Using the ideas from the temperature dependence of the vapor pressure we can make a new equation that relates the vapor pressure, P1, at one temperature, T1, to the vapor pressure, P2, at another temperature, T2, we find the following (after some algebra).
\[\ln \left({P_2\over P_1}\right) = {\Delta H_{\rm vap}\over R}\left({1\over T_1} - {1\over T_2}\right)\]
This is the Clausius‐Clapeyron Equation. There are many forms in which you might find this equation, however they are all equivalent.
Note: This also means that if you plot ln(P) vs 1/T the graph will be a straight line with a slope equal to ∆Hvap/R.
Derivation of the Clausius-Clapeyron equation assume that the vapor behaves as an ideal gas such that its molar free energy (chemical potential) is simply given as
\[G = G^{\circ} + RT\ln\left ( {P \over P^{\circ}} \right )\]
This video works through an example in which you are given the normal boiling point and heat of vaporization of a substance and asked what the vapor pressure of this substance would be at a specific temperature. This question requires use of the Clausius-Clapeyron equation described above.