and
t, it
follows that
dG
- -S at ~ 6 dG
=
V,
—
(n
dP
dG
and
the direction in which the volume decreases at constant
T
and
P.
'
+n
)(i>3ii 3
+
v t Hi
—
(/j^j
—
18-1
y 2lu 2 ).
All the lettered points in Fig. PI 8-1
separates the plane into two regions: on the
lie in
left
a
one plane. The
line
wave has the speed
v
CD and
T and pressure P, we = e If we go to a slightly different equilibrium dT and pressure P + dP, then the new degree of
reaction equilibrium exists at temperature
must have dG/de state at
T
and P; whereas from Eq. (18-13), we see that an increase of pressure at constant temperature causes a reaction to proceed in absorbed at constant
PROBLEMS
dt
When
629
=
at
temperature
reaction will be
ee
+
T
+ de
e
e .
e
.
and the change
in
dG/dt during
this process
is
zero.
Therefore,
-£®*+i(»*+s*-« de
deFig. PI 8-1
Solving for
rfe
=
&,,
we
get
dS ' d (
A
rr
o-(r/ot-
on the right the speed v'. Show by Lagrange's method that the time for the to travel the path APB is a minimum when v/v' = sin tp/sin
wave
dV/ di Jn
(18-10)
d*G/dtr
,
Recognizing
that, at
thermodynamic equilibrium, d£>
=
2" t/S
The
or
entire system
is
thermally insulated.
If the final
T/ and that of the liquid is Tf, show by Lagrange's method that the condichange of the universe to be a maximum is that Tf = Tf, 18-3 Consider a homogeneous mixture of four ideal gases capable of undergoing the reaction is
dO e
we
/r.p
(18-11)
\de / 3 ',/-
(dQ/deh.p
get
(d^/de 2 )^
tion for the entropy
(18-12)
viAi
(dF/deV./.
3P/r
(a
2
G/ae
+
V2A2
"33^3
+
ViAi.
(18-13)
2 j
7..
How many
thermodynamic equilib. ium, d 2 G/de 2 is positive. Equation (18-12) therefore states that an increase of temperature at constant
(a) (b)
components are there if one starts with: Arbitrary amounts of A\ and At only. Arbitrary amounts of all four gases.
pressure always causes a reaction to proceed in the direction in which heat
(c)
ci
Since
G
is
a
minimum
at
is
.
temperature of the metal
moles of A\ and
vi
moles of
A-> only.
630
I
Icat
18-4
(CaCOj), a
and Thermodynamics
Consider a system composed of a solid phase of calcium carbonate solid phase of calcium oxide (CaO), and a gaseous phase consist-
C0 CaC0
ing of a mixture of
being present
2,
3
631
Heterogeneous Systems
three constituents
Consider a system consisting of a pure liquid phase in equilibrium gaseous phase composed of a mixture of the vapor of the liquid and with a inert gas that is insoluble in the liquid. Suppose that the inert gas (somean
amounts. These are the substances which
times called the foreign gas) can flow into or out of the gaseous phase, so that
CaO vapor,
vapor, and
initially in arbitrary
all
are present in a limekiln, where the reaction
18-8
the total pressure can be varied at will. How many components arc there, and what (a)
CaCO,
^ CaO + CO
the variance?
is
Assuming the gaseous phase to be a mixture of
(b)
ideal gases,
show
that
;
=
g"
RT(
+
\n p),
takes place. (a) (b)
How many
components are there, and what is the variance? Assuming the gaseous phase to be a mixture of ideal gases, show that
where g"
=
the molar Gibbs function of the liquid, and
Suppose a
(c)
pCnOpCO,
is
P
pressure from
j,
and/> refer to the
vapor.
to
more
little
P
+ dP,
foreign gas
added, thus increasing the
is
Show
at constant temperature.
that
pCaCO,
d
(c)
CaC0
If solid
components are
3 is
there,
introduced into an evacuated chamber,
and what
is
ammonium
Solid
hydrosulphide
trary
where v"
is
the molar volume of the liquid, which
NH HS ^ NH, + 4
(a)
How many
(b)
Assuming that the gaseous phase
there,
(d)
there
is
no foreign gas
to
and what
is
In fr<\
the variance?
is
a mixture of ideal
gases,
If solid
NH4HS
K.
when
components are
placed in an evacuated chamber,
is
how many
and what is the variance? components are there in a system composed of arbitrary amounts of water, sodium chloride, and barium chloride? 18-7 At high temperature the following reactions take place: 18-6
p,
there,
+
CO,
= 4=
;~^
CO + H
2
0.
(P
(b)
-
How many
components are there
is
the vapor pressure
when no
there
is
sufficient air
foreign gas
above the water
very small amounts of several solutes
G =
is
PB =
present.
4.57
make
to
is
Mono
+ Wi +
Ma«2
+
•
•
•
,
and
the subscript zero refers to the solvent,
if
H
2
0.
=
gh
+ RT\n x
k
.
we (a)
CO
Arbitrary amounts of C, 2, and H>. Arbitrary amounts of C, CO2, 2> CO, and
H
the
(Gibbs equation),
Po)
2CO
with (a)
P and
til
Mk
+ Hj
is
mm, show
that,
the total pressure
equal to 10 atm, p = 4.61 mm. 18-9 The Gibbs function G of a liquid phase consisting of a solvent and
where
start
total pressure
How many
C and CO2
where the show that
final state
In the case of water at 0°C, at which
(e)
/•XII.HS
(c)
a
show where Po
=
is
HgS.
that pKll,pll.K
practically constant.
is
Integrating at constant temperature from an initial state where
partial vapor pressure
components are
"dP = RT -£,
the variance?
