3.D.2.a. Vaporization Curve

Consider the coexistence curve between the liquid and gaseous phases of a pure substance.  This is indicated as the vaporization curve between the triple A and critical point C in Figure 3.4.

 

If we neglect the volume changes of the liquid phase, we have

        Δ v lg = v g v l v g = RT P

where the last equality assumes the vapor to be an ideal gas.

The Clausius-Clapeyron equation thus simplifies to

        ( dP dT ) coex = Δ h lg TΔ v lg PΔ h lg R T 2                               (3.15)

where Δ h lg  is the molar latent heat of vaporization.  Treating Δ h lg  as a constant, we get, along the vaporization curve,

        dP P = Δ h lg R dT T 2

        lnP= Δ h lg RT +const

        P= P exp( Δ h lg RT )                 (3.16)

where P  is the pressure for T .

Exercise 3.2

Find the molar heat capacity c coex  along the vaporization curve.

Answer

        c coex =T ( s T ) coex

With T, P as independent variables, we have

        ds= ( s T ) P dT+ ( s P ) T dP

On the vaporization curve, we have

        dP= ( dP dT ) coex dT PΔ h lg R T 2 dT

so that

        ( ds ) coex =[ ( s T ) P + ( s P ) T ( P T ) coex ]dT

                        [ ( s T ) P + ( s P ) T PΔ h lg R T 2 ]dT

Thus

        c coex = c P + ( s P ) T PΔ h lg RT

Using

        ( s P ) T = ( v T ) P = R P

where the last equality is valid for an ideal gas, we have

                c coex = c P Δ h lg T                        (2)

which means for T< Δ h lg c P , we have c coex <0  so that the vapor gives off heat as T is raised.