We derived, on the previous page, the Clapeyron Equation:

It now remains to show how it may be applied to the most commonly encountered phase boundaries:

For the boundary between solid and liquid phases, the entropy of transition, ΔS_{trs}, may be replaced by ΔH_{trs}/T (This is because on the phase boundary the two phases are in equilibrium, so the heat transfer, which is equal to the enthalpy change accompanying the process, occurs reversibly and may be equated with TΔS. Rearrangement gives the above substitution.) We thus obtain, for the melting (fusion) transition:

Since the change in volume upon melting is small, this phase boundary is always steep. Note that, if the enthalpy and volume changes of fusion change so little with temperature that they can be treated as constant, the above expression may be integrated as follows:

which gives

as the approximate equation for the solid liquid phase boundary.

At the liquid-vapour boundary, we pursue a similar approach, by initially substituting ΔH_{vap}/T for ΔS_{vap}. This gives the Clapeyron equation for the liquid-vapour phase boundary as:

The volume change upon vaporisation is much greater than that upon melting. Thus this phase boundary is much less steep than the solid-liquid phase boundary. It follows from this that the reciprocal, dT/dp , would be much larger for vaporisation than fusion, and hence that the boiling point is much more sensitive to the external pressure than the melting point.

It is possible to manipulate this equation by the introduction of some approximations. Since the molar volume of a gas is much, much greater than the molar volume of a liquid, we may approximate ΔV = V_{m,g }– V_{m,l }» V_{m,g} i.e. neglect the molar volume of the liquid since that of the gas is so much greater. Further, we can make the approximation that the gas behaves perfectly, which means we can make the substitution V_{m}(g) ≈ RT/p. These two approximations turn the exact Clapeyron equation into:

We may now rearrange the above expression, using the property that dp/p = d ln p (a general property in calculus. This property is clearly related to the familiar result that integrating 1/x with respect to x gives ln x.)

Rearrangement yields the Clausius-Clapeyron equation, which gives the temperature dependence of the vapour pressure of a substance:

If we make the assumption that the enthalpy of vaporisation is independent of temperature over the range of interest, then the expression may be integrated very simply to give an equation relating vapour pressure and temperature:

The solid-vapour boundary may be treated in exactly the same way, though the enthalpy change of vaporisation must be replaced with the enthalpy change of sublimation.