(NH 4 HS) is mixed with arbiamounts of gaseous NH 3 and H 2 S, forming a three-constituent system of two phases, undergoing the reaction 18-5
v
how many
Using the relation
V=
(dG/dP)r, show that
632
I
Icat
Heterogeneous Systems
and Thermody
H=
Using the relation
(b)
G — T(dG/dT) r show
II
which means that there
(a)
.... AJ =
in water,
1
=
g"
+ RT In
-
v")
dP
in x at constant temperature,
= RT d \n
Assuming the vapor to behave
(c)
like
(1
initial state
x
= In
P
is
an
0,
^
P = P
=
,
In (1
-
*)
show
that
=
x,
+ |1 (Px - p Px
and regarding v" temperature from
P = Px
,
and derive
Justify neglecting the last
P ~
=
Po(l
term on the
-
the vapor pressure of the
For an infinitesimal change
-/" AT = -s" dT + R (b)
(e)
let
x stand for the mole
Substituting for
=
R
In (1
show h'"
T
in x at constant pressure,
In (1
—
-
x)
dT
+ RTdla
x) the value obtained
that (a) reduces to
h"
dT + RTdln
(1
-
Substituting for
Taking show
//,-,
ice.
x.
The
+ RT In
is
x).
(1
show
-
that
x).
from the equation
dissolved in water,
and the
equation of phase equilibrium
(1
R
change in x
+ R In In
(1
—
show that
-
h'
(1
is
- x),
-
at constant pressure,
dT
x)
x) the
(a)
into account that x
+ RT d In
(1
show that
- x).
value obtained from the equation
reduces to
dT = RTdln
«
1
(1
and
-
x).
calling
h"
that the depression of the freezing point
.«, = AT
A
Consider the system of Prob. 18-10, and
of phase equilibrium,
—
and show that
right,
(Raoult's law),
x)
dT = -s" dT
T
fraction of the sugar. (a)
(/;)
§"
infinitesimal
h"
Po 18-11
—.
mole-fraction of sugar in solution.
For an
of phase equilibrium,
fusion
P*
—
molar Gibbs function of pure ice, molar Gibbs function of pure water,
).
dilute solution.
or
(a)
= = =
-s'
ideal gas
to a final state x
the vapor pressure of the pure liquid, and
(d)
g' ,/'
-*).
as constant, integrate the preceding equation at constant
an
where
and x "
calling //"
RT"-
very small amount of sugar
equilibrium with pure
in
i =
solution.
1
is
-*),
(1
is the molar Gibbs function of water vapor, g" the molar Gibbs function of pure liquid water, and x the mole fraction of the sugar in
For an infinitesimal change
A
18-12 solution
is
where g'"
{v
and
1
and the
in equilibrium with
g'"
(b)
into account that x
is
pure water vapor. Show that the equation of phase equilibrium
is
Taking
Zn,./:,,
no heat of dilution. very small amount of sugar is dissolved
A
18-10 solution
=
«
h" the latent heat of vaporization l v , show that the elevation of the boiling point is (c)
that
,
633
RT-
—
:
X.
— is
ti
the heat of
APPENDIX A PHYSICAL CONSTANTS
Symbol
Constant Electronic charge Electronic rest mass
Speed of
light in
e
1.602
m
9.109
vacuum
One electron volt One atomic mass unit One liter atmosphere
Nv
-
3
Mb
M* mp R m A\A 0"
1 1
liter
9.273 X 10- 2I erg/Oc -24 5.051 X lO erg/Oe 10-" g 1.673 8.314 J/mole deg 5.670 X 10- f crg/s cm 2 1.602 X 10- 12 erg 1.660 X 10~ 24 g 1.013 X 10" J
X
•
'
eV
1
C
6
k
Aa
Stcfan-Boltzmann constant
10-' 9 10
h
Universal gas constant
X
X 10-' 8 g 2.998 X 10 cm/s 6.626 X 10-"erg-s 1.381 X 10-' erg/deg 6.023 X 10" mole-' 9.649 X 10 C/molc
c
Planck's constant
Bollzmann's constant Avogadro's number Faradav's constant Bohr magneton Nuclear magneton Proton rest mass
Rounded value
u
atm
•
deg 4
APPENDIX
B
Ricraann Zcta Functions
RIEMANN ZETA FUNCTIONS
To
obtain Eq.
(3),
expand
=
f{x)
(— W <
x*
x
<
it)
in a Fourier series:
=
oo
an
x*
Hence,
Setting
.v
-
7r
v 4
K IT
=
= k
—H
=—
Jo JO
'
5
* 4 cos B*«fc
/
= -(-1)-— -
Jo
-
8it
!
V)
(-1)" —
Lt_
n-
cos nx
=
1-
71
L=
=
n2
7T
0)
12
c* Substituting Eq.
(2)
and solving
for
w=
n
To
obtain Eqs.
(1)
and
(2)
6
rr l
it
I 1
(3)
90
expand ihc function
(2),
/(A')
=
.V
2
(-7T
<
X
<
r)
in a Fourier series: f
= ) =0
/(*)
a» cos nx,
71
ao
=
1 .v
I
2
dx
=
Jo
Hence,
x1
2 -
/
7T
Jo
2 .v
cos
=
gives Eq. (1),
;u-rf.v
= ( — 1)"
— 4
(-1)"
= and
j
n2
f-
4
3
Setting x
— 3
/•* r»
setting
2-» a-
=t
gives Eq. (2).
C. R. Wylie, Jr., Advanced Engineering Mathematics, 2d ed., pany, New York, 1960, Chap. 7. t
=
1
McGraw-Hill Book Com-
*
.
Eq.
"
(-')" cos nx.
>
«,
l
— gives
)
£, t
««> -
V
2
1
71
48
"
i- 48 y i f M, H
air 2
5
(-1)"-'
—
— n*
gives
tt"
y
(-I)"
n*
(3).
4
637
BIBLIOGRAPHY
TEMPERATURE Herzfeld, C. M.,
and
F. G. Brickwcddc:
Publishing Corporation,
New
"Temperature,"
vol. Ill, pt. I,
Reinhold
York, 1962.
CLASSICAL THERMODYNAMICS Buchdahl, H. A.: "Concepts of Classical Thermodynamics,"' Cambridge University Press,
New
York, 1966.
Callcn, H. B.:
Pippard, A.
