must
be
identical
with ;j,
G(,V,p,t)
and (iO) becomes
A3)
is equal to 'h for s singlecomponentsystem I or G for an ideal gas, sec B1) below. Gibbs free energy per particle,G'/A'. If more than one chemical species is present, A3) is replacedby a sum ov
Thus
the
chemical
poiemiaf
all species:
A4)
Gibbs FreeEnergy
The
becomes
identity
ihermodynamic
xda = dV
+

pdV
A5)
^e/INy,
and E) becomes
shall
We
the
develop
G ss: YJ^jfif
that
ts
reacting
\302\243njdNj
(!6)
+ Vdp.
adz
of chemical
the property equilibria by exploiting with respect to changes in the distributionof in a t, p. No new atoms comeinto the system
theory
a minimum at
molecules
~
\302\253
dG
constant
themselves reaction; the atoms that are presentredistribute snucics to another molecular species.
ti \302\273f
Omipurhtw
iiiutupfri
Let
F.
with
sttf
its
is Jitiacut
wluit
molecular
one
from
:ihuuJ
\\\\vi
rd.ttions
iwo
= p(iVrT,K)
{cFfdN)tty
A7)
and
*=
$G/dN)tiP
We found
lM
in F.18)
for an ideal
(IS)
fi{T,p).
gas
MN.r.F) Tlog(WKnQ) , so
that
ti(N,z,l')
is
not
and therefore
of N
independent
She iaiegra!of{S7). Thai is, f is not direciiy proportional lo N number of particles is increased. Instead,from F(t,V,N)^
But
she
Gibbs
free energy for
C(r,/\302\273,N)
the
\302\273 F
if ilie
system
we cannot is kept at
+
PK
gas
potential
in
the form
cotislanl volume
ihc
as
B0J
is
=
N/V
 l] +
Arr[!og(p/rny)
==
p/r.
Vr
PO
,
NrIog(;j/THQ)
of ihe ideal gas Saw
JV;i(r,f) .is
 I].
\302\273.
by use
F =
write
F.24),
Ni[\\og(NJVnQ) ideaJ
A9)
We readily
identify
in B1)
the
chemical
us
/((t,p)
= i
lo\302\243(pJxnQ)
,
B2)
9; Gibbs Free Energy
Chapter
Reactions
Chemical
and
G \302\253 in sit.V) in We see [hat N appears by reference to the result iVjih.pV unavoidably but not in /j(t,p) in B2). The chemical potential is the Gibbs free energy per particic, A9), but il isnol ihcHctnihoHz free energy Of course, we are free lo wrile p its cilher pet panicle.
A9) or
122},as is convenient.
IN
EQUILIBRIUM
We may write the
REACTIONS
equation of a chemicalreactionas v,A,
+
v2A2
+
= 0
\342\226\240  \342\226\240
+
v,A,
B3)
,
B4)
species Hj
in
the
reaction
equation.
species, and the Vj are the coefficients Here v is the Greek lettermi. For the
of the reaction
Clj = 2HC1 we lave
+ =
Ai
the chemical
Aj denote
the
where
 Clj;
A,
H,;
Aj
=
fj=l;
Vi=l;
HC1;
\302\273j=2.
B5)
of chemical
discussion
The
conditions
of constant pressureand
energy isa minimumwith
The
differential
of
respect
constant
presented
In
temperature.
in the
to changes
for
equilibrium
proportions
reactions
under
the Gibbs free of the reaclants.
G is
dG
Here//j
equilibria is usually

 Z fj^i
+
adz
Vi!p.
B6)
is the chemical potentialof species j, asdefined by \\is = {BGJdNj)sp. At = 0;then B6) reduces pressure dp = 0 and at constant temperaturefa
to
07)
rfC1/,/W,.
j
The changein
the
potentials of the zero.
Gibbs
free
re\302\243tctbnts\302\273 Xti
energy comlit^rtufii
in
a
reaction
G
depends
is &i cxtrcmum
on the chemical tjo uu musl miu
for
Equilibrium
The change coefficient
dN} in
the
the chemical
v,m
of species/
of molecules
number
equation
]>>/.;  0.
We
may
Ideal
Gases
is proportionalto the dN} in
Write
the form
\"
JNj dfif indicates
where in
how many
\302\253
,
rjd$
B8)
timesthe reactionB4)
takes
place.
The
change
dG
becomes
B7)
dG
In equilibriumdG
=
0, so
\302\273
dft
vjfi
I
B9)
that
C0)
This is the conditionfor
pressure
and
for Ideal
0 when
utilize
We
F.48)
nj
is the
concentration

which
is
the
depends internal
Uul ihcmull
when
p and
genera!
the constituents
equilibrium condition acts as an ideal gas.
potential of speciesj as
of species j

logc;) ,
C1)
and
\302\253QJZ/int)
C2)
,
on ttie temperalure but not on ttie concentration.HereZ/int) function, F.44). Then C0) can be rearrangedas partition 5\302\273gnj
*
of
rflogMj
cj a
the
of
form
useful
we assume that each lo write the chemical fij
where
constant
Gases
simple and
We obtain a Y^Vjfj
of matter at
temperature.*
Equilibrium
=
transformation
in a
equilibrium
is moregencrat; onceequilibrium
t arc specified..
=
\342\226\240
$>,!(*<:,
is reached,
,
ihe rcaclion
C3a)
does nol proceedfurihi:r,
Chapter 9: Gibbs
Free
and Chemical
Energy
Rcactia
C3b) The lefthand sidecan berewritten
as
C3c) side can
the righthand
and
be expressedas C3d)
Here
the equilibrium
called
K(t},
constant, is a function only of the
temperature.
Wiih{32)\\vehave
s
K{t)
free energy
internal
the
because
C4)
nil\302\253/'
is Ffim) =
TiogZj{int).
From
and
C3c,d)
we have
C4)
C5)
FkVj
law of mass action.The result
as the
known
concentrations of the reactanlsis in
of any one
the conceniraison
concentration of one or moreofthe the equilibrium
To calculate
a consistent on our
choice of the
without
happen l\\2
^
a conscious
2H, ilie simplest
is
It
effort
not on
the
of the dissociatedpanicles(here
2H)
of
the
ground
state
of the
A
change
equilibrium
it is
essential
to choose in
partition
function
The
depends
Zji'mi)
different
need
zeros
properly the energy or free
for the energy
to arrange ibis, but it docs not part. For a dissociation reaction such clioosethe zero of the internal energy
diilicuh our
procedure is to
of each compositeparticle(here
C4),
eigenstales.
to give
be related
reaction.
in ihe
difference
as
must
reactants
different
in the
energy of each reactant.We
internal
zero of the energy
a change
reaclants.
other
the value of each
because
consistency
alone.
temperature
force
constant K(x}in
of the
zero
the
way
here
of the
a function
reaciam will
product of the
the indicated
that
says
H2 a!
molecule} res!.
to coincide
we piacc
Accordingly,
composiie particle at
~\302\243fl,
with the
where
is \302\2433
energy the energy the
energy
Equilibrium for
required
in
to be
laken
is
and
positive.
Examplei Equilibrium
of
atomic
rcacuon Hj
for the
aciioil
mass
hydrogen imo
particle into its constituents
the composite
to dissociate
reaction
the
Ideal Gases
atomic hydrogen
and molecular hydrogen. = 2H ttz ~ 2H = 0
or
The siatcmcnl of ihe for
the
dissociaiion
law
of molecul
is
C6)
denotes ihc concenmuion of nlolecutar
Here [llj] atomic
of
li
h>drogen.
hydrogen,
and
fjt]
i
foltovsihai
l37) [h^tpo17^'
lhai is, the relume concentration proportional lo the square rooi of K is
equilibriumcon:.lani
given
logK
in
of the
terms
of
atomic
ihe
concent
at a given is inversely temperature hydrogen ration of molecular hydrogen. The equilib
by
=
log^tHj)
internal free energy of

2IognQ(H) 
H,. per
molecule.
Spin
F(U2)/i, factors are
C8)
absorbed
in
F(H,}.
is H,, the of energy is laken for an H atom at rest. The more lightly bound is Kt leading of Hj in ihe 10 a higher more negative is F{Ht), and ihe higher proportion al absolute zero. eV per molecule, mixture. The energy to dissociate Ht is 4476 be said lhai ihe dissociationof molecular into atomic hydrogen is an It may hydrogen of dissociaiion: The En associated wiih ihe decomposition example entropy gain entropy [he It is believed of Hj into two independent in loss particles compensates binding energy. reaction in not The that most the is H and of Ht: intergalactic hydrogen space present as in the direction of H by the low values of ilie concentration of Hj. equilibrium is thrown Here
[he zero
Hjdrogcn
is very dilute
Example:
pi! and ihe limitation of water.
in
intergalactic
space.
In
liquid
waicr
ihe ioni^lion
process
09}
H2O*~>H+ + OH\" eeds
to
o.ximalcly
txicnt. Al room tcinpcraluic she coiiccmraikm product
a slight by
the reaciion
cquilibfium
i
D0}
where ihc jonac ^nefcasco
aGGin&
oy
lo
as required
can
ions
decrease
of
z\\t\\
iictu
ionizaiion
be increased
\342\200\224
fOH
^=
\021
of H* ions is
wai^r
\302\273^ntJ ihe concentration of C3i 1 tons wtll decrease constant. Similarly, ihe concentration [H*][O!!'] * a hase lo ihe water, and ihe H concentration will adding
by
siaie oi wafer
The physical process
prolon
product
is more
ions ate
H*
suggests\342\200\224ihc
of H2O molecules.This
with groups*
per liicr. inpurcualcr\302\243H*l donor. The concentration
moles
in
lo aa as a
to llic
the
accordingly.
ihe
is said
mainiain
\"
of OH
arc given
concentrations
An acid
!0\021molr''.
Chemical Reactions
Free finersy and
Gibbs
9;
Chapter
protons,
affect ihe
not sigiltficantry
does
than ihe equation
complicated
not bare
associated
are
but
of
ihe reaction
letmsof
ihe
vulidily
equation. it
convenient
is ofien
lo express theacidily
oraikalinii) otasolulionin
pti,
defined as
pH s log10[H+].
The
pH
concentration in The
base ten of iht; hydrogen ion concentrais ihe negative of the togarilhm = liier The of water 7 because !CT7mair'. oCsolution. is per pH pure [H*] acidic sotuiions have pH near 0 oi even have an negative; apple may pH  3. lo is basic. has a of 73 it 75; slightly plasma pH
solution
ota moles
slrongesi blood
Human
Kttictic
Kxati\\ple2
nB,
action.
modelof/nms
AB. We suppose that t!ie concentrations
moieculc
nAB denote
is
AB
of A,
where the me constant C describes rale constant D describesthe reverse atoms A and B. in thermal equilibrium = Oand so that du.a:di
the
B combine to form a collision of A and B. Let ha, formed in a biaiomic of nAB is B, and AB respectively. The rale of change Suppose
formation
a function
of
temperature
derived earlier by Suppose
\"
The
by
o[AB
some
ptolon.
505A958).
catalytic
in
a collision
of A
result
is consistent
=
law
ihe
its component ace constant,
D3)
,
D/C
with ihe
B, and
with
process, the ihemiat decay of AB into the concermaiions of all consiiiutents
only. This
action
of mass
that we
thermodynamics.
AB is noi formed
dominant
one surrounding
standard
A and
atoms
that
nAB
formed
D!)
principally
by
the bimolecular
collision of
A
and
B, but
is
process such as
species present A [cvie* is given
is
most by
molecules surroundlikely it*1 4HjO, acomplc*of4 water P(oc_ Roy. Soc tLondon) A147. M. Eigcn and L. Dc Macycr,
,
\342\226\240 .
Here E is ihe
So long as ttie
catalyst which intermediate
to its original slate AE ts so short lived that
is returned
product
up as AE, ihc ratio iiaiiu/iUb in equilibrium direct above. process A + B<AB treated
is lied the
actually The
proceeds, ihc equilibrium must in equilibrium of the direct equality
of detailed
be
the
and
ai ihe
end
IdealCases
for
Equilibrium
seeond step. quantity of A
of the
no significant
us if AB were formed what route Hie reaction by same. The rates, however, may differ. inverse reaction rate's is culled the principle must
No
be
the same
in
rnaiier
balance.
Comment: Reactionrates. The law of mass action expresses ihc condition satisfied by ihe concentrations once a reaction aboui how fasi lias gone to equilibrium, li iclis us nothing AH as it proceeds, ihc reaction proceeds. A reaciion A + B = C may bat evolve energy before the reaction can occur A and B may have 10 negotiate a potential as in barrier, is called the activation energy. Only moleculeson the high Figure 9.2. The barrier heigh! will not be able to get end of their will be able to read; others distribution energy energy over the potential hill. A catalyst speeds up a reaction by offeritig an alternate reaction path a lower energy of activation, but it does not change the equilibrium with concentrations.
Schematic
AH measures ihe energy evolved in the reaction the equilibrium concentration ratio [A][B]/[C]. The barrier to be negotiated is the height of the potential energy the reaction it determines the rate at which the can proceed, and takes place.
Figure 9.2
The quantity
nnd determines activation before
reaction
coordinate
Chapter 9: Gibbt FreeEnergy
Reactions
Chemical
and
SUMMARY
1. The
Gibbs free energy
G3 is a minimum
thermal
in
2. (cG/3r)H,= a;
=
= W,,(r,ri
4
ofmass
law
for a
action
of
the
pV temperature and pressure.
V;
= p.
(SG/SN),.P
chemical reaction is that IK'
a function
iff +
at constant
equilibrium
tfG/cph.,
3. C(r,p,W) The

U
 ^w.
alone.
temperature
PROBLEMS
/. Thermal expansion near absolutezero,
(a)
the
Prove
three
Maxwell
rela
relations
,
(dV/di)P
=
{dts/dp),
,
(aVldN)p = +(ap/ap)^ , (Qj/cr)jV
two
omit
(lie help Coefficient
of
subscripts
of D5a) tjicrniiti
approaches
D5c)
appear similarly in D5b) and D5c).It is commonto that occur on both sides of theseequalities, Show wilh (b) the third law of thermodynamics tjiat the volume Coeffi
should
subscripts
those
~Ea/dN)x.
D5b)
D5a) should be written
Strictly speaking,
and
=
D5a)
and
expansion
zero as t
* 0.
2.
Thermal
ionization
in
hydrogen
c +
wheree isad
H*ftH,
electron on a proton H*. (a) of the reactants satisfy the relation of an
the
that
Show
of atomic
the; formation
Consider
of hydrogen.
the reaction
the
as
eicutron,
concentrations
equilibrium I
D7)
[e][H+J/[HJs\302\253acxp(//T),
/ is
where
(he energyrequited to tontze
refers to the electron.Negiecithe
not aSTed
and
electrons
concentration
atomic
of
spins
and
the
~
{im/2zh2K11 this assumption does
hydrogen,
particles;
tiQ
is known as the Saha equation.Ifall the arise from the ionization of hydrogen the then protons atoms, of protons is equal to that and the of the electron Electrons,
concentration is
The result
result.
final
the
by
given
]
D8)
0]~[H]\"V'3\302\253p(//iT).
problem arises in
A similar
thermal ionization
semiconductor
of impurity
The
is
electronic
excited
[H(exc)] andT 
with
[e]
/ is
a simple
the ionization
proportional
energy, the square root of the
to !
electrons to the system, then (He
state, conditions
for
ts not
this
that
shows
which
concentration.
atom
(b) Let [H(exc)]denotethe first
/,
Here
problem.
concentration
elccjron
If we add excess will decrease.
C)
^1 and not
involves
\"Boltzmannfactor\" hydrogen
the
I
exponent
B) The
ujth
connection
of electrons.
are donors
that
atoms
in
physics
No lice lhat: A)
hy
adsorption
which atthe
of
concentration
equilibrium
the is \\l above ground surface of the Sun,with
5000K.
of protons
concentration
H atoms in the state. Compare
[H]
=
1023 cm~
3
 I
3,
of donor
hnization
impurities in semiconductors.
\\ A
pentavalent
impurity
atom in crystalline silicon introduced in place ofa tetravalent the role of in free space, but with silicon acts likea hydrogen atom e2/e playing e2 and an effective mass m* playing the role of the electron mass m in the state of the of *hc kmizalion energy and radius o( the ground description the diclctltic consta.t\\t free silicon electron. For impurity atom, and alsofor the e ~ 11.7 and, 0pproxima!e!y, m* =0.3 m. U there are 10\" donors p^r cmJ, K. 100 electronsat cent ration of conduction estimate the con
(called
4.
a donor)
monomer
polymers
4
the
Consider
Siopotymergrowth.
of linear
A'mer
made
= (A'
up
of
chemical
identical
f l)mer.
of a sotution basic reaction step is
equilibrium
units. The
Let K^ denote liieequilibrium
constant
for
Chapter
9: Gibbs free
this reaction,
Reactions
Chemical
and
Energy
(a) Show from the law
mass
of
\342\200\242 [\342\226\240 \342\226\240]
lhat the concentrations
action
satisfy
+
[N
from the
(b) Show
1]
=
[If \"/K.KjKj
Iheory of reactionsthai
\342\200\242 \342\200\242 \342\226\240
K.v
ideal
for
gas
D9)
conditions
(an ideal
solution):
where iVmer
MN is
,
^Bnti2/M,^rm
wQ(W)
the mass of the Nmer molecule,and
molecule,
(c)
Assume
ratio [N
concentration
4
N t]/[N]
\302\273 j,
at
so
thai
room
FN
E1) is the
nQ{N) =
Hq{N
free energy +
if there
temperature
1),
of one
Find
is zero
the free
basic reaction siep: that is, if AF = FKi.l ~ Fs ~ fj =0. = in a bacteria! Assume as for ammo acid molecules ceil. The 10aocm~3, [I] molecular weight of the monomer is 200.(d) Show that for the reaction to go in the direction of long molecules we need AF < This ~0.4cV, approximately. condition is not satisfied in Nature, but an ingenious is followed that pathway simulates the condition. An elementary is given by C. KiUel, Am. discussion J. energy
Phys.
in the
change
40,60A972).
for the expression equilibrium, (a) Find a quantitative = = n+ n~ in the particleantiparticle concentration n reactionA+ 4 A\" = 0. The reactants may be electronsand positrons; protons Let the mass of and antiprotons; or electrons and hoies in a semiconductor. the either particle be M; neglect of the particles. The minimum energy spins A\" is ATake the zero of the energy scaleas the release when A* combines with with no particles energy present, (b) Estimate n in cm\023 for an electron (or a 300 that K. with a A such A/t = 20. The hole is hoie) in a semiconductor T \302\253* viewedas the antiparticic to the electron. Assume that the electronconcentration
5. Pavtkteantipartkk thermal equilibrium
is equai to
the
hoie
concentration;
assume
aiso
titat
the
particles
are in
of (a) to let each particle have a spin of classical regime, (c) CorrectIheresult Particles that have amiparticfes are usually fermions with spins of \\.
the 3.
10
Chapter
Transformations
Phase
PRESSURE
VAPOR
276
EQUATION
the Coexistence Curve,p Versus
Derivation of Point Triple
278
t
284
Latent Heat and Enthalpy Model
Example:
System
WAALS
DER
VAN
284
for GasSolid
285
Equilibrium
287
OF STATE
EQUATION
Mean Field Method
CriticalPoints
288
the
for
Gibbs Free Energy
van
der
the
van
of
2S9
Gas der Waals Gas Waals
291
Nucieation
Fe
from
agnet
295
298
OF PHASE TRANSITIONS
THEORY
LANDAU
302
Ferromagnets
Example:
First
294 ism
Order
302
Transitions
305
PROBLEMS
1.
2. Calculationof 3. Heat 4.
and
Energy,
Entropy,
for
dTjdp
GasSoltd
7. Simplified
Note: tn
305 305
Water
Order
the
305
305
Equilibrium
305
Equilibrium
of 6. Thermodynamics First
der Waals Gas
of Vaporization of Ice
5. GasSolid
8.
of van
Enthalpy
the
Superconducting
306
Transition
Model of the SuperconductingTransition Crystal
first section
307
307
Transformation
s denotes c/iV,
the
entropy
per atom.
In
the
section
on fcr
Chapter 10: Phaie Transformations
PRESSURE
VAPOR
of pressure versus
The curve

