Example 18.9
Calculate the constant-volume heat capacity of 1 mole of gaseous water at
727°C and 1 bar pressure. Compare this to a value of 33.0 J/(molK) at 727°C
and about 1 atm pressure (where 1 atm 1.01325 bar) and comment on the
difference.
Solution
The three vibrational temperatures for H 2 O are 2287, 5163, and 5350 K. (See
Example 18.5.) Our expression for CVbecomes
640 CHAPTER 18 More Statistical Thermodynamics
CV (6.02
1023 mol^1 )1.381
10 ^23
K
J
3
2
3
2
(^)
2
1
2
0
8
0
7
0
K
K
2
exp
2
1
2
0
8
0
7
0
K
K
^1 exp
2
1
2
0
8
0
7
0
K
K
2
We g e t
CV(8.314
m
J
ol
)(3 0.658 0.154 0.137)
CV32.8
mo
J
lK
This is a very good agreement with experiment! The variance comes from the
fact that H 2 O is certainly not acting as an ideal gas, even at these tempera-
tures and pressures. At higher temperatures and lower pressures, predicted
values are much closer to experiment. Finally, notice how much the vibra-
tions contribute to the heat capacity. The lower-energy vibration, which has
the lower value ofv, contributes much more than the two higher vibrations
combined.
The point of this section is that statistical thermodynamics can derive ex-
pressions for thermodynamic properties of molecules. Many computer pro-
grams are available that use the expressions in Table 18.5 to calculate thermo-
dynamic properties of molecules, given their energy levels (which can be
determined spectroscopically or theoretically). Application of these equations
gives the physical chemist a powerful tool for understanding the thermody-
namic properties of molecules.
18.9 Equilibria
Classical thermodynamics is very useful when applied to chemical or physical
processes that are in a state of equilibrium. How well does statistical thermo-
dynamics apply to equilibrium?
Let us assume a balanced, gas-phase equilibrium:
(^) AA (^) BB (^) CC (^) DD (18.55)
where A and B represent reactants, C and D are the products, and (^) A, (^) B,
(^) C,and (^) Dare the molar coefficients of the balanced chemical reaction.
JQPJ
(^)
5
1
1
0
6
0
3
0
K
K
2
5
1
3
0
5
0
0
0
K
K
2
exp^5
1
3
0
5
0
0
0
K
K
^1 exp
5
1
3
0
5
0
0
0
K
K
2
exp^5
1
1
0
6
0
3
0
K
K
^1 exp
5
1
1
0
6
0
3
0
K
K
2