"Thermodynamics," John Wiley & Sons, Inc., New York, 1960. Thermodynamics," Cambridge University Press, New
B.: "Classical
York, 1957,
STATISTICAL MECHANICS
Guggenheim, Inc.,
New
E. A.: "Boltzmann's Distribution
Law,"
Interscicnce Publishers,
York, 1963.
and D. L. Turcotte: "Statistical Thermodynamics," AddiCompany, son-W'esley Publishing Inc., Reading, Mass., 1963. MacDonald, D. K. C: "Introductory Statistical Mechanics for Physicists," .lohn Lee,
J.
F., F.
W.
Sears,
Wiley & Sons, Inc., New York, 1963. Ter Haar, D.: "Elements of Thermostatistics," Holt, Rinehart and Winston,
New
Inc.,
York, 1966.
THERMAL PHYSICS "Entropy and Low Temperature Physics," Hutchinson & Co. London, 1966. Fast, J. D.: "Entropy," McGraw-Hill Book Company, New York, 1962. Fay, J. A.: "Molecular Thermodynamics," Addison-Wcslcy Publishing Company,
Dugdale,
J. S.:
(Publishers), Ltd.,
Inc.,
Reading, Mass., 1965.
640
Heat and Thermodynamics
Landsberg,
P. T.:
"Thermodynamics," Interscicnce
Publishers, Inc.,
New York
1961.
Morse,
M.: "Thermal Physics," William A. Benjamin, Inc., New York. 1964. "Fundamentals of Statistical and 'ITiermal Physics." McGraw-Hill Book
P.
Company, New York, J.:
Law
"Third
1965. of Thermodynamics," Oxford University Press. Fair
Lawn
ENGINEERING THERMODYNAMICS
Reynolds,
Company,
W. C: "Thermodynamics," McGraw-Hill Book Company, New York
1965.
Zemansky, M. W., and H. C. van Ness: "Basic. Engineering Thermodynamics," McGraw-Hill Book Company, New York, 1 966.
CHEMICAL THERMODYNAMICS Caldin, E.
Chemical Thermodynamics," Oxford University Press
Fair
Denbigh, K.: "Principles of Chemical Equilibrium," Cambridge University
Press,
F.:
N.J., 1958.
York, 1955.
Guggenheim,
E. A.:
"Thermodynamics," Interscicnce
Publishers, Inc.,
New
York,
1957.
Kirkwood,
and
G.,
J.
Book Company,
I.
Oppenheim: "Chemical Thermodynamics," McGraw-Hill
New York,
1961.
Reid, C. E.: "Principles or Chemical Corporation, New York, 1960.
Thermodynamics," Rcinhold Publishing
Wall, F. T.: "Chemical Thermodynamics,"
W. H. Freeman and Company, San
Francisco, 1958.
LOW-TEMPERATURE PHYSICS Atkins,
K. R.: "Liquid Helium," Cambridge University Press, New York, 1959. J. G.: "Helium Three," Ohio State University Press, Columbus. Ohio,
1960.
Din,
&
Sons, Inc.,
New
York,
F.,
Lane, C. T.: "Superfluid Physics," McGraw-Hill Book Company, New York, 1962. Lvnton, E. A.: "Superconductivity," John Wiley & Sons, Inc., New York, 1962. McClintock, M.: "Cryogenics," Rcinhold Publishing Corporation, New York, 1964.
Xewhouse, V.
Schmidt, E.: "Thermodynamics," Dover Publications, Inc., New York, 1949. Van Wylen, G. L: "Thermodynamics," John Wiley & Sons, Inc., New York, 1959.
Daunt,
John Wiley
York, 1966.
Obert, E. P.; "Concepts of Thermodynamics," McGraw-Hill Book New York, 1960.
New
Physics,"
Mendelssohn, K.: "Cryophysics," Interscience Publishers, Inc., New York, 1960. Mendelssohn, K.: "Quest for Absolute Zero," McGraw-Hill Book Company, New
N.J., 1961.
Lawn,
C: "Low 'Temperature
Jackson, L.
1962.
Reif, F.:
Wilks,
641
Bibliography
and A. H. Cockett: "Low-temperature 'Techniques,"
lishers, Inc.,
New
Interscience
Pub-
York, 1960.
Garrett, C. G. B.: "Magnetic Cooling,"
Harvard University
Press,
Cambridge,
Mass., 1954,
Gopal, E. S. R.: "Specific Heats at New York, 1966.
Low
Temperatures," Plenum
Press,
Inc.,
Hoare, F. E., L. C. Jackson, and N. Kurti: "Experimental Cryophysics," Butterworth Scientific Publications, London, 1961.
L.:
&
Sons, Inc.,
New
Dover Publications,
Inc.,
"Applied Superconductivity," John Wiley
York, 1964. Scurlock, R. G.
New
:
"Low Temperature Behavior
of Solids,"
York, 1966.
Shoenbcrg, D.: "Superconductivity," Cambridge University Press,
New
York,
1952.
White, G.
K..:
"Experimental Techniques in Low-temperature Physics," Oxford
University Press, Fair
Lawn,
N.J., 1959.
ANSWERS TO SELECTED PROBLEMS
Chapter
1
/>'[/'
1-1
(a)
P(V-nb);
.
-
1
(b)
P(V-
nb)
P'V
= 1
1-2
w (A)
% M
nB'/V'
„
>
C
M'
pv\ „«
Curie's equation.
W = 0-0 —-r
PV
-
0+ C9P
— = —C
Woiss' equal ion.
1
- nliO, ideal-gas equation.
1-3
419.57°K.
1-4
(a)
4.00°K.
1-5
(a)
4.00°K.
Chapter 2
2-4
(a)
2-5
418 atm.
501 atm;
X
(b)
2-11
5.06
2-12
21.3 Hz; 33.8 Hz.
2-13
3.36
X
P» v
nB'/V
I0«dyn. 10 6 dyn.
48.8°C.
.
644
Heat and Thermodynamics
Answers
Chapter 3 3-3
929
-0.34 (A)
Chapter 6 J.