EQUATION
by
together
appropriate conditions can of a system is a portion phase
and under
or solid
a liquid
in
energy
another
one
with
interact
of matter at constant substance. The curve is a real gas in which the atoms
of the
isoihcrmsof
the
We
or molecules associate
free
the
a quantity
for
volume
temperature is determined calledan isotherm. consider
phase,
A
that is uniform in composition.
Two phases real
ofa in
atoms
the
in
and
solid
we say
As
There
are isotherms at
in
isothermsfor
and gas
soiid
which
holds also
equilibrium
liquidgas
low icmperaturcsfor for
the
which
coexist. Everything equilibrium
solidgas
solidliquid equilibrium. and
on a
coexist
may
vapor*
section of an isotherm only
the the
phase\342\200\224exists, only a single phase\342\200\224the the pressure. There is no more reason to cali this phase a gas than we avoid the issueand callit a fluid. Values of the critical temperature
great so
no
fluid
temperaiure
a liquid,
if
isotherm lies below a critical temperaturerc. Above
of the
temperature
how
the
other.
coexist and
the
Liquid critica!
phase.
gas
liquid for
and the
each
with
boundary between them. An isotherm pV plane in which liquid and gas coexist 10.1, part of the volume contains Figure
a definite
show a region in
gas may
equilibrium
with
coexist,
may
for severalgasesare Liquid and gas
iO.t.
in Table
given
matter
extent of an isotherm a to from zero pressure infinite along they coexist at most only pressure; of atoms, fixed number section of the isotherm. For a fixed and temperature the there will be a volume above which all atoms present are in gas phase. at room bell an sealed A small drop ofwater placedin evacuated temperature jar will A
water
of
entirely. the
evaporate the
from
atoms
relations
are
by
suggested
for
the
the entire
along
the bell jar
filled
with
gas at
H2O
some pressure.
may already saturated with moisture There is a concentration of water, however,above which into a liquid drop. The volume vapor will bind themselves
exposed
The thermodynamic
conditions
coexist
never
leaving
entirely,
evaporate drop
will
to air not
Figure
JO.i.
conditions
equilibrium
for
of two
the
coexistence
of two
systems that are
in
phases are the
thermal,
diffusive,
Vapor Pressure
IO.I
Figure
is constant, but Efiei^
is only
and its
the s^nclc
vapor are in
of a
isotherm
Pressurevolume
such that liquid temperature is, t < tc. in the twophase
and
Gas
+ gas
Liquid
Liquid
gas phases
region of
Hquid
reyj
gas
at a
may coexist,that 4 gas the pressure
may change. At a given lempcrature a Jjtjuiu v*iluc of tltc ptcssurc for Vvh^cri at we move the Jf this pressure equilibrium.
volume
some of the gas is condensed to liquid, but remains unchanged as Jong as any gas remains.
down,
piston
pressure
Table 10.1 Critical temperatures T,.
of
gases Tt, in
K
in
He
5.2
H2
Nc
414
N.
Ar
151
210
Oj
Kr
Xc
289
CO,
K
33.2
126.0
1543 647.1
7
the
.W.2
Equatio
10: Phase
Chapter
Transformations
and mechanical contact. Thesecondilions are for
or,
=
18\\
Ml
=
the
phases
pressure
temperature
=
the general point in the px plane alone is stable, and if liquid phase
At a
Metastablephases
may
may
phase
have
a lower
by
occur,
be the pressure for
which that
+
dp;iQ
the
divides
It is a
pl
=
and
liquid
.
A) phases.
gas
in the
species
Note
that
the
two phases must be
B)
/ijl
two ns
coexist: If /i, < ng the gas phase alone is stable.
phases <
Pi
do not
or superheating.
supercooling
A
metastable a
chemical potential.
temperaturez0.Suppose Pa
p2;
a transient existence, sometimesbrief, sometimes long,at another and ntore stable phaseof the same subslance
Derivationof the CoexistenceCurve, Let pQ
=
for which
temperature
has
fit
potentials are evaluatedat the common of the liquid and gas, so that
/l,(p,T)
the
t2:
The chemical
coexist.
common
and
Pi = Pg
/Jj,;
where the subscripts / and g denote the chemical potentials of the same chemical if
=
tj
and gas,
liquid
*j
equal
that
+ rfi.Thecurve
p, t plane into a
the
p
two two
in thep.T
Versus
phases, phases
r
liquid and also coexist
gas, coexist at the at the nearby point
plane along which the two
phase diagram, as
given
in
Figure
phasescocxist
10.2
for H2O.
condition of coexistencethat C)
D)
dt).
We
relationship betweendp a series expansion of each sideof D) to
C) and
Equations
make
D) give a
and
dx.
obtain
.
E)
ion
Cot
ojthc
Figure
10.2
relationships
the
jiB in
and
Phase diagram of H.O. The chemical poienmls /t,. ;i,. solid, liquid, and gas phases ate of the
shown.The phase
here
boundary
bciwcen
ice
and water is not cxacily vertical; the slope is actually negative,although large. After very Iniemalionat Critical Tobies.Vol. 3. and P. \\V Proc. Am. Acad. Sci.47, 4i A912) Qridgman, forms of ice. see Zemamky, for the several p.
100
100
0
200
Temperature,in In
the
limit
as dp
300
375.
400
\302\260C
and dz approach zero, F)
by
C)
which
and
E). This
result may
is the differential
be rearrangedto give
curve equation of the coexistence
or
vapor
pressure
curve.
The derivatives of the chemicalpotentialwhich
in terms ofquantities
accessible
to
measurement.
occur
In the
may be
expressed treatment of the Gibbs
in G)
Chapter 10: Phase Transformations
free
With
9 we
in Chapter
energy
the definitions s
v
the volume
for
relations
the
found
per moleculein
and entropy
1 (cG\\
V
5 =
V/N,
o/W
(9)
each
we have
phase,
(dp\\
)JJ Then
for
G)
becomes
dp/dt
01) ~
Here
sa
st is
moleculefrom
of entropy
the increase the
gas, and
the
to
liquid
the
of
ihc
is
t\\
vg
we transfer one increase of volume
when
system \342\200\224
from the liquid to the gas. It is essential to understand thai the derivative dp/dz in (I!) ts not simply taken from the of state of the gas. The derivative refers to the very equation of i in which the and and special interdependent change p gas liquid continue
of the system
we
when
to coexist.The number varied, subject only The be
of to
of molecules
numbers
quantity
added
~
sa
to the
one molecule
transfer
the
s, is
when
decrease
added
in the
the
system
as
vary
Here
constant.
a
moiecule
is
volume
the
Ns and
outside
in the
is transferred
reYersibly
the
of
temperature
from
molecule
one
transfer
(Va
the
are
system
from
constant.
process, the
must
that
the liquid
(If heat is
temperature
to the gas.)The quantity
of
will
heat
transfer is l1Q ~
by virtue
will
gas phases, respectively. liquid related directly to She quantity of heat
sysfern to
the
A\\
phase
and
to the gas, while keepingthe
not addedto
=
+ Na
N, in
in each
molecules
\" TE*
of the connectionbetween
5'}'
and
heat
A2) the
change
of entropy in a
reversible process.The quantity L
=
tEs

A3)
of the
Dcritaiion
defines She latent heat of
Coexistence
vaporization, and is easily
measured
Cur
by
elementary
calonmetry.
We let A*\302\253
the change
denote
to
she
We
gas.
of volumewhen
combine
A1), U3),
A4}
igv,
is transferred
molecule
one
from the
liquid
and A4) to obtain
A5)
ClausiusClapcyron equationor the vapor pressure equation of this equation was a remarkableearly The derivation of accomplishment Bo!li sides of and are determined thermodynamics. (!5) easily experimentally, the equation has been verified to high precision. We obtain a particularly useful form of A5} if we make two approximations: that volume the (a) We assume by an atom in the gas I'j: occupied vg \302\273 is known
This
phaseis replace
as the
much
very
larger
in the
than
liquid (or solid)phase,so that
pressure
atmospheric
(b)
so
that
may
Av by vg:
A6)
may
the ideal written as
that
assume
We
i'9/i;( ^
be
!03, and the
approximation is very
gas law pVg
=
Av
With these
{16)
vt/yr
Avsva= At
we
S
Ngz
applies
to the
good.
gas phase, A7)
zip.
approximations the vapor pressureequationbecomes
of temperature, molecule. Given L as a function curve. this equation may be integrated to find the coexistence the the heat L is independent of temperature over latent if, in addition, Thus the of interest, we may take L = l.,Q outside integral. range temperature when we integrate (IS) we obtain where
L
is the
latent
heat per
r /dp
(lv
A9)
10: Phase
Chapter
=
logp
Traasfortnat
\342\200\224Z0/i+
where p0 is a constant.We
one
where
LQ as
defined
to one
instead
refers
If Lo
nioiecuie.
) =
constant;
,
p0CXp(L0/T)
the latent
heat of vaporization of
mole, then
A'cta, where No is t!ie Avogadro constant. For water the latent heat at the iiqufdgas transition is 2485J g~' at O'Cand 2260 J g~' at lOO'C.a substantial variation with tcmpcraiure. as !ogp The vapor pressure of water and of ice is plottedtn Figure !0.3 R
is the
versus 1/T.
gas constant,
R =
The curve is linear over
substantial
Crit
ca! p
with
consistent
regions,
the
Jin.
\\ id
wa
er
X 103 \342\200\224 1 atm
pressure
Vapor
ofwaier
and of
The vertical scale is dashed line b a straiglif
\\
\342\226\240
102
\"
is 1/T. The
ice
iine.
S
I
\\
10
Si*\"
vc
\\ s \342\200\242Ice
V
1
\\
1.5
2.0
2.5
3.0 3.5 4.0 lO'/T,
in
K~\302\273
4.5
5.0
2
Tcmperamre, Figure
4
3 in
K
Vapor pressure versus iemperaiurcfor 4He. After H. van Dijk Research offheNauonal Bureau of Standards 63A, \\2
10.4
eial.. Journal of A959)
104, vapor pressure of 4He, plottedin Figure of temperatures between I and 5 K. The phasediagram of 4He at low temperatures was shown in Figure 7.14. Notice tUat the liquidsoHd eoexistence curve is closely horizontal below 1.4K. We infer from this and (I!) [hat the entropy of the liquid is very nearly equal to the entropy of the solidin this region. It is remarkable that the entropies a normal should be so similar, because liquid is much more disordered than a of a normal liquid is considerably higherthan that solid, so that the entropy
result
approxmiate
is widely
used
in
the
B0). The
measurement
10: Phase
Chapter
Transformations
of a normal solid. But
3He,
the
slope
is a
4He
the
of
liquid. For another quantum liquid, curve is negative at low temperatures
quanhim
liquidsolid
entropy of the liquid is lessthan the entropy solid has more accessible statesthan the liquid! Liquid 3He a Fermi gas, has a relatively low entropy for a liquid it approximates because has a low enSropy when t \302\253 which a large proportion generally zF because of the atoms have Sheir momenta ordered the Fermi into sphere of Chapter 7.
(Figure
region the
in this
and
7.15),
of the
solid. The
Triple
point.
The
poin!
triple
of
t
is that
a substance
point p,, t,
in
She
p~z
all three phases, vapor, liquid,and solid, are in equilibrium. plane = = solid Here ng /i( /js. Consider an equilibrium mixture ofliquidand phases in a volume enclosed that somewhat larger than occupied by the mixture alone. The remaining volume will contain in the vapor, equilibrium wish only bo!h condensed phases,and at a pressure equal to the common equilibrium at which
The
of
of
pressure
vapor
both
temperature
point
iriple
at
substance
She
This pressure is the triplepointpressure. is not identical wish the melting temperature
phases.
somewhaton pressure; triple under common equilibrium the
the
For
the
water
defined
that
such
Latent
the
pressure of the two condensedphases. is 0.01 K above the atmospheric temperature = 273.16K. The Kelvin scale is O.Oi\302\260C = T, vapor
point, of
triple
water is exactly273,16K; seeAppendix
ofa
Tile latent heat
and enthalpy.
heat
phase to she gas phase,is equal
the liquid
of the
temperature;
melting
pressure
point
triple
two
phases
difference
of
enthalpy.
The
H
s
at
constant
V
f
pV
= dU f
=
On the coexistencecurve jig
L Values
of//
at coiisnint
are
\302\253 tAa
tabulated;
dV
f pdV
~
//,. Thus
\302\273 At/
+

phases, where H is
=
Hr
1
difference
When
called
the
we cross the
applies:

ihey are found
\342\200\224
Vtlp.
\302\273 Ml
=
B2)
,
jt,)dN
at constant
pressure:
T
+
pdV
(}tg
pAV
from
latent heat is also equal to the
coexistence curve,the thermodynamic identity Tito
the entropy
t times
B.
as
transformation,
phase
to
The pressure. the two between is dH
differential
is the
lemperaSure
point
temperatures depend meiSing temperature
Melting
pressure.
atmospheric
pressure Ha
 llt.
by integrationof the
B3} heat
capacity
of the
Derivation
Coexistence
Curve, p
Ver
B5)
jc,,
Example: Model system for a solid in equilibrium
gassotidequilibrium.
We
construct
a simple model can derive ihe easily a to apply liquid.
io de
as in Figure 105. vapor Roughly the same model would of N atoms, each bound as a harmonic oscillator of freImagine the solid to consist u to 3 fixed ol force. The binding center oleach atom in ihe ground siaicis frequency energy that is, ihe energy of an atom in its ground to a free atom at rest. \302\243fl; state is \342\200\224 co referred \342\200\224where The energy states of a single oscillator arc ntioi h a or zero is r.o, positive integer in o:.e we suppose itial each ntom can oscillaiconly (Figure 10.6). For the sake of simplicity dimension. The result for oscillators in [hrce dimensions is left as a problem. ol a single oscillaior in llie solid is The Junction pariiiion
describe
pressurecurve
Z, =
for
ihis
with
a gas,
We
model.
\302\243\302\253p[(n/10>
= expOWt)
eo)/r]
= \302\243\302\253p(,,Aa./T)
B6) The Ziee energy
F, is
F* = u* ~ The Gibbs free
energy
in the
Gs
Tff,

B7)
tlogZ,.
solid is, per atom,
\302\253
Vt

to, +
pt',
=>
Fs
+
pvx =
//,.
Figure
10.S
wiih
aionis
pressureis a energy ol the in
than
aioms
in a
Aloms
in the
in
cqmSibn
gas phase. The equilibri ol temperature.
function
The
solid phase is
in [he
atoms
ihe
solid
phase, bm the cmiopy of
gas lo be lends
in
higher
the
at may
temperature
high
be
in ihe
iIk
atoms
all ot
gas.
\\
gas phase. . The
equilibrium configmaiion is dfiietiiiiacd of the iwo ciTccls.A! low cotmtcrpkiy
lanpemSurc most of
lo
;src in ll
most of
ihe a
y
ihe
10: Phase Transformation
Chapter
(u. The lor of frequency ssumed 10 be % below that
in!he
o(a
gas phase.
Ground Male < aloui
bound
The pressure
volume
i>i
per
in
in ihe
gas phase: c, \302\253 vr
that
o[
solid phase is much
[he gas with than
smaller
but Uie which il is in contact, atom in [he the volume vt per
\342\226\240
neglect the term absolute activity is
If we
the
is equal to
solid
the
atom
pv, we have
(or
the chemical
potential of
,/t) =
[hesolidp, S
We
to be
' with
equilibrium
p
K
we
insert
nQ from
V
=
gas
phase,
and we
take Ihe spin of
C0) \302\253q
The gas is in
B9)
gas approximation to describethe zero. Then, [rom Chapter 6,
the ideal
make
atom
whence \342\200\236
exp(logZJ
the
f
the solid
t\\
'\"a
when
inQexp(E0/t)[l
).f
=
or ).\342\200\236
 exp(/1<0/t)].
C1)
C.63J:
C2)
=
(j^j
The
mode!
simplest
interactions
the
he was
below,
liquidgas phase transition is that ofvander Waals,\\vho gas equation pV = Nx to take into accounlapproximately between we atoms or molecules. By the argument that of a
ideal
the
modified
led to a modified equation of stateof the (p
van
the
as
known
 Nb) = Nx ,
+ N2a/V2)(V
der Waals
equation of state. Tliis is written
molecules, and the constant /' is a measureof their of (Figure 10.7). We shall derive C3) \\vitli..tlie help We shall then trcal ~(SF/DV)liN. in order to exhibitthe liquidgas For an ideal gas we have, from
The hard core the
b is
had
gas
at
repulsion
To
we
now
add a
volume
NT{log[)fQ(V
correction
a is
between
two
repulsion
range
= properlics of the model relation p
general
V,
be treated
can
distances but
the
the volume per molecule. We replace ~ of instead Nb). Thus, C4),we N/{V
this
in
constant
for
the
C4
+ i].
NT[log(na/n)
in C4) by
=
atoms
F.24),
free
approximately as
volume
V
~
the concentration
therefore
F
short
tlic
the ihcrmodynamic
short
not the
available
N
for
transition.
=
F(idealgas)
if
C33
the a, b are interactionconstantsto be defined; of the long range attractivepart the of mieraeiion
a measure
give
form
V. The
volume
of State
Equation
OF STATE
DER WAALS EQUATION
VAN
Waats
Der
Van
when
Sb, n
=
N/V
have
 Nb)/N] + intermolecular
C5}
1}. attractive
forces.
Figure 10.7
The
iiucraaion
energy
between
The repulsion plus a long range aiiracHou. short range repulsion can be described that each molecule by saying approximately
has a hard, impeneirable
coie.
10: Phase Transformation*
Chapter
Mean Field
Method
There exists
a
taking
the effect
of weak long range
particles system.The most gases and to ferromagnets.Let of two atoms separated a distance
of a
known
widely tp(r)
the average value of at r = Ois
is n,
gas
the
atom
interactions
potential
for
the
among
method are to energy of interaction
the concentration
r. When
method,
of the
applications the
denote
by
the
the mean field
called
method,
approximate
simple
account
inio
of atoms in
ihe total interactionof all other
on
atoms
C6)
where 2a denotes
value
the
of
ihe
We exclude the
convention.
useful
integral \\dVip{r). The factor of two hard core sphereof voiume b from
is
of integraiion. In writing C6) we assume that the concentration ihe volume accessible to the moleculesof the gas. That ihroughout we use the mean value of n. Tins assumption essence is the of the mean
volume constant
approximation. of concentration
in
we
language
energy
say
the
\"
molecules.
interacting From
regions
thaS
it follows
C6)
of a
that
The
factor
exact
s MJ =
is
Heimholu
free
of a
energy
F(vilW) =
The
pressure
*=
obiain
the
the
free
C7)
N2a/V.
it arranges
problems;
only once in
N(N  I), which
We add C7) to C5) to
change the energy and
\\BNna)
is counted
molecules
of bonds
number
is,
V by
volume
j is common to selfenergy
\"bond\"between two
is
\342\226\240
the interactions
gas of N moleculesin AF
n
field we ignore the increase concentration uniform of strong attractive potential energy. In modern mean field method neglectscorrelationsbetween
assuming
By
a
the
we
der
v;m
approximate
Waais
that an
interaction
the total energy.The as N2.
approximation
for the
yas:
.Vi{log[\302\273u(l'r
Sh)/N~\\
+ 1}
 Nza/V.
0$)
is
C9) '\342\226\240^\302\253'\027^5F
Critical Pawls for
the
Figure
suggests
pressure be
used
of intermolecular near ihe boundary Y. The van dci Waats argument lhal ihcse forces contribute art internal Nxa!Vl which Is lo be addedlo ihe
10.8
forces that of a volume
Direciions
on molecules
ad
as ihe
pressure
in
gas taw.
the
1
Q
o o
o
of volume Khas N not b. The volume molcculcSi V S'b. Intuition by molecules is occupied suggests that iHis fece volume should be used in V. llic gas law in place of the coiiiaincr Volume
V
Figure
each
O 0
O o
Q
(p
Waals
10.8
and
coniainer
of volume
o
\302\251
der
Hie
10.9
+ N2a/V2){V
equatum of
~ Nb)
\302\253 A'l
staSe. The terms in
a
D0)
.
and
b arc
interpreted
in
10.9.
Points
for She van
ne the
quantities
der
Gas
Waals
pc ^ a!21b2\\
Vc
s
3Nb;
xc
^ 8a/27b.
D1)
Chapter
10:
Phase
Trans/or,
. 
0.95tt
P/Pc
0
Figure
In termsof
these
der Waals
Tile van
10.10
critical
the
van der
the
quantities
equation of
3
Waals equation
\\(V
f7HJAH
This equation is plotted in temperature tc. The equation
stal
Courtesy of R. Cahn.
temperature.
Figure may
1010 be
l\\
8t
3/
3t/
for
written
becomes D2)
near the temperatures in terms of the dimensionless
several
variables ps
9sVfV.\\
pjPc\\
t
e t/tc ,
D3)
D4)
This si!
gases
result
!ook
ts
known alike\342\200\224if
as the they
correspondingstates. In termsof p, V, t, the van der Waals equation. Valuesof a
law of obey
Free
Gibbs
usually obtained
and b are
Realgasesdo one
At
same p,
at the
substances
not
the
obey
P~V
curve
phases.
and
coincide,
At a
horizontal
corresponding
local
the
Here
and
tc the critical
Above
tt
of the
van
der
G
gives
N.
of pressure
function
as
a function
we
cannot
by
gas
of
Nt{log[\302\273u(K
V,
t,
N;
It is
want
Nb)/N]
the natural
put G
conveniently
(9.13).
the characteristics With G = F + pV,
of
exhibits
pressure.
instead ofvolume.We
obtain ;i(t,p)as G(z,p,N)/N
relation
Waals

^
\342\200\224^
This equation p, t, Unfortunately
Vet
Gas
constant

=
pc,
\342\226\240
Waals
der
the !iquicigasphase transition at have from C8) and C9) the result
G(x,VtN)
=
respectively.
exists.
free energy of the van
The Gibbs
liquid
9 = 1;i 1. We call critical volume,and criticaltemperature, p = 1;
by D4) if
pressure, separation
Energy
of the
@i
no phase
Free
Gibbs
and minimum
maximum
a
f has
constant
at
V
is no separation between the vapor and of inflection point
satisfied
are
conditions
ihe subsiances.
there
aThese
states of
point, the curve of p versus
horizontal point of inflection.
tc. States of two
pc and
observed
Gas
M'aah
(let
van
to high accuracy.
equation
the critical
point,
called
Vy x are
the
to
fitting
by
of the
Energy
variables
into an
G(z,p,N)
+ 1],
analytic
because
we
D6)
for G are form
we can
as
/i that determines the phase coexistence
temperature
results of numerical calculationsof G versus plotted pare 10.! 1 for temperatures below and at ihe critical temperature. At any the lowest branch represents the stable phase; the otherbranches
represent
unstable
=
ft,
in Figure
a
then
fig. The
phases.
The pressure
at which the branchescrossdetermines
gas and liquid; this pressure is calledthe equilibrium !0.12. for G versus t are plottedin Figure vapor pressure. V < Vs In which only ihe 10.13 shows, on a pV diagram, the Figure region and the V > V2 in which liquid phase exists region only the gas phase exists. between and The value or The phasescoexist of V2 is determined Vx V^. Kj by the condition that /i](r,p)== }ia{x,p) along the horizontal line between Vx and the
transition
between
Results
V2 This
will occur if the
shaded area belowthe Jine
is equal
to the
shaded area
10;
Chapter
Phase
Ttansfo
0.40
t\342\200\224

rasure
,Vapor[
/ \\
Figure 10.11
der
Waals
(a) Gibbs free of stale:
equaiion
(b) Gibbs
free energy
versus
stale;i = %..
above
the
To see
line.
have
difference
dG
=
versus
pressure
i = 0.95tc. Courtesy
pressure
for van der
of
for van R.
Cahn.
Waals equation of
ihis, consider dG
We
energy
Vdp at
\302\253 ~adx
+
Vdp +
constant i and constanttola!number
of G between V\\
and
D7)
pdN. of
particles.
The
V1 is
G,G,=
fWp,
D8)
j
1 ~p

