Page 161 - Essentials of physical chemistry

Gibbs’ Free Energy and Equilibria 123

3
We can use the density of water at 258C of 0.9970480 g=cm and the molecular weight of
18.01528 g=mol to calculate the molar volume of water at 258Cas:

3
3
(18:01528 g=mol)=(0:9970480 g=cm ) ¼ 18:06862 cm =mol ¼ V 0 :
H 2 O
Then we can integrate the Gibbs–Duhem equations to find the partial molal volume of water.

m
V ð 1 ð m V H ð 2 O ð
1 1 3 1
n 2 dV 2 , so m (1:77)m 2 þ 2(0:12) dm:
dV 1 ¼ dV H 2 O ¼
n 1 n H 2 O 4
0 0 18:06862 0
V
This looks complicated but just do the integral term by term and insert the numbers.

3
3 (1:77) 2 m 2 þ 2(0:12)m 2 1 m
4 3 2
, that is the right side of the

V H 2 O 18:06862 mL=mol ¼
1000 g

18:01528 g=mol 0
equation is to be evaluated between the limits of 0 and m for the general treatment of any molal
concentration of NaCl. Moving the volume of pure water to the right side and carefully evaluating
the other formula at the upper limit of m and the lower limit of 0 leads to
" #
3
1:77 m 2 þ 0:12m 2
2 3
55:50844
V H 2 O ¼ 18:06862 (as cm =mol). Note that the partial molal volume of the
water actually contracts with increasing amounts of NaCl. Combining these results and rounding the
numbers to fewer places considering the limited significant figures in the original polynomial we
find (finally):
3 2 1 2
V H 2 O ¼ 18:069 0:0159m 2 0:00216m and V NaCl ¼ 16:62 þ 2:66m 2 þ 0:24m :
Aside from some esoteric calculation, what does this mean? First, it means that electrical inter-
actions within the solution cause departures from the usual sum-of-blocks total volume of the
solution. This is a warning that on the molecular scale electric effects become important in liquids.
Second, notice that the partial molal volume of the NaCl solute gets larger as ‘‘m’’ increases since
that brings in the charged particles to the solution. On the other hand, the partial molal volume of the
water decreases as more charged ions are in solution; the water ‘‘shrinks’’ on a molar basis as the
NaCl is added. Third, we can calculate the total volume of the solution using the moles of water and
the moles of NaCl by multiplying the moles by the formulas we have obtained for a given value of
‘‘m.’’ Use of the Gibbs–Duhem equation now allows us to maintain our naive idea of just
multiplying the molar volume formulas by the number of moles and adding to get the total. This
sort of study of solution behavior is currently an area of considerable research, although the
approach uses large and complicated computer simulations rather than the relatively simple example
we have shown for a binary solution.

CHEMICAL POTENTIAL FOR OPEN SYSTEMS
Historically, one of the main contributions of J. W. Gibbs to the development of thermodynamics
was his extension of G ¼ H TS to open systems. This is an important consideration for on-
stream processes encountered by chemical engineers. We have already introduced the concept of
the chemical potential, m ¼ (G i =n i ), in two previous applications in this chapter; first in the
i
0
treatment of gas species m(T, P) ¼ m þ RT ln P and then again in the discussion of the
Gibbs phase rule. So far the treatments referred to closed systems and it seemed that m is just
i