(a)
-179.1.
(A)
-1.13.1 at 300°K;
-180
-338 J at 1°K, -517 J at 1°K.
J at 300°K;
Chapter 4
6-6
A Kr
6-7
1
X 10-* a® + mW/cm cleg.
3.75
4-21
2.3
4-25
0.0046 deg.
4-26 4-27
7.16 g.
4-28 4-29 4-30
1360°K. 5750°K.
4-31
13 min.
6-8
2.41
6-9
k'
86 lamp at 77°K.
V
(7730°K); !0»
X
=
10 16
X
(7.73
10° °K); 10 c
V
(7.73
X
10' °K).
atoms/s.
A{w)/AV.
6-10
3.25
s.
6-11
1.85
X
10-
mmofllg.
4
1.50
X
"'
10
A0 2
Chapter 7
.
•
cm 2
1.38
7-5
.
(«)
362
J;
362
(A)
362
J;
0.
(c)
-12.2
J.
-12.2
J;
J;
97 percent.
Chapter 9
6,
same on both
5P /4. Vr on left =
sides
and equal
to 8
;
P, same on both sides and equal
(A)
8 V„/5;
Non-quasi-static,
5-2
76 cm.
5-3
125 cm.
5-4
1.7 g.
5-5
Three times the
Vj on right
damped
=
12 Ku/5.
8.33 J/deg;
9-6
(«)
0;
9-7
(a)
1340 J/deg; 1310 J/deg;
9-9
0.0300/?.
(A)
to
(c)
(a)
2.32.
fr)
Chapter 5
(A)
9-10
(a)
6.3 J/deg;
9-11
(a)
0;
9-12
277 J/deg.
-1120 -1210 (A)
X
1.83
(A)
9-13
-16.0
9-14
1.5
9-15
(a)
4Ko/9;
(/)
0;
(,g)
(A)
nCp
In (63/8)
5.80 J./deg;
(c)
()
5.80 J/deg.
190 J/deg. 90 J/deg.
J/deg;
J/deg;
1.39 J/deg; 10- 5 J/deg.
(f)
3.80 J/deg.
J./deg.
oscillation.
aim;
300°K;
0.0570 J/deg. 3 7'„/2;
(A)
«CP
to
In (21/4)
-
-
21 To/4;
(d)
(19/4)«C.r
;
to
,iR In (27/8);
2nfl In (27/8).
initial value.
ft 3 ;
1020
(b)
0.221
5-12
(A)
500
5-13
(A)
571°K.
5-14
(a)
10.2
5-17
to
-9.5dcg/km.
5-18
1.41.
5-19
(a)
1.18
5-20
(a)
0.966
5-21
1.27.
5-24 5-25
320 m/s.
5-26
V
1.23 V.
4-13
5-11
20.25°K.
J,
-0.198
to
5-1
645
J.
3-7
4-7
Problems
Diatomic.
5-27 5-28
3-6 3-11
to Selected
9-26
About 40U°K.
fclb.
ft.
fcjj
s; s;
Chapter 10 614°K;
(b)
(A) (A)
(f) (
3530°K;
()
107.5
k.I.
X
10-1
2
10-2
(A)
10".
~10";
to
~10";
(«0
je'/AT- 10 8
69 em. 47.1
cm.
1.27.
80 percent He; 20 percent Ne.
Chapter 11 11-6
(a)
-87.5
11-7
(a)
1.11 kJ;
11-8
-50,6 J; to -36.9; -3.96 U; (c ) 5.07 k.I'. -0.436 deg; +0.099 deg; +3.63 deg. J;
(A)
(A)
(d)
0.684 deg.
i«C o r
;
646
Heat and Thermodynamics Answers to Selected Problems
Chapter 12 12-3
(a)
12-4
37°K;
(A)
12-6
70 percent. 63.6 kJ/mole;
12-7
(a)
12-8
(a) /v-.-t
12-9
P =
69.5 kJ/mole;
132kJ/mole; 195°K; (A)
=
(A) l
su
20.6°K.
=
7.2.
600.8°K.
12-17
{c)
6.7 percent.
(A)
428 atm/deg;
12-19
(rf)
0.77 percent.
12-20
(c)
8.26
=
(a)
0.826 cm/s;
15-6
(a)
-0.00796.1;
5.80 kJ/raolc;
=19.?
deg;
l
=118
VA /R
dee.
21.9 m/s.
(b)
1-62 m/s.
(A)
(A)
1.10
0.0313
(A)
-0.0318.1;
0.0685
to
-0.0363
0.140.1.
J;
X
10~ 9 erg.
J.
J.
Chapter 16
Chapter 17
(a)
3.6 J;
13-4
(A)
-0.900, -0.336, +1.40, 4-1.37 deg/m.
13-7
-214 -188
-3.18
(A)
J;
6.78 J;
(c)
-1.21 deg.
(d)
kJ. kJ.
to
lOOmdcg. -6.67 mV. (b) -33.8 mV; -101 +0.152 mV/deg.
13-23
(a)
6000°K;
(b)
9.84 erg/cm';
13-24
(a)
160 erg;
(A)
120 erg;
0.25, 1.00, 25.0,
(c)
J; out.
3.28
0)
40 erg;
0:
(d)
dyn/cml
(rf)
Chapter 14
J=i = (a)
14-14
4.04°K.
s.
10.4 J;
14-15
(a)
-0.264
14-16
(a)
89
10.4 J;
(A)
J;
(A)
(<,-)
0.3°K.
kOe
(A)
4 kOe.
14-17
(a)
1.03°K
(A)
0.0031 5°K.
14-19
(a)
2.00°K
(A)
0.1 4°K.
Chapter 15 15-1
19 percent.
15-2
0.588 J/g
•
0.315 atm;
17-5
(a)
17-6
94,600 J/mole.
17-7
39 percent;
17-8
100 percent.
(A)
0.613;
/i/s.
13-1
14-13
1250,1.