0.95
nqb =
!
LiqutaS^
\302\273bic

G/Nrc
k
i
t
'
i
0.990
0.988
0.986
0.984
\342\226\240
't/t, (a)
Figure
versus temperature of A. Manoliu.
Gibbs free energy
10.12a
of slale a! p
=
0.95 pe. Courlesy
for
van
der Waals
= p
nqb
equalio
1.0
=
1

\\
\\

\\oas
2.70
0.90
I.OO
1.10
1.05
t/t,
(b) Figure
of slate
10.12b
a! the
Gibbs crilical
free energy pressure
pc.
versus temperature for van
der
Waais
cquatiO!
Chapter
Id: Phase
Transformations
T
~
conslan
 ^Liquid Cocx
\\
c
fir
Stetice
X
<\302\243
i
a
as
\\ s.
v.t
10.13 Isotherm of van lemperafure belowliie critical Figun e less
than
above
lliegas phase exisls. Between in stable equilibrium lies along the
line
and is
liquid
and
liquid
and
volumes
bul ihe When
equals
the volume
/^{r,p)along
Hie
we require Nuclcatiou.
of the
magnitudes
j.tg
Let
V
that
and
V2
coexistence
is available.
shaded
areas,
areas are
equal, line
coexistence
horizontal
=
Vx
an inhomogencousmixture of two phases. The gas phases coexist. The proportion of liic be such lhai ihe sum of fheir gas phases must
integral isjust ihesum ofthe Ihe
volumes
only
Kj
Ihe system
volumes
For
phase exisls; for
!hc liquid
Vt only
a
der Waalsgasat lempcralure.
one

G?(t,/>)
drawn
in
and one positive. Gj(t,p) and [is{x,p}=
negalive
the
figure.
In equilibrium
/*,. Aji
\342\200\224 \342\200\224be
;i,
ftg
ihe
chemical
potential
difTerence between
the vaporsurroundinga smallliquid and the liquid in bulk (an infinitely droplet is if A/i iarye drop), positive, Ihe buik liquid will have a lower free energy than the gas and thus the will be more stable than the gas. However,the liquid surface the free free of a liquid drop is positiveand tends lo increase energy Al the of'he small radii the surface can be dominantand energy liquid. drop can be unstable the change with respect to the gas.We calculate in Gibbs drop of molecules is the concentration R forms. freecnergy when a drop of radius If\302\273, in the liquid, AG
=
G,

= G9
{
+ 4nR2y
D9)
Ferromagtutism
the surface
y is
where
drop
will
grow
free energy per unit Gt <
when
An
Gr
0 =
=
ti&G/dR
This
is the
fend
to
critical radius
R tlie
larger
drop
will
to
tend
E1)
that
because that,
drop
will
energy.
too, will
At
lower
free energy.
the
must that energy barrier (Figure10.14) fluctuation in order fora nucleusto grow beyond in E1) D9):
The free
(&G)C
to
free
the
Jower
will
spontaneously
grow
R the
At smaller
a drop
of
because
spontaneously
evaporaie
E0)
\302\2532y/H,A/i.
nuclcation
for
when
+ SnRy ,
4nR2iii&n
Rc
The liquid
is attained
of AG
maximum
unstable
tension.
surface
or
area,
If we
assume [hat
express
Aji as
the
vapor
=
for
r
the vapor pressure tn of the bulk liquid (R ~*
Kcrro
300 K
at
water
and p
E2)
like an
behaves
p is
pressure
by a ihcrni.i! substitution of by
overcome
R,. is found
(lenPW/nMl*I}.
Aft
where
be
ideal gas, we can use Chapter5
,
tlog{p/peq)
the
gas
co).
phase
and
use
~
We
y
pe
72erg
the
equilibrium
cm\022 to
vapor
estimate Rt
x !0~6cm.
= l.J^tobel
ism
magnet
means a magnetic has a spontaneous magneticmoment,which field approxithe mean moment even in zero applied magneticfield.We develop defined as the of ihe magnetization, approximation SO the temperature dependence each moment niagnelic magnetic per unit volume. Tlie centra! assumptionis that alom experiences an effective fieid BE proportional to she magnetization:
A ferromagnet
BE
where
I is
a conslans. We
take
the
\302\253 AM
external
E3)
,
applied
field as
zero.
Chapter
10:
Phase
Tram/or
barn
Critical *\"
or
t for growH,
3n nuclei
cortdcnsut
fx
'
1
1
L
ondensation
\\
,
nuclei

__
evaporate
\342\226\240'\342\226\240\342\226\240
1
Figure 10.14 of
function
Excessfree
drop
radius
energy
of drop
relative lo
gas, as
R, both m reduced units. The
gas
is
because the iiquid has the lower free energy for but the surface energy ofsmaii dropscreates an energy of nuclei of the liquid barrier ihat inhibits the growth fluctuations eventually may carry nuclei over ihe phase. Thermal
supersaturated this
curve
as drawn,
barrier.
a system with
Consider
a concentration h of magneticatoms,
magnetic moment fi. In Chapter3 we magnetization in a field B:
and of
M
In the mean field
.
.
approximation
an
found
each
exact
of spin
E3)
Me . \342\226\240\"\342\226\240_\342\226\240_\342\226\240
this
.
for a
becomes,
n/nanh(/dMA)
for the
result
= HjitanhOiB/t).
,
.
j
E4)
ferromagnet, '..
'.
E5)
Fcrromu^cth,
1.0
0.8
0.6
0.4
0.2
0
soiuiion of \302\243q. for the Graphical E6) reduced magnetization ,,i as a function of temperature. The reduced magnetization is defined as m = M/n/t. The lefthand side of Hq. E6) is piolicd as a straight line in 10.15
Figure
The righthand side is tanh(m/0 and is three difierenl of for values the reduced plotted t = r/n;i'^ *= i/tt. The three curves temperature to the temperatures 2x,,zt, and Q.5rt. The correspond with unit slope.
versus
curve for
=
i
in
the
2 intersects
as appropriate for ihe is no externai applied magneiic
m s= 0, t =
(or this
it) is
murks
ictnpetalure for
Curve
lo
tangent
i =
0.5 is in
tmersccts ihc straight 1
\342\200\242 0 the
magneiic
we write it temperature
iiie
tn f
s
line
moments
are
Figure
and
10.15. The
M
versus
shall
We
M.
=
left
of
sides
and n/i. As
0.94
l.so
that all
absolute zero.
see that
E6)
ianh{Hi/r). this
equation
separately
intercept of the two curvesgives
temperature of interest.The critical
curvesof
up at
iined

in
region s= \302\273i
whence
z/n[i2X,
We plot the right in
ferromagnciic in at about
moves up to
hi
as
the suatght Sine m at ihc origin; ilic onset of fcrroiiiagnetisiu.The
solutions of this equation M exist in the temperature range between0 and v To solve E5) terms of the reduced magnetization ??i \302\253 and the reduced M/h/i
nonzero
with
region (iherc curve for / = i
paramagnetic The fieid).
intercept
a transcendental equation for
al
in only
iine
straight
x obtained
in ihis
temperature
ts
way reproduce
f
the
=
I,
;is functions of \302\273t, value of m at Ihe = or zt h/i^L The
roughly the featuresofthe
10;
Chapter
Phase
Transformation
of Figure 10.16 Saiuration magnetization as a nickel function of temperature,together with ihc theoretical curve fci spin \\ on ihe
mean fieid theory.
resuiis, as
experimental
in
shown
magnetizationdecreases
smoothly
OF
THEORY
LANDAU
Landau gave a systematic transitions
consider
free
to
applicable
at
systems
F =
energy
with
minimum
\302\243,
system,
electrons bonds
have
can
a certain
be indepaideally
mean
field theory
the
It isnot helpful
system
can
to
all possible
consider
by a single order
be described
might be the magnetization in a ferromagneti in a ferroelectric system, the fraction of polarization in a superconductor, or the fraction of neighborAB
in ait
value
alloy c,
=
specified,
equilibrium the order parameter the Landau iheory we imagine Ihat \302
AB. In thermal cc{c).
In
and we consider the LandauTree
FL(\302\243.x)
(\"unction
energy
=
the energy and entropy are taken when the value \302\243 no! necessarily c0. The equilibrium specified where
of phase transi
xi, which
Greek
bondstotola! will
to what variables?
dielectric
superconducting
of the
variety of systems exhibitingsuch transitions.We their Heimholtz constant volume and temperature, so that \342\200\224 in The ta is a minimum big questionis,a equilibrium.
suppose here that
parameter the
TRANSITIONS
a large
respect
variables. We the
U
at z
zero
PHASER
formulation
for nickel As t increases the = rc, called the Curie temperature
10.16
Figure to
E7)
order value
has
parameter
\302\2430{t)is
the
value
the
of
Landau
Transitions
of Phase
Theory
makes FL a minimum, at a given the actual Hdmholtz t, and f(r) of the system at i is equal lo thai minimum:
c, that
F(t) Plotted as a function than one minimum.
\302\273
S
Fl($0,t)
of
for \302\243
r, ilic
constant
The lowest of these transition another phase i is increased. of
function ferroelectric
are
systems
exam
wellbehaved function something
this.
We
that \302\243
it
can
not be
lowest
In a
state.
equilibrium the
becomes
more
as
minimum
Landau (\"unction is an even applied fields. Most ferromagneticand ferro
pies of of
should
that
of
absence
the
in \302\243
the
have
for which Ihe
to systems
ourselves
E8)
Co
Landau free energy may
minimum
restrict
* \302\243
determines
first order We
if
FL(\302\243r)
free energy
taken
also
that
assume
be expanded
for
For
granted.
F 1(^,1) is a
in a
sufficiently
power series in
\302\243\342\200\224
function of
an even
as \302\243,
assumed,
entire
The
dependence of
temperature
coefficients
g6 These
g2,gx,
ga\\
FL(\302\243,x)
is
are matters
coefficients
in the
contained for
expansion or theory.
experiment
changes sign at ^(x) example of a phase transitionoccurswhen a temperature i0, with y4 positive and the higher terms negligible.For simplicity
The simplest take
we
linear
g2(t)
in i: ~
<72(r)
the
over
these
Witli
The
temperature
The
which
of interest,
F0)
and we take g4
as
in that
constant
range.
idealizations,
form F0)
certainly
range
temperature
 to)* .
(r
fails is not
and cannot be accurateover a very wide temperature range, on tembecause such a linear dependence at low temperatures consistent
equilibrium
has the
with
value of
the is \302\243
law.
third
found
it
at the
minimum of
FL{$
;t)
with
respect
roots
f~Q
and
?
= (to
 r)(a/g4).
F3)
Phase
Chapter
iff;
With a
and ga positive, the root c,
energy
function
Transformations
=
above i0;
at temperatures
F1)
0 corresponds
J=Xt) =
The other root, c,2 energy
of
FdZ'J)
as
of F(r) with
in Figure
ihe
for \302\243:
temperature
minimum of the free
here the HelinhoHzfreeenergy
is
F4)
0g(l).
t) corresponds to the minimum of the free below t0; here the Helmhoitz is free energy
temperature
of
a function
Figure 10.18,and shown

(a/gj(to
at temperatures
function
The variation
~
to the
10.17. The variation ihree is shown in temperatures representative is of the equilibrium value of \302\243 dependence is shown
in Figure
10.19.
Our model describesa phasetransition
elergoes
to
continuously
Figure for an
zero
as the
in
which
temperature
the value of the order paratnis increased to t0. The entropy
10.17 Temperature dependence of (he free energy of the second order. ideatized phase t ransition
Figure
10.18
free energy
Landau
function
As the reprcscniaLivc temperatures. the equilibrium value of \302\243 gradually posiiion
of the minimum
Figure
10.19
temperature, curve is not realisticat use Of Eq. F0): the third thatdf/rfr
increases,
polarization
a secondorder
>0asi*0.
at \302\2432
;is tic fined
by
t0 the
of the tree energy.
Spontaneous for
versus
temperature drops below
low
law
versus
phase transition. The because of the temperatures of thermodynamics
requires
Chapter
10: Phase
\342\200\224
Transformations at t
is continuous
dFfdz
temperature f0.
a
Such
Transitions
nonzero
a
with
latent
The real
second order transitions;
is no
there
latent
heat at the
transition
definition a second order transition. order heat are called first we transitions; a remarkable world contains of diversity arc ferromagnels and superexamples is by
transition
them presently.
discuss
= rQ,so that
best
the
superconductors.
Example:
Landau which
In the mean field approximation, ferromagnets the satisfy To show moment a field consider a n atom of theory. this, 3, pin magnetic magnetic we shall set equal to ihe tijean field >M as in E3). The interaction energy density is Ferromagnets.
V(M) =
j iscommon to selfenergy
ihe factor
where
F6)
\302\261
g{M) = constant in Ihe
regime
in
FL{M) ~
At
transition
Ihe
M
which
\302\253
n/j.
Thus
constant 
temperature
\\M2(/.'~
the coefficient of M*
with
First Order
Transitions
A
latent
lion order
at
heat
constant transitions
in metals transformations
the
discussion
density is given approxi
F7)
,
fursciion
\342\200\224A+

entropy
M2J2nn2
the free energy
i0 in agreement
The
problems.
per
lermsof
vanishes,
unit
volume
higher
is
order.
so ihat
F9}
\302\273}i2>.,
following
F8)
E6).
transiphase transition. The liquidgas In ihe physics of solids first transition. is a first order pressure and in phase transformaare common in ferroelectric crystals a first order iransition describes and alloys. The Landau function
characterizes
a first order
when the expansion coefficientg*
is
negative
and
gb Is
positive.
We
consider
first
l(K20
Figure first
Landau
free energy
at representative function has equal minima For t below rf the as shown. \302\243
order
versus
function
transition,
i1
temperatures.
Order
Transitions
m a At
xc
at a is minimum finite absolute is a there iatger values of ^; as r passes through tc in the position of the absolute disconimuouschcinge minimum. The artows mark the minima. the Landau
The extrema of
this
function
are
given
by
at
=
\302\243
0 and
o
the roots
FigureJ0.20: G1) 
Either
\302\243
0 or
G2) At with
transition
the c,
~
0 and
temperature
with the
rc the
root c^O.
free energies will
The
value
of
be
xc will
equal
for
not be
the phases
equal to r0,
Chapter
Phas
10;
T
and the
order parameter\302\243 (Figure
weui earlier, where \302\243 transformation
show
may
those
differ from
results
These
xc.
at continuously as in supercooling
zero
to
hysteresis,
hysteresis exists in
t0 
coexistencecurve in the px planebetween ClausiusClapeyron equation: dp
L is
the two 2.
The
the latent heat and
van
heat L
order
transfor
but
supcrsaturation,
no
is the
An
two
phases
must
satisfy tiic
L
_
der
Waais
volume
difference per
~
pV is the
the energy
'
is a minimum
of state
(P +
atom between
necessarily
V +
enthalpy.
is 
N2a!V2){V
Nb) =
Nx.
function
energy
and entropy
not value \302\243,,
H
where
equation
free
Landau
 Hlt
~ H,
.
4. In the
first
A
phases.
latent
3. The
tf.
or
1. The
where
continuously to zero at phase transition treated
transition.
order
a second
rrt in K
order
second
the
in
~
not go
does
10.21}
20
40
_60
are
the
taken
when
thermal
with respect to
when \302\243
parameter has the specified The function Fl equilibrium value \302\2430. the order
the
system
is
in
thermal
equilibrium.
A first
5,
phase transition is characterizedby
order
a
Intent
heal
and
that
the
by
hysteresis.
PROBLEMS
/,
dcr
van
ofthe
entropy
and
energy,
Entropy^
(c)
the

enthalpy//
//(r./>) arc
results
given
2. Calculationof equation
U =
INi.
= U
+ pV
=
Nt

jNr +
to first order
near p

~
the
for
G6)
terms o, h.
\\V;ials correction
the vapor
from
Calculate
G5)
2N2ajV\\
2NuPfx.
der
van
~ 1 atm
G4)
i%
NhP
the
in
heat of vaporizational
wmcr. The
G3)
N2.ii!V.
+ N2bx/V
water.
for
(IT{dp
of elT/dp
value
the
en
+ 1}.
Nb)JN]
N[log[nQ(V
H{i,V)
All
Show
is
energy
the
thai
Show
(a)
gas is
Waats
o(b) Showthat
IVaah gas.
of van der
enthalpy
Hquidvypor
100Xis 2260
Express
Jg\021.
pressure equaof in
equilibrium the result
keivin/atm.
3. Heat of vaporization Hgal of vaporization
ice
4.
Gassolid
we
let the
at
ice.
of
the
latent
soiid move in \302\273 haS)
heat
the
range
the heat
Jmol\"'
of vaporiza
of interest.
of the exampleB6}C2) in dimensions,
three
vapor
per atom
pressure
is tQ
which
(a) Show that in
the
is
\342\200\224
\\i.
the gassolid equilibrium under the exof the solidmay be neglected over the temthe ofthe be cohesive Let e0 solid,per atom. energy
Consider equilibrium. that the entropy assumption
GaS'SoU'd
temperature
a version
Consider
equilibrium.
osciSiators in the
(b) Explain why extreme
vapor over iceis3,SSmm
of water
pressure
~l\302\260C.
high temperatureregime(t
5.
The
mniHgai OX. Estimatein
and 4.58
~2=C
10: Phase
Chapter
Transformatio
Treat the gas as idea!and
of ihe
energy
system
minimum
of
ihe
+ F,
number
total
the
the
independent
container,
thai the
(a) Show
volume
of the much
total HeimhoHzfree
is
F = F, where
that the
the approximation
Make
monatomic.
accessible to the gas is the volume V of Smaller Volume occupied by the Solid,
free energy
\"
+
that in ihe
G8)
(b) Find
constant,
is
Ng
to N^show
respect
 I] ,
N,r[!og(N,/l'il0)
N = N,
of atoms, with
+
Ufa
the mini
equilibrium condition 09)
(c) Find the equilibrium
vapor
6.
of the
Thermodynamics
pressure.
superconducting transition, (a) Show
th
2/i0
SI units for Be. Because Bc decreases with side is negative. The superconducting phase
temperature,
increasing has
(SO)
[i0 ih
.
in
that
the
lower
entropy:
it
the
right
is the
more
ordered phase. As t ~+ 0, the entropy'in both phases will go to zero, consistent r? with the third law. What for the sliapc of the curve of Bt versus docs this imply = = this ihe SIiow llutt result hits 0 and hence (b) At r = xtt we have Bt a^ aN. following consequences: A) The two free energy curves do not cross ;if tt but are the same: as shown in Figure 10.22.B) The two energies merge, Usfr,.) = heat with the transition at r \342\200\224 associated tt. U.vW C) There is no Intent What is the latent heat of the transition when out carried in a magnetic field,
at r < i{7 (c) related
that
Show
Cs and
CN, the
heat capacitiesper unit
volume,
are
by
(81)
Figure S.iS is a than
linearly
dominated
plot of Cj'T
with
by Cs.
T1
vs
decreasing
Show that
and
r, while Hiis
implies
shows
that
Cs decreases
Cs decreasesas yz.
For
t
much \302\253
tc>
faster
AC
is
_
0.2
X. Normal
.1
Superconductor
*STC=1.180K
0.5
lure.
Temper;
K
of Experimental values of the Tree energy as a function in the superconducting state and in the normal stale. Below the transition is lower temperature T, = 1.180 K (he free energy in !he transition slate. The (wo curves at the supcrcondtiding merge heat is second order {(hereis no laient tempcra(ure, so thai the phase transition a! Tc). The curve and of transiiion in zero is measured FN is magnetic field, Fs normal slaie. measured in a magnetic to in (he field suftkien! pu! the specimen
Figure 10.22
temperature
Tor aluminum
Courtesy of
N.
E. Phillips.
model of the superconducting transition. TheBc(i}curves that iutve shapes close to simple paraboias.Suppose
7, Simplified superconductors
=
Bt(i)
that
Cs
linear in r,
as for
Assume
and plot
calculate
heat
8.
crystal
of
order
two.structures,
siable form
low
of the
than linearly as (Chapter 7}. Draw
gas
the i dependencesof
and ihe latent First
Fermi
of
the transition.
temperalure
substance.
separated atomsat
form
by a and
he
then
(83}
also t * 0. on the resultsof
the two
heat capacities,
Cs(rc)/Cv(r(}= 3. crystal that
can existin
'hat and /?. We suppose the /J structure is the stable high
the
energy
6 to
Problem
the
energy
Cs is
that
Assume
entropies,
Consider.'!
If the zero of the infinity,
two
Show that
transformation.
denoted
I
most
d/r(}2].
faster
vanishes a

