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Chemical thermodynamics                  25

                dsuniverse  = dssystem + dssurroundings               (2.20a)
                dsuniverse = 0 for reversible (equilibrium) transformations   (2. 2 0b)
                dsuniverse>O for irreversible (spontaneous) transformations  (2.20c)

                The second law of thermodynamics cannot be proved. It is believed
              to  be  valid  because  it  leads  to  deductions  that  are  in accord  with
              observations and experience.  The following exercise provides an ex­
              ample of such a deduction.
                Exercise 2 .4. Assuming the truth of the second law of thermodynam­
              ics, prove that an isolated ideal gas can spontaneously expand but not
              spontaneously contract.
                Solution.  We  will  consider a  unit  mass of the gas.  If the gas is
              isolated, it has no contact with its surroundings; therefore,  dssurroundings
              =  0 , and

                            d su ni verse  = dssystem  +  dssurroundings =  ds gas   (2.2 1 )
                   i
              Also,  f   the gas  s   i s olated ,  dq = dw = O ;   therefore, from Eqs. (2.5) and
                            i
              (2. 6 ) ,   du = O.  If du =  O ,  it follows from Joule' s   law for an ideal gas (see
              Note  I  in this chapter) that dT=  O .  Hence, the gas must pass from its
              initial state  I )   to its final state (2) isothermally.
                        (
                                             s
                To obtain an expression for d s ga •  we can follow any reversible and
              isothermal path from state I to state 2, and evaluate the integral
                                            f  dqrev
                                        ga
                                      ds s =   2
                                             I  T
              For an ideal gas (see Exercise  . 2 8)
                                         2

                                     dqrev   dT    dp
                                     --   =  c  -  -  R -
                                       T    P  T   p
              where R  is  the  gas  constant  for  a  unit  mass  of the  gas  and  c the
                                                                        P
              specific heat at constant pressure of the gas. Therefore,





              or,
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