Page 157 - Essentials of physical chemistry

Gibbs’ Free Energy and Equilibria 119

2 2
Thus, multiplying by (V =V ) we can cast this into easily measurable quantities and we find

qV qV qV qV
T
2 2
qT qT V TVa qT qP
P P P T
:
(C P C V ) ¼ 2 ¼ ; a , b
qV V b V V
qP
T
TVa 2
:
b
(C P C V ) ¼
This general expression is most useful for liquids since (C P C V ) is usually very small for solids.
Example 3
Let us calculate C P C V for water at 208C. The challenge is to sort out the units. First, we need the
3
molar volume of water at 208C using the density at 208C of 0.9982063 g=cm and the molecular
3 3
weight of 18.01528g=mol, which yields (18.01528 g=mol=0.9982063 g=cm ) ¼ 18.04765cm =mol ¼
6 3
18.04765 10 m =mol. Use a and b from Table 6.4.
3 3 6 3 2
Note that 1 cm ¼ (0.01 m) ¼ 1 10 m and we need to recall that 1 Pa ¼ 1N=m .
3
TVa 2 (293:15 K)(18:04765 10 6 m =mol)(0:206 10 = C) 2
3

(C P C V ) ¼ ¼ 4 6
b (4:591 10 =10 Pa)
1
Note a common problem in data tables is that the power of 10 is shifted, so here [10 ( C) ] means
3
that the number in the table has been multiplied by 1000 and the units are reciprocal degrees
4
1
centigrade. Similarly [10 (MPa) ] means that the number in the table has been multiplied by 10 4
2
and the units are reciprocal megapascals. We also flipped reciprocal pascals into m =N in the
denominator of the expression. Thus, we find that
TVa 2
¼ 0:489032 Nm=mol K ¼ 0:489032 J=mol K ¼ 0:116882 cal=mol K:

b
(C P C V ) ¼
While this is a small value compared to R ¼ 8:314 J=mol K ¼ 1:987 cal=mol K, it is not a small

number meaning that C P and C V are noticeably different for liquid water. We did this as an exercise
showing the use of two of the HUGA equations and an interesting application of units. This is one of
the topics usually shown in a graduate course in thermodynamics but in our approach of only
covering a few topics well we have shown an application of some of the HUGA equations and there
will be others. Knowing the HUGA equations and how to use them is roughly half of all
thermodynamics.
OPEN SYSTEMS: GIBBS–DUHEM EQUATION FOR PARTIAL MOLAL VOLUMES
The relationship we are about to describe is due to the work of Pierre Duhem (1861–1916) a French
physicist who translated Gibbs’ work into French and was in his own rights a prolific author of
thermodynamic studies. So far the applications of thermodynamic (except for the on-stream
ammonia synthesis discussed above) have been for what are ‘‘closed systems’’ where it is possible
to enclose the ‘‘system’’ with a boundary and separate it from the ‘‘environment.’’ Many of the
synthetic applications in chemical engineering are carried out with on-stream processing rather than
in a batch reactor, a system in which a continuous flow of reactants is processed and continuous
product flows out of some sort of reaction chamber. While most laboratory synthesis is carried out in
batch fashion, there are also static phenomena, which depend on adding an arbitrary amount of one