# Clausius-Clapeyron Equation Example Problem

## Predicting Vapor Pressure

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[PT1,vap/PT2,vap] = (ΔHvap/R)[1/T2 - 1/T1]
Where:
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