-2.29°C, 306 atm.
to
Chapter 13
14-4
15-5
16-2
12-18
(a)
0.1,0.9,9; 223 m/s;
65.4 kJ/molc.
31.2 kJ/molc; /„•
m
12-11
13-12 13-14
(«)
1.21.
/R = 137 deg; /,„/* 0.0463 atm/deg.
12-16
13-8
(a)
15-4
25.4 kJ/mole.
l
(*)
(b)
T=
34.9 arm;
15-3
deg.
1.44°K.
431°K.
81 percent.
17-10
709 J/deg.
17-11
0.00129 deg-'.
(c)
61,200 J/mole.
647
INDEX
INDEX
Abbot, C. G., 109 Abraham, B. M., 510 Absolute entropy, 240 Absolute zero, 209, 503
Boltzmann's constant, 1 52 Boorsc, H. A., 543, 546 Borisov, N. S., 532
Born, M.,
Absorptivity, 98
Adiabatic demagnetization, 461 Adiabatic wall, 5 Affinity, 596
1
98
Bose-Einstein condensation, 510 Bose- Einstein statistics, 273
applied to photons, 422 Bots, G. J.
C,
Allen, J. F., 512
Brickwedde,
Amagat, Ambler,
Bridgman,
E. II., Ill
E., 481
Andres, K., 300, 328
Ash me ad,
I,.,
511, 527
459
J.,
Atkins, K. R., 381, 522, 525, 640
Atomic magnetism, 442 Avogadro's number, 147 Babb,
S.
F..,
Jr.,
M.
Baker,
D., 119
11.
I.,
371
Bardeen,
J., 547 Bcattic-Bridgeman equation, 34, 139 Blackbody, 98
Briilouin function, 452 Brillouin's equation, 50, 452, 507
Brownian motion, 146 Buchdahl, H. A., 639 Buckingham, M. J., 380 Burton, E., 464
Cailletet, L. P.,
Caloric, 71 Calorie, 86
Capacitor, 403
Boiling point elevation, 633
Careri, G., 529
I..,
102
503
Caldin, E. F., 640 Callen, 11. B., 412, 639
Blaekbody radiation, 98, 420, 431 Blaekman, M., 313, 317 Blaisse, B. S., 485 Bohr magneton, 443 Boltzmann,
563
270
373
Bagatskii,
639
P. VV., 34, 289, 412,
Brillouin, L.,
Andronikashvilli, E.
516
F. G., 344, 472,
Calorimetry, 72, 85
Caraihcodory, 197 Carbon thermometers, 472 Carnot,
S.,
197, 215
652
He at and Thermodyn Index
Carnot cycle, 197, 208, 222 Casimir, II. B. G., 468, 547 Catalyst, 565 Cazin, A., 117 Cell, 43, 399
Cerium magnesium
nitrate,
Degeneracy, 253 Degree of dissociation, 579 Degree of ionization, 579 Degree of reaction, 568, 577 dc Haas, W. J., 87, 461, 465, 468 458
de Klerk, D., 466, 476, 511, 518, 521 Denbigh, K,, 640
Chase, C. E., 381
Chemical equilibrium, 26 Chemical potential, 574 Christie, D. E., 285
Denenstein, A., 526 Depression of the freezing point, 633 Desormcs, C. B., 126, 215
Equilibrium, thermal, 4,
5,
27
thermodynamic, 26 Equilibrium constant, 587 Equipartition of energy, 161, 266
Garfunkcl,
Estermann, I., 160 Eutectic, 614
Garrett, C. G. B., 640
W.
Everitt, C.
F.,
Gcballe, T.
Exact differential, 38 condition
for,
282
Expansivity, thermal, 37, 297, 328 Extensive quantity, 46
Diathermic wall, 5 Dielectric, 403
Clausius theorem, 215, 245 Clement, J. R., 471
Fairbank, W., 380, 532, 554
Diesel cycle, 176
Fast, J. D., 639
Dietcrici equation, 139, 391 Dillingcr, J. R. 524
Feher,
Cockett, A. H., 640 Coefficient, of performance, 182
of volume expansion, 37, 297 Coherence length, 548 Collins helium liqucfier, 343, 503
Complexions, 255
Components, number Composite system, 66
of,
622
Compound
system, 65 Compressibility, 37, 302
J.,
136
Cryohydrate, 616 Cryohydric point, 616 Cunsolo, S., 529
Curie constant, 46 Curie's equation, 46, 455
Dal ton's law, 557 Daniell cell, 43 Daniels, J. M., 466 Daunt, J., 477, 523, 640 Davy, Sir Humphry, 79 Deavcr, B. S., 554 Debye, P., 314, 460 Debyc's heat capacity, 314
W., 612
J.
460, 461, 465, 476, 503
Gibbs free energy, 280 Gibbs function, 280, 570 magnetic, 450 Gibbs function change in a reaction, 597 Gibbs-Melmholtz equation, 329 Gibbs' paradox, 563
Fermi level, 323, 421 Fermi temperature, 324
Glatzcl,
Dissipative effects,
Feynman,
Goodman,
1
93
Doll, R., 554
Film, 42
Dugdale, J. S., 386, 639 Dulong, P. I.., 307, 314, 315 Durieux, M., 471
Fine, P.
J.
G.
M,
130
Elastocaloric effect, 434 Elcctrocaloric effect, 405
C,
J.,
296, 301
B. B.,
J.,
542 640
380, 481, 516, 547
Graham, G. M.,
law of thermodynamics, 79 for composite systems, 82
GrjzSnvold, F.
for simple systems, 81
Grilncisen
sound, 517
297, 301
Grayson-Smith, H., 464
C, 296 gamma, 334, 386
Guggenheim, Gusak, V.
E. A., 241, 253, 365, 639, 640
G,
371
basic, 20 in
low-temperature physics, 470
secondary, 20 standard,
Ford, R.
Energy
Forrez,
253
Fleming, P. J.,
G,
Haldanc,
1
524 532
J.,
Hall,
Entropy flow, 241 Entropy principle, 231 Entropy production, 242
Fowler, R.
