c05 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come
80 ENTROPY AND THE SECOND LAWinvolving the energy and the entropy used more by engineers than by chemists; this
is called the Helmholtz free energy,A=E−TS.PROBLEMS AND EXAMPLE
Example 5.1 The Standard Entropy of Silver
TheCp/Tvs.Tdata set for silver from 15 to 300 K isTCp Cp/T lnT
15.0000 0.6700 2.7100 0.0447
30.0000 4.7700 3.4000 0.1590
50.0000 11.6500 3.9100 0.2330
70.0000 16.3300 4.2500 0.2333
90.0000 19.1300 4.5000 0.2126
110.0000 20.9600 4.7000 0.1905
130.0000 22.1300 4.8700 0.1702
150.0000 22.9700 5.0100 0.1531
170.0000 23.6100 5.1400 0.1389
190.0000 24.0900 5.2500 0.1268
210.0000 24.4200 5.3500 0.1163
230.0000 24.7300 5.4400 0.1075
250.0000 25.0300 5.5200 0.1001
270.0000 25.3100 5.6000 0.0937
290.0000 25.4400 5.6700 0.0877
300.0000 25.5000 5.7000 0.0850TheCp/Tvs.Tcurve for silver is given as Fig. 5.2. Select from packaged software
(for example, SigmaPlot©C) or write a short program of your own that will enable you
to integrate the data set for silver to find the standard entropySat 300 K. The problem
is simplified by the lack of phase transitions in solid Ag over the temperature range,
including the melting and boiling points, which are well above 298 K. TheCRC
Handbook of Chemistry and Physics, 2008–2009 (89th ed.) value for the standard
S^298 Ag at 298 is 42.67 J K−^1 mol−^1.Solution 5.1 SigmaPlot©Ccontains a macro that carries out integration under curves
that are displayed as a smooth function. First plot your function, then execute
Tools → macro → run → compute. Be sure to designate your plot be-
low the macro. The SigmaPlot output for this integration is 42.2076 over the interval
from 15 K to 298 K.
The result is pretty close to the standard entropy in the handbook, but it lacks
a contribution below 15 K. This problem is usually handled by the Debye method
(Problem 5.7), which assumes a third-power equation leading toCp=AT^3 = 0 .67 J K−^1 mol−^1