Recall that the vapor pressure of a liquid increases with a rise in temperature. However, this dependence is not linear.

To illustrate, the vapor pressure of water at 50 °C is 0.122 atm, while at 100 °C, it is 1 atm. Vapor pressure curves sharply upward with increasing temperature, resulting in an exponential curve.

In comparison, when the natural log of vapor pressure is plotted against reciprocal temperature, a straight line is obtained, and its equation is called the Clausius–Clapeyron equation.

Here, *R* is the ideal gas constant; *c*, a constant characteristic of the liquid, is the y-intercept; and the slope of the line is equal to the negative of the molar heat of vaporization over the gas constant.

The equation allows the calculation of the molar heat of vaporization from the experimental measurements of equilibrium vapor pressures and temperatures.

For example, suppose the natural log of ethanol vapor pressure plotted as a function of reciprocal temperature gives a straight line with the slope of negative 4638 kelvins.

The equation for the slope of the line, along with the value of *R*, gives the molar heat of vaporization of ethanol as 38 560 joules per mole.

If the molar heat of vaporization of any liquid and its vapor pressure at one temperature are known, the equation’s two-point form can be used to calculate the liquid’s vapor pressure at a different temperature.

Take the example of water, whose enthalpy of vaporization is 40.7 kilojoules per mole. If the vapor pressure of water at 373 kelvins is 1 atm, what will be its vapor pressure at 383 kelvins?

To solve, use the two-point form of the equation and substitute the given values of vapor pressure, enthalpy of vaporization, the two temperatures, and the gas constant to get the vapor pressure of water at 383 kelvins as 1.409 atm.

The rise in vapor pressure from 373 kelvins to 383 kelvins is 0.409 atm, which clearly indicates that an increase in vapor pressure as a function of temperature is a non-linear process.