Page 118 - Physical chemistry eng

5.4 CALCULATING CHANGES IN ENTROPY 95

1.35 bar to a pressure of 3.45 bar. The entropy change for this process is obtained
using Equation (5.23) with the value of b = 1>T from Example Problem 5.4.

T f P f T f P f T f
C P C P dP C P P f
¢S = dT - Vb dP = dT - nR = dT - nR ln
T T P T P i
3 3 3 3 3
T i P i T i P i T i
T T 2 T 3
-5
-8
-2
800. a18.86 + 7.937 * 10 - 6.7834 * 10 + 2.4426 * 10 b
K K 2 K 3 T
= 2.50 * d
T K
3
450.
K
3.45 bar
-1
-1
-2.50mol * 8.314 J K mol * ln
1.35 bar
= 27.13 J K -1 + 69.45 J K -1 - 37.10 J K -1 + 8.57 J K -1
-19.50 J K -1
= 48.6 J K -1

EXAMPLE PROBLEM 5.6
In this problem, 3.00 mol of liquid mercury is transformed from an initial state charac-
terized by T = 300. K and P = 1.00 bar to a final state characterized by T = 600. K
i
i
f
and P = 3.00 bar.
f
–3
–1
b
r
a. Calculate ¢S for this process; = 1.81 10 –4 K , = 13.54 g cm , and C P,m
–1 –1
for Hg(l) = 27.98 J mol K .
b. What is the ratio of the pressure-dependent term to the temperature-dependent
term in ¢S ? Explain your result.
Solution
a. Because the volume changes only slightly with temperature and pressure over the
range indicated,
T f P f
C P T f
¢S = dT - Vb dP L nC P,m ln - nV m,i b(P - P )
f
i
T T i
3 3
T i P i
600. K
-1
-1
= 3.00mol * 27.98 Jmol K * ln
300. K
200.59 gmol -1
-4
-3.00mol * * 1.81 * 10 K -1 * 2.00 bar
6
10 cm 3
-3
13.54 gcm * 3
m
5
* 10 Pabar -1
= 58.2 J K -1 - 1.61 * 10 -3 J K -1 = 58.2 J K -1
b. The ratio of the pressure-dependent to the temperature-dependent term is
–5
–3 10 . Because the volume change with pressure is very small, the
contribution of the pressure-dependent term is negligible in comparison with
the temperature-dependent term.
As Example Problem 5.6 shows, ¢S for a liquid or solid as both P and T change is
dominated by the temperature dependence of S. Unless the change in pressure is very
large, ¢S for liquids and solids can be considered to be a function of temperature only.