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                214
               Chapter 7                  TABLE 7.1
               One-Component Phase Equilibrium
               and Surfaces
                                         Enthalpies and Entropies of Fusion and Vaporization a
                                                                                                              THE
                                                     T nmp    fus H m    fus S m  T nbp    vap H m    vap S m    vap S m
                                         Substance
                                                      K     kJ>mol  J>1mol K2   K     kJ>mol   J>1mol K2  J>1mol K2
                                         Ne           24.5   0.335     13.6     27.1    1.76     65.0      64.8
                                         N 2          63.3   0.72     11.4      77.4    5.58     72.1      73.6
                                         Ar           83.8   1.21     14.4      87.3    6.53     74.8      74.6
                                         C H 6        89.9   2.86     31.8     184.5   14.71     79.7      80.8
                                          2
                                         (C H ) O    156.9   7.27      46.4    307.7   26.7      86.8      85.1
                                             5 2
                                           2
                                         NH 3        195.4   5.65     28.9     239.7   23.3      97.4      83.0
                                         CCl 4       250.    2.47      9.9     349.7   30.0      85.8      86.1
                                         H O         273.2   6.01     22.0     373.1   40.66    109.0      86.7
                                           2
                                         I 2         386.8  15.5      40.1     457.5   41.8      91.4      88.3
                                         Zn          693.    7.38     10.7    1184.   115.6      97.6      96.3
                                         NaCl       1074.   28.2      26.2    1738.   171.       98.4      99.4
                                         a    H and   S are at the normal melting point (nmp).   H and   S are at the normal boiling
                                                                                           vap m
                                           fus
                                                                                      m
                                                                                   vap
                                              m
                                                   fus m
                                         point (nbp).   S THE  is the normal-boiling-point   S value predicted by the Trouton–Hildebrand–
                                                                            vap m
                                                   vap m
                                         Everett rule.
                                               H      is usually substantially larger than   H  .   S  varies greatly
                                              vap  m,nbp                             fus  m,nmp  fus m,nmp
                                                                                                           3
                                         from compound to compound, in contrast to   S  . Amazingly,   H for  He be-
                                                                                 vap m,nbp          fus
                                                                                       3
                                         tween 0 and 0.3 K is slightly negative; to freeze liquid  He at constant T and P below
                                         0.3 K, one must heat it.
                                             Although H O(g) is not thermodynamically stable at 25°C and 1 bar, one can use
                                                       2
                                         the experimental vapor pressure of H O(l) at 25°C to calculate   G° of H O(g). See
                                                                        2                       f  298  2
                                         Probs. 7.67 and 8.36.
                                          7.3           THE CLAPEYRON EQUATION
                                         The Clapeyron equation gives the slope dP/dT of a two-phase equilibrium line on a
                                         P-T phase diagram of a one-component system. To derive it, we consider two infini-
                                         tesimally close points 1 and 2 on such a line (Fig. 7.5). The line in Fig. 7.5 might in-
                                         volve solid–liquid, solid–vapor, or liquid–vapor equilibrium or even solid–solid equi-
                                         librium (Sec. 7.4). We shall call the two phases involved a and b. The condition for
                                                                 b
                                                            a
                                         phase equilibrium is m   m . No subscript is needed because we have only one com-
                                                                                                          b
                                                                                                    a
                                         ponent. For a pure substance, m equals G [Eq. (4.86)]. Therefore G   G for any
                                                                            m                       m    m
                                         point on the a-b equilibrium line. The molar Gibbs energies of one-component phases
                                                                                               a
               Figure 7.5                in equilibrium are equal. At point 1 in Fig. 7.5, we thus have G m,1    G b  . Likewise,
                                                                                                      m,1
                                                                                                       b
                                                                              b
                                                                        a
                                                                                     b
                                                                 a
                                                          b
                                                    a
                                                                                               a
                                         at point 2, G m,2    G m,2 , or G m,1    dG   G m,1    dG , where dG and dG are the in-
                                                                                     m
                                                                        m
                                                                                                       m
                                                                                               m
               Two neighboring points on a two-  finitesimal changes in molar Gibbs energies of phases a and b as we go from point 1
               phase line of a one-component              a      b
               system.                   to point 2. Use of G m,1    G m,1  in the last equation gives
                                                                           a
                                                                        dG   dG  b m                        (7.13)
                                                                           m
                                             For a one-phase pure substance, the intensive quantity G is a function of T and P
                                                                                            m
                                         only:  G   G 1T, P2,  and its total differential is given by (1.30) as  dG
                                                m
                                                                                                              m
                                                      m
                                         10G >0T2  dT   10G >0P2  dP.  But the equations in (4.51) give 10G >0T2     S m
                                                                                                    m
                                                  P
                                            m
                                                                                                         P
                                                                 T
                                                           m
                                         and 10G >0P2     V .  So for a pure phase we have
                                                           m
                                                     T
                                                m
                                                     dG   S  dT   V  dP   one-phase, one-comp. syst.        (7.14)
                                                                      m
                                                              m
                                                       m
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