Maxwell Relations
Section 4.2: Fundamental Relations for Closed Simple Systems
4.3 Additional Useful Thermodynamic Identities
We have asserted without proof that (∂U/∂V)T,nvanishes for an ideal gas. We can now prove this assertion and can obtain a formula that allows evaluation of this derivative for nonideal gases and liquids. We convert Eq. (4.2-3) to a derivative equation by nonrigorously “dividing” bydV, converting the quotients to partial derivatives, and specifying thatTandnare held fixed. The process is mathematically indefensible, but gives the correct derivative relation:
∂U
∂P
T,n
T ∂S
∂V
T,n
−P ∂V
∂V
T,n
T ∂S
∂V
T,n
−P (4.3-1)
We apply the Maxwell relation of Eq. (4.2-18) to the first term to obtain ∂U
∂V
T,n
T ∂P
∂T
V,n
−P (the thermodynamic
equation of state) (4.3-2) The relation shown in Eq. (4.3-2) is called thethermodynamic equation of state. For an ideal gas,
T ∂P
∂T
V,n
TnR
V P (4.3-3)
so that
∂U
∂V
T,n
P−P 0 (4.3-4)
It is now necessary only to specify thatPV nRTto define an ideal gas.
E X A M P L E 4.7
Show thatPis proportional toT in an ideal gas ifV andnare constant, using Eqs. (4.3-1) and (4.3-4).
Solution For an ideal gas
T ∂P
∂T
V,n P ∂P
∂T
V,n P
T
At constantVandn, 1 PdP 1
TdT
ln (P)ln (T)+ln (constant) P
T constantat constantV andn
E X A M P L E 4.8
For a gas obeying the truncated virial equation of state, PVm
RT 1+ B2 Vm show that
∂U
∂V
T,n
RT2(dB2/dT ) Vm2 Solution
P RT
Vm + RT B2 Vm2 From Eq. (4.3-2)
∂U
∂V
T,n T
∂P
∂T
V,n
−P
If bothVandnare held fixed, thenVmis held fixed;
T ∂P
∂T
V,n RT
Vm + RT Vm2
B2+TdB2 dT
∂U
∂V
T,n RT
Vm + RT Vm2
B2+TdB2 dT
−RT
Vm −RT B2 Vm2
RT2 Vm2
dB2 dT
Exercise 4.5
Find the value of (∂U/∂V)T,n for 1.000 mol of argon at 1.000 atm and 298.15 K, using the truncated virial equation of state.
The partial derivative (∂U/∂V)T,n is an important measure of the deviation of a system from ideal gas behavior. It has the same units as pressure and is known as the internal pressure.If the internal pressure is positive, the energy increases as the volume increases and work is done against the attractive intermolecular forces. The internal pressure is a measure of the net cohesive forces of the system and can have a large positive value for liquids.
To derive a useful equation for the internal pressure, we begin with Eq. (4.3-2) and apply the cycle rule to the partial derivative on the right-hand side of the equation:
∂U
∂V
T,n
−T ∂P
∂V
T,n
∂V
∂T
P,n
−P
∂U
∂V
T,n
T α
κT−P (4.3-5)
whereαis the coefficient of thermal expansion andκTis the isothermal compressibility.
E X A M P L E 4.9
Evaluate the internal pressure of liquid benzene at 298.15 K and 1.000 atm.
Solution
∂U
∂V
T,n
(298.15 K)
1.237×10−3K−1
9.67×10−10Pa−1 −101325 Pa 3.81×108Pa
This internal pressure is equal to 3760 atm.
Exercise 4.6
a. Evaluate the internal pressure of liquid water at 25◦C and 1.000 atm.
b. Calculate the gravitational force per unit area on a column of water 100 m in height. Explain how the internal pressure relates to the fact that in a giant sequoia tree, sap can be brought to a height of nearly 100 m in the tree whereas barometric pressure can raise it only to about 10 m against a vacuum. What can you say about the attractive forces between the sap and the walls of the vessel containing it? If a gas bubble appeared in the sap, what would happen?
An equation for (∂H /∂P)T,n that is analogous to Eq. (4.3-2) can be derived in a similar way. We convert Eq. (4.2-11) to a derivative equation:
∂H
∂P
T,n
−T ∂S
∂P
T,n
+V ∂P
∂P
T,n
T ∂S
∂P
T,n
+V Using the Maxwell relation of Eq. (4.2-22), we obtain
∂H
∂P
T,n
−T ∂V
∂T
P,n
+V (4.3-6)
E X A M P L E 4.10
Show that for an ideal gas (∂H /∂P)T,n0, using only the equation of state,PVnRT, and Eq. (4.3-6).
Solution
∂H
∂P
T,n −T
∂V
∂T
V,n V −T
∂
∂T nRT
P
P,n +V −nRT
P +V 0
Exercise 4.7
a. Find an expression for (∂H /∂P)T,nfor a gas obeying the truncated pressure virial equation of state:
PVmRT+A2P
whereA2is a function ofT. It has been shown that the second pressure virial coefficientA2 is equal toB2, the second virial coefficient.
b. Evaluate (∂H /∂P)T,n for 1.000 mol of argon at 1.000 atm and 298.15 K. Data onB2and dB2/dT are found in Example 4.3.
