Applications of Maxwell’s Thermodynamical Relations
Discipline Course-I Semester-II Paper No: Thermal Physics : Physics-IIA Lesson: Applications of Maxwell’s Thermodynamical Relations part1 Lesson Developer: Dr. Vinita Tuli College/ Department: ARSD College, University of Delhi
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Applications of Maxwell’s Thermodynamical Relations
Table of Contents Chapter: Maxwell’s Thermodynamical Thermodynamical Relations Introduction
Applications of Maxwell’s Relations
Summary
Exercise/Practice
References/Bibliography
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Applications of Maxwell’s Thermodynamical Relations
Introduction Out of several properties of a thermodynamic system some of these like internal energy and entropy cannot be measured. So thermodynamic relations can relate these properties with those that can be measured like pressure,temperature, compressibility etc. In thermodynamic relations un-measurable properties can be written as partial derivatives involving both intensive and extensive variables. A thermodynamical relations is a rule which is obtained by a simple thermodynamic reasoning and applies to most of the systems. The usefulness of the above relations lies in the fact that they relate quantities which seem unrelated. They help us to link data obtained in various ways or replace a difficult measurement by another one. They can also be used to obtain values of one variable from the calculations of another variable. These relations are very general and immensely useful as they simplify analysis of thermodynamic systems. The most convenient way to derive these relations is to use partial di fferentiation.
Value Addition How to Remember Maxwell’s Relation’s : Note : The first four Maxwell’s thermodynamical relation need not be memorized. These can be easily written by remembering the word SP or TV i e (sportive). In this, SP TV represent entropy, pressure, temperature and volume. In deriving the first two relations, S is written in the numerator on the left hand side of the equation and the rest of the quantities P, T and V are written (i) in the clockwise order and (ii) in the anticlockwise order. In the clockwise direction, the right hand side expression in with a + sign and for the anticlockwise direction, it is –ve. These equations are S P …(i) = V T T V
S V …(ii) = T P T P Similarly to write the other equation, S is written in the denominator of the right hand side of the equation and the other quantities are written (i) in the anticlockwise direction and (ii) in the clockwise direction. These equations are T P …(iii) = S V S V
T V = P S S P
…(iv)
Applications 1. Specific Heat Equation The specific heat at constant pressure is given by
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Applications of Maxwell’s Thermodynamical Relations Q T P
C P =
S T P
= T
( Q T S )
the specific heat at constant volume is
Q T V
C V =
S T V
= T
if entropy S is taken as a function of T and V and since dS is a perfect differential,
S S dT dV T V V T
dS =
S S T S V = T P T V T P V T T P
S S S V T T = T V T T P V T P
T
S P = V T ( from Maxwell’s second relation) T V
S S P V = T T T P T V T V T P
T
P V T V T P
C P C V = T
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…(1)
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Applications of Maxwell’s Thermodynamical Relations (a) For a perfect gas :
The equation of state for one mole of a perfect gas is PV = RT R P T = V V
R V = T P P equation (1) takes the form
R R V P
C P C V = T
=
=
R 2T PV R 2T RT
C P C V = R
(b)
Van der Waals :
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Applications of Maxwell’s Thermodynamical Relations
The equation of state for a Van der Waals gas is
a P 2 (V b) =RT V
where a and b are constants. From the Van der Waals equation, we get
a RT P V 2 = (V b) Differentiating the equation with respect to T at constant volume, we have
R P T = (V b) V and differentiating with respect to T at constant pressure we have
0
2a V
= V T P 3
R V (V b) T P (V b) RT
2
2a R V RT T (V b)2 V 3 = (V b) P
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Applications of Maxwell’s Thermodynamical Relations R (V b)
V T = RT 2a P (v b)2 V 3
P V and in equation (1), we have T V T P
Substituting the value of
R R V b V b RT 2a (V b)2 V 3
T
C P C V =
R
=
2 2a (V b)
1
V3
RT
Neglecting b in comparison to V as b is much smaller than V
C P C V
R
= 1
2a
V 2
V 3 RT
R
= 1
2a
VRT
2a = R 1 VRT
1
Expanding binomially the above equation and neglecting higher powers of a as a is very small in comparison to V , we have
C P C V = R 1
2a VRT
2. 2 To Show that C P CV TE V
