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 solutions vapor pressures at different temperatures to the heat of vaporization. The Clausius-Clapeyron equation is expressed byln[PT1,vap/PT2,vap] (ΔHvap/R)[1/T2 - 1/T1]Where:ΔHvap is the enthalpy of vaporization of the solutionR is the ideal gas constant 0.008314 kJ/K ·molT1 and T2 are the absolute temperatures of the solution in KelvinPT1,vap and PT2,vap is the vapor pressure of the solution at temperature T1 and T2 Step 1: Convert  °C to K TK  °C 273.15T1 14.7  °C 273.15T1 287.85 KT2 52.8  °C 273.15T2 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] -2.305take the antilog of both sides 10 torr/PT2,vap 0.997PT2,vap/10 torr 10.02PT2,vap 100.2 torr Answer The vapor pressure of 1-propanol at 52.8  °C is 100.2 torr.