Equation of phase equilibrium, 610 Equation of reaction equilibrium, 580 Equation of state, 28, 34
Frankel, R. B., 466, 476 Free-energy change in a reaction, 597 Free expansion, 115 Freezing mixture, 615
Equilibrium, 4 chemical, 26, 557 600, 626
11.,
Frandsen, M.,
J. B. S.,
E.,
183
532
calorimetric definition, 73
512 Fourier effect, 410 Fourth sound, 522 effect,
J. P.,
H.
Heat, 73, 80
517
Fountain
Franck,
Ilabgood, H. W., 369 Ilafner, E. M., 130
Enthalpy, 275 magnetic, 491 Entropy, 214
mechanical, 26
285
S. R.,
Gortcr, C.
412
First
Elevation of boiling point, 633 Energy gap, 547
of,
J. J.,
C, 332 G, 373
Fixed points, 11
Ehrenfest, P., 198, 395
displacement
Gladun,
Gopal, E.
creeping, 522
First
Einstein's heat capacity, 314
level,
Gilvarry,
Goldman,
R., 272, 529
First-order phase transition, 346
536
Gibbs,
F.,
Disorder, 239 Displacement of equilibrium, 600, 626
Edwards, D., 381 Effusion, 146
Cooper, L. N., 547 Corak, W. S., 326, 542 Creeping film, 522
W.
Gibbs' theorem, 559
Duthie,
Compressibility factor, 115
Cronin, D.
G,
472, 534
II.,
Giauque,
Din, P., 363, 640 Direction and entropy, 240
adiabatic, 295, 302
Critical field of a superconductor, Critical state, 29, 368
639
J. A.,
326
P.,
437 Fermi-Dirac statistics, 273, 324, 547 Fermi energy, 323
;
Clusius, K., 296, 301
Fay,
M.
Gas thermometer, 14 526
Dewar,
Clement, M., 126, 215
401
cell,
Fusion, 31, 372
Clapeyron, E., 215 Clapeyron's equation, 349, 502 Clark, A. L., 130
Sir James, 503
Fuel
653
of reaction, 590
thermodynamic definition, 78 Heat of fusion, 372 Heat of sublimation, 353
241
386 1
at absolute zero, 353, 360
17
Freezing point, depression Friedman, A. S., 11? Frohlich, H., 533
Heat of vaporization, 361
of,
633
law of corresponding
states for,
at normal boiling point, 366
Heat
capacity, 82
of conduction electrons, 325 Debye theory of, 314 definition, 83
Einstein theory of, 313
367
654
I
Icat
and Thermodynamics
Heat capacity, of gases, 122 measurement of, 84
Index
Inversion curve, of hydrogen, 339 of nitrogen, 337
of paramagnetic subsystem, 457
Irradiance, 99
of reacting gas mixture, 602
Irreversible process,
of superconductor, 542
Isenthalps, 336
Heat conduction,
Heat engine,
1
91
380
E.,
Kramers,
II.
G,
Kuridt's tube, 136 1
Kunzlcr,
92
Kurti,
of hydrogen, 339 of nitrogen, 337 Isentropic process, 219
Isotherm, 6 Isotope effect in superconductivity, 534
IT calorie, 87
Helmholtz, H., 79
461, 463, 465, 476, 484, 503, 641
Lagrange multiplier, 155, 257 Lambda transition, 371, 377 Landau, R., 509, 517 Lande, A., 198 I.andc splitting factor, 443 I.andsbcrg, P. T., 198, 640 Lane, T., 527, 641 Langcvin's equation, 507 Laplace, P. S., 134 Lasarew, 15. G, 523 Law of mass action, 586 Lee, J. F., 639
Helmholtz function, 279, 570 C, 386 Henry, W. E., 453 Hcrzfeld, C. M., 639 Heterogeneous systems, 607
Jackson, L.
Higher-order phase transitions, 377 Hill, R. W., 345, 388, 467
Jones, R. V., 300 Joule, J. P., 79, 115
Lenz,
Hilsch, P., 536
Joule coefficient, 115
Leverton,
Hilsch, R., 536
W.
Joule
Lewis,
VV.,
Hcltemes, E.
G.
S.,
G,
526, 641
James, B. W., 299 Jensen, H. G, 130 Johnston, H. L., 112 Jones, G. O., 512
116
Hoare, F. E., 641 Holborn, L., Ill Hudson, R. P., 466, 481, 511, 518, 521 Hull, G. W., 472 Hydrostatic system, 28
effect,
115
Joule-Kelvin coefficient, 337 Joule-Kelvin expansion, 182, 335 Joule-Thomson expansion, 182, 335
S.,
466
temperature,
P., 343,
111, 503, 522,
1
7,
209 of,
231
503
Kellers, C. F., 380 Kelvin temperature,
scale,
207
Keyes,
23
maximum
temperature population, 490 Inversion curve, 337 of helium, 340
Kinetic theory, 146 Kirby, R. K, 300, 360 Kirchhoff's equation for sublimation, 349,
of,
338, 340
353 Kirchhoff's law, 99
Kirk, A., 503
Kirkwood, Kittel,
J. G.,
640
G, 486
Knudsen
effect, 15,
Little,
W.
A., 554
nuclear, 479
Magnetization, 45, 446 Magnetocaloric effect, 461
Mass
action, 586
Matthias, B., 533, 534
Maxwell, E., 381 Maxwellian distribution, 157, 273 Maxwell's demon, 271 Maxwell's equations, 283 Mayer, J. R., 79 Mazzoidi, P., 529 Mechanical equilibrium, 26 Mechanical equivalent of heat, 86
Mechanical
stability,
Megh Nad
Saha, 594
572
Mehl, J. B., 527 Meincke, P. P. M., 297, 301 Meissner effect, 540 Meixner, J., 412
Membrane
equilibrium, 559 Mendelssohn, K., 523, 641
Mendoza,
E.,
215
Microstates, 255
R. L., 385 J. II., 467 Misencr, A. D., 513 Mole fraction, 558 Mills,
Milner,
Monomolecular film, 42 Morse, P. M., 447, 640 Mullcr, G.