Bt0[l
of
scale
density
is
taken
eilher
is the
a slrucmre semperalure
as the
1/@) at r
==
0
stale of uill
be
Chapter 10:
Pha:
negative. The phasestable
velocity of sound ve U?@). to lower values phase, corresponding
thermal excitationsin
phase.The !he energy.
the
larger
Soft
free
by the
is
$
thermal
excitation,
tend
systems
the
of
the
~rc2r'i/3Oi;3/i3,
by
given
velocity of
all phonons.(b)
Show
value of U{0);thus
amplitudes than in the a the entropy and the lower
stabie at high tempera!ures,hard
to be
low.
in the
the lower
/J phase is lower than vx in the clastic moduli for /?, then the
have larger the larger
will
phase
in
the (a) Show from Chapter 4 that phonons in a solid at a temperature much at
systems
the
have
\342\200\224 0 will
If the
<
Ux@)
a
t
at
free
energy
density
icssthan the Debye temperature
Debye approximation that at the transformation
v
with
will
a finite
actual
is
transformation
defined
real solution if t'p phase
<
transformations
as the
trans
thermal energy
(85), U
refers to unii volume.
that
must
the
that
L = 4[U,@)and
(84)
This example is a simplified model in solids, (c) The latent heat of
Show
(84)
as the
iv
system through the transformation.
In
taken
temperature
>v.').
There be ofa classof
contributed
t/,@
be
latent
to carry the heal for this model is supplied
(85)
11
Chapter
Mixtures
Binary
310
SOLUBILITY GAPS OF
ENTROPY
AND
ENERGY
MIXING
wrth Interactions NearestNeighbor Example: Binary Alloy of Structures Mixture Two Solids with Different Crystal Example: Low 3Hc\"He Mixturesat Temperatures Example: Liquid for Simple Solubility Gaps Phase Diagrams PHASE
EQUILIBRIA
AND
SOLID
BETWEEN
311 318 319 320
321
LIQUID
322
MIXTURES
Advanced Treatment: Eutectics
325
SUMMARY
330
PROBLEMS
330
Potentials in TwoPhase Equilibrium Energy in 3He4He and PbSn Mixtures
1. Chemical 2.
3.
Mixing Segregation
Coefficient
of
Impurities
4. Solidification Range of Binary 5. Alloying of Gold into Silicon
Alloy
330
330 331 331
33i
11:
Chapter
Mixtures
Binary
of materials science,and large parts of applications chemistry are concerned of with the biophysics, properties multicomponentsystems
and
Many
phases in coexistence.Beautiful, unexpected,and important occur in such systems. We treat the fundamentals of the subject
or more
two
have
effects
physical
in this chapter, with
Mixturesarc
of
systems
Mixtures
ternary and quaternary mixtures.If molecules, A
to form
three
and
mixtures
are called
four constituents
\"oii
expression
phases, such as oil and their and water do not niix\" means [hat
or
more
mole
and not
arc atoms,
constituents
when its constituents are intermixedon an A mixture is heterogeneous phase, as in a solution.
a single
two
contains
with
the
species.Birjury
is homogeneous
mixture
scale
situations.
an ailoy.
is called
mixture
the
simple
different chemical
or more
two
two constituents.
only
from
drawn
examples
GAPS
SOLUBILITY
have
that
distinct
The
water.
atomic
when
it
everyday
does not form
mixture
a single homogeneousphase.
Thepropertiesof
differ
mixtures
solidification
and
melting Heterogeneous
Consider
mixtures
may
a goldsilicon
from
the properties
ofmixturesareofspeciai at lower temperatures than
properties melt
alioy: pure Au
melisat
lO63cCand
substances. The
of pure
interest. Heterotheir
pure
constituents.
Si at
I404X,
at 37OCC. This is not but an ailoy of 69pet Au and 31 pet Si melts (and solidifies} ihc result of ihe formationof any lowmelting AuSi microscopic compound: a two mixture of almost phase investigation of the solidified mixtureshows pure
Au
side
by
side
with
almost
pure Si (Figure I I.I}.Mixtures
with
such
because and they are of practicalimportance precisely of their lowered melting points. What determines whether two substances form a homogeneous or a heteroare in equilibrium mixture? What is ihe composition of ihe phaseslhat heterogeneous can be of mixtures with each osher in a heterogeneous mixture?The properties at a fixed semperaiure will underslood from the principle that any system evolve to the of minimum free energy. Two subsiances wiil configuration is the configuradissoive in each oilier and form a homogeneous mixtureif that will free energy accessible to the components. The subsiances of iowest configuration
properties
are common,
Gaps
Solubility
SO/tm When a mixture of 69 pci Au and 31 pet Figure II.I Heterogeneousgoldsilicon alloy. Si is melted and then solidified, the mixture Au a into segregaies pure phase of almost \302\253iih a almost coexistent of Si aboui phase (Sight phase) pure {darkphase).Magnified is that of the lowestmelting AuSi mixture, the 800 times. The composinon given eutectic a later socalled mixture, concept explained in she text. Photograph courtesy
ofStephan Justi. a
form
side by side is
phases
then we say I hat A
if [he
tnixlure
heterogeneous
the
mixture miMure
hclerogeneous
free energy of
combined
tower ihaii the free energy exhibits will
of
the
Uvo
the
separate
mixture:
homogeneous
a solubility
melt at
gap. a lower [cmperalurethan
the
separate
free energy of the homogeneousmeltis lower than the two combined free of solid energies separate phases. ihis we assume for simplicity that the external Throughout chapter be and we sel pV = 0. Then volume changesdo not neglected, may
substances
work,
and
if I he
the appropriate
Ilian the Gibbs free We
discuss
compounds
the
energy
free energy is G.
We
tire
will usually
HelmhoUz
free
the
pressure involve
F rather
energy
simply speak of
com
free
energy.
welldefined binary mixtures of constituents Ihat do no! form with each other. Our principal interestis in binary Consider alloys.
Chapter H: Binary
Altitun
or more
two
have
concerned
are
biophysics,
science, and large parts of chemistry with the properties of multicomponent systems
materials
of
applications
Many
phases
coexistence.
in
Beauiiftil,
that
and important
unexpected,
Ihe fundamentals physical systems. in this chapter, with examples drawn from simplesituations. We treat
in such
occur
effects
and
of
the
subject
SOLUBILITYGAPS are
Mixtures
or more
of two
systems
molecules, A
is homogeneous
mixture
Binary species. four constituents
and
three
are
constituents
an alloy. when its constituents
is called
mixture
the
chemical
different
constituents. Mixtureswith ternary and quaternary mixtures.If the
have only two
atoms,
arc intermixedon an
a Single phase, as in a solution. A mixture two water. contains or more distinct phases, such as oil and \"oil and water do no! mix\" that means their mixture expression a single
melting
solidification
and
mixtures
may
a
an
of 69
alloy
from
the
properties
side
by
alloy:
pet Au
side
are common,
properties
of their loweredmelfing What heterogeneous
determines mixture?
and
evolve dissolve configuration of
with
they
a
two
compound: phase
mixture
pure Si (Figure 11.1).Mixtures
almost
and
AuSi
are
of practical
importance
microscopic
of almost with
such
precisely because
points.
whether two substances form a homogeneousor 3 heteroWhat is the composition of the phasesthat are in equilibrium
be other in a heterogeneousmixture?Theproperties ofmixtures can wiil from the principle that at a fixed temperature any system to the configuration of minimum free energy. Two substanceswill in each other and form a homogeneousmixture if that is the configurawill lowest to the free energy accessible components. The substances
with each understood
substances. The
of pure
of mixtures are of special interest.Heterolower temperatures than their constituents. Si at 1404\302\260C, pure Au melts at 1063\302\260C and pure 31 pet Si melts (and solidifies} at 370\302\260C This is not
of the formationof any iowme!ting investigation of the solidified mixture shows Au
form
at
melt
the result
pure
everyday
does iiot
properiies
Consider goldsilicon but
The
it
phase.
homogeneous
The properties of mixtures differ Heterogeneous
atomic
is heterogeneous when
to form
scale
mixtures
are called and not mole
Solubility
10/mi
H.I Figure jure 11.1 Siisn is melted
Heterogeneous goldsiliconalloy. When a mixture of 69 pet Au and 31 pet and ihen solidified,!hemixiure into a phase of almost pure Au segregates a codxisieni of almost Si aboui wiih (dark (lighi phase) phase pure phase). Magnified 800 limes. The composition given is !hal of ihe towesimelting AiiSi mixiure, ihc eulectk: mixture, a concept explainedtater socalled in ihe texi. Photograph courtesy
ofSiephanJusii. mixture if the combined free energy of the two separate is free of the lower than the by side phases homogeneous mixture: energy we say that the mixture exhibiis a solubility gap. then A hcierogeneous mixture will melt at a tower temperature than the separate comif the free energy of the homogeneous melt is lowerthan ilie substances free of the two separate solid phases. combined energies Ihis we assume for simplicity that the external pressure Throughout chapter and we set pV ~ 0. Then volume changesdo not involve be neglected, may a heterogeneous
form
side
and
work,
than
the
appropriate
We
compounds
discuss
free eneryy
free energy G. We
the Gibbs
binary wiih
each
mixtures
will
is the Hetmlioltzfree energy usually
of constituents
other. Our
simply
speak
rather
F
of the free
energy.
that do not form weildefined
principal interest is in
binary
alloys.
Consider
Chapter Hi
a mixture
Binary
Mixtures
of
aloms
JVA
the composition of the system
We express
x the sysiem
Suppose
per atom
in
\302\273
1
A'b/N;

Suppose
further
curves shape.
B}
jVa/N.
an
average
free energy
with
C)
two
homogeneous
Any
respect
to
two
separate
x
<
at
points,
of this shape are common,and mixture
in
derivative
second
ihe
x, < is unstable
\302\273
form shown
the functional
lias
that/(.\\)
we can draw a line tangent to the curve at
cause this
x
x of B aioms;
= F/N.
contains a range in which
energy
the fraciion
iermsof
forms a homogeneous solution,wiih
/
Free
B.Thetotal
by
given
this curve
NB atoms of substance
A and
ofsubsiance
is
atoms
of
number
in the
we
wilt
Figure
11.2.
Because
d2f/dx2 is negative, x = xx and x \342\204\242 x^. see
later what
composition range D)
xp
phases of
may
composition x, and x^. We
is that ihe average free energy per atomof the mixture segregated a the i\" line and in ihe on the Thus the straight given by connecting points [}. point a lower has free energy than entire composition range D) the segregatedsystem shall
show
the homogeneoussystem.
Proof
i
The
free
energy
of a F
where ,V,
Nfi are
and
=>
the total
segregated mixture of the two phasesa and NJix,)
+
NfJ\\xfi)
numbers of atoms in
j$ is
E)
. phasesa
and
ji, respectively.
These numbers satisfy the relations
which may
be
solved
for .V,
and Ny.
0)
Gaps
Solubility
Free energy per alom as a function for a of composition, a aiom of system gap. tf the free energy per has a shape such that a tangent can be drawn homogeneous mixture touches the two x and that curve at diftereiit /?, (lie composition points the two points is unstable. Any mixture with a range between in this two phases with the composition range will decompose into composition _v, arid ,\\f. The free energy of the two phase mixture is It. given by the point / on the straight line, below the point tl.2
Figure
with
E) we
From
a solubility
obtain
fjix)
(S) JV
for
the
free energy
straight line through
the
the point i
in
thefx
points
of the
two
plane. a and
system.
phase
If we
/?. Thus/
set in
.v.
(he
=
is linear in x and is a v,c see (hat the line docsgo A'3 or.v^, result
This
between
interval
on the straight lineconnectinga and
p.
.y4 and
xfi
is given
by
Binary Mixtures
II:
Chapter
We have not yet
made useof
(he
that
assumption
(he straight
line is tangentto
3 and /J, and therefore our result holds for any straight line points two points 2 and/Jin commonwith/(.v). Bui fora given vaiue of x, (he lowest free energy is obtainedby drawing (he lowest possible straight line that has (wo points in common on opposite sides of a. The lowest wiih/(v), possible line the is shown. The and straight (wopointtangent x3 compositions x? are
f{x)
the
at
that has
the limits of the solubility of (he system. gap Once (he system has reachedits lowest free (he (wo phases must be in energy, to diffusive with both atomic species,so thai their chemical equilibrium respect satisfy
potentials
/*a> = We show
point tangent with as in Figure 11.2.
the
two
/jB
are
(9)
Pb*
given
by (he
of the two
intercepts
plot at x
edges of thc/(x)
vertical
*=
0
and
a ~
1,
MIXING
OF
ENTROPY
AND
ENERGY
jja and
i that
Problem
in
=
Pb\302\273
/*\302\253;
The Heimhoitz free energy F ~ U ~ to has contributions from Ihe energy and from the A and B on We treat the effect of mixing two components entropy. both terms. Let uA and \302\273a be the energy per atom of the pure substancesA and B, referred
atoms
to separated
at
Tlie
infinity.
energy
average
per atom of the
constituents is u
=
(uANA
+ vsyn)/N
which defines a straight line in
the
per
mixture
atom
separate homogeneous
of
the
homogeneous
constituents. is
niixture
u~x

nA
4

(\302\253B
A0)
uA)x ,
Figure 11.3. The average energy be may larger or smallerthan for the
plane.
In (he example of Figure11.3,(he energy than the energy of the separate larger
of
the homoge
constituents.The
of mixing. If (he re term in the free energy is negligible,asat 1 = 0,a positive mixing mixture not will that a is stable. mixture means such energy homogeneous Any in the the \342\200\224ia then separate into two phases. But at a finite temperature term free energy of the homogeneous the tends to lower mixturealways free energy. a contribution, the ofa called mixturecontains entropy of mixing, Theentropy of the separate components. The mixing that is not present in (he entropies the different arises when of atoms species are interchangedin position; entropy of such interthis a different state of (he system.Because operation generates
energy
excess
is called
\342\200\224
the energy
and Entropy
Energy
11,3
Figure
in a
sysicm
Energy with
per atom
a positive
as a function
of composition
energy.
mixing
of Mixing
A
simple
a solubiliiy gap may occur is thai of a in which the system energy per atom of the homogeneous mixilire Him of ihe separate phases, so that is greater than 1 s. The 0 for att c mposi ing e rgy i e bci differ een the u[x) curve and the straight line. example
for which
states a mixture has more accessible and hence the mixturehas the higher entropy. In C.80) we calculated the mixing entropy A, ^B,. to find
changes
two
tlie
than
erM
of
separate
substances,
a homogeneous
alloy
(ID
as
plotted
in Figure
11.4. The
that the slope at the ends of the
N dx which
goes
to
+ co
as x
curve of aM
range
composition

X)
the important property is vertical. We have
x has
versus

bgx
+ 0 and to \342\200\224 co as
 ioj
x *
I.
A2)
Chapter II:
Binary
Mixtures
=
da^/elx
Ffgure
11.4
Mixing entropy.
interchange of two
atoms
Tn
species leads to a new
system. The logarithm of liie number of mixing
Consider
is the
slate
in this
related
slates
way
of the is the
entropy.
now the
quantity
u[x)
which
of two constii uenis an
mixture
any
of different
\342\200\224 X
free energy
per atom without

(a
the

A3)
ffjtf
mixing
entropy
contribution.
The
is usually nearly the same for (he an, nonmixing part of the entropy,
entropy
curves
dependence
to fo(x),
~zaM/N in
Figure of/0(.v)
we obtain at
11.5. In drawing itself,
because
the for
Energy
Free energy temperatures.The curve Figure 11.5
per
atom
fQ is the
versus composition, free energy per atom
und
Entropy
of Mixing
at three without
the
a parabolic mixing entropy contribution. For ilHistraiion composition dependence is assumed, and the temperaturedependence of/0 is The tliree solid curves represcnl the free energy neglected. including the mixing for the temperatures 0.8 rM, 1.0 tM, and 1.2 rM, entropy, where there is a solubility rw is the maximum temperature for which gap.
our
The
this is
argument
construction
separation
phase
at 0.8 rw
is apparent.
follow irrelevant. Three importantdeductions
of the/(x)
from
the
curves:
(b)
f{x) turns up at both ends of the composition contribution. infinite range, entropy slope of the mixing Below a certain temperature rM there which is a com position range within than is ihe the second derivative of the fo[x) curve stronger negative second it derivative of the positive taM contribution,thereby making values of x. to draw a common tangent to f(x) at two different possible
(c)
Above Ty
(a)
At
all
finite
temperatures
because
the
of the
curve
has
a positive
second derivative at al! composilions.
Binary Mixtures
11:
Chapter
We conclude that the AB system with below the solubility gap temperature tM.
widens with composition
only
range
solubility of A
in
B and
as t * of B
in
The
0.
At
A,
a result
finite
any
energy
can reach the
a
gap
edgesof
there
temperature
earlier
obtained
exhibii
of the
will
range
composition
the gap
but
temperature,
decreasing
mixing
positive
in
the
finite
is a
3. The
Chapter
below Positive is that the mutual solubility is limited only tw. We now discuss three examples. mixing energies arise in different ways.
new result
Consider an alloy A^jB, with nearestneighbor interact'ionSi in than the attractive interinteraction between unlike atoms is weaker interaction between as bonds. There are like atoms. For simplicity we speakof the interactions be ihe potential three different bonds: energies of AA, AB, and BB. Let uAA) uAB and uBB each bond.These binding energies will usually be negativewith respect to separated aloms. We assume the atoms are randomly distributed among the lattice sites.The average of ihe bonds surrounding an A alom is energy Binary
Example:
alloy
(he attractive
which
=
uA
where (t  x)is ihe propoition mean field approximationof

A
of A and x is the
=
A
The total energy is obtained by summing the average energy per atom neighbors,
The
factor^
can be written
ip[(l
Figure
mixing
II.5.
proportion of B.This for B atoms,

+
X)HAD
result
is wiiucn
in ihe
over
both
A5)
N\"UD.
atom
types. Ifeach
atom has
p nearest
is
+
xJUAA
2jcA
aiises because eachbond is sharedby
 *Kb + the
two
atoms
A6)
*3\302\253db]
it
The
connects.
result A6)
as
u =
is the

A4)
tO. Similarly,
Chapter
\"a
,
xuAa
+
x)uAA
energy.
On this
ip[(l
model

the
x)uAA
mixing
+
xum]
energy
as a
+ uM.
function
A7)
ofx
is a
parabola, as
in
Energy
A solubility gap
occurs whenever (/'//dx1
< 0, that
= 2P[fAB
^r

and
Entropy
of
Mixing
is, when
i(\302\253AA
+
B0)
O3
From A2),
x{l 
N dx2 The
sign holds
equal
for
,x =
$. Wilh
T*i

