Clausius-Clapeyron Vapor Pressure Calculator

How the Clausius-Clapeyron Equation Is Used

The Clausius-Clapeyron equation relates the vapor pressure of a liquid to temperature, governed by the enthalpy of vaporization. It’s used to predict boiling points at different pressures and to measure vaporization enthalpies.

Clausius-Clapeyron Equation: ln(P2/P1) = −(ΔH_vap / R) × (1/T2 − 1/T1)

Where:

• P1, P2 = vapor pressures at temperatures T1 and T2 (same units)
• ΔH_vap = molar enthalpy of vaporization (J/mol)
• R = gas constant = 8.314 J/mol·K
• T1, T2 = temperatures in Kelvin

Worked Example — Water at High Altitude: Water boils at 100°C (373 K) at 1 atm (101,325 Pa). What is the boiling point at 0.75 atm (Denver altitude)?

Given: ΔH_vap(water) = 40,700 J/mol

• ln(0.75/1.0) = −(40,700/8.314) × (1/T2 − 1/373)
• −0.2877 = −4,895 × (1/T2 − 0.002681)
• 1/T2 = 0.002681 + 0.0000588 = 0.002740
• T2 = 1/0.002740 = 365 K = 91.8°C

Water boils about 8°C lower in Denver — matching the altitude boiling point formula result.

ΔH_vap Reference Values:

• Water: 40,700 J/mol (44,000 at 25°C)
• Ethanol: 38,600 J/mol
• Acetone: 31,300 J/mol
• Benzene: 30,800 J/mol

Applications: distillation column design, vacuum evaporation in food processing, pressure cooking optimization, pharmaceutical lyophilization (freeze-drying).

Reading the result

The equation is a two-point relationship, so it needs one vapour pressure and its temperature as a reference before it can tell you anything at a second temperature. That reference is usually the normal boiling point, where the vapour pressure is 1 atm by definition, which is why the boiling point is such a convenient anchor for any liquid.

Because the relationship is exponential, small temperature changes move vapour pressure a lot. Water at 100 °C exerts 760 mmHg; at 90 °C it is only 526. That steepness is the reason a pressure cooker works at all, and the same reason a mountain kettle boils at a temperature too low to cook a decent pasta.

Measuring the heat of vaporization yourself

Turn the equation around and it becomes a measurement technique. Record vapour pressure at several temperatures, plot ln(P) against 1/T, and the points should fall on a straight line of slope −ΔHvap/R. Multiply the slope by −R and you have the enthalpy of vaporization for a liquid nobody tabulated. The linearity of that plot is also a useful check: real curvature means ΔHvap is changing appreciably over your temperature range, and the two-point form is no longer trustworthy.

That assumption is the main limitation here. The derivation treats ΔHvap as constant and the vapour as an ideal gas, both of which hold well below the boiling point and fail near the critical point.