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Teaching entropy                                              37


                      p  2                    1-2: Adiabatic compression
                                              2-3: Isothermal heating
                                              3-4: Adiabatic expansion
                                   T H        4-1: Isothermal cooling



                                               3


                                 1
                                            T L
                                                             4
                                                                    V
              Fig. 3.4 The operation of the Carnot cycle on a p-V diagram.

              The infinitesimal work done due to the expansion of a unit mass of the ideal
              gas is δW¼pdv. Substituting p¼R g T/v, where v denotes the specific vol-
              ume, and integrating Eq. (3.10) yields

                                      ð 3
                                                         v 3
                                            dv
                                 Q H ¼  R g T  ¼ R g T H ln              (3.11)
                                       2     v           v 2
              where R g is the gas constant. In a similar manner, the amount of heat rejected
              by the cycle per unit mole of the gas is found as follows.

                                                   v 4
                                       Q L ¼ R g T L ln                  (3.12)
                                                   v 1
              For the adiabatic compression and expansion processes, the first law is
                                                                         (3.13)
                                         dU ¼ δW
              Substituting dU¼c v dT and δW¼pdv into Eq. (3.13) leads to

                                                      dv
                                                                         (3.14)
                                   c v dT ¼ pdv ¼ R g T
                                                       v
              Dividing both sides of Eq. (3.14) by T and integrating over the adiabatic
              compression process gives

                                          T 2       v 1
                                      c v ln  ¼ R g ln                   (3.15)
                                          T 1       v 2
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