x)
these results{19)yields
M\302\253ab

iO'AA
+ ^a)]
B2)
as die lower
of the temperature for a solubility limit gap. are many reasons why mixed bonds may be weaker than ihe bonds of the sepafaic constituents.If the constituent in radius, the difference introduces atom* of an alloy differ clasticstrains that water molecules raise the energy. Water and oil \"do not mix\"' because There
carry a large water
strong
electric
molecules.
as
the
moment
dipole
This attraction
weaker
oiloil
Example: Mixture of two
that leads to a strong electrostatic attraction which are only is absent in wateroil bonds,
between about
as
bonds.
solids
with different
crystal structures.
Consider a homoge
of gold is the facesilicon. The stable crystal centeredcubic structure in which nearest equidistant every atom is surrounded by twelve of silicon is the diamond structure in which structure every neighbors. The stable crystal aiom is surrounded by only four equidistant nearest neighbors, if in pure Au we replace a wiih the small fraction xof the atoms by Si, we obtain a homogeneousmixture Au^.Si., 1 fee crystal structure fraction x of the of Au. Similarly, if in pure Si we replace a small aioms by Au, we obtain a homogeneous mixture Au, ,5^, but with the diamond crystal siructure of Si.There are two different free energies, one for each crystal structure (Figure range, or else pure Au and 11.6).The two curves must cross somewhere in the composition curve consists of the lower Si would not crystallize in different The structures. equilibrium a sotubility with a kink at the crossover point. Such a system exhibits of the two curves, in the on side of the curves shown either crossover The figure are schegap composition. to the in the actual extends so close Au~Si system the unstable range schematic; edges of the from x = 0 to x = 1. that it cannot be representedon a fultscale extending plot diagram homogeneouscrystalline
mixture
of gold and
structure
Chapter
11;
Binary
Mixture
\\/
Figure
11.6
Tree
energy versus
homogeneousmisiuresfor mixture
crystallize
free energy curves
which
in ^ilfcrem
composition for cryslallinc [he [wo constituents of the
crystal
are involved,
one
Two
structures.
for each
differe
crystal structure
Different crystal structures for the pureconstiiucmsarean cause or solubility important in crystalline solid mixiures. Our a/gument to mixtures of ihis kind, provided gaps applies the two structures do not transform coniiriuously into each other wilh changing composicomposition. This when is a tacit assumption in our discussion, an assumption not always satisfied the two crystal structures are closely similar. The other we make throughout assumption this is that no stable compound formation occur, should in the presence of comchapter the behavior of the mixture be more formation compound complex, may
vs^~7cz:\":\"s'.\".r:'~ Exampk: mixture
with
Liquid
SIU*
a solubility
..,.,...
~
\342\226\240\342\226\240\342\226\240 \342\200\224..\342\200\224 ..... ... \342\226\240\342\200\242\342\226\240\342\226\240>\342\226\240\342\226\240\342\226\240\342\226\240\342\226\240\342\226\240\342\200\224
He m*.Uuivi
gap is the
at W
miMuniof
liquid mix
tanpcraiures.
The
ilie two iiefium
isotopes JHeand JHe, atoms
moat
interesting
;.,,
of
ocmii fcimjon^ unti of the 'aHer bosons, 1 lie re js a soluoiltty u\302\273io sn the mtx turc oclow 0.S7rCj ii1/ in i igure 11.\027. 1 Ins property ss utilised m the Iicliuitt cJj'tiiion refftccr^tor have a must be positi\\e to The origin of the solubility gap. {Chapter 12). Tito mi.\\ing energy low temperais tht! folio\" ing: 4Hc aloiiisarc bosons.At suliieiently positive mixing encray temperatures almost jli \342\226\240*! le afoins have state orbii;tl of the sysicm, vvherc occurs) the ground they
tiie Toruicr
30
20
10
40
Phaie
Diagrams for
50
60
.70
SimpleSolubility
80
Gaps
100
90
\342\226\240\" Atomic
Figure
11.7
Liquid
percent
mixtures of
JHe and
He
pure
\342\226\240'He
4Hc.
kinetic energy. Almost trie entire energy of the mixture is contribuicd by t!ie which are fermions.The of a degenerate Fermi atoms, energy per atom gas increases v,ilh concentration 7. This energy has a negative secondderivative as n1'*,as in Chapter
kinetic
zero
3He
Pltase
Diagrams
Tor Simple
Solubility Gaps
dependence of solubiiily gaps,as in the 11.8. The two compositions xx and xf arc plaited horizontally, Figure The .v^ and xf branches merge at the vertically. corresponding temperature maximum temperaturet,m for which a solubility gap exists. At a given temperatemperature, overall composition falls within the raoge enclosed mixture whose by any of actual curve is unstable the as a homogeneous mixture.The phasediagrams
A
phase
mixtures
form of
diagram
represents
with solubility
the temperature
according gups may be more complex,
(he free energy relation/(.v),
but
the
underlying
principles
to
the
aclual
are ihesamc.
II;
Chapter
Binary Mlxtur
Slabk
Decomposilion
/
\\
1
1 1 1
Uns
Figure 11.8 Phase diagram gap. A homogeneous mixture i if
temperature
the point
curve. The mixture
will
curve
boundary
system
PHASE
(*,i)
a solubility
be unstable stability boundary
tlic
ai
form two
separate phasesof the the intersections of the stability boundary line
for
with
temperature
r. The
stability
calculated quantitatively
a parabolic
BETWEEN
EQUIUBRIA
with
x will
below
Tails
shown here was
of Figure lt.5,
system
oCcomposition
then
given by curve with the horteontal compositions
a binary
for
the
for
fo{x).
LIQUID
AND SOLID MIXTURES a small
When
of
the
fraction
phenomenonis mixtures.
of a
solid that forms We
the homogeneous liquid mixturefreezes,
is almost always
readily
understood
consider
a simple
different
from
that
composition
of the liquid.
The
liquid and solid model, under two assumptions; (a) Neitherthe from
the free
energies for
Phase
solid nor she liquid has a solubility
constituenl
is
A
The free energies
melting
temperature
ta or pure
Semperature tb of pure constituentB. and ta ra. Tor the solid and fL(x) Tor the liquid, are
between fs{x)
atom,
per
(b) The
gap.
Solid Mixtures
Between Liquid and
the melsing
Jhan
lower
a SemperaSure
consider
We
Equilibria
11.9a. The two curves intersect at some comLeSus draw a commonlo boSh aS .\\ ~ xs posision. jangenS curves, touching/j = and fL a! x xL.We can define three composision ranges, each with differcnS shown
in Figure
qualitatively
internal
equilibria:
x < xL, the
(a) When
in
system
is a
equilibrium
homogeneous
liquid.
a system in equilibrium consistsof two phases, solid phase of composition xs and a liquid phaseof composition xL. x > xs the system in equilibriumjs a homogeneous When solid. (c) a arc The and of and a so!id in compositions xs xL liquid equilibrium phase of the temperature dependent. As ihc temperature decreasesthe free energy solid decreases more rapidly tlKll of the liquid. The Ungctitiai points in than Figure 11.9amove to the Icfi, Tliis behavior is rcprcscnScdby a phase diagram stinihlr to the earlier representationof the equilibrium curves for composition
(b)
When
mixtures
xL<
with
xs, the
x <
11.9b the curve
In Figure
separation.
phase
for
xL
is called
the
curve. Hquidus curve; the curve for xs is the soltdus have been determined experimentallyfor vast numbers The phase diagrams of binary mixtures. Those for most of the possible binary alloys are known.*
For
Figure phase diagrams are more complicatedthan for a 11.9b, simple system, germaniumsilicon. When is lowered in a binary liquid mixturewith the the temperature phase of Figure lj.9b, solidification takes placeover a finite diagram temperature a liquid with the range, not just at a fixed temperature.To see this, consider is lowered, initial composition xiL shown in Figure 11.10.As the temperature \342\200\224 of the solid formed is given solidification begins at t composition x,. The is changed. 'hat of the In the so the remaining liquid by xtsi composition the example xiS > xiL, so that the liquid moves towards lowervaluesof x, where if solidification is lower. The temperature has to be lowered temperature the of the moves solidification is to continue. The composition along liquid = at t The solid formed the curve until solidification is compleied tA. liquidus metal
most
homogenizeafterward
in
for
a long
slandard
Constitution
iabutatlons
of
binary
arc
by
solids
many
M.
in
atomic
solid
may homoge
temperature remains is too slow, and the
if lhe diffusion
\"
indcnniieiy.
Hansen,
of binary alloys, firsi second supplement, alloys,
Constitution
Edioti,
for
''frozen
The
equilibrium.
particularly
diffusion,
time. But remains
\342\200\242
The
atomic
by
homogeneity
R. P.
and is not in
in composition
is nonuniform
high
the
alloys
was drawn
which
Coitsilxatlon of binary supplement,
allays,
McGrawHill,
McGrawHill, 1969. .
McGrawHill.
1965; R
A.
1958; Shunfc,
11.9
Figure
ihis example
Phase
equilibrium
neither phase
cxhibhs
btiwccn
liquid
a solubiliSy
and
solid
mixtures.
In
gap. We assume
the free energies for iie two plxiscs; The upper figure (a) shows The curves xL ihc lower figure (b) shows ihc corresponding diagram. phase and xs in She phase diagram are called ihe liquid us and She solid us curves. = 940cC and The phasediagram is She GcSi phase diagram, wish TCt
tA
x < xlt.
<

I412\"C
7\342\204\2425i
324
Phase
Figure
Mosi
11.10
Scmperalurc,
but
liquid over
highernwiliiig
consiiluenl
lowermching
consliSucnt
solidification
lemperalure
Advanced Treatment;
mixiuresdo
a finite
not
temperature
liquid
at a
range
from
and Solid
Mixtm
sharp t,
'o ta.
The
first, thereby enriching the liquid phase and thus lowering She of ific liquid. precipilaSes
in ihc
down
tower meltingtemperatureof the system: a mixture of 69pet An and solidification
Liquid
solidify
Eutecltcs. There are many
liquid phase remainsa
compositions
Between
Equilibria
to
temperatures
constituents.
31 pet
starts at a
binary
The
Si starts to
in which
systems
the
below the alloy is such a
significantly go!dsi!icon
solidify at
higher temperature.When
370\302\260C.
we
At
plot
other
the
of alloy oflhe onset of solidificationas a Function composition, obtain Mixtures with two we the twobranch liqutdus curve in Figure 11.11.
temperature
solidification minimum temperature liquidus branchesare calledcutectics.The is She eutectic is the eutectic SemperaSure,where She composition composition. is a two phase solid, wiih The solidifiedsolid at the eutecticcomposition nearly pure gold sideby side wiih nearly pure silicon, as in Figure 11.1.In the solidAuSi mixture shcre h a very wide solubility gap. The low mching point oFthe for the free occurs eutectic composision becausethe homogeneous energy melt is lower than the free energy of the two for at solid, temperatures phase or above She cutectie temperature. Such behavior is common among systems thai exhibita solubilitygap ill the solid but not in the liquid. The behavior of eutccjj'cs can be understood from the free energy plotsin Figure11.12a. We for the solid as in Figure 11.6, assuine_/^(.\\\")
11:
Chapter
Mixtures
Binary
1600 404\302\260
1400
1200
U
1063\302\260
1000
E
/
/ \\
/ \\ 0\302\260
1/
31
Pure
11.11
Figure
Au
Euieciic
and
range
Figure
but
II.12a
below
10
iempctaiure indicaie
37OX
complcic
80
90
100
Pure Si
silicon
percent
go!dsiljconalloys.The
iogeihcr ai ihe euieciic daia poinls ai ihe mixiurc docs noi
to difTcrcnt
corresponding
temperature
diagram of
ihe experimcnial
composiiion
constituents.
phase
60
50
40 Atomic
branches ihai come line
30
20
10
0
Hquidus
iwo
consisisof
T, = 37O;C.The horizonial ihai ihroughoul the eniirc
iis solidificalion
unlit
ihe
euieciic
crystal structures a and ^ for the two pure confor a temperature above the cutectic
is constructed
the melting
iemperature of cither consihuent,
so
that
to energy of the liquid reachesbelow the common tangent phase curves. We can draw two new twopoint tangents tltat give even lower free energies. We now distinguishfive different ranges: composition is a (a) and (e). For x < xaS or x > x^, the equilibrium state of the system solid. In the first range the solid will have the crystal structure a; Homogeneous in the second the structure is range ($. (c). For xlL < x < XpL, the equilibrium state is a homogeneousliquid. and For is in a liquid (b) (d). x^ < x < xaL or x^L < x < Xp$, phase equilibrium with a solid phase. the As is lowered, faS and fa decreasemorerapidly than/L, temperature and of the the range H.12b homogeneous liquid becomesnarrower. Figure shows the corresponding the two curves. solidus phase diagram, including tUe free
the solid
Figure 11.12 Free energies(a)and sysiem.
At
to
the
theeulecltc
common
t, Jhe
temperature
tangent
to f^
energy fL
a homogeneous
A mixture
meits at a
above
(he
and fps, as in
tangent,
of Jhe liquid phase is tangential
free energy
which fL touches the tangent is the iies
(bj in a
diagram
phase
Figure
eutcctic
although
11.13.
The
composition At x < xt, the
composition.
fL may be beiowthe free energy
at
free of
solid.
of composition equal to
single temperature,just like
the
a
eutectic
pure
composition
substance.
solidifies
The solidification
and
of
Chapter 11: Binary
Mixtures
FiCe energies
Figurel!.!3
compositions
away
From
and
ends
at
the eutecttc
and
ends
at
a higher
in
a euseclic
system at t
, andati
< xr.
starls at a
higher temperature at the starts eulectic temperature. Melting temperature the
euieciic
composition
temperature.
The minimum properly of the utilized. The AuSi eutectic plays
melting
a large
temperature
role
in
of eutectics
semiconductor
is widely device
tech
welding of electrical contact wires madeofgoldtosilicon devices. Leadtin exhibit a euieciic (Figure alloys at a i83\302\260C solder below that of pure tin, to 11.14) give melting temperature 232;C. is to whether a sharp melting temperature or a melting range According comor a different citlicr the exact cutectie compositionB6pet lead) desired, Salt the because of the low is on ice melts ice composition employed. sprinkled
technology:
eutectic
the
cutectie
temperature
permits
2L2\"Cof
low temperature
the H2O~NaC!
eutecttcat 8.17moipet
NaO.
The in character, solfdus curves of eutectic systems vary for the greatly die ioclt PbSti system (Figure Il.M) die solid phases in equilibrium with contain :tn appreciable fraction of tltc minority const [merit, and this fraction in other increases with decreasing systems this fraction may be temperature, small or may decrease with or both. The AuSi system decreasing temperature, with is an example: The relative concentration of Au in solid Si ill equilibrium of only an AuSi melt reaches a ma\\imum value 2 x !G~6s.ound i 300\" C, and it drops off rapidly at lower temperature. In our discussionof the free energy curves of Figures 31.12 and 11.13we assumed the lite composition tltat at which the liquid phasefree energy touches
10
\020
20
pure Sn
30
40 Atomic
50
60
percent
lead
70
90
100
pure Pb
10/tm Figure
of [he Jackson.
11.14 PbSn
diagram of the PbSn s> stem, after Hatlicn.{b) Microphotngrapf; of J. D. Hunt and K. A. magnified about S0Otimes.Courtesy
{a} Ptiasc eutmic,
S29
11:
Chapter
Mixtures
Binary
curves lies between the
solid phase
tangent
to the
In some
systems this point lies outsidethe
and
were
fL
i
iti Figure
interchanged
i
interval,
Such
.\\2a.
and
compositions
xlS
as
and/t
if either/aS
arc caiied
systems
xfiS.
or/flS
peritectic
systems.
SUMMARY
1.
side by side is lowerthan
phases
separate
the combinedfreeenergy
gap when
a solubility
exhibits
mixture
A
free
the
of the
energy
two
of
homogeneous
mixture.
2, The in
the alloy
For
position.
3. The mixing
for
energy
uM
for
p nearest
4. The
5. Mixtures
minimum
we have
A} _IBI,
px(l

is
interactions
nearestneighbor =
species are interchanged
of different
atoms

j(uAA +
uBB)]
.xL versus
t
x)[uAB
,
neighbors. is
Hquidus
equilibrium for a solid
when
arises
entropy
mixing
the
curve
composition
with a solid.
a
for
The solidusis the compositioncurve
phase in equilibriumwith
a
in
phase
liquid Xs
i
versus
liquid.
two branches to the liquidus curve solidification temperature is called the
with
The
eutectics.
called
are
eutectic
temperature.
PROBLEMS
L Chemicalpotentialsm twophase
[eniials ;iA and /jB of phase mixtureare given with 1. liquid
the vertical Mixing 3He4He
energy
die two by
the
intercepts
B of
and
oFthe twopoint
edges oFthe diagramat x
=0
x
and
the chemical an equilibrium
tangent
mixtures
in
in
the
Similarly,
solubility
oFPb
112
Figure
andPbSn mixtures. The phasediagram of 3He Figure 11.8 shows that the solubility
> 0.
potwo
\342\200\224 1.
in 3He~*He
finite (about 6 pet) as r residual Figure.11.14 shows a finite remains
that
Show
equilibrium.
atomic species A
PbSn
in solid
phase
oF
liq
in 4He
diagram
of
Sn with decreasing
t.
do
What
such
residual
finite
solubilities
about
inipty
the
Form
of the
Function
u(.x)?
3.
Segregation
In
this limit the
1to A, wish A' \302\253 Let B be an impurity the oF can be as linear Free nonmixing parts expressed energy both solid phases. of x, as fQ(x) = /0@) + x/0'@),for and functions liquid Assume thai the liquidmixtureis in equilibrium with the solid mixture. Calculate concentration the coefficient. ratio k ~ xs/xL, called the segregation equilibrium For k \302\253 then a and substance be many systems may I, purified by melting and partial resolidificatioti, discarding a small FractionoFthe meit.Thisprinciple used in the purification of materials,as m the zone is widely of semirefining \342\200\224 = = \302\243 ! T 1000K. Give a numerical value for eV and semiconductors. for/os' /Dt'
4.
of impurities.
coefficient
of a
range
Solidification
binary
alloy.
Consider the solidificationofa
binary
of the that, regardless diagram of Figure.' 11.10.Show B in component initial the melt will always become fully composition, depleted ion the id i Seas the time remnant That sol the last of the meit solidifies. is, by will not be complete until the has dropped to TA. temperature
alloy
5.
wish
Alloying
the phase
of gold
hto
silicon,
(a) Suppose a
and onto a Si crystal, subsequently diagram, Figure 11.11,estimate how
silicon crystal. the estimate
The
for
densities
800\302\260C.
of Au
heated deep
and Si
1000A
to she
400\302\260C
gold
layer From
will
of Au the
is evaporated AuSi phase
penetrate
are 19.3and 2.33gcm\"\023.
into the (b)
Redo
12
Chapter
Cryogenics
COOLING BY EXTERNAL IN
AN
Gas Liquefactionby
WORK
334
ENGINE
EXPANSION
for
Effect
337
Effect
JouleThomson
the
Example: JouleThomson
van
der
Waals
Helium,
Pumped
Cooling:
to 0.3
K
HeliumDilution Refrigerator;
341 342
Miilidegrees
DEMAGNETIZATION:
ISENTROPIC
QUEST
333
339
Liride Cycle Evaporation
Gas
346
ZERO
FOR ABSOLUTE
NuclearDemagnetization
348
SUMMARY
350
PROBLEMS
350
1. Helium
2. Ideal
as a
van
der
Waals
359
Gas
35i
Carnot Liquefier
3. Claude
Cycle
4. Evaporation
Helium
35!
Liquefier
352
Cooling Limit
for 5. Initial Temperature
Demagnetization
Cooling
352
physics and techiioiogyofthe productionoftow temperatures. the physical principles of the most important cooling methods,
is the
Cryogenics discuss
We
lowest
the
to
down
cooling
a gas
of
temperatures.
principle oflow temperature generation
The dominant
by kiting it do work be a conventional may
against
is the
lOmK
to
down
during an expansion. The
a force
gas; the free electron gas in a semiconatoms dissolvedin liquid 4He. The force semiconductor; or internal to the gas. Below be external against which work is done may 10mK the dominant cooling principle is the iscntropic demagnetizationof a
gas
employed
or
the
ihe cooling methods chain lhat starts cooling
discuss
We laboratory
to the
Household
cooling
evaporation
liquid helium
below its
COOLING
BY
In the
method boiling
EXTERNAL
EXPANSION
AN
in
the
they occur
in which
order
by liquefying helium and
a
in
from
proceeds
1 ;iK. sometimes lowest laboratory temperatures,usually lOmK, and automobile air conditioners utilize the cooling appliances
same
IN
gas of 3Hc
substance.
paramagnetic
there
virtual
that
temperature,
is used
in
the
to about
for
laboratory 1
cooling
K.
WORK
ENGINE
isentropic expansion
of a monatomicidealgas
lower pressurep2, the temperaturedropsaccording
from
to a
pi
pressure
to
(i)
by F.64). temperature working
process
and Ti = 300K; then the temSuppose p, = 32atm; p2 = iatm; will drop to T; \342\200\224 75 K. We are chiefly interested in helium as the and for helium A) is an excellentapproximation if the cooling gas, is reversible.
The problems in implementingexpansioncoolingarise of actual expansion processes.The problems irreversibility of good low temperature lubricants. by the nonexistence and cooling cycles follow Figure 12.1.The compression
the
from are
partial
compounded
Actual expansion
expansion parts
of
itisng
by
Work in on
External
Heal
{Expansion
ejection
Expansion
ngine
Working\342\200\224
volume
gas ts Simple expansion refrtgeraior.A working is the heat of into the compressed; compression ejected environment. The compressed room temperature gas is heat counleriiow further in the exchanger. It then precooled to a does work in an expansion engine, where it cools volume. Afkr extracting temperature below that of the working hea{ from the working the gas returns to the compressor volume,
FEgure
via
{he
I2.I
heat
exchanger.
Ens
Chapter 12:
Cryogenics
separated.The compressionis
the cycle arc
exchanger by contact
flow heat
temperatureof the
in
cooling
the
on
the
expansion
of the
design
The
work
done the The total
by
plus the work
She
=
W
{Ul
on the
performed
important
fij and
the expansion
work p{Vl gas.
the displacement
plus
gas
into
flowing
difference
enthalpy
Vx
a given
to
refer
mass of
is the energy U2 the the gas against pressurep2
with
engine
the gas
of
the
gas
work
The
is the difference
For a monatomic ideal work
the
of
boih
to move
V% required
p2
by ihe engine
extracted
exchangeris as
is the
engine
expansion
where
leaving
energy
im
requirements
cooling
gas: The iota!energy
and output
compressor,
by
the
iruernat energy U^
engine is the
heat
the
via
compressor
expansion engine.
extracted
the input
between
the
the
reduces
The design of Sheheat
engine.
counter
in a
precooled
return gas stream at the low cooled to itslowesttemperature a low friction turbine.The cold gas extracts
The heat exchanger greatly
exchanger. imposed
usually
temperature
gas is then
load and then returns to
the cooling
from
room
near
room
above
cold
the
with
The
load.
engine,
expansion
heat
as
the
or
at
performed
The hot compressedgas is cooled to temperature. by ejecting heat into the environment. The gasis further
 {Uz +
4 PlVt)
gas
U
j=
pV =
and
Nr
=
PlV1)
Hs
~~
B)
H2.
Nr, hence H =
The
\\Nx.
engine by the gas is 
\302\273
W
N{t,
C)
r2).
The countefHowheat exchangeris an enthalpy device: it is an exchange expansion engine which extractsno externalwork. Most use expansion engines to prccool the gas closeto its Hquefiers gas It is impractical to carry She expansion cooling to temperature. liquefaction of of a liquid the point liquefaction: the formation phase inside expansion
enginescauses
mechanical
is
a
usually
JouleThomson
liquefiers
usually
temperatures,
with
multiple
The principle of
to
the
electron
potential potentials.
cooling by
gas
in
or
more
stage
expansion expansion
of an
below.
exchangers. isentropic
semiconductors.
When
ideal gas
electrons How
is applicable
from
a
semi
high electron concentration into a semiconductor she the electron gas expands and does work concentration, against barrier between the two substances that equalizesthe two chemical is used electronic cooling, called the Peltiereffect, The resulting
semiconductor wuh electron
heat
final liquefaction
Helium and hydrogen engines at successive tem
discussed
stage, two
eontain
The
difficulties.
operating
with
a lower
Gas Liquefaction by
down
to about
So 135K Gas
195 K quite routinely;in by the
Liquefaction
units
multistage
a
12.2.The work
~plt(Vj doneon
Here
dVt
the
is
in
gas
pushing
it through
is negative
The overall processis at constant expansion valve acts as an expansion = 0 in If B), we have H\\ ~ li2 in the
11
\342\204\242
ideal
\\Nr,
so
that
down
\342\200\224in
ts
r2
of
al! gases.
causes
significant
will condense.
gas
lower
pressure
p2,
the displacement and {he expansion valve
work
a
engine
sec this,
that extracts effect.
JouleThomson
There is
the expansion.
the
gas on the downstreamside. To
enthalpy.
At are
interactions
the
between
difference
{he
+p2(\"/2 recovered from the and dV2 is positive.
work
displacement
temperatures
pressure p, is forcedthrough
valve into space with
an expansion
called
constriction
as in Figure
Gas at
is simple.
implementation
practical
Effect
JouleThomson Effect
Intcrinolccular attractive interactionscausethe condensation icmpcratures slightly above the condensation temperature that work strong enough against them during expansion of the cooling of the gas. If the coolingis sufficient, part This process is JouleThomsonliquefaction.
The
JouleThomson
achieved.
been
hnvc
the
notice that the
zero work. With For an ideal gas
zero coolingeffect
for
an
gas.
gases a small temperaturechange work done by the molecules duriiig expansion. in real
2,2
The JouleThomson
through an expansion value. If be a temperature change during
done against initially
will cool
the
below
The sign
the
eflccl.A
the the
gas
is pushed
notlflieal. ihere will because of work expansion
gas is
forces.
If the
on JouleThomson expansion
internal
of the temperature
temperature is inversion temperature, riB,, the gas
intermolecuhtr
a certain
of
because
valve
Expansion
Figure
occurs
12:
Chapter
Cry
Liquefaclion dala
U.I
Table
n.
K
CO,
195
cm
112
902 77.3 20.4
o,
N,
H, \342\200\242He
4.SS
JHe
3.20
the
The las
liquid. Jrti
T( ano\" oot
T,,.,
Tt,
K
(jas
lo
for
U/mol
304
B050)
25.2
191
A290)
155 126 33.3 5.25 3.35
umn,
measured
em'/mol
223
66
6.82
67
621
5.57
205
0.90
28.1 34.6 28.6
51 B3)
0082
320
0.025
50.8
n walls
pressure.
mosphcric
der Waais
effect for
have
ran
\302\253=
JWt
gas, where a and
corrections caused by corrections
opposite
the
short
is
0.14
for
because its lrjple us LNG give daia
quamilics for shipping we of air. For helium,
signs. The
the critical
far
li'aab
gas.
+
{S2fV){bx
b are
positive
range repulsion and
initial such
poinl
fuel.
boih
c
Liquid for ihe
temperature. an expansion
for common
== lab
in
found
gases
A0.75)
that
D)
constants. The the
= 2/rt,
tempera!urc, defined by A0.46). lemperature. iln, is the inversion
We
 2a)
tola! correciion changes tinv
where xc
0.7!
12.1.
Table
JouleThomson
a van
8.7
'
H for
45
be la'ken up
can
thai
on the depends change during a JouleThomsonexpansion All gases have an inversiontemperature below which TIn, above which heals (he it cools, gas. inversion temperatures
Example:
314
893
of natural gas, which is liquefied in huge and niirogen are separaied iu lhe liquefaction ei isotope 4Hc and for 3He.
in
Mite
34.4
Carbon dioxide solidifies
are listed
wall
8.18
tndkai
Atl/V,
V,.
AH,
K
long sign
range at
last
two
atiraction.
arc the The correc
terms
the temperature
E)
at fixed For t < iin, the enthalpy here in expansion the work done against the increases; temperature increasesas the volume In a process at consiant enihafpy attraciive interactions between molecules is dominani. this increase is compensated by a decrease of the \\Nt ierm, that is, by cooling the gas. For The
temperature
Gas Liquefactionby i a
ioIccuIcspenetrate
farther
lhe repulsive
JoutcThot
s because now the anl: ai lhc higher
fixed lemperamre inio
the
work
done by the
lempcraluie
the
regio
wilh litjueficrs the JouleThomsonexpansioniscombined heat exchanger, as shown in Figure J2.3.The combination is a cycle in 1895 to called a Lindc cycle, aficr Carl von Linde who used such air starting from room temperature.In our discussion we assume that liquefy is ihe same the expanded the heat at from exchanger gas returning temperature as the compressed it. We neglect any pressure differencebetween gas entering the output of the heat exchanger and the pressure above the liquid.
Linde
cycle.
In gas
a counlerflow
To
and
from
comprcs
Figure by
12.3
The Lindecycle.Gas
combining
a countcrflow
JT expansionvaJv.
Liquefied
gas
JouleThomson
expa
Iieatexchanger.
Figure
as a
Performance of helium inpui pressure,
12.4
fund
ion ofihe
for various values of
ihc
inierna! refrigeration load available
ii^uciicrand
ihc
ihe
heal
through
See Problem Plenum,
still
exchanger A. J.
1971.
p. 1S7.
K
if
gas boiledoff
rather
Croft
cunes give at 4.2
coid helium
3. Afier
The solid
in
than
boiled
QiM (he
= toad
curvesgive tfDUl
A&mWLdcryo&mcs