where T is the absolute temperature E is the isothermal elasticity of the gas, is the coefficient of volume expansion and V is the specific volume of the gas. From Maxwell’s second thermodynamical relation
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Applications of Maxwell’s Thermodynamical Relations S P V = T T V from the definition of C P and C V we have
Q T P
C P =
S T P
= T
Q T V
C V =
S T V
= T Taking S f (T , V ), we have
S S dT dV T V V T
dS =
dividing the above equation by dT and then carrying out the process at constant pressure we have
S T = P
S T S V T T V T V T P
But from the maxwell’s relation , we have
S P V = T T V
S S P T = T T P T V T V
T
V T P
P V T V T P
C P = CV T
…(2)
taking the general equation of state for a gas as P f (V , T )
P P dT dV T V V T
dP =
Dividing the above equation by dT , we have
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Applications of Maxwell’s Thermodynamical Relations
dP dT
P P V T V V T T P
=
At constant pressure dP = 0 the above equation becomes
P P V T = V T V T P
Substituting the value in equation (2)
2
P V C P C V = T V T T P
…(3)
2
V P C P C V = T T P V T
…(4)
modulus of elasticity at constant temperature is given by
P V T
E = V
and the coefficient of volume expansion is given by =
V V T P 1
Substituting these values in equation (4) we have
C p C V = TE 2V
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…(5)
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Applications of Maxwell’s Thermodynamical Relations 3. Clausius-Clapeyron’s Equation (First Latent Heat Equation)
The equation
dP dT
L T (V2 V1 )
was derived by Clapeyron using Carnot’s reversible cy cle.
Therefore, it is also called Clapeyron’s eqauation. Whenever there is a change of state, from a solid state to liquid state or from liquid state to vapour state, the temperature remains constant as long as the change takes place. This temperature (either melting point or boiling point) depends upon the pressure and is a characteristic of each substance. The above relation tells us how melting or boiling points vary with pressure. Later Clausius also obtained the same relation by using the Maxwel l’s thermodynamic relations. Derivation : From second thermodynamical relation, we have
S P V = T T V Multiplying both sides by T , we get
S P = T V T T V
T We know that
T S = Q Substituting this in the above equation, we g et
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Applications of Maxwell’s Thermodynamical Relations
P Q V = T T V T
…(6)
Q V represents the quantity of heat absorbed or liberated per unit change in volume at T constant temperature. As there is a change in volume due to the heat absorbed at constant temperature, the heat represents the latent heat used when a substance, changes from solid to liquid (melting) or liquid to vapour (boiling) state when the temperature remains constant, during the change of state. If L is the quantity of heat required to change the sate of a unit mass of the substance, V2 and V 1 the corresponding specific volumes (Volume per unit mass) then
Q = L V = V2 V 1 Dividing the above equations we get
L Q V = V V T 2 1 L
Therefore we have
V2 V 1 dP dT
=
P T V
= T
L T (V2 V 1 )
…(7)
Where dT is the change in melting point or boiling point due to a change in pressure dP . This is Clausius-Clapeyron equationt or first latent heat equation.
Effect of Pressure on Boiling Point of Liquid When a liquid converts into vapour state, we know that there is always an increase in P volume. Implies that V2 V 1 ; T and L are always positive quantities, therefore is T V +ve. This shows that boiling point of the liquid is raised by increasing the pressure. This is known as elevation of boiling point of the liquid.
Effect of Pressure on Melting Point of Solid When a solid melts, there may be an increase in volume, as in case of wax (such substances are called wax type substances) or there may be a d ecrease in volume, as in the case of ice. (such type of substances are called ice type substances).
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Applications of Maxwell’s Thermodynamical Relations
In melting of wax type substances, volume of the substance always increases.
(V2 V 1 ) = +ve Which implies
dP dT
= +ve
Hence the melting point of the substance is raised with the increase in pressure. In case of ice
(V2 V 1 ) = – ve hence melting point of such substances is lowered with increase in pressure. This is known as depression in melting point. Substances which expands on melting, have their melting points raised with the increase of pressure and the substance which contract on melting have their melting points lowered with the increase of pressure.