B.,
536
S.
O., 317
Lundstrom, V., 317 Lynton, E. A., 641
Nabauer, M., 554 Neganov, B. S., 532 Negative Kelvin temperature, 487 Negciptcmp, 489 Nernst, W., 498
F, G, 117 Kikoin, A. K., 523
International Table (IT) caloric, 87 Inversion, 337
M. Yu., 532 Lindemann's equation, 376 Lindenfeld, P., 473
Liburg,
Lundquist,
Intensive quantity, 46
of a magnetic ion subsystem, 456 International practical temperature scale,
130
537
Magnetic temperature, 462 Magnetism, 45 atomic, 442
I.orcntz's law, 95
Internal energy, 76 1 1
F.,
Lounasmaa, O. V., 345, 380, 388
Kelvin's thermoelectric equations, 41 Kerr, E. C, 381
of a gas,
11.
Lombardo, G., 437 London, F., 510, 547, 553 London, H., 547
Kcesom, P. H., 296, 318 Keesom, W. H., Ill, 542
Increase of entropy, principle Inexact differential, 38 Integrating factor, 82
419
VV.,
Liquidus, 614
H,
Katz, L., 130
partition function, 263
D., 481
Liquefaction of gases, 338
Kapitza,
Ideal gas, 119
T
Linear expansivity, 41 Kaeser, R.
1
Lee,
Leupold, H. A., 543, 546
Kamerlingh-Onnes, 539 Ice point,
472
J. E.,
N„
G
lelmholtz free energy, 279
Him,
380, 465, 516
of helium, 340
67
Heat radiation, 101 Heat reservoir, 89 Heer, C. V., 477 Heinglen, F. A., 304 Helium, solid and liquid, 382 1
Kqjo,
655
438
McClintock, M., 641 MacHonald, D. K. G., 639 MacDougall, D. P., 465, 503 Macrostate, 255
Magnetic enthalpy, 491 Magnetic Gibbs function, 450 Magnetic intensity, 45 Magnetic ions, 445 Magnetic polarization, 446, 479 Magnetic quantum number, 443
Nernst-Simon statement of the third law, 489, 501, 508 Nernst's equation, 594 Ncsmeyanov, A. N., 360 Newhouse, V. L., 641 Newton, Sir Isaac, 134
Normal boiling point, 13 Normal freezing point, 20 Normal melting point, 13 Norma! sublimation point, 13
656
Heat and Thermodynamics
Nuclear demagnetization, 484 Nuclear magnetism, 479 Nuclear magneton, 479 Obert,
640
E. F.,
Olszewski, K., 503
O'Neal, H. R., 546 Onsager, I.., 243, 529 Onsagcr's reciprocal relation, 243 Oppenheim, 1., 640 Optical pyrometer, 425 Osborne, D. W., 510, 516, 527 Otto,
J.,
Inde
Porous plug experiment, 335 Pound, R. V., 490, 493 Powell, R. L., 95
Satterthwaitc, C. B., 326, 542
Stone, N.
Saturation curve, 30
Stretched wire, 40, 396
Pressure of radiation, 431
Schmidt,
Pressure-volume diagram, 29 Purcell, E. M., 490, 493
Schneider,
Pyroelcctric effect, 405
Schricffer, J. R.,
Partial entropy, 561 Partial pressure,
Partington,
557 184
J., Jr.,
Partition function, 258
magnetic, 450
Pcarlman, Pellam,
N,
J. R.,
Peltier effect,
318, 327 511, 518, 521, 522
410
Penetration depth, 551 Perlick, A., 296, 301
Perpetual motion,
1
79
Persistent electric current, 553 Persistent flow of He, 527
Peshkov, V. P., 518 Petit, A. T., 215, 307, 314, 315 Pfaffian differential form, 82, 203 Phase, 335, 566, 572 Phase equilibrium, 610
Phase
rule, 612,
622 N. E., 546 Phonons, 509 Photon gas, 420 Phillips,
Piezoelectric effect, 406
Pippard, A. B., 198, 548, 639 Planck's radiation equation, 425 Plumb, H. R, 137 Pohl, R. O., 437 Polarization, dielectric, 405 magnetic, 446, 479 Pollack, G. L., 374
Population inversion, 490
F.,
Surface energy, 397 Surface film, 42, 396 Surface pressure, 43
344
Surface tension, 42
Scars, F.
Sutton,
Radford, L. E., 466 Radiant emittance, 98
W., 117, 198, 639 Second law of thermodynamics, 177 Caratheodory statement, 198 Clausius statement, 184, 248 Kelvin statement, 178, 248 Kelvin-Planck statement, 178, 497
Radiation pressure, 431 Ramsey, N. F., 490, 493
Planck statement, 178 Second-order phase transition, 395
Taylor, R. D., 381
Rankine
Second sound, 516 Secbcck coefficient, 413 Scebcck effect, 409
critical,
Reaction equilibrium, 580 Refrigerator, 179
Seguin, 79
ideal gas, 17, 209
Seidel, G., 296
Kelvin, 207
Regelation, 394
Sclden, R. W., 524
magnetic, 462
Regenerator, 168 Reid, E., 640
Semipermeable membrane, 558 Shapiro, K. A., 522
maximum
Quinell, E.
1(
547
385
Superconducting solenoid. 464, 552 Superconducting transition metals, 539 Superconductivity, 532
Scurlock, R. G., 641
II.,
472
Ill
Paramagnetic solid, 45, 445 Parks, R. D., 554
Schottky's equation, 457
Scott, R. B.,
Quasi-static adiabatic process, 124 Quasi-static process, 52
640 G., 369
E., 71,
W.
466, 476
Sublimation, 31, 353
536
J.,
Schuch, A.
Otto cycle, 173 Overall coefficient of heat transfer, Overton, W. C, 303
Saur, E.