ihe
Hia, ihe
is placed
by lhc load
oiT into
Liude cycle,
of 1 aim and
pressure
ouiput
temperature.
The broken
coefficient
liquefaciion
input
an
by the
operating
litjueficrs for
inside ihe
is relumed
liie atmosphere. (C. A. Baiiey, ed.),
Evaporation Coaling;
The
comhimtiion
hc;\302\273
fraction
is
X
Constant
liquefied.
=
Hla
lhe input
of
heat
the
and
output
under lhe pressurepoal.
peraturc

+
enthalpy ihat the
suppose
J)H9Ui'
\302\243!
lhe enthalpy
tfHl) is
exchanger
consimu
K
\302\2436}
mo!c of llou, = H(Tin.pBJare the enlhalpies per both at lhe common pressures, upper temperature
H(TiMpa) and
gas at
combination;
to 0.3
requires lhat
cniluilpy =
Here
Hie
enter
//;\342\200\236 IWii,
is ;i
valve
exchangerexpansion
arrangement. Let one moleof gas
PumpedHelium,
From
F)
per moleof liquid
its
at
boiling
lcm
the fraction
we obtain
17)
called
the
coefficient.
liquefaction
when
lakes place
Liquefaction
>
//\342\200\236\342\200\236, Hia;
> H{Tia,pJ.
HiT^J the
Only
JouleThomson
will
take place.
with
rapidly
If
Figure 12.4 shows lhe experimentally. from them Tor helium. The liquefaction
known
are
G)
calculated
coefficient
coefficientdrops
numerator
lemperalure of lhe heat exchangermaHer. at this temperature cools the gas, liquefaction
expansion
The three enthalpiesin liquefaction
(8)
at the input
enthalpies
the
is, when
thai
Tiat
increasing
decrease of the
of the
because
denominator. To obtain useful liqueinversion ;. > 0.!, input temperatures below onethirdof the liquefaction, say For are this usually required. temperature many gases requires precedingof and the engine. The combination of an expansionengine gas by an expansion is invariably a Linde engine cycle is called a Claude cycle. The expansion 12.1preceded by another heat excitauger, as in Figure in
Coofing:
Evaporatfon from
Starting
evaporation latent
and
G)
heat
liquid
the
Pumped Helium, lo 0.3 K helium, the of !hc liquid
simplest route lo lower temperaturesis
helium, by cooling of the of vaporization liquid Iteltum
The heat extractioncauses
ihc
:itonu'c
ihc
forces
of the
increase
that
cnuscJ
Tltomson cooling tlte initial initial state is a liquid.
further
cooling:
helium staie
is
pumping is extracted
work
io liquefy in a gas, while in
lieiium
away
along
is done the
first
evaporation
vapor,
with the
f
j
y '
vap,..
against the interpi;>ce.
hi JouL
cooling
the
17:
Chapter
Table
Cryogenics
12.2
3Hc reach
Tempera turds,
in
which
kctvifi.at
1 he
vapor
of 4He and
pressures
values
specified
p(lorr)
The
lowest in
helium
0.79
0.28
0.36
vacuum
1.27
1.74
2.64
0.47
0.66
1.03
179
cooling of
by evaporation
technology
pressure
vapor
gas and
0.98
liquid helium 14). As the ternperalure drops,the (Chapter (Table 12.2} and so docsthe raie ai which can be extracted from the liquid helium
accessible
tempcralure
isa problem equilibrium
0.66
drops
its heat ofvaporization
bath. cooling
Evaporation
Helium
in
such a5
Dilution
classical
K io
of
pressure
refrigeration
principles
0.0! K. is
dominated
evaporation refrigeratorin a We saw in Chapter 7 that
cooling
everyday ait
conditioners.
Militdegrees
Refrigerator:
Once the equilibriumvapor 0.6
cooling principle
household refrigerators and freezersant! in is in the workingsubstance. difference only
devices The
dominant
ts the
lose their by
very
clever
dropped to
I0\023
torr,
range utility. The temperaiure
dilution
helium
the
3He has
liquid
from
which is an
refrigerator,
disguise.* quantum are bosons, while
3He atoms are fermions. is not important at temperatures appreciably higher distinction This of \"fie, 2.17 K. However,the two than transition the temperature superfluid Below as altogether different substances at lowertemperatures. behave isotopes like 0.87 K. liquid 3He and 4He are immiscibleovera wide composition range, in Chapter 11 and is shown in the phase oil and water. This was discussed *He
of 3He4He mixtures in diagram the range labeled unstable wil!
atoms
11.7.
Figure
A mixture
decomposeinto
two
with
in
composition
whose
phases
separate
area. that are given by the two brandies of the curve enclosing compositions 3He phase. 3He phase floats on top of the dilute The concentrated 3He to about 3He in As T \302\273 the concentration of the dilute drops 0, phase 6 pet, and the phaserich in 3He becomes essentially pure 3He. Consider a liquid
*
For good reviews, sec D. S. Belts.Contemporary 36, 181A968);for a general review or cooling
Physics techniques
9.97 {1968): IC. 1 K see W.
below
Lounasmaa, Repts. Prog, Phys. 36, 423 A973); O. V. Lounasmaa, below t K, AcademicPress,Hew York, 1974. A very elementary
methods Scientific
American
221,26
(t%9).
.
\342\226\240
Wheatley.
Am1
J. Huiskamp
Experimental accoun!
and principles
Phys. O. V. and
Is O. V. Lounasmaa, .
12.5
Figure
dilution Cooling principle of ilm helium wiih a JHc4He nmiure. When from ihc pure ]He fluid and
Hlc is in equilibrium mixiure, sHe evaporaics
3He4He
mixture wiih more than 6 pet 3Hea*
range, near the bottomof Figure11.7.At
atoms have condensedinto
these
refrigerator. Liquid is added io the absorbs heat in ihc
4He
a temperature
temperatures
in the almost
millidcgree all the 4He
Their entropy is negligible which then behave as if they were of the mixture. If the 3He concenalone, as a gas occupying the volutne present the excess condenses into concentrated liquid 3He and exceeds concentration 6pct, heat If concentrated liquid 3He is evaporatedimo the 4He latent is liberated. the latent heat is consumed.The principle rich of evaporation phase, cooling can again be applied: this is the basts of the heliumdilution refrigerator. To to obtain see how the solution of 3He can be employed refrigeration,
comparedto
that
of
the
the
remaining
ground
state
orbital.
3He atoms,
the equilibrium between the concentrated3Heliquidphase the and tile lliai JHc:4lic nilio of dilute gaslike plliise (Figure 12.5).Suppose with the dilute phase is decreased,as by dilution pure *He. In order to restore 3He aiomswil! the equilibrium from the concentrated concentration, evaporate
consider
iHc
3He liquid.Coolingwill
result.
be a cyclic process the 3He4He mixturemust again. separated is tile different common method Tile most distillation, equilibrium using by 12.6 shows a schematic 3He and *Hc of 12.2). (Table Figure vapor pressures on these principles. The diagram is highly built diagram of a refrigerator In in actual refrigerators titehcat exchangerbetween oversimplified. particular, An alternate chamber and the still has an elaborate multistagedesign. the mixing of 4He method* to separate the. 3He4He mixture utilizes tile superfluidity of reasons it is less commonly used, below 2.17 K. For a variety, practical
To
obtain
although
Us
performance
is excellent.
.
\342\226\240 .
Chapter
12:
Cryogenics
3He pump
loop
Key: Liquid \342\226\240 I 'lie
Dilution Refrigerator:
Helium
AtitliJegre
Hdium dliulion refrigerator. Prccooledliquid a mixing 3He enters chamber a( (he tower cud of the assembly, wlicrc cooling takes place by ihe quasiof the 3He atoms into the denser JHcJf1cmixed cvaporaiion underneath. phase 12.6
Figure
The quastgas of JH atoms dissolvedin liquid *He then diffuses through heat exchanger into 3 still. There the JHe is disiilledfrom the 3Hc4Hc
a countcrfiow
mixture
a useful 3He evaporation and circulation selectively, and is pumped olf.To obtain heat must be added to the still, 10 raiseUs temperature to about 0.7 K, at which
rate,
vapor pressure is ssiH much smaller. Thus, the 4He does not a nearly stationary appreciable extent; ihe *Hc moves riirough JHe is returned to ihe system and is condensed background of 4Hc.The pumpedoff in a condenser that is cooled to about I K by contact with a pumped 4He bath. The constriction below the condenser takes up the excess pressuregenerated by the the *He
temperature
lo any
circulate
circulation in rhe
first
pump over ihe pressure in siill. ihcn in the counter/low
still. The liquified JHe is cooled further, heat exchanger, beforereentering tlic miung
the
chamber.
In the convendilution refrigerator has a low temperaturelitnft. conventional evaporaiioii this limit arose because of the disappearance of refrigerator the but the quasigas phase of 3Hepersists down to t = 0, However, phase, gas ihe heat of quasivaporizationof JHe vanishes to x2, and as a proportionally Ihe rate heat removal from the mixing chamber vanishes as i1. TI'S result, low device;* limit is about 10 mK, In one representative temperature practical was capable of a temperature of 8.3 mK has been achieved:ihe same device
The helium
removing 40/AVat 80mK. ihe
design
there is
SmK
below
Temperatures
of Figure
I
2.6,
no needto cooi the
mixingchamberdrops removed from The
dilution
the
off the
3He
incoming its
below
single shot operation. If, 3He supplyafter some time of opeiation,
be ncltievcu by
can
we shut
sleady
itself, state
and
value,
of the
ihe temperature until
has
3He
all
in
been
chamber.
refrigerator
is not the
oniy cooiingmethodin
the
inillikelvin
known the peculiar propertiesof JHe.An alternate method, in Figas Pomcranchsik cooling, utilizes the phasediagramof 3He,as shown and between Figure 7.15, with its negative liquid slope of the phase boundary ant! solid 3He, The interested reader is referredto the reviews by Huiskamp Lounasmaa, and by Lounastnaa, citedearlier.
that utilizes
range
'
N.
H.
Pcnnings,
84, 102A976}.
R. de
Bruyn
Ouboicr,
K. \\V. Tacoois.
Phjiica 8
SI. !0! A976).
and
Physiea
B
DEMAGNETIZATION:
ISENTROPIC
QUEST FOR ABSOLUTEZERO 0.01
Below
doniimim
K the
cooling process
is the isciitropic(adiabaiic)
dcm;ig
paramagnetic substance. By this process, temperaturesof I niK have been attained with electronic paramagnetic systems and j /(K with nuclear systems. The method dependson tlie fact that at a fixed temperaparamagnetic temperaturethe of a system of magnetic momentsisloweredby application of a entropy slates are to because accessible the system fewer magnetic field\342\200\224essentially small. when ilic level splitting is large than when the level is splitting Examples of the dependence of the entropy 2 on tlie magnetic field were given in Chapters of a
iictizatioii
3.
and
We first apply a will
magnetic field
a value
attain
without
will
then flow
remain
t2 into
to the
appropriate
reducedto B2
the
changing
the
tj
the spin
specimen
system only
of
value
entropy
means
which
unchanged, <\342\226\240< When
at constant
Bt
temperature ij. The spin excess If the magnetic field is then Bj/tj. the of spin system,the spin excess
that B2/z2
will
is demagnetized
from
the
system
#i/ri
equa'
12.7
can
vibrations, as in
of interest the
the temperatures
At
Bi,
isentropically,entropy
of lattice
entropy of the will be usually negligible; thus the entropy ofthe spin system during isentropie demagnetization of the specimen. Figure
\302\253
HBz
lattice
essentially
constant
\\Latttce
1
Total
w
Spin
Lattice
Time\342\200\224
Before Time
Figure
12.7
cooling
of
Time\342\200\224\342\200\242
Before Time
New
equilibr:
at which
magnetic field js removed
field
removed
demagnetization the total entropy of the S in of the lattice should be small entropy with the entropy of the spin system in order to obtain significant the lattice. During
isentropie
specimen is constant. The comparison
equilibrium
at which
magnetic is
New
Fig
vibrations
initial
.
is
Quest for
Demagnetization:
lsentropic
Zer
Absolute
as a function of Icmpcralure.assum Entropy fora spin \\ sysiem of field 100 Bx gauss. The specimenis magnetiz magnetic Thu cxlcrna! ntagnctit insulated isothermaiiy ihcrmaMy. along ah, and is then field is 1 timed on a reasonable off along/>c. Ill order to keep the figure sculc llic initial temperature tlie field are lower woi and external than magnetic Tj
12.8
Figure
inicrna!
an
used
in
The
random
practice.
steps
out in the
carried
cooling processarc shown
field is
applied at temperaturetx
the
the
with
giving
surroundings, {At? ~ 0)
insulated
and the field
with
the
specimen
isothermal
in
thermal
in good
removed; the specimenfollows
6c, ending up at temperature t2. The thermal contact is broken helium provided by gas, and the thermal
gas with The
the contact
constant ai
is
t,
by removing the
a pump.
population
of a
magnetic subievelis a function
is the magnetic momentof a ofthe
contact
path ab. The specimenis ihen
path
entropy
12.8. The
Figure
population
spin.
distribution;
The
spinsystem
hence the
only
entropy
of
ntB/x,
is a
spin entropy is a function
function
only
m
where only
oimBjx.
localinteractions is theeffeclive field that corresponds to thediverse among temperature r2 reached in an spins or ofthe spins with the lattice, the final
If SA the
isenlropic demagnetization experimentis
(9} rt the initial temperature.Results which as CMN, Figure IZ9 for the paramagnetic salt known . . magnesium nitrate, where
B
is the
initial
field and
are
denotes
shown
in
cerous
Final
Figure 12.9
removed
as
cniirciy, and
fields
iiiiiial
field
magaclic
bul
0.6
0.5
in K
temperature,
Bf versus
experiments ihe magnetic
In fhese
nil rale.
After
Final
0.4
0.3
0.2
0.1
0
final
field
was
not
indicated values. The were idcnliait in all inns. S. Still and J. H. Milncr.
m tfic
only
icmpcraiurcs
unpublished results 61\" J. 6, by N, Kuril, Kuovo Cimcnio (Supplement)
cilcd
1109A957).
The processdescribed so into a cyclicprocess thermally
is
far
a single
shot
disconnecting,
by
demagnetized working substance from at t,, and repeating the process.*
converted
process. It is easily in one
way
or
it to the load, reconnecting
the
another, the
reservoir
Nuclear
Dcmagnelizalion
nuclear
Because much
are
weaker
paramagnet.
\342\200\242
C.
similar
lhan 100 limes lower with a nuclear paramagnet in The initial temperature of the nuclearstage
temperature
C. B.
V. Hctr.
Rosenblum,
lhan
arc weak, nuclear magneticinteractions a to reach electronic interactions. We expect
moments
magnetic
W.
Barnes,and
A. Slcyerl.
and i.
Daunt.
J.
G.
A.
Barclay.
Rev. Scj, insi. 25. IGS8 j954); 17, 3S! A977).
OHgcnics
electron
with
an
a
nuclear
spin
W. p. PraH,
S. S.
t\\'uc!ear Demagnetization
Iniiial magnetic
0.6
in
Held
KG
1
Initial
B/T'm
\\QS
G/K
Nuclear demagnctizaflons of copper skirting from 0.012 K and various fields. After M. V. Hobdcn imd N. Kuril. Phil. Mag.
Figure
I2J0
in
nuclei
the
metal,
1.1902!1959).
must be lower than in an cooling = = lfwestartatB SOkGandT, 0.01 10 percent decrease on magnetization is experimen!
over
This is
to
sufficient
ihe
overwhelm
T2 ss 10\027K.The
temperature
electron spincooling experiment. * 0.5, and the entropy K,then/fiB/*87\\ of the maximum spin entropy. a final lattice and from (9) we estimate nuclear
first
oui by Kurli and coworkers on Cu nuclei at about 0.02 K as attained by electron stage
the
B\302\261
moments
of
=
3.1
the
and
I0~6K. reached in this experimen!was 1.2 so line of Ihe form of (9): 7\\ = TC.1/B) Bin wilh ^auss, of the magnetic mointeraction field gauss. This is the ciTcctive
Cu nueiei.
nuclei
Temperatures
cooling
first
The
cooling.
demagneltzallon
load was
The motivation conduction
at
below
the
electrons
temperature
1//K
for
have been
using
help
of the
first
nuclei
ensure
in a
metal rather than
rapid thermal
contact
stage.
achieved in experiments in
the system of nuclearspinsitself,
results
The
x
in an insulatoris that
of lattice
from a
starting
metal,
lowest temperature in Figure 1210fil a llial
was carried
experiment
cooling in
particulatly
in
which
experiments
the
combinations of cooling experimentsand
that were
nuclear
resonance
magnetic
experiments.*
SUMMARY
1. The
dominant
two
a gas
of
cooling
of the
principles
by letting
work
do
it
of low temperaturesarc ihe a force during an expansion
production
and the iscntropicdcmagncii/atioii of a
against
substance.
paramagnetic
is done work cooling is an irreversible process in which interatomic forces in a It is used as last the attractive against gas. cooling
JouleThomson
2.
stage
in
gases.
lowboiling
liquefying
3. In evaporation cooling the work is also doneagainst but starting from the liquid phaserather than the
the 5.
dilution
helium
of magnetic
system
moments,
an
when
external
moments may magnetic nuclear moments, temperatures
using
devices
cooling device in
which
4He.
lowering of the
utilizes the
The
strength. By
evaporation
household
of cooling
laboratory
gas of 3He atomsdissolvedin
demagnetization
Isentropic
is an
refrigerator
gas is the virtual
basis
the
forces,
Using different
phase.
gas
working substances, evaporation coolingforms and cooling devices, automobile air conditioners, (in the range 4 K down to SGmK).
4. The
interatomic
the
magnetic
temperature
of
field is
reduced
be electronic or nuclearmoments. in the microkelvin range may
a in
be
achieved.
PROBLEMS
L Helium for
as a
helium
Use
of the
and
liquid
data
in
6
it as such
in
helium Table
for \342\200\242See,
example,
.
a
der
van
Waals
tlte liquefaction coefficient X gas. Select the van der Waals
a way that Tor one and that 2a/b is the
12.!. Approximate the Hout
849A970).
(a) Estimate
H'aafcgas.
by treating
coefficientsa volume
der
van

