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58 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy

To obtain Equation (3.37), both sides of Equation (3.36) have been divided by dT, and
the ratio dV/dT has been converted to a partial derivative at constant P. Equation (3.19)
has been used in the last step. Using Equation (3.9) and the cyclic rule, one can simplify
Equation (3.37) to

0V 2
a b
0P 0V 0T P
= C + Ta b a b
V
C P V = C - T
0T V 0T P a 0V b
0P T
b 2 b 2
C = C + TV or C P, m = C V, m + TV (3.38)
P
m
V
k k
Equation (3.38) provides another example of the usefulness of the formal theory of
thermodynamics in linking seemingly abstract partial derivatives with experimentally
available data. The difference between C P, m and C V, m can be determined at a given tem-
perature knowing only the molar volume, the isobaric volumetric thermal expansion
coefficient, and the isothermal compressibility.
Equation (3.38) is next applied to ideal and real gases, as well as liquids and
solids, in the absence of phase changes and chemical reactions. Because b and k
are always positive for real and ideal gases, C - C 7 0 for these substances.
P
V
First, C - C V is calculated for an ideal gas, and then it is calculated for liquids
P
and solids. For an ideal gas, (0U>0V) = 0 as shown in Example Problem 3.3, and
T
0P 0V nR nR
Ta b a b = Ta ba b = nR so that Equation (3.37) becomes
0T V 0T P V P
C - C = nR (3.39)
P
V
This result was stated without derivation in Section 2.4. The partial derivative (0V>0T) =
P
Vb is much smaller for liquids and solids than for gases. Therefore, generally
0U 0V
C W ca b + Pda b (3.40)
V
0V T 0T P
so that C L C V for a liquid or a solid. As shown earlier in Example Problem 3.2, it is
P
not feasible to carry out heating experiments for liquids and solids at constant volume
because of the large pressure increase that occurs. Therefore, tabulated heat capacities
for liquids and solids list C P, m rather than C V, m .

The Variation of Enthalpy with Pressure
3.6 at Constant Temperature
In the previous section, we learned how H changes with T at constant P. To calculate
how H changes as both P and T change, (0H>0P) T must be calculated. The partial
derivative (0H>0P) T is less straightforward to determine in an experiment than
(0H>0T) P . As will be seen, for many processes involving changes in both P and T,
(0H>0T) dT W (0H>0P) dP and the pressure dependence of H can be neglected
T
P
relative to its temperature dependence. However, the knowledge that (0H>0P) T is not
zero is essential for understanding the operation of a refrigerator and the liquefaction of
gases. The following discussion is applicable to gases, liquids, and solids.
Given the definition H = U + PV , we begin by writing dH as

dH = dU + P dV + V dP (3.41)
Substituting the differential forms of dU and dH,
0H 0U
C dT + a b dP = C dT + a b dV + PdV + VdP
V
P
0P T 0V T
0U
= C dT + ca b + PddV + V dP (3.42)
V
0V T