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To deduce the second latent heat equation of Clausius given as
C2 C 1 =
dL dT
L T
where C1 and C 2 are the specific heat of a liquid and its vapour and L is the latent heat of the vapour. For a change of state from liquid to vapour we get L S2 S 1 = T
…(8)
Where S1 and S 2 are the entropies of the liquid and vapour states Differentiating equation (8) with respect to T, we have
dS2 dT
dS1 dT
= L
1
T
2
1 dL
T dT
Multiplying the above equation by T, we get
L dL dS2 dS T 1 = T dT dT dT
T
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Applications of Maxwell’s Thermodynamical Relations
C2 C 1 =
dL
L
…(9)
dT T This is known as Clausius second latent heat equation. Problems based on second latent heat equation i.
Calculate the specific heat of saturated steam given that the specific heat of water at 100°C = 1.01 and latent heat of vaporization decreases with increase in temperature at the rate of 0.64 cal/K. Latent heat of vaporization of steam is 540 cal.
Sol. Here
C 1 =1.01, C 2 = ? T = 373 K
dL dT
= – 0.64 cal/K
C2 C 1 =
dL dT
C 2 = C 1
L T
dL dT
L T
= 1.01 (0.64)
540 373
= – 1.077 cal/g ii.
If L = 800 – 0.705 T, show that the specific heat of steam is negative.
Sol. Given :
L = 800 – 0.705 T
dL dT
= – 0.705
At normal boiling point of water i.e., at 100°C = 373 K, the value of L would be L = 800 – 0.705 × 373 = 537 The specific heat of water in liqui d state, C 1 = 1 Using second latent heat equation
C2 C 1 =
dL dT
L T
= 1 (0.705)
537 373
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Applications of Maxwell’s Thermodynamical Relations = – 1.14467 Thus, the specific heat of saturated vapour pressure at 100°C in negative
5.
To prove that for any substance, the ratio of the adiabatic and isothermal elasticities is equal to the ratio of the two specific heats.
Isothermal elasticity of the gas is given by
P V T
E T = V
Adiabatic elasticity of the gas is given by
P V S
E S = V
E S E T
P V V S = P V V T P V S = P V T
P T T V S S = P S S V T T Taking the first term of the numerator to the denominator and the first term of the denominator to the numerator
E S E T
S T P V T S = T S P V S T
…(i)
from the first four Maxwell’s equations we have
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Applications of Maxwell’s Thermodynamical Relations
V S = T P P T
T P V = S V S T V = P S S P
S P V = T T V Substituting these values in equation (i), we get
E S E T
V T = V S
P P S V P P T V
Taking the first term of the denominator to the numerator and the second term of the numerator to the denominator
E S E T
S V V T P P = S P P T V V
Multiplying and dividing the equation by T
S T P = S T T V T
Q T P = Q T V We know
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Applications of Maxwell’s Thermodynamical Relations Q T = C P P Q = C V T V
and
E S E T 6.
=
C P C V
=
…(ii)
Using Maxwell’s thermodynamical relations, to prove that the ratio of the adiabatic to the isobaric coefficient of volume expansion is 1
(1 ) Sol.
Adiabatic coefficient of volume expansion is given by
S =
V V T S 1
...(i)
Isobaric coefficient of volume expansion is given by
P =
V V T P 1
…(ii)
Dividing the above two equations 1 V S P
=
V T S 1 V
…(iii)
V T P
V T S = V T P Bringing the first term in the numerator to the denominator =
1
T V V S T P
Using Maxwell’s relation
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Applications of Maxwell’s Thermodynamical Relations T P V = S S V Substituting in the above equation
S P
=
1
P V S T V P
Splitting the first term of the denominator =
1
P T V T V S V T P
Taking the second term of the den ominator to the numerator and multiplying by T
S T V = P V T T V T P T
We know that
P V T V T P
C P C V = T and
S T V
C V = T
Substituting these values in the above equation we g et S P
S P
=
=
C V
(C P C V )
1
C P C V
1
Dividing the numerator and denominator by
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Applications of Maxwell’s Thermodynamical Relations C V S P 7.
