Page 157 - Modern physical chemistry
P. 157

7.7 Correcting the Calorimetric Entropy to the Ideal Gas State   149

             Here temperature Tl must be low enough so that (7.23) is a good approximation.
                If we assume that the third law applies to the given pure material, then the entropy at
             o K is zero. The entropy at very low temperatures is then given by increment (7.24). At tem-
             peratures up to the lowest first-order transition, one adds the amount given by (7.21). At
             the transition, one adds amount (7.22). Then up to the second first-order transition, one
             adds the amount given by (7.21) for the pertinent temperatures. And so on. The result is
             known as the calorimetric entropy 8CJJJ.or for the final temperature and amount of material.
                When the final phase is gaseous, the entropy change on transforming it to the ideal
             gas state is generally added.

             Z 7 Correcting the Calorimetric Entropy to the Ideal Gas State

                In an ideal gas, the molecules do not interact appreciably except during collisions.
             At a given temperature, this condition is approached as pressure P is reduced to zero.
             The change in 8 can be determined for taking the actual gas down to zero pressure and
             then taking the ideal gas form back up to the initial pressure.
                Consider one mole of gas at a given temperature T.  With its entropy 8  a function of
                                             dS=( ~~ 1                               [7.25]
             P and T,  we have

                                                       dP.
             But for the Gibbs free energy, equation (5.82) is

                                            dG=VdP-8dT.                              [7.26]
                                       -a~:r =( :~ 1 =-( ~~ 1·                       [7.27]
             From this, we obtain



                In describing the actual gas, we employ Berthelot equation (3.73). From it, one gets
                                         av)
                                        (       R  R 27  P  Tc 3                     [7.28]

                                         aT  P = P + P  32 Pc  T3 '
             while the ideal gas equation yields
                                              (~~)p  ;                               [7.29]



                Using (7.28) when lowering the pressure to PI' and (8.29) for raising it back to P
             gives us
                    P
              I:lS = -f ' (R + R 27 ~  ] dP _ fP.  R  dP = fP. R 27 ~  dP = 27 R ~ T~  P   [7.30]
                                                                3
                                    3
                                                               Tc
                                   Tc
                    P  P  P  32 Pc  T3      P P       P  32 Pc  T3     32  Pc  T3  p'
             Letting 1'* be zero, so the actual gas is ideal there, leads to
                                                         rc 3
                                           S. -8= 27 R~  •                           [7.31 ]
                                            1     32  Pc  T3
                Here 8 j  is the entropy of the hypothetical ideal gas at pressure P, 8 the entropy of the
             real gas at pressure P, T the absolute temperature, P the critical pressure, and Tc the crit-
             ical temperature for the given gas. When P equals 1 bar, 8 j  equals tf, the standard entropy.
   152   153   154   155   156   157   158   159   160   161   162