J.,
cycle, 171
Raoult's law, 632 Rayfield, G. W., 529
G
Rcif, F.,
529
Shepard,
Resistance thermometer, 21 Restricting equations, 621
Reversible adiabatic surface, 200 Reversible cell, 43, 399 Reversible process, 191 Reynolds, VV. C, 640
Riemann
zeta function, 315, 636
Rinkel, 130
Roberts,
I..
Robinson,
D., 472, 481
N. H., 466
Rodd, 271
C,
11.,
472
Shoenberg, D., 641 Simon, F. E., 373, 387, 461, 463, 465, 476, 498, 503 Simon and Glatzel equation, 373 Simpson, O. C., 160
G, 80
Soft superconductors, 536
Solar constant, 109 Solidus, 615
Rollin, B. V., 522
Rose, F.
437
Shirley, D. A., 466, 476
Slater, J.
F.
I.,
Sherman, R.
481
Rossini, F. D., 117 Rotons, 509 Riichhardt's method, 128 Rudnick, I., 522 Rumford, Count, 79
Sackur, O., 266 Saha's equation, 595 Santini, M., 529
Spencer, H. M., 123 Stark efTect, 447 Statistical
mechanics, 146, 251
Steam engine, 171 Steam point, 11 Stefan,
J.,
657
102
Stefan-Boltzmann constant, 102 Stcfan-Boltzmann law, 102, 428 Stern, O, 160
J., 549 Swcnson, C. A., 386, 387 Swim, R. T., 303 Sydoriak, S. G., 472
Temperature, 7 absolute zero, 19 Celsius, 19
29
inversion, 338, 340
negative Kelvin, 487
superconducting transition, 532 Temperature-entropy diagram, 218 for carbon dioxide, 221 for helium, 345 for
hydrogen, 344
Tencrz, E., 317
Tcr
Ilaar, D., 639
Tetrode,
II.,
266
Thermal conductivity, 91, 93 Thermal efficiency, 167 Thermal effusion, 438 Thermal equilibrium, 4, 5, 27 Thermal expansivity, 37, 297, 328 Thermal radiation, 97 Thermal stability, 583 Thermocouple, 22
Thermodynamic Thermodynamic Thermodynamic
equilibrium, 26 potential, 570
probability, 255 Thermoelectric phenomena, 409
Thermoelectric power, 409
Stirling engine, 168
Thermoelectric refrigeration, 41
Stirling refrigerator,
Thermometer, 9 carbon, 472
179 Stirling's approximation, 256 Stoichiometric coefficients, 577
CMN,
476
658
Heat and Thermodynamics
Thermometer, gas, 14 germanium, 473
Vinen,
W.
529
F.,
Virial coefficients,
1 1
resistance, 21
Volume
thermocouple, 22 vapor pressure, 471
von Minnigerode, G., 536 Voroncl', A. V., 371
Thermometric function, 9 Thermometric property, 9
Vortex rings in He, 528
Third law of thermodynamics, 497, 541 Third sound, 525
Thompson, Thompson,
Thomson
B., 79 K..,
Washburn,
field
of a superconductor, 537 Throttling process, 182, 276, 336 Tisza, L., 510, 517
Tolman, R. C, 412 Townsend, P., 549 Toxcn, A. M., 550 Triple point,
Wasscher,
W., 117
E.
D., 380, 516
J.
Water-gas reaction, 588 Weinstock, B., 510 Werntz, J. H., 524
Westrum,
E. F., 296
Wexler, A., 326 White, D., 112
1
White, G. K., 300, 328, 641 Wiedemann and Franz ratio, 94 Wien's law, 428 Wiersma, E. G, 461, 465, 468 Wilhelm, J., 464
data, 32
of water, 11, 31, 32 Triple point cell, 12
Trouton's rule, 368 Turcotte, D. L., 639
Turner, I.. A., 198 Two-fluid model of liquid helium, 510 Tyndall, J., 102
Type Type
Wall, F. T, 640 Waller, I., 317 Walters, G. K., 531
532
effect, 41
Threshold
expansivity, 37
superconductors, 539, 542, 552 II superconductors, 539, 550 I
Wilks,
J.,
640
Wire, 40
Woods, S. B., 130 Woolley, H. W., 344
Work,
51
adiabatic, 74
of a hydrostatic system, 53
Unattainability of absolute zero, 498 Unavailable energy, 236 Universal gas constant, 113
of a magnetic
Wroblewski,
Wu,
Van Van Van
der VVaals equation, 35, Dijk, II., 471
1
39
Itterbeck, A., 517
van Lacr,
P. H.,
542
Van Ness, H. C, 640 Van Wylcn, G. J., 640 van't HofT isobar, 591 Vapor pressure, 31
formula
for,
360
measurement
Vapor
of,
359
connection with entropy constant, 357 Vaporization, 29, 361 F, L.,
34
63
C.
S.,
S. F.,
503
481
Wylie, C. R.,
Jr.,
636
Yang, C. N., 481 Yates, B., 299
Young's modulus, 41
Zeeman
effect,
445
Zemansky, M. W., 640
pressure constant, 353
Verwiebe,
solid,
of a reversible cell, 61 of a wire and a surface film, 60
Zero-point energy, 1 9, 324, 385 Zero-point entropy, 501 Zeroth law of thermodynamics, 6 Zimmerman, W., 527 Zuslandsumme, 258
»-
Robert Boyle 1627-16'Jl
James
1'. Joule ISIS 1889
Benjamin Thompson Count 17931814
Rumford
Rudoij Claurius 1822-1888
Nicolas Lionard Sadi Carnol 1796-18)2
Guslav Robert 18241SS7
Kirchhoff
J.
WiUardCibbi
1839-190)
WaUheT Nernst 1
8 fit- 1 911
/^\\
William Thomson 1824-1907
Lord Kelvin
Clerk Maxwell 18311879
/•
^
////Ac Kamerlingh
Onnes
1853-1926
Constantin Cantheodory 1873-1950
Max
Planck
1858-1947
Albert Einstein 1879-1955
f
>
H O O <
i
3
556.
rt )
7
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72807
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