=
1
…(iv)
1
Using Maxwell’s thermodynamical relations prove that the ratio of the adiabatic to the isobaric pressure coefficient of expansion is
( 1) Sol. Adiabatic pressure coefficient of expansion is
S =
1 P
P T S
...(i)
Isochoric pressure coefficient of expansion is
V =
1 P
P T V
…(ii)
Dividing (i) by (ii) we get
P P T S = 1 P P T V 1
S V
…(iii)
P T S = P T V
Taking the first term of the numerator to the denominator =
1
T P P S T V
Using Maxwell’s relation and substituting in the above equation
V T = S P P S
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…(iv)
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Applications of Maxwell’s Thermodynamical Relations S V
=
1
V P S P T V
…(v)
Splitting the first term of the denominator we get =
1
V T P T S T P P V
Taking the second term of the denominator to the numerator S
T P = V P T P T V
And multiplying by T we get
S T P = V P T T P T V T
using
P V T V T P
C P C V = T And we get
S V
S V
=
=
=
C P C P C V
C P C V C P 1 C P 1
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…(vi)
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Applications of Maxwell’s Thermodynamical Relations
Summary A thermodynamical relations is a rule which is obtained by a simple thermodynamic reasoning and applies to most of the systems. The usefulness of the above relations lies in the fact that they relate quantities which seem unrelated. They help us to link data obtained in various ways or replace a difficult measurement by another one. They can also be used to obtain values of one variable from the calculations of another variable. These relations are very general and immensely useful as they simplify analysis of thermodynamic systems. The most convenient way to derive these relations is to use partial differentiation. They can also be derived from other methods.
Exercise 1.
2.
Derive Maxwell’s quantities.
thermodynamical
relations
connecting
the
thermodynamic
Show that
P V T V T P
C P C V = T
where symbols have their usual meaning. 2.
Use Maxwell’s relations to obtain C P CV R for an ideal gas where C P and C V are specific heats at constant pressure and constant volume. And show that for a Van der Waal’s gas
C P C V = R 1
2a VRT
Where symbols have their usual meaning. 3.
prove the relation.
E S E T
4.
=
C P C V
Derive the Clausius-Clapeyron equation given by
dP dT
=
L T (V2 V 1 )
and explain the effect of pressure on (a) boiling point of a liquid (b) melting point of a solid.
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Applications of Maxwell’s Thermodynamical Relations
6.
Prove the relations
S P = V T T V
S P = T P T P
OBJECTIVE QUESTIONS 1. (a) (b) (c) (d)
Four thermodynamic potential are given by: Pressure, volume, temperature and internal energy function. Pressure, volume, internal energy and Helmholtz function. Internal energy function, Helmholtz function, enthalpy and Gibbs function none of these
2.
Specific heat of saturated vapour pressure is (a) zero (b) positive (c) negative (d) sometimes positive sometimes negative
3.
From Maxwell’s thermodynamic relations
E S E T
?
(a) 1 (b)
1
(c) (d) 2 4.
The Clausius-CLapeyron equation is given by: dP L (a) dT T (V2 V1 ) (b)
dP dT
L L(V2 V1 )
dP
TL(V2 V1 ) dT (d) none of these (c)
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Applications of Maxwell’s Thermodynamical Relations
5.
The value of C P C V for an ideal gas is given by (a) TE V 2 (b) T 2 E V (c) TE 2V 2
2
2
(d) TE V 6.
Rice takes longest time to cook (a) in submarine 100 m below the surface of the sea. (b) at the sea level (c) at Simla (d) at Mount-Everest
7.
Paraffin wax contracts on solidification. The melting point of wax will (a) increase with pressure (b) decrease with pressure (c) no change with pressure (d) decrease linearly with pressure
References: 1. 2. 3. 4. 5. 6.
Thermodynamics, kinetic theory and statistical thermodynamics by F.W. and G.L. Salinger. Fundamentals of Statistical and Thermal Physics by F. Reif. A Treatise on Heat by M.N. Saha and B.N. Srivastava. Thermal physics by C. Kittel/H. Kroemer. Heat and Thermodynamics by M.W. Zemansky and R.H. Dittman. Thermal Physics by S.C. Garg, R.M. Bansal and C.K. Ghosh.
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Sears
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