Clausius–Clapeyron Calculator
Predict the vapor pressure of a liquid at a new temperature from a known vapor pressure, the two temperatures and the enthalpy of vaporization.
How it works
ln( P₂ / P₁ ) = −( ΔH_vap / R ) · ( 1/T₂ − 1/T₁ ) → P₂ = P₁ · e^that
The Clausius–Clapeyron equation links a liquid’s vapor pressure to its temperature through the enthalpy of vaporization. Because vapor pressure climbs steeply — exponentially — with temperature, a modest warming can multiply it several times over. Given one measured vapor-pressure/temperature pair and the enthalpy of vaporization, the equation predicts the vapor pressure at any other temperature, which is how boiling points at altitude and distillation conditions are worked out.
Worked example
Water boils at 373.15 K where its vapor pressure is 1 atm, and ΔH_vap ≈ 40,700 J/mol. At 350 K: ln(P₂/1) = −(40,700 / 8.314) × (1/350 − 1/373.15) = −0.868, so P₂ = e^(−0.868) = 0.42 atm — which is why water boils below 100 °C on a mountain.
Frequently asked questions
Why must temperatures be in kelvin?
The equation uses reciprocals of absolute temperature (1/T). Celsius has an arbitrary zero, so using it would give nonsensical reciprocals. Always convert to kelvin first.
What does the enthalpy of vaporization control?
It sets how sharply vapor pressure responds to temperature. Liquids with strong intermolecular forces have a large ΔH_vap and a steeper vapor-pressure curve than volatile liquids with weak forces.
Is ΔH_vap really constant with temperature?
It decreases slightly as temperature rises, so the two-point equation is an approximation. Over a modest temperature range it is accurate; over a very wide range a temperature-dependent ΔH_vap gives better results.
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This calculator is for educational purposes. Double-check important results before acting on them.