The Clausius-Clapeyron Equation
To make use of the above equations, an expression for the saturation vapour pressure is required. Consider the co-existence of pure water vapour and a plane water surface such that thermal equilibrium prevails. Let the subscripts and refer to the vapour and liquid phases respectively and represent the internal energy. The first law may be stated as:
but since the equilibrium occurs at constant temperature ( ) and pressure , one may write:
or where is the Gibbs free energy for saturated water vapour.
What this means is that the Gibbs free energy is unchanged during the passage between liquid and vapour phases or
where and are the Gibbs free energies of vapour and liquid respectively.
Now consider equilibrium occurring at a slightly different pressure and temperature for which . Using eq.(11) then implies that . The Gibbs free energy
for a gas at temperature and pressure is usually defined by so that is given by:
but the first three terms on the RHS sum to zero by virtue of eq.(10) and so:
Applying this to the relation with gives:
and on using eqs.(7) and (8) together with the assumption (i.e. the volume of vapour is much greater than the volume of the equivalent mass of liquid water) reduces to:
Finally, the perfect gas equation for water vapour ( ) can be used to to eliminate above giving:
which is the Clausius-Clapeyron equation.