M. Chapcllier,
Hi(q
M. Goldman,
AH
4 f(rin
V.
mole 2Nb is the actual molar
actual inversion temperature. denominator in G) by setting 
H. Chau
xliH)
A0)
,
and A.
.._...\342\226\240
Abragara,
Appl.
Phys41,
where All is the latent heat of vaporizationof liquid how this helium. (Explain arises if one treats the as an ideal gas). The approximation expanded gas /. as a function of the molar volumes Yin and resulting expression gives Vatll. Convert to pressures by approximating the l\"s via the ideal gas law. (b) Inaert numerical values T = 15 K and compare with for 12.4. Figure Carnot liquefier. (a) Calculatethe work mole of a monstomic ideal gas if the liquefy Assume that the gas is suppliedat roomtemperature
2. Ideal
one
pressure p0 at whidi the liquefied 7\\ be the boiling temperature of the
of vaporization.Show
that
under
gas gas
these
\\VL thai
iiqticfier To.
be required to operated rcversibiy. the same and under
would
is removed,
1 atmosphere. typically and A// the latent at this pressure, conditions
Let
heat
A1)
To derive A1) assumethat the gas is first cooled at fixed pressure p6 from To the fixed between to Tfc, by means of a reversible refrigeratorthat operates upper temperature Tb ~ To attd a variable lower temperatureequalto the gas ~ 1\\.After reaching Tb the Initially temperature. T, = To, and at the end T, at the lower the latent heat of fixed extracts temperavaporization refrigerator of temperature Tb. (b) Insert To = 300 K and values for Tb and AH characteristic liter of liquid helium. helium. Reexpress the result as kilowatthours per Actua! helium liquefiersconsume5 to lOkWh. liter. 1 mols\021 in which cycle helium Hqucfier. Considera heliumliquefier enters the Lrnde stage at T(o = 15 K and at a pressure pla = 30 aim. all the in liter hr\"'. Suppose that liquefied (a) Calculate the rate of liquefaction, helium is withdrawn to cool an externalexperimental the releasing apparatus, load in boiledoff helium vapor into the atmosphere. Calculate the cooling the the it is this watts sufficient to evaporate heliumat rate liquefied. Compare if the liquefier is operated as a closedcycle with the load obtainable cooling the apparatus into the liquid collectionvesselof the refrigerator by placing the heat so lhat the still cold boiledoff helium gas is returnedthrough liquefier, and ex(b) Assume that the heat exchangerbetween exchangers, compressor ideal that return is the (Figure 12.1) sufficiently expanded gas expansion engine the same temperature Tc as the that leaves it with pressure is at essentially pout compressedgasenteringit with pressure pc. Show that under ordinary liquefier must extract the work operationthe expansion engine
3.
Claude
of gas
Te 
TJ ,
A2)
Chapter
12:
per mole
Cryogenics
of compressed gas. HereTin,
pin,
pBUt, and
X
have
the
same
meaning
Undo cycle sectionof this chapier. Assume the expansion engine between ihe and operates isemropically pressuretemperature pairs {pc,Tc) Estimate (Pia>Ti(l). From A2) and the given values of (pia,Tia),calculate(pc,Te). (c) the minimum compressor power required to operate the iiquefier,by assuming the compression is isothermal from to pc at temperature that Tc ~ 50\"C. poai Combinethe result with {hecooling loads calculated under (a) into a coefficient of refrigerator for both the modes of operation. Compare with performance,
as in ihe
Carnot
iimit.
cooling limit. Estimate the lowest temperatureTmia that can if the cooling load is 0.1 W 4He evaporation cooling of liquid = and the vacuum I02filers\021. Assume (hat the pump has a pump speed S helium vapor pressure above the boiling is equal to the equilibrium helium lo TBliJ1, and assume that ilic helium gas warms vapor pressurecorresponding the to roorn and expands accordingly before it enters tip temperature ptunp. Nota: Tlte molar volume of an ideul and atmospheric at room gas icmpcramrc pressure G60torr) is about 24 liters. Repeat the calculation for a mtjch smaller heat load (I0~3 W) and a faster puinpA0J is defined liter s\"\021). Puntp in speed 4. Evaporation be achieved by
Chapter
14.
temperature far demagnetizationcoaling, Considera paramagnetic field ofiOOkG with a Dcbyc temperature {Chapter 4) of 100K. A magnetic or lOtcsia is available in the laboratory. the temperature to which the Estiniate that salt must be prccoolcd other means order cooling by in significant magnetic process. Take may subsequently be obtained by !he isentropic demagnetization 5. Initial salt
to be I Bohr ion the magnetic momentofa paramagnetic By signifimagneton. 0.1 of the initial we to understand temperature. cooling may cooling
significant
13
Chapter
Statistics
Semiconductor
ENERGY
FERMI
BANDS;
LEVEL;
355
AND HOLES
ELECTRONS
358
ClassicalRegime of
Law
362
Action
Mass
362
intrinsic Fermi Level
363
/rTVPE AND Donors
Fermi
SEMICONDUCTORS
P'TYVE
363
and Acceptors
364
Semiconductors
Extrinsic
in
Level
365
Semiconductors
Degenerate
368
Impunly Levels
Occupationof Donor
369
Levels
Example:
Gallium
SemiInsulating
Arsenide
373
pn JUNCTIONS
ReverseBiased
pn
NONEQUIUBRIUM
SEMICONDUCrORS
Abrupt
377
Junction
QuasiFermi
Drift and
Flow:
379
Diffusion
Example:Injection
381
Laser
Example:
Carrier
379 379
Levels
Current
372
Through an Impurity Level
Recombination
383
SUMMARY
385
PROBLEMS
387
1. Weakly
2.
Intrinsic
387
Doped .Semiconductor and
Conductivity
Minimum
Conductivity
387
3. Resistivity and Impurity Concemraiiou 4. Mass Action Law for High Electron Concentrations
387
5. Electron
387
and
Hole
Concentrations
in InSb
6. Incomplete lonizationof Deep Impurities 7. Builtin Field for Exponential Doping Profile
8.
Einstein
Relation
for
High
Electron
Concentrations
387 3S7
388 388
13:
Chapter
Ha
ht n,
Carrier
^
conduct v\302\273'ilti
conccnlration
= cffcciivc =
effective
38S
Lifetime
388
Pair Generation
ElectronHole
iiiiiiui^zny
388
Laser
9. Injection
10. Minority 11.
Statistics
Semiconductor
[tin xind valence b jfiu\302\243 \\ elects oeis snd
of holes
hoics^ donors
*inu
3cccplovs.
I he
^
c^uantiiEn
conccnlnU/on
for condudion
quamum
coiKcntralion
for
ctccirons;
holes.
densities of states for the conduction In the semieonductoi tileralurc n,. and % ate called ihe effective and valence bands. Notice iKal we use ;i fo( tin; chemical potential or Fcimi level, and we use Ji foi
caiiicr mobilities.
ENERGY
The is
FERMI LEVEL;
BANDS;
ELECTRONS
HOLES
AND
of the FermiDirac disiributionto eiecirons in semiconductors application central to the design and operation of all semiconductor and devices,
to much of modern electronics.We
of semiconductorsand
treat
devices
semiconductor
of the physics thermal physics. of the physics of
those
below
thus
aspects are parts of
that
is familiar wjlh the basic ideas as 'n texts on solidslatephysics the crysialline soiids, treated in and on semiconductor the We assume the devices cited references. general of bands and of conduction electrons and hotcs. Our principal by concept energy aim is to understand the dependence of the altimportant concentrations of conduction electronsand of holesupon the concentration and the impurity that
assume
We
eleclrons
the
reader
in
temperature. A
semiconductor
band t
=
and
0 al!
is a
system
with
electron
orbitats
grouped
into two
energy
energy gap (Figure 13.1). The lower band is the valence at the upper band is the conductionband.* In a pure semiconducior valence band orbitats are occupied and alt conduction band orbitals
bands separatedby
an
are empty. A full band cannot carry any current, so that a pure in a semiconductor at r = 0 is an insulator.Finite conductivity
semiconductor
follows either
in the conduction from the presence of electrons,catled conduction electrons, orbitats in the valence band, called notes. band or from unoccupied Two different electrons and holes: mechanisms rise to conduction give of electrons from the vatencc band to the conduction band, Thermalexcitation of or the presence that change the balance between the number impurities of electrons available to fill them. of orbitats in the valence band and the number and the energy the band We denote valence the energy of the top of by \302\243,., rence of the bottom of the conductionband by e{. The differed
is the energygap of the semiconductor.Fortypical eV. 0.1 and 2.5 electronvolts.In silicon,e, ^ 1.1 \342\200\242 We tieai both bands of bands wtih additional
as single
for out (imposes it bands; gaps wiihtn each gioup.
does
semiconductors Because
t ^
noi mailer thai
Eg
is between
1/40 eV at boih
may
room
be groups
Chapter 13: Semiconductor
Statistics
Empiy
band
Conduction
Energy
atr =0
gap
Filled
air =0
t3.1
Figure
Energy
Air = conduct the
structure of
band
orbilajs occur in
a puic semiconductor
bands
\\vhjch
we usually
havec,, \302\273t. Substances 2.5 eV are usually insulators. Table 13.1 gives semiconductors, together with other properties Let
nt
ihe
denote
concentration of
insulator.
exlerid
gap.
energy
temperature,
or
through the crystal. Gallorbitaisuplothe top of the valence band are filled,and ihe the bands is called ion bantl is empty. The energy interval between
The electron
the
pure semiconductorthe ii, =
crystal is electrically neutral. Most semiconductorsas usedn\\ devices impurities that may become thermally
gap of more than later.
electrons and two
about
for selected
gaps
energy
needed
of conduction
concentration
holes. In a
with a
wiH
nh
the
con
be equal:
B)
\302\253*.
if the
\342\226\240temperature.
positively
tiiat
Impurities
charged
in the
give
an
have
ionized
electron
been
In the
inteniionaiiy
doped with
semiconductor
to the
crystal
process) are called donors.Impurities
(and that
at room become accept
Bands; Fermi
Energy
data
structure
Band
13.1
Table
Lml; Elec
Energy
Q
ions and
c Idea
at
gaps
300 K
itions
effective

olcs
li
free
1300K
\",.
eV
'
2.7 x 10
Cc
0.67
1.0 x !O 9
GaAs
1.43
Let
4.6 X
!he
\302\253j+be
'
10
10\"
0.58
11.7
0.35 0.71 0.42
0.07
0.073
15.8
13.13
12.37
0.39
0.015
band (and become negatively of positively
concentration
charged acceptors. An
ts called
1.06
0.56
17.S8
the
in
charged
acceptors.
of negatively
concentration
to
vacuum
ilia
10\"
6.9 x 10'* 6.2x 10\"
the valence
are called
process)
relative
cicctr on mass
x to'a
1.5 x
4.9 x 10 '
from
electron
an
1.1 x 52
4.6 x 10 '
0.18
of the
c
t.[4
1.35
masses,
constants
in/ftn
c
Si
,,p InSb
units
in
Dielectric
fstatcs
Dcnslty
i concentr
antun
=

/[/
charged donors and
The difference C)
na~
donor concentration.The electrical neutrality
the net ionized
the
na~
condition
becomes
=
which
specifies The electron
distribution
the difference concentration
[i is
the chemical potentialof the
electrons. In semiconductor theory the Fermilevel. called Further, almost
level
always
reserved
is designated
of a meia!
which
in
ihe
limil
r *
by ef we
for ihe
or
by
at
The
electrons.
To \302\243.
theory
temperature.
confusion
avoid
as cf our
and
with
the
the Fermi
Fermi
energy
which stands for tile Fermi
previous
usage
to
is always potential is the character fi
chemical
semiconductor
e refers
subscript
electron and hole mobilities,and
designated
any
E) /
electron
the in
0, we shall maintain
chemicalpotential
D)
no~~t
betweenelectronand hole concentrations. may be calculated from the FermiDiracdis
exp[(e where
\342\200\224
Hj*
6:
of Chapter
function
=
An
of the
level
letter /j for the
;i and
Given
the distribution
The
t, the number of conductionelectronsis obtained function /,(e) over all conduction band orbitals:
of
number
[l \302\243
 /.(*)]
= I
ftU).
V)
VI!
VB
is overall valenceband
the summation
summing
is
holes
*\\ =
where
by
Here
orbitals.
we have
introduced
the quantity
at energy e is unoccupied.We say a hole\"; that is the distribution [hen/h(e) function for holes just as f\302\243t) is the distribution function for electrons. Comof with shows that the involves hole occupation probability Comparison (8) E) \342\200\224 e where the electron y. y. occupation probability involves c p. = = The concentrations and nh nt NJV NJV depend on the Ferm't level. But what is the value of the Fermi level? It ts determined by the electrical \342\200\224 = as An. This is an neutrality requirement D), now written nh{y) nt{y) the must for to solve we determine the functional implicit equation y.; equation
which
that an orbital probability the unoccupied orbital is \"occupiedby the
is
ne{y) and
dependences
nh{y}.
Classical Regime We
assume
that
by the
defined
regime
concentrations are
and hole
electron
both
requirements
that
\302\253 1
fr
and
fh
in
as
\302\253 I,
the
classical
in Chapter
This will be true if, as in Figure 13.2, the Fermi level lies insidethe energy and ts separated from both band edgesby energies that large enough 
exp[(\302\243c
To few
satisfy times
inequalities
are
satisfied
(9) both larger (9)
\342\200\224
place
n)
(gc
than
in many
/i)/t]
\302\253 1;
and
(/t

exp[0*

eu)
to
have
ej/t]
\302\253 1.
be positive
6. gap
(9)
and al
least a
The a semiconductor is callednondegenerate. and limits on the electron and hole concentrations
t. Such upper
applications.
/J.E.)and /h(g)reduceto classical
With
distributions:
(9)
the
two
occupation
probabilities
Classical
':.\"\\~'\\
I
Regi
Conduciion I
band
13.2 Occupancy oforbiials as a finite temperaiure, to the Fermiaccording The conduclion and valence bands may be represented Dirac disSribution function. numbers in terms of temperaturedependenteffedive Nc, Nc of degenerate orbiials The located aS the iwo band edges e,, \302\243\342\200\236. n(, n( arc ihe corresponding quantum
Figure
=
We
use
F)
and

exp[(e
A0) to
write
=>
the
total
number
cxp[(,,

of conduction
A0) electrons in the
form
N,
A1)
Statistics
13: Semiconductor
Chapter
where we define
N, \342\200\224 \302\243 is \302\243c
Here
band
the
A2)
ej/r].
conduction electron referred to
of a
energy
\302\243exp[~(s Cfl
conduction
the
ec as origin.
edge
The expression for
lias the maihematical fomt of a partition function a similar conduction band. In Chapter3 we evaluated sum denoted there by Zlt and we can adapt that rcsuil lo the prescuJ problem with an for modificalion band siructure effects. Because of the approximate rapid decrease of cxp[{\302\243  e()/i] as c increases above its minimum value for
one
at ee,only
Nc
in the
electron
ihe
of orbitals
distribution
a
within
range
of a few
above
x
cc really
evaluation of the sum in A2). The orbitalshigh in the band make a negligible contribution. The important is that near the band edge point the electronsbehave very much like free particles. Not only arc the electrons of the semiconductor, but the energy mobile,which causes the conductivity distributionofihe orbitalsnear the band edge usually differs from that of free in the particles only by a proportionality factor in the energy and eventually matters
sum
in ihe
for Z%.
We can arrange for a suitable proponionaliiyfactor by use of a device we calculated the called the densifyofsfates effective For free particles mass. in C.62), but for zero spin. For particlesof spin  the partition function Z\\ result is larger by a factor of 2, so that A2) becomes
Nc = Zj
=
this gives
Numerically,
NJV = 2.509 x where
Tis
1019 x
dependenceas this
Nt A3),
formally
for but
ihe same temperature a by proportionality facior. We exhibits
semiconductors
actual differs
by wriiing,
in magnitude in
io
analogy
A3),
Nc *= 2(mSz/2nti2K'2V
where Experimental
is more
(H)
.
G/300KK'2cm\023
in kelvin.
The quantity
express
A3)
2(mx/2nh2)i!iV.
2nQV=
me*
is
values
than
called
the
arc
given
a formality.
effective mass for denshyofbtates in Table 13.1. The introduction
In the theory
of
electrons
A5)
, electrons.
Experi
of effective
masses
in crystals
it
is
shown
that
Classical Regime
Ihe dynamical behavior of electrons and
forces such as electricfields, free electron
the
from
Ihe densityofstaJes We
the
define
that
is
holes,
the influence of external effective wilh masses different
under
of particles
dynamical massesusually masses, however. mass. The
quantum
coticetilralion
for ??\302\243
eleclrons
conduction
from
different
are
as
NJV 
nc =
A6)
By A0 the conductionelectronconcentration
=
ne
NJV
becomes
A7)
Jo the assumption The earlier assumption (9) is equivaleni that n,. \302\253 nc, so that the conduction electrons act as an ideal gas. As an aid to memory, we may level at ;i. IVanriiuj: think as wiili the Fermi of Ne arising from N( orbilalsat \302\243\302\243, In is invariably called the effective density of the semiconduclor literature nt
statesof
conduction
the
Similar
reasoning
band.
gives the
number of holesin
valence
ihe
band:
 e)/t]
wiih
the
definition
A9)
We define
the quantum concentrationnv
e
where
wk*
concentration/^
is
llns
NJV
dcnsityofsJaJcs
s
for
holes
as
2{in^z/2jihi)m.
effective
mass for
holes.
By
(IS)
ihe
hole
s NJV is
nh
\342\200\224 aBexp[\342\200\224(/(
\342\200\224
e,)
B1)
gives the carrier concentrationin
A7), this
Like
positionof
ceniraiion and the
edge.
In
of the
valeuce
is
of the
independent
'V'k = 'WexpffE,.where the and
energy gap
the common
concentration
t^
Fermi level so longas the
concentrations
Then
classical regime.
in the
valeuce band effective density of states
Action
The productn^nh are
the
called
con
quantum
to the
band.
of Mass
Law
is
nu
(he
of
terms
relative
level
Fermi
literature
semiconductor
(he
the
s=
\302\2439
ec

eff.
In
O/r]
s=
ncnuexp(~
we have
semiconductor
a pure
B2a)
,
eJz)
value of the two concentrations is called the of the semiconductor. By B2a),
\342\200\224
>ibt
ut.
carrier
intrinsic
B2b)
The Fermi level independence of the retains its value even when ne <\302\243 nh, as atoms,
impurity We
then
may
both
provided
product in the
concentrations
means
n^iij,
this
that
product
presence of electrically charged remain in the classical regime.
B2a) as
write
B2c)
The mass
of
value
action
the
depends
product
law of
only on
the temperature.This result
semiconductors, similar to
chemical
the
mass
action
is
the
law
(Chapter 9).
Intrinsic FermiLevel For
an
intrinsic
sidesof A7)
and
semiconductor
eB
= e^
nh
and
we
may
equate
the righthand
B2b):
neexp[(ec Insert
ne =
\342\200\224 and
eB
divide
 ^)/r] = by n,.exp(
(vO\022exp(\302\243/>*).
\342\200\224\302\243(/r):
ej/2r].
B3)
and Acceptors
Donors
logarithms to obtain
We lake
=
/i
{U,
+
+ \302\243,.)
= l(cc
t Iog(u,/u,)
of A6) and B0). The Fermilevel (he middle of lhe forbidden gap, but by use
is usually
that
amount
Pure
an
semiconductor lies near the exact middle by an
intrinsic from
displaced
B4)
3rlog(\302\273ifc7\302\273i,*),
small.
AND pTYPE SEMICONDUCTORS
wTYPE and
Donors
for
+ e,) +
Acceptors
are an
semiconductors
idealization of Httle
Semicon
interest.
practical
to usually have impurities intentionally added in order semiconthe concentration of either conductionelectrons A more with conduction electrons than holes is called \302\273itype; a semiin devices
used
Semiconductors
or holes.
increase semiconductor
n and electrons is called ptype.The letters p and in carriers. Consider a silicon positive signify negative majority crystal atoms. which some of the Si atoms have been substituted by phosphorus hence each P has Phosphorus is just to the right of Si in the periodic table,
semiconductor with
fit
into
the
Si it replaces. valence band; hence a Si crysial
filled
than the
more
electron
one
exactly
more holes than
do not electrons P atoms wiil contain
extra
These some
with
more conduction electronsand, by the law of muss action, fewer holes than \302\253 Si crystal Next consider aluminum atoms. Aluminum is just to the left pure of Si in the periodic fewer than the Si table, hence Al lias exactly one electron it replaces. As a result, Al atoms increase of holes and decreasethe the number of
number
electrons.
conduction
Most impurities in the same columnsof the periodic behave in St just as P and Al behave. What matters
electrons
relative
from
Impurities
Similar reasoning For the presentwe may
enter
assume
that
is the
to Si not the total number of other columns of the periodic table will
and
can be appliedto other lhat
assume
band or
the conduction
each acceptor
fill
one
and Al will
number of valence
electronson behave
not
for
semiconductors,
donor
each
as P
table
the
atom.
so simply.
example
GaAs.
atom contributes one electronwhich hole also in the valence band. We
atom removes one electron,either
the
from
valence
are called the approxifrom the conduction band.Theseassumptions all impurities when ionized are either approximation of posiimpurities: fully ionized A\". donors D+ or negatively charged acceptors positively charged electrical The condition D) told us that neutrality
band or
An
=
nt
\342\200\224~ nfc
nj+
\342\200\224
na~.
.
.
.
B5)
J3; Semiconductor
Chapter
Becausenh
=
Statistics aciioti law,
mass
the
m,V\302\273*from
equation for
we secthat
B5)
to a
leads
quadratic
nc\\
B6)
V\".^!^1.
root is
The positive
=
i{[{A'O2
\302\273*
~
because
and
nh

