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August 18, 2010 11:36      9in x 6in     b985-ch03     Elementary Physical Chemistry





                                             The Second Law of Thermodynamics              27

                            3.3.4. Isolated Systems

                            Variation of heats is difficult to determine, and most often one calculates
                            entropy changes of isolated systems, where dq = 0. Thus, for an isolated
                            system, Eq. (3.18) takes the form
                                                      ∆S isolated ≥ 0                  (3.19)




                              Comment: The results obtained so far, which relates entropy to heat
                              change divided by temperature, was derived for a particular system —
                              an ideal gas. What assurance do we have that the efficiency criterion
                              applies to other systems? It does! There is a theorem (not to be
                              developed here) that proves that all systems operating reversibly
                              between the same temperatures have the same efficiency.Thus,if
                              we derive the entropy for one system, we have it for all.





                            3.4. Determination of Entropy
                            Note: We consider in this section only reversible changes.


                            3.4.1. Entropy change in Phase Transitions (solid–liquid,
                                  liquid–vapor, solid–vapor)

                            i) At constant T

                                                    ∆S =    dq/T = q/T                 (3.20)

                            ii) At constant T and P, q P =∆H and ∆S =∆H/T              (3.21)

                            Example 3.1.
                                                        ◦
                                   H 2 O(s, 0 C) = H 2O(l, 0 C)  ∆H =6.01 kJmol −1
                                          ◦
                                          ∆S = 6010J mol −1 /273.15 K = 21.99J K −1  mol −1
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