Clausius–Clapeyron Equation (Two-Point Form)

Integrate the Clapeyron equation with the two honest simplifications, that the vapour is ideal and that $\Delta H_{vap}$ does not change over the interval, and this two-point form falls out. It is the workhorse for turning two boiling points into an enthalpy of vaporisation, or one boiling point into all the others. A liquid boiling at 373.15 K under one atmosphere and at 354.75 K under half an atmosphere has $\Delta H_{vap} = R\ln 2 / (1/354.75 - 1/373.15) = 41.5\ \text{kJ/mol}$, which is water to within a couple of percent.

Run it forward on water from 100 °C down to 80 °C with $\Delta H_{vap} = 40.7\ \text{kJ/mol}$ and it predicts 48.2 kPa. Steam tables say 47.4 kPa. That 1.7% gap is not an arithmetic error, it is the cost of holding $\Delta H$ constant across 20 K, and it grows fast as you widen the interval. Near the critical point $\Delta H_{vap}$ collapses towards zero and the equation fails outright.

The unglamorous mistake is temperature units. Both temperatures go in as absolute values, because they appear as $1/T$ and the reciprocal of a Celsius reading is meaningless. This page takes any temperature unit and converts, but if you are working the equation on paper, convert to kelvin first. The unexpected use, incidentally, is in the kitchen and on mountains: the same equation, run backwards, tells you that at 3000 m water boils near 90 °C, which is why high-altitude cooking directions exist.