Clausius-Clapeyron Equation Example Problem

The Clausius-Clapeyron equation is a relation named for Rudolf Clausius and Benoit Emile Clapeyron. The equation describes the phase transition between two phases of matter that have the same composition.

Thus, the Clausius-Clapeyron equation can 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. 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.

This example problem demonstrates using 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/T₂ - 1/T₁]
Where:
ΔH_vap 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
P_T1,vap and P_T2,vap is the vapor pressure of the solution at temperature T₁ and T₂

Step 1: Convert °C to K

T_K = °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/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.