numvana

Clausius-Clapeyron Equation Calculator

Estimate how a substance's vapor pressure shifts between two temperatures from its molar latent heat of vaporization, via the integrated Clausius-Clapeyron equation.

Vapor pressure at T2 (same unit as P1)
70.6047
Pressure ratio (P2 ÷ P1)
0.6968×

How it works

The integrated Clausius-Clapeyron equation estimates how a pure substance's vapor pressure changes between two temperatures, assuming its molar latent heat of vaporization L stays roughly constant over that range: ln(P2/P1) = −(L/R) × (1/T2 − 1/T1), so P2 = P1 × exp[−(L/R) × (1/T2 − 1/T1)], where R = 8.314 J/(mol·K) is the molar gas constant and both temperatures are in kelvin.

L — the energy needed to turn one mole of the substance from liquid to vapor — is something you look up for your specific substance (it varies by substance and isn't invented or stored by this calculator); the result comes out in whatever pressure unit you entered P1 in, since only the ratio P2/P1 is computed from the formula.

Because L is treated as constant, accuracy degrades the further T2 is from T1, or near the substance's critical point where the real latent heat drops sharply — this is a good estimate over a modest temperature range, not an exact equation of state.

FAQ

Where do I find the molar latent heat of vaporization for a substance?

It's a published physical property specific to each substance (for example, water's is commonly cited around 40.7 kJ/mol near its boiling point) found in chemistry reference tables or a substance's safety data sheet — this calculator runs the formula on whatever value you supply rather than storing or looking one up itself, since that figure varies by substance and even somewhat by temperature.

Why does the result get less accurate over a large temperature range?

The derivation assumes L doesn't change between T1 and T2, but in reality latent heat decreases as temperature rises toward the substance's critical point. Over a few tens of degrees the approximation is usually good; stretched across a much wider range, or close to the critical point, the constant-L assumption breaks down and the estimate drifts from the true vapor pressure.

Related calculators