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Chemistry LibreTexts

Clausius-Clapeyron Equation

Skills to Develop

  • Apply the Clausius-Clapeyron equation to estimate the vapor pressure at any temperature.
  • Estimate the heat of phase transition from the vapor pressures measured at two temperatures.

The Clausius-Clapeyron Equation

The vaporization curves of most liquids have similar shape. The vapor pressure steadily increase as the temperature increases. A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the pressure \(P\), enthalpy of vaporization, \(\Delta{H_{vap}}\), and temperature \(T\) are related,

\[P = A e^{(- \Delta H_{vap} / RT)} \tag{1}\]

where

  • \(R\) (8.3145 J mol-1 K-1) is the gas constant and
  • \(A\) is an unknown constant.

This is known as the Clausius-Clapeyron equation. If \(P_1\) and \(P_2\) are the pressures at two temperatures \(T_1\) and \(T_2\), the equation has the form:

\[\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right) \tag{2}\]

The Clausius-Clapeyron equation allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

Example 1: Vapor Pressure of Water

The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1. Estimate the vapor pressure at temperature 363 and 383 K respectively.

SOLUTION
Using the Clausius-Clapeyron equation (Equation 1), we have:

\[P_{363} = 1.0 \;exp \left[- \left(\dfrac{40,700}{8.3145}\right) \left(\dfrac{1}{363\;K} -\dfrac{1}{373\; K}\right) \right]\]

\[ = 0.697\; atm\]

\[P_{383} = 1.0 \;exp \left[- \left(40,700/8.3145 \right)\left(\dfrac{1}{383\;K} - \dfrac{1}{373\;K} \right) \right]\]

\[ = 1.409\; atm\]

Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process.

Discussion
We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization various slightly with temperature.

The Clausius-Clapeyron equation applies to any phase transition. The following example shows its application in estimating the heat of sublimation.

Example 2: Heat of Sublimation of Ice

The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice.

SOLUTION
The enthalpy of sublimation is \(\Delta{H}_{sub}\). Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form:

\[\Delta H_{sub} = \dfrac{ R \ln \left(\dfrac{P_{273}}{P_{268}}\right)}{\dfrac{1}{268 \;K} - \dfrac{1}{273\;K}}\]

 \[= \dfrac{8.3145 \ln \left(\dfrac{4.560} {2.965} \right)}{ \dfrac{1}{268\;K} - \dfrac{1}{273\;K} } \]

  \[= 52370\; J\; mol^{-1}\]

Note that the heat of sublimation is the sum of heat of melting and the heat of vaporization.

Discussion
Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization

Example 3: Heat of Vaporization of Ethanol

Calculate \(\Delta{H_{vap}}\) for ethanol, given vapor pressure at 40 oC = 150 torr. The normal boiling point for ethanol is 78 oC.

SOLUTION

Recognize that we have TWO sets of \((p,T)\) data:

  • Set 1: (150 torr at 40+273K)
  • Set 2: (760 torr at 78+273K)

\[ \ln p = \dfrac{-\Delta{H_{vap}}}{RT} + c\]

Substituting into the above equation twice produces:

\[ \ln 150 = \dfrac{-\Delta{H_{vap}}}{(8.314)\times (313)} + c\]

and

\[ \ln 760 = \dfrac{-\Delta{H_{vap}}}{(8.314)\times (351)} + c\]

Subtract these two equations, to produce:

\[ \ln 150 -\ln 760 = \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right]\]

\[-1.623 = \dfrac{-\Delta{H_{vap}}}{8.314} \left[ 0.0032 - 0.0028 \right]\]

Solving for \(\Delta{H_{vap}}\) :

\[ \Delta{H_{vap}} = 3.90 \times 10^4 \text{ joule/mole} = 39.0 \text{ kJ/mole}\]

Contributors

  • Chung (Peter) Chieh (Chemistry, University of Waterloo)

  • Albert Censullo (California Polytechnic State University)