3.5 The Third Law of Thermodynamics and Absolute Entropies 147
3.41 Find the value ofS◦mfor solid silver at 270.0 K from the
following data:
T/K CP, m/J K−^1 mol−^1 T/K CP, m/J K−^1 mol−^1
30 4.77 170 23.61
50 11.65 190 24.09
70 16.33 210 24.42
90 19.13 230 24.73
110 20.96 250 25.03
130 22.13 270 25.31
150 22.97Assume that the Debye formula can be used from 0 K to
30.0 K. Since there are an odd number of data points that
are equally spaced, you can use Simpson’s rule.3.42 The following are heat capacity data for pyridine.^10
T/K CP, m(s)/ T/K CP, m(s)/ T/K CP, m(l)/
JK−^1 mol−^1 JK−^1 mol−^1 JK−^1 mol−^113.08 4.448 151.57 63.434 231.49 120.67
21.26 12.083 167.60 68.053 239.70 122.23
28.53 19.288 179.44 71.756 254.41 124.54
35.36 25.309 193.02 76.467 273.75 127.93
48.14 33.723 201.61 79.835 293.96 131.88
64.01 40.748 212.16 84.446 298.15 132.74
82.91 46.413 223.74 94.328 307.16 134.55
101.39 50.861 231.49 101.25
132.32 58.325A value of 8278.5 J mol−^1 is reported for the enthalpy
change of fusion at the normal melting temperature of
231.49 K.
a.Assuming that the Debye formula can be used between
0 K and 13.08 K, find the absolute entropy of solid
pyridine at 231.49 K.
b. Find the absolute entropy of liquid pyridine at
231.49 K and at 298.15 K. Use a numerical technique
to approximate the integrals needed.
3.43 For each of the substances in Table A.7, calculate the
entropy change of vaporization at the normal boiling
temperature and compare your result with Trouton’s rule.
Which of the substances qualify as “normal” liquids?
3.44 For each of the following substances, calculate the entropy
change of vaporization at the normal boiling temperature
and compare your result with Trouton’s rule. Which of the
substances qualify as “normal” liquids?Substance Boiling ∆vapHm/kJ mol−^1
temperature/Kpropane 231. 020. 133
n-butane 272. 624. 272
n-pentane 309. 227. 594
2-methyl butane 301. 027. 074
2,2-dimethyl propane 282. 623. 634
hexane 342. 231. 912
2-methyl pentane 333. 432. 119
3-methyl pentane 336. 432. 400
2,2-dimethyl butane 322. 830. 422
2,3-dimethyl butane 331 29. 790Summary of the Chapter
Kelvin’s statement of the second law of thermodynamics is that heat put into a system
that undergoes a cyclic process cannot be completely converted into work done on the
surroundings. Clausius’ statement of this law is that heat cannot flow from a cooler to
a hotter body if nothing else happens. The mathematical statement of the second law
was shown to be a consequence of the Kelvin statement. It asserts thatS, the entropy,
is a state function if we definedSdqrev
TIt was shown from the second law that in any reversible process the entropy of the
universe remains constant, whereas in any irreversible process the entropy of the uni-
verse must increase.(^10) F. T. Gucker and R. L. Seifert,Physical Chemistry, W. W. Norton, New York, 1966, p. 445.