3.D.3. Liquid-Vapor Coexistence Region

Consider the phase diagram in the P-v plane depicted in Fig.3.7. [cf. Fig3.4-6].

Along an isothermal T 0 < T C , the straight line segment AB inside the vapor-liquid coexistence region indicates a mixture of the vapor and liquid phases.  Here, both phases are at the same temperature T 0  and pressure P 0  as given by a point on the vaporization curve.  The molar volumes of the pure vapor and liquid phases are given by v g = v A  and v l = v B , respectively.  The state D of molar volume v D  thus represents a mixture of phases obeying the lever rule

        v D = x g v g + x l v l                         (3.24)

where x indicates a molar fraction.  With the help of the constraint x g + x l =1 , we can rewrite (3.24) as

        ( x g + x l ) v D = x g v g + x l v l  

Þ            x l x g = v g v D v D v l                     (3.25)

Metastable States

According to eq(2.179), mechanical stability requires

κ T = 1 v ( v P ) T >0   Þ        ( v P ) T <0              (a)

 

Consider again the isotherm T 0 < T C  in Fig.3.7.  If we extend the gaseous state beyond point A into the coexistence region (see dashed line), mechanical stability can still be maintained as long as condition (a) is satisfied.  However, since the free energy is not a minimum, the system is only metastable. These states are called supercooled (vapor) states.  Similarly, metastable states resulting from the continuation of the liquid state past point B into the coexistence region are called superheated (liquid) state.

 

It is possible to extend the superheated liquid states into the negative pressure region.  In which case, no wall is required to contain the system.

 

Note that at the critical point C, the molar volumes of both phases are equal.  Since the relevant symmetries are also the same, these phases become indistinguishable.  Also, there are no metastable states.

Law of Corresponding States

The liquid-vapor coexistence curve in the T-ρ plane can be plotted in terms of reduced quantities T T C  and ρ ρ C , where the subscript C denotes a critical value.  Guggenheim found that such curves more or less coincide for a large number of pure substances (see Fig.3.8).  This is an example of the law of corresponding states, which postulates that all pure classical fluids obey the same equation of state involving reduced quantities.

The curve in Fig.3.8 is given by Guggenheim as the solutions to the following equations

        ρ l + ρ g 2 ρ C =1+ 3 4 ( 1 T T C )                           (3.26)

        ρ l ρ g ρ C = 7 2 ( 1 T T C ) 1/3                             (3.27)

Heat capacities

In the coexistence region, the total internal energy of a mixture (point D in Fig.3.7 ) is

        U tot = n g u g ( v g , T 0 )+ n l u l ( v l , T 0 )                               (3.28)

where n i  is the number of moles of phase i substance present and u i ( v,T )  is the molar energy of the substance when it is in the pure phase i with molar volume v and temperature T.  Dividing by the total number of moles, we obtain the molar total internal energy

        u tot = x g u g ( v g , T 0 )+ x l u l ( v l , T 0 )                                (3.29)

so that the molar heat capacity at constant volume is

        c v = ( u tot T ) v D

                = x g ( u g T ) coex + x l ( u l T ) coex + ( u l u g ) coex ( x l T ) v D          (3.30)

where we've used d x g +d x l =0 .

The following steps are required to related (3.30) with directly measurable quantities.

1.         From u l = u l ( v l , T 0 ) , we have

( u l T ) coex = ( u l T ) v l + ( u l v l ) T ( v l T ) coex

                = c v l + ( u l v l ) T ( v l T ) coex                    (3.31)

        Similarly,

                ( u g T ) coex = c v g + ( u g v g ) T ( v g T ) coex          (3.32)

2.                ( u l u g ) coex = h l P 0 v l ( h g P 0 v g )

=Δ h lg + P 0 Δ v lg

=[ T 0 ( dP dT ) coex + P 0 ]Δ v lg                (3.34)

3.         Finally, to calcualte ( x l T ) v D , we begin with

               v D = x l v l +( 1 x l ) v g

        so that

0= ( x l T ) v D v l + x l ( v l T ) coex ( x l T ) v D v g +( 1 x l ) ( v g T ) coex

        i.e.,          ( x l T ) v D = 1 Δ v lg [ x l ( v l T ) coex +( 1 x l ) ( v g T ) coex ]

                                        = 1 Δ v lg [ x l ( v l T ) coex + x g ( v g T ) coex ]          (3.36)

Putting everything into (3.30) gives

        c v = x g [ c v g + ( u g v g ) T ( v g T ) coex ]+ x l [ c v l + ( u l v l ) T ( v l T ) coex ]

                +[ T 0 ( dP dT ) coex + P 0 ][ x l ( v l T ) coex + x g ( v g T ) coex ]                  (3.37)

        = x g { c v g +[ ( u g v g ) T T 0 ( dP dT ) coex + P 0 ] ( v g T ) coex }

                + x l { c v l +[ ( u l v l ) T T 0 ( dP dT ) coex + P 0 ] ( v l T ) coex }                    (3.37a)

Now, from

        du=TdsPdv

we have

        ( u v ) T =T ( s v ) T P=T ( P T ) v P                      (3.38)

 

Also, using the Maxwell relation

( x y ) z = ( x y ) w + ( x w ) y ( w y ) z                              (2.8)

we can write

        ( P T ) coex = ( P T ) v + ( P v ) T ( v T ) coex                       (3.39)

 

Putting (3.38-9) into the terms included in one of the square brackets in (3.37a), we get, in the coexistence region,

        ( u v ) T T 0 ( P T ) coex + P 0 = T 0 ( P T ) v T 0 ( P T ) coex

                                                = T 0 ( P v ) T ( v T ) coex

Hence, (3.37) simplifies to            

        c v = x g { c v g T 0 ( P v g ) T ( v g T ) coex 2 }+ x l { c v l T 0 ( P v l ) T ( v l T ) coex 2 }                (3.40)

with all quantities on the right side measurable.

 

Since the isotherm is also an isobar in the coexistence region, c P  is infinite for a mixture in the coexistence region.  Thus, adding heat at constant pressure to a mixture in the coexistence region only converts some liquid into vapor.  The temperature remains constant until the entire mixture is turned into vapor.