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THE TEMPERATURE DEPENDENCE OF ENTROPY       143



                and hence

                                                     T 2
                                           S = C p ln                           (4.17)
                                                     T 1
                (b) C p is not independent of temperature (over larger temperature
                ranges). We employ a similar approach to that above, except that C p is incorporated
                into the integral, yielding


                                          S 2      T 2
                                                     C p
                                            dS =       dT                       (4.18)
                                                     T
                                         S 1      T 1
                which, on integration, yields Equation (4.9).

             Worked Example 4.7 What is the increase in entropy when warming 1 mol of chlo-
             roform (III) from 240 K to 330 K? Take the value of C p for chloroform from the Aside
             box on p. 142.

                                                  H

                                              Cl
                                                Cl  Cl
                                                 (III)

             We start with Equation (4.9), retaining the position of C p within the  The ‘T’ontop and
             integral; inserting values:
                                                                          bottom cancel in the
                                          330 K             −2            second term within the
                                             91.47 + 7.5 × 10 T
                   S = S 330 K − S 240 K =                      dT        integral.
                                        240 K        T
               Rearranging:
                                           330 K  91.47
                                   S =              + 7.5 × 10 −2  dT
                                         240 K  T
             Performing the integration, we obtain

                                S = [91.47 ln T ] 330 K  + 7.5 × 10 −2  [T ] 330 K
                                              240 K              240 K
             Then, we insert the variables:

                                         330 K           −2

                            S = 91.47 ln        + 7.5 × 10 (330 K − 240 K)
                                         240 K
             to yield
                                     S = (29.13 + 6.75) JK −1  mol −1

             so
                                         S = 35.9JK  −1  mol −1
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