We can now obtain a useful relation betweenCPandCVfor systems other than ideal gases. Equation (2.5-11) is
CPCV+ ∂U
∂V
T,n
+P ∂V
∂T
P,n
(4.3-7) The CV term represents the energy change due to an increase in temperature that would occur if the volume were constant. The (∂U/∂V) (internal pressure) term repre- sents the energy absorbed in raising the potential energy of intermolecular attraction, and the P(∂V /∂T) term represents work done against the pressure exerted by the surroundings.
We now use the thermodynamic equation of state to write CPCV+
T
∂P
∂T
V,n
+P−P ∂V
∂T
P,n
CV+T ∂P
∂T
V,n
∂V
∂T
P,n
(4.3-8) We apply the cycle rule, Eq. (B-15) of Appendix B, in the form:
∂P
∂T
V,n
∂T
∂V
P,n
∂V
∂P
T,n
−1 (4.3-9)
which gives
CPCV−T ∂P
∂V
T,n
∂V
∂T
P,n
2
(4.3-10) We obtain
CPCV−T ∂P
∂V
T,n
∂V
∂T
P,n
2
CV+ TV α2 κT
(4.3-11)
whereαis the coefficient of thermal expansion andκTis the isothermal compressibility.
Sinceα(which is occasionally negative) is squared and sinceκTis always positive,CP
is never smaller thanCV.
E X A M P L E 4.11
Calculate CV,m for liquid water at 0.00◦C and 1.000 atm. The constant-pressure heat capacity is equal to 75.983 J K−1mol−1. The coefficient of thermal expansion is equal to
−68.14×10−6K−1at this temperature (this is one of the few cases in which this quantity is negative). The compressibility is equal to 5.098×10−10Pa−1. The molar volume is equal to 18.012 cm3mol−1.
Solution
CP,m−CV,m (273.15 K)
18.012×10−6m3
−68.14×10−6K−1 2 5.098×10−10Pa−1
0.04481 J K−1mol−1
CV,m75.983 J K−1mol−1−0.04481 J K−1mol−1 75.938 J K−1mol−1
Exercise 4.8
a. Find the value ofCV,mfor liquid water at 25.00◦C and 1.000 atm. The coefficient of thermal expansion is equal to 2.07×10−4K−1, the molar volume is equal to 18.0687 cm3mol−1, and the compressibility is equal to 45.24×10−6bar−1.CP,mis equal to 75.351 J K−1mol−1. b. At 3.98◦C liquid water has a maximum density and the coefficient of thermal expansion
vanishes. What is the value ofCP,m−CV,mat this temperature?
c. Show that Eq. (4.3-11) leads to the expression for an ideal gas obtained in Chapter 2:
CP,m−CV,mR (ideal gas)
d. Calculate the value of the ratioγCP,m/CV,mfor liquid water at 25.00◦C and 1.000 atm and compare it to the value of the same ratio for argon gas at the same temperature and pressure.
There are a number of organic liquids that have fairly large coefficients of thermal expansion. In these cases the difference betweenCV,mandCP,mis somewhat larger than is the case with water in the preceding example. For almost all solids the difference betweenCP,mandCV,mis quite small.
E X A M P L E 4.12
Calculate the value ofCV,mfor liquid benzene at 20◦C and 1.00 atm. The density of benzene at 20◦C is 0.8765 g cm−3. The value of the isothermal compressibility in the appendix applies at 25◦C. Assume that you can use this value at 20◦C.
Solution From Eq. (4.3-11)
CPCV+ TV α2 κT
From the appendix, the value ofCP,mfor liquid benzene is 135.31 J K−1mol−1. The molar volume is
Vm
78.11 g mol−1 0.8765 g cm−3
1 m3 106cm3
8.912×10−5m3mol−1 CV,m135.31 J K−1mol−1
− (293.15 K)(8.912×10−5m3mol−1)(1.237×10−3K−1)2 9.67×10−10Pa−1
93.96 J K−1mol−1
Exercise 4.9
The constant-pressure specific heat capacity of metallic iron at 298.15 K and 1.000 atm is equal to 0.4498 J K−1g−1. The coefficient of thermal expansion is 3.55×10−5K−1, the density is 7.86 g cm−3, and the isothermal compressibility is 6.06×10−7atm−1. Find the constant- volume specific heat capacity at 298.15 K.
We can now obtain two additional relations for the heat capacities. From Eqs. (2.5-8) and (2.4-4),
CP ∂H
∂T
P,n
(4.3-12) CV
∂U
∂T
V,n
(4.3-13) We take Eq. (4.2-11) for a closed system and convert it to a derivative relation, speci- fying thatPis fixed:
∂H
∂T
P,n
T ∂S
∂T
P,n
+V ∂P
∂T
P,n
(4.3-14) The derivative ofPwith respect to anything at constantP is equal to zero, so that
CP ∂H
∂T
P,n
T ∂S
∂T
P,n
(4.3-15) Similarly,
CV ∂U
∂T
V,n
T ∂S
∂T
V,n
(4.3-16)
Exercise 4.10
Use Eq. (4.3-15), Eq. (4.3-16), and the cycle rule to show that CP
CV κT κS
(4.3-17) whereκSis theadiabatic compressibility,
κS −1 V
∂V
∂P
S,n
(4.3-18)
P R O B L E M S
Section 4.3: Additional Useful Thermodynamic Identities