^
\302\253 \302\273* l([(AnJ
Most
the
often
+ V]1''2
 An],
nr or
either
is much
ft*
[(AnJ +
an ntype
=
An*]\"*
=x
A/i
+ nffAn
ptypc semicondiiclor
Ji. ^ The
majority
n?/\\An\\
proportionalto Level
Fcrmf
By use
An

the
B9)
ln*l\\bt\\.
B7) becomes
nh
An;
** n^/An
and B7)
is negative
\302\253 n,;
while
extrinsic
the
carrier
mmoriiy
C0)
becomes
{A/i +
nk ^
\302\253 n,.
h;V1A*i1
limit
^
B8)
C1)
jAffj.
is nearly
concentration
equal to
is inversely
jAnj.
in Extrinsic
of the massaction
having to calculate
the
solving
m +
carrier concent ration in
the magnitude of An,
/ij,by
B7)
+ (hj&nJ]1'2
^[i
&n is positive and
semiconductor nt
In a
B8)
nt.
In
than n,:
larger
be expanded:
then
can
B7b)
extrinsic semiconductor. The squarerootsin
defines an
Condition
B7a)
\302\273
\\An\\
This
,
An}
compared to the intrinsic con
is Urge
concentration
doping that
so
concentration,
+
have
we
An
+ V]1/J
A7) or
Semiconductor law Fermi
concentrations without Tlic Fermi level is obtained from n, or
we calculated level
B1) for/c
first.
the carrier
Scum
Degenerate
Figure various
the
13.3
The Fermi icvet in
doping
band edges.
A
sniaii
silicon
as a function
The Fernifleveis
concentrations. decrease
of liic energy
of lempcraturc,
are expressed relative
gap
wiih
icmperaiure
for
Ic
has
been negieaed.
now use B7)
We may
to
find
ft as
a function of
temperature and doping level
13,3 gives numerical results for Si. With decreasing Figure eiiher Fermi level in an extrinsic semiconductorapproaches She valence band edge.
ths
temperature
An.
the
cr
conduction
DegenerateSemiconductors one
When quantum
of
concentration,
carrier.
the carrier we
The calculation
is increasedand approachesthe use the classicaldistribution no longer A0)
concentrations may
of the carriorconcenir.iiion now
Fermi gas in Chapter 7. The sum is written equal to [he number of electrons, states times the distributionfunction: of the
N
over as
follows
all
occupied
an
iiiteura!
the
quan
for
iliiit
treatment
orbitais, which n over the density yf
IS:
Chapter
Statistics
Semiconductor
where for free
llie panicles of mass \302\273j
is
of stales
densily
C4)
Thai is, \302\251(\302\243)(& is ihe make ihe by
in
n,V;
Lei x
of orbitals
number
in
the
transition to conductionelectronsin \342\200\224 oblain by m,*; and \302\243 by e ec. We
s (e 
and
er)/r
ij
=
~
(fi
et)/t. We
interval
energy
+ (\302\243,e
we
semiconductors
use the definition A6)
of
To
ck).
iV
replace
obtain
J(c to
C6)
The integral /(;;)in When
ee

p.
as the
is known
C6)
\302\273 x
we
have
\302\273 1, \342\200\224ij
FermlDirac integral. so
that

cxp(x
jj)
\302\273 1.
In
this
limit
C7)
the
result
familiar
for
the ideal
gas.
several limes electron concentrationrarely exceeds the quantum concentration nc.The deviation between the value of/i from C5) and the approximation then can be expanded into a rapidly C7) converging = series the r of ratio power njn^ calledthe JoyceDixon approximation:*
In semiconductors the
\342\226\240 \",/\"\342\200\236
C8)
\302\253\342\200\224iSf) t3.4
Figure
compares
the exact
relation C6) witll
the
approximations
C7)
and
C8). \342\200\242
W.
B. Joyce
1.483S6
and B. W.
x lO\"*:^,
Dixon.
App!.
 4A2561x
Phys. Lclt.
1Q6.'
31,354 A977).
If the
right side
of C8) is
wrine
Slmicomlitct
Dtgcncnitt
7
6
5
4
3 2
I i)
above
conduciion
band
edge Er. The
JoyccDixon approximation
C8).
=
0
(.
1
2
3
4
5
6
 i,)h
dashed curve reprcsenisihc firsi
icrm
of ihc
When
Sta
Semiconductor
13:
Chapter
longer small comparedto n(,
neisno
law must be modified.In Problem4
ask
we
Ihe
ihe
of the
expression
io show
reader
mass action
ihat
C9) If
the
Itself
gap
here
will
on
depend
Impurity
carrier concentrations, the value
on the
depends
localized
P
atom
be used
n, to
concentration.
Levels
The addition of impuritiesto a conduction or valenceband into
as
of
Ihe
where the
gap,
energy
from
orbitals
some
moves
semiconductor
the
orbiials now appear
in a silicon crystal.If the electron to the Si conductionband, the atom as a positively charged ion. The positiveion attractsthe electrons in appears the conduction an electron band, and the ton can bind just as a proton can bind an electronin a hydrogen atom. the However, binding energy in the phosphorous
iis extra
released
has
consider
We
states.
bound
semiconductoris severalordersof the energy is to be by square paniy
column
V donors
donor
corresponds
in
the binding
of the static dielectric constant, and for 132 gives the ionizationenergies Si and a Ge. The lowest orbital of an electron bound to \342\200\224 = to an energy level Asd below the edge of the st st
effects. Table
mass
of
because
because
mostly
lower,
magnitude
divided
conduction band (Figure13.5). There
is
one
set
orbitals
of bound
for
every
donor.
argument applies to 'he valence band, as in Figure
A parallel from
set ofbotmdorbitals as
Aej. Ionization
ionization
an
wiih
6rrteV.
For
zinc,
T:.Mc 13.2 column
energy
Asa ~
\342\200\224
il!
the
all
for
most
acceptor,
!uiiu..iio.i encr^e*ofcuhmm acceptors
in Si
arui
Ge,
are
listed
VI donors
column
important
\302\243,,,ofthesameorder
Ea
in Si energies for column III acctipcors
In GaAs the ionizationenergies
closeto
holes and acceptors.Orbiialsare split off For each acceptor atom there is one 13.5. in Table
except oxygen are =
&\302\243a
24meV.
V
in mcV
Ace
11
AI
13.2.
cp OIS
Ga
In
16
49
45
57
65
12.7
10.4
10?
iOS
11.2
Some
of
Occupation
'
\"'
!'
At,
Letch
Donor
\302\261
Donor
13.5
Figure
and acceptor
impurities generate orbitals deep inside the forbidden ionization multiple orbitals corresponding to different of Donor
Occupation A
level
donor
can
Hence there are [wo
different
occupationsof these level is occupied one
two
by
spin.
opposite
gap,
energy
with
sometimes
states.
Levels
be occupied
As a
in Uic
levels
impurity
orbitals
electron,
result, the
by an electronwith with
orbitals are
not
the donor
either
spin
nt
spin down.
energy. However, the
the same
independe
up or
of
each
other;
Once
the
cannot bind a second electron with
occupation probability
for
a
donor
level
is not
but function, by a function given by the simple FcrmiDiracdistribution is vacant, the the orbital treated in Chapter 5. We write that donor probability so that the donor is ionized, in a form slightly different from E.73):
the origin to singly occupied donor orbital relative that of the energy. The probability the donor orbital is occupied by an electron, the is is so thitt donor neutral, given by E.74):
Here
Ed
is
the
energy
of a
Statistics
Semiconductor
13;
Chapter
In the ionized conditionA\"
Acceptors require extra thought. each of the chemical bonds between
ihe
atom
acceptor
of
and the
the
acceptor,
surrounding
semiconductor atomscontainsa pair of electrons with There antiparaliel spins. is only one such state, hencethe ionized contributes one condition term, only exp[(/i  cJ/t], to the Gibbs sum for the acceptor, lit the neutral condition A the
of
one electron
acceptor,
the missing electron may
is missing
the
from
up or
haveeithcrspin
bonds.
surrounding
spin down, the
is representedtwice in the Gtbbs sum for the acceptor, by Hence the thermalaverageoccupancy is '
A
condition
neutral
The
exp[{^ ~
2 +
A,
the
with
\"\021+2
Efl)/i]
exp[(Efl
orbityl
acceptor
a

Because
neutral condition term 2 x J ~ 2.
\"}
{~
$x]\"
unoccupied,
occurs
with
probability
______
The
of
value
From
An
\342\200\224
== nd+
or D2)
D0)
_ __________
is the
na~
difference
D3)
of concentrations of D*
arid
A\342\204\242.
we have
D4)
D5) The
condition
neutrality
This expressionmay of
functions
the
position
represent the positive
four
visualized
be
in
and all negative
posicivechargesequal
by a
Fermi
of the terms
be rewritten
may
D)
D6);
logarithmic
level
(Figure
the two
charges.The actual total
the
negative
as
plot of n\" and n* 13.6}.
The
solid lines representthe Fermi
level
as
four dashed
occurs
sum
func
lines of
aii
where the total
charges.
holes can be neglected;for electrons can be neglected. If one of the two impurity nu~ \024* \"i 'he be the can neglected, species majority carrier concentrationcan be calculated in The closed formConsider an ittype semiconductor with no acceptors. For
~
nd+

^
jio~
\302\273
nh
as
in
Figure
13.6, the
Occupation of
Figure
13.6
intersection approximation
with
will A7)
the
be
13.6 is now given by the intersection point of the n* the interdonor concentration is not too high, the the straight portion of the incurve, alongwhich approxiWe rewrite this as
in Figure
point
neutrality
curve
of Fermi teve!and eteciroi coniaining both donors and acceptors.
determinaiion
Graphical
sciniconducior
an n1ypc
DonorLetch
?!e curve,
on
holds.
U ihe
cxp(/
\302\243i)/t]
=
\302\2537>
(\302\273A)\302\253PfcA);
(\302\273>Jexp[(\302\243,
tj/t]
= njn*
,
D8)
s
nc*
Is
electron
the

neexp[~(et
the
with
be present
would
that
concentration
the Fermi level coincided
Eli)/i] = ;i,exp(A^/t)
donor
level.
~
nd* to
the
in
~
Here
if
conduction
band
~
the donor
ec
Aed
. {49} Ej is
iontzacton energy. insert
We
set
D4) and
into
D8)
nt
obtain
E0)
+
nf3
is a
llus
=
shallow
weak that
donor
8\302\273j\302\253 \302\273/, the
A
for
v
\302\253 i.
x ~
With
large and
=? 1 +
xI1*
we
8\302\273d/He*

ionization.For Table 13.2, so
the
for
that il.4 pet of the donorsremain
the subjectof
Problem
+
ix2
;tj(l
gives P
=s 0M5nc
ne*
expanded by use of
\302\253
2it//ne*
parentheses
example,
that
\\x
is sufficiently
doping
\342\226\240\342\226\240\342\226\240
E3)
,
obtain
^ \302\253\342\200\236 \302\273\342\200\236
The secondtermin
E2)
I}
close to nc.ffthe
may be
root
square
+
is
solution
+ (S\",,//)/)]1'3
V([l
nt* is
levels,
E1)
i\302\273rfn,*.
quadratic equation in n,; ihe positive \302\273\342\200\236
For
=
\\n,nt*
in
the first

order departure from complete
300 K, we = from D9). If \302\253j Si at
unionized.
E4)
2
The
have
Ae,j
0.0!nf,
limic
Eq.
of weak
^
t.74r E4)
from
predicts
ionization is
6.
gallium anaiM*.: Could pure GaAs be prepared, it would have catikt concciUfdtioita! room temperature 10'cm\023. Wjih such a tow of\302\273,< !O art conceuiraiion of carriers, bs; closer @~\" less than a nicial, the conductivity would an as would be useful than to a sciniconductor. G;iAs insulator conveniiona! [
an intrinsic
StuMitiiHtuting
pn
However, ic with
is possible
concentra
high
near tnerinste carrier Cm3) of oxygen impuriiy levels near the middle to achieve
lions (tOi5IO17
in GaAs by
concentrations and
Junctions
chromium
doping
two impu
together,
have iticir of ihe energy gap. Oxygen enters an and is a donor in GaAs, as expecicd from the posiiion of O in the periodic Cable relative to As; the energy is an acceptor v> i'h an level* is about 0.7 eV belowts. Chromium level about Q.S4eV below energy et. a GaAs Consider boih and chromium. The ratio of liiS two crysiai doped with oxygen conceniraiions is not critical; anything with an O:Cr raiio between abouc 1:10and 10:1 will do. If the conccitiraiions of all olher are small compared with those of O and impurities Cr, (he position of ihc Fermi level will be governed by ihe equilibrium betweenelm ons on O and holes on Cr. The of Figure construction t3.6 applied io tliis system shows that over the indicated concentration raiio rangedie Fermi is pinned io a range between level t.5i above the O fevel and l.5r below (he Cr level. With ihe Fermi level pinned near the imrinsic middle of the energy gap, the crysiai must act as nearly Gallium arsenide doped in this way is called semiinsulating GaAs and is used extensively to to10 il cm\\ substrate as a tiijjhresisiivity for GuAs devices. A similar [!0a prodoping procedure is possible in inP, with iron ihe taking place of chromium. impurities
ttiat
As site
pn JUNCTIONS
Semiconductors
are almost never uniformly doped. An underan semidoped requires understanding of nonuniformiy of structures called pn junctionsin which the doping semiconductors, particularly to ntype within 'he samecrystal.We consider ton from /vtype changes with posi! a semiconductor at .v = 0 which the doping changesabruptly crystal inside from a uniform donor concentration nd to a uniform acceptor concentration as in Figure i3.7a. This is an exampleof a p~i\\ junction. More complicated na, a device structures are made up from simple bipolar transistor h.is junctions: used
m
devices
understanding of devices
iwo closely spacedpn junctions, ofthe sequence or n~pn. pi\\~p in the builtin electrostaticpotential Vbi, even step p~n junctions contain a With no externally absence of an externally applied voltage 13.7b). (Figure the are in of diffusive junction applied voltage, the electrons on the two sides of tlic two which means that the chemical potentials(Fermilevels) equilibrium, within Fenni level ihe band sides are the same. Because ihe posiiionof the level forces :i of the Fermi siructurc depends on the localdoping,constancy shift
in
the
shift
is
eVN.
electron The
energy
potential
bands stop of
in
ihe
crossing
height eVbii%
required to equalize the total chemical intrinsic chemicalpotentialsare unequal,
an
junction example
potential
as
discussed
of
two
(Figure 13.7c). of the potential systems
in Chapter
5.
The step
when ihc
A pn junclion. (a) Dopingdislribulion.!l is assumed 13.7 the that from doping changes abruptly ntype to ptype. The two levels arc usually different, doping (b) Electrosiaiic poteniiat. The Figure
buitlin two
voltage sides
wilh
concentrations, shifted
relaiivc
generate the
diffusive between ihe equilibrium electron concenlralions as wet! as hole level must be (c) Energy bands. Becauseihe Fermi Vbl
ealabtishes
differenl
to each buitlin
other, (d) Spacecharge
voltage
and
to
shift
the
dipolc
energy
required
bands.
to
We assume that
the two doping concentrationsnd,
nondegenerate range, as
defined
Hf
are
If the
donors
the p
side, then the
\302\253 nd
\302\253 \302\273c;
\302\253 \302\273( na
the n
side and
on
the
from \302\261
side
n
Jij,
\302\253 \302\273t.
E5}
the acceptorsfully
on
ionized
electron and ho!econcentrations satisfy =z vd\\
ne
one on
extrinsic but
by
ionized
fully
in the
lie
tta
and
the other
on thep band
The conduction \302\273\342\200\236.)
s:
nh
E6)
na,
side. (We have droppedthe on the
energies
i\\
p sides
and
superscripts
follow
from
A7): E7)
\302\253\302\253/*xlbgl^/nj;
H~ xlo&{nJnc) =
ccp = by
eVtf

For doping concentrationsnd A
is 0.91
eV in silicon
in electrostatic
step

=
E8)
~
must
satisfy
the
at room
temperature. to shift
is required
potential Poisson
to
other.
each
find eHi band
the
The
= es ~9.2t,
edge
electrostatic
energies
on
potential
equation
~
(SI)
Hfl=s0.0inL.,we
0.0inrand
E9)
.
Tlog(W>,2)
\302\243CJ. e\302\253
the two sides of the junction relative
tlogK1/\"^).
Hence
B2c).
which
;1
= ~~ ,
F1)
of the semiconductor. space charge density and e the permittivity varies. must be whenever In the vicinity of the junction
where
p
is the
(Figure 13.7d}. concentration is less Positive
space than
the
charge donor
on
the
n
concentration,
side
means
indeed,
that
as the
the electron
conduction
Statistics
Semiconductor
13:
Chapter
raised relative to the fixed Fermi decreaseof the electronconcentration ie. band edge ts the
Take Then
ofthe
origin =
ec(x)
electrostatic \342\200\224
and
e
\302\243c(~a))
ne{x) =
ThePoisson
potential at
x = co.soihai
d.x i/x^
(/.x \\il\\J
t
dx =
f
{
x =
r
/
i6d\\
, ,1
e
J
0:
F5)
]
\\dxj interface
F3)
^
J
^
AM2
oc) Integrate uiih the initial condition
the
F2)
to obtain
_
At
ntcxp[e
] Jdtpfdx
by
an exponential
becomes
A7)
^ Multiply
predicts
A7)
is
F1)
equation
level,
0 we assume that'
F6) Vn is
where
the n side.
part of
that
the builtin electrostaticpotentialdropthat
The exponential
on the righthandsideofF5)can
be
on
occurs
and
neglected,
we obtain
E for the
v
ofthe
component
[peii,/c)(K,
electric
field E
F7)
*/e)Vn
= dipjdx. at the interface.Similarly, FS)
\302\243=[{2eMB/\342\202\254)(^t/e)]\022,
where
ihepstdcTlie we
potential drop part of ilie builtinelectrostatic be the same; from this and from two \302\243 fields must
I; is that
that Vn
on
occurs V
Vp
=
Vbi
find
(\302\243^')\022
F9)
rseBiasedAbrupt is the Sameas if On
field E
The
lite
from
to a
junction
the
Junction
been depleted
electrons had
side all
iitype
p~n
distance
, G0)
(Vw2r/f)
no
with
at
depletion
>
\\x\\
theory as a measureofthe
layer into
on the
p side,
The totaldepletion
width
1f
=
we assume
 4.25 \302\243
\302\273o
wK
nd =
i0licm\023;e
and w
x 104VcmwI
ReverseBiased Let
of penetration
depth
a voltage
V
be
4.70
the
to a
applied
p side n
to the
p side
in bulk contains
side
law. As
action
holes,
and
increased
\302\273 side,
contains a very a very
by the
are
2z/e
== 1 volt,
we lind
iO^5cm.
pn junction,
holes
and low
which
of such sign that that
means
voltage
V
raises
p side the
cond uction
drive
will
the
of
concentration
little
current
approximately
applied voltage,Figure
ihe same 13.S.
The
as
if
builtin
the
field
at
a
electrons
side to the p side.But conduction electrons, and
The distributions
flows.
is at
potential
from the ;i
low concentration of holes,consistentwith
a result, very
potential
x

Vbi
Junction
pn
Abrupt
=
~ l(ko;aiid
negative vohagerelativeto the n side, energy of the electronson the p side.This from
used in semiconductor device of the space charge transition
n side.
the
Similarly,
wa is
distance
The
\\vK.
the
the ifie
mass
of electrons, voltage
were
the interface is now
given by
G3)
Chapter
13: Se
..\342\226\240