Clausius–Clapeyron Equation Example Problem

Predicting Vapor Pressure

Vapor forming in a beaker
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The Clausius-Clapeyron equation may be used to estimate vapor pressure as a function of temperature or to find the heat of the phase transition from the vapor pressures at two temperatures. The Clausius-Clapeyron equation is a related named for Rudolf Clausius and Benoit Emile Clapeyron. The equation describes the phase transition between two phases of matter that have the same composition. When graphed, the relationship between temperature and pressure of a liquid is a curve rather than a straight line. In the case of water, for example, vapor pressure increases much faster than temperature. The Clausius-Clapeyron equation gives the slope of the tangents to the curve.

Clausius-Clapeyron Example

This example problem demonstrates how to use the Clausius-Clapeyron equation to predict the vapor pressure of a solution.


The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C.
Heat of vaporization of 1-propanol = 47.2 kJ/mol


The Clausius-Clapeyron equation relates a solution's vapor pressures at different temperatures to the heat of vaporization. The Clausius-Clapeyron equation is expressed by
ln[PT1,vap/PT2,vap] = (ΔHvap/R)[1/T2 - 1/T1]
ΔHvap is the enthalpy of vaporization of the solution
R is the ideal gas constant = 0.008314 kJ/K·mol
T1 and T2 are the absolute temperatures of the solution in Kelvin
PT1,vap and PT2,vap is the vapor pressure of the solution at temperature T1 and T2
Step 1 - Convert °C to K
TK = °C + 273.15
T1 = 14.7 °C + 273.15
T1 = 287.85 K
T2 = 52.8 °C + 273.15
T2 = 325.95 K
Step 2 - Find PT2,vap
ln[10 torr/PT2,vap] = (47.2 kJ/mol/0.008314 kJ/K·mol)[1/325.95 K - 1/287.85 K]
ln[10 torr/PT2,vap] = 5677(-4.06 x 10-4)
ln[10 torr/PT2,vap] = -2.305
take the antilog of both sides 10 torr/PT2,vap = 0.997
PT2,vap/10 torr = 10.02
PT2,vap = 100.2 torr


The vapor pressure of 1-propanol at 52.8 °C is 100.2 torr.