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.

### Problem:

The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C.

Given:

Heat of vaporization of 1-propanol = 47.2 kJ/mol

### Solution

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[P_{T1,vap}/P_{T2,vap}] = (ΔH_{vap}/R)[1/T_{2} - 1/T_{1}]

where

ΔH_{vap} is the enthalpy of vaporization of the solution

R is the ideal gas constant = 0.008314 kJ/K·mol

T_{1} and T_{2} are the absolute temperature of the solution in Kelvin

P_{T1,vap} and P_{T2,vap} is the vapor pressure of the solution at temperature T_{1} and T_{2}**Step 1** - Convert °C to K

T_{K} = °C + 273.15

T_{1} = 14.7 °C + 273.15

T_{1} = 287.85 K

T_{2} = 52.8 °C + 273.15

T_{2} = 325.95 K**Step 2** - Find P_{T2,vap}

ln[10 torr/P_{T2,vap}] = (47.2 kJ/mol/0.008314 kJ/K·mol)[1/325.95 K - 1/287.85 K]

ln[10 torr/P_{T2,vap}] = 5677(-4.06 x 10_{-4})

ln[10 torr/P_{T2,vap}] = -2.305

take the antilog of both sides 10 torr/P_{T2,vap} = 0.997

P_{T2,vap}/10 torr = 10.02

P_{T2,vap} = 100.2 torr

### Answer:

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