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)

Flip the sign inside the bracket and you get (ΔH_vap / R) × (1/T1 − 1/T2), which is the same equation with one fewer minus sign to lose. That is the form the result panel prints.

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 (373.15 K) at 1 atm (101,325 Pa). What is the boiling point at 0.75 atm, which is roughly 2,400 m of altitude?

Given: ΔH_vap(water) = 40,650 J/mol

• ln(0.75/1.0) = −(40,650/8.314) × (1/T2 − 1/373.15)
• −0.2877 = −4,889 × (1/T2 − 0.0026799)
• 1/T2 = 0.0026799 + 0.0000588 = 0.0027387
• T2 = 1/0.0027387 = 365.1 K = 92.0°C

That is about 8°C lower than at sea level. Denver itself sits lower, at 1,609 m and roughly 0.82 atm, where water boils near 94.5°C. The calculator runs this same inversion for you and reports it under “Solving the other way”, at four reference pressures.

ΔH_vap Reference Values (at the normal boiling point):

• Water: 40,650 J/mol, rising to about 44,000 at 25°C
• Ethanol: 38,560 J/mol
• Benzene: 30,720 J/mol
• Methanol: 35,210 J/mol

Those are the same four values the dropdown offers, so the numbers in the panel and the numbers here cannot drift apart. The water entry is the one to watch: quoting 40.65 kJ/mol and then working at 25°C introduces about 8% of error before you start, because the heat of vaporization is not actually constant.

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 vapor 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 vapor 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 vapor pressure a lot. Water at 100°C exerts 760 mmHg; at 90°C the measured value is about 525. 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 decent pasta.

Run that ten-degree step through this calculator and it returns 530 rather than 525. The gap is not a bug, it is the constant-ΔHvap assumption showing itself over a single ten-degree interval near the boiling point. The Antoine equation is fitted rather than derived and lands on 525.3, much closer to the measured figure. Use Clausius-Clapeyron for the physics and the ΔHvap it gives you; use Antoine when you want the number right.

Measuring the heat of vaporization yourself

Turn the equation around and it becomes a measurement technique. Record vapor 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 vapor as an ideal gas, both of which hold well below the boiling point and fail near the critical point.