Page 40 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:36      9in x 6in     b985-ch03     Elementary Physical Chemistry





                                             The Second Law of Thermodynamics              25


                            Similarly, for the adiabatic transition from the lower (T 1,V ) to the higher
                                                                               1
                            isotherm (T 2 ,V 1 )weobtain
                                             ln(T 2 /T 1 )+(γ − 1) ln(V 1 /V ) = 0    (3.12b)

                                                                     1
                            which is equivalent to
                                             ln(T 1 /T 2 )+(γ − 1) ln(V 1 /V ) = 0    (3.12c)

                                                                     1

                               Obviously, comparing (3.12a) with (3.12c) shows that V /V 2 = V /V 1
                                                                                        1
                                                                                2


                            or V /V = V 2 /V 1 . Substituting this result in Eq. (3.10b), shows that
                               2
                                  1
                                                    q 1 = −RT 1 ln V 2 /V 1           (3.12d)
                               Finally, substituting this expression in Eq. (3.8) and using (3.10a) yields
                                                      ξ =1 − T 1 /T 2                  (3.13)
                               Note that since ξ = −w/q 2 =(q 1 + q 2)/q 2 =1 + q 1 /q 2 it follws
                            that q 1 /q 2 = −T 1 /T 2 and thus, q 1 /T 1 + q 2 /T 2 = 0. Replacing this by the
                            differential form dq rev/T (emphasizing the reversible nature of the process)

                            yields, upon integration along a closed contour,  dq rev /T =0, which is
                            another indication that dq rev/T is an exact differential.


                            3.3.2. Irreversible Process
                            What is the efficiency when the process is irreversible?
                               Let us first assume that the efficiency is greater for the irreversible
                            than for the reversible case. Denoting the irreversible efficiency as ξ and
                                                                                        ∗
                                                      ∗
                            the reversible as ξ, and since ξ >ξ,we musthave
                                                       ∗  ∗
                                                    −w /q > −w/q 2                     (3.14)
                                                          2
                               Consider now two Carnot cycles, the starred and the non-starred one,
                            coupled together (Fig. 3.3). Let us, for simplicity, adjust the engines so
                            that −w  ∗  = −w. Dividing both sides of (3.14) by −w ,1/q ∗ 2  < 1/q 2
                                                                              ∗
                                                                                 ∗
                            and therefore, q 2 <q . Hence, since the w’s are the same, q >q 1.This
                                               ∗
                                                                                 1
                                              2
                            means that heat, in the amount of q − q 1 , is transferred from the lower
                                                            ∗
                                                            1
                            temperature reservoir to the higher temperature one, in violation of the
                            Clausius Principle. So the efficiency of an irreversible change cannot be
                            greater than that of a reversible change.
                               Can the efficiency of an irreversible change be less than that of a
                            reversible change? Yes, there is no prohibition! Since ξ =1 − T 1 /T 2 for
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