# 23.4: The Clausius-Clapeyron Equation

#### Evaporation

In Section 23.3, the Clapeyron Equation was derived for melting points.

$\dfrac{dP}{dT} = \dfrac{ΔH_{molar}}{TΔV_{molar}}$

However, our argument is actually quite general and should hold for vapor equilibria as well. The only problem is that the molar volume of gases are by no means so nicely constant as they are for condensed phases. (i. e., for condenses phases, both $$α$$ and $$κ$$ are pretty small).

We can write:

$\dfrac{dP}{dT} =\dfrac{ ΔH_{molar}}{TΔV_{molar}} = \dfrac{ΔH_{molar}}{T} \left[V_{molar}^{gas}-V_{molar}^{liquid} \right]$

as

$V_{molar}^{gas} \gg V_{molar}^{liquid}$

we can approximate

$V_{molar}^{gas}-V_{molar}^{liquid}$

by just taking $$V_{molar}^{gas}$$. Further more if the vapor is considered an ideal gas, then

$V_{molar}^{gas} = \dfrac{RT}{P}$

We get

$\dfrac{1}{P} .\dfrac{dP}{dT} = \dfrac{d \ln P}{dT} = \dfrac{ΔH_{molar}^{vap}}{RT^2} \label{CCe}$

Equation $$\ref{CCe}$$ is known as the Clausius-Clapeyron equation. We can further work our the integration and find the how the equilibrium vapor pressure changes with temperature:

$\ln \left( \dfrac{P_2}{P_1} \right)= \dfrac{-ΔH_{molar}^{vap}}{R} \left[\dfrac{1}{T_2}-\dfrac{1}{T_1} \right]$

Thus if we know the molar enthalpy of vaporization we can predict the vapor lines in the diagram. Of course the approximations made are likely to lead to deviations if the vapor is not ideal or very dense (e.g., approaching the critical point).

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 vaporization curves of most liquids have similar shapes. 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 vapor pressure $$P$$ and temperature $$T$$ are related,

$P \propto e^{- \Delta H_{vap} / RT} \label{1}$

where

If $$P_1$$ and $$P_2$$ are the vapor pressures at two temperatures $$T_1$$ and $$T_2$$, then a simple relationship can be found:

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

This is known as the Clausius-Clapeyron Equation and 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.

Alternative Formulation

Note the order of the temperatures in Equation \ref{2} matters as the Clausius-Clapeyron Equation is often written with a negative sign (and switched order of temperatures):

$\ln \left( \dfrac{P_1}{P_2} \right) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_1}- \dfrac{1}{T_2} \right) \label{2B} \nonumber$

Example $$\PageIndex{1}$$: 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 $$\ref{2}$$), we have:

\begin{align} P_{363} &= 1.0 \exp \left[- \left(\dfrac{40,700}{8.3145}\right) \left(\dfrac{1}{363\;K} -\dfrac{1}{373\; K}\right) \right] \nonumber \\[5pt] &= 0.697\; atm \nonumber \end{align} \nonumber

\begin{align} P_{383} &= 1.0 \exp \left[- \left( \dfrac{40,700}{8.3145} \right)\left(\dfrac{1}{383\;K} - \dfrac{1}{373\;K} \right) \right] \nonumber \\[5pt] &= 1.409\; atm \nonumber \end{align} \nonumber

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 can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation.

Example $$\PageIndex{2}$$: 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:

\begin{align} \Delta H_{sub} &= \dfrac{ R \ln \left(\dfrac{P_{273}}{P_{268}}\right)}{\dfrac{1}{268 \;K} - \dfrac{1}{273\;K}} \nonumber \\[5pt] &= \dfrac{8.3145 \ln \left(\dfrac{4.560}{2.965} \right)}{ \dfrac{1}{268\;K} - \dfrac{1}{273\;K} } \nonumber \\[5pt] &= 52,370\; J\; mol^{-1}\nonumber \end{align} \nonumber

Note that the heat of sublimation is the sum of heat of melting (6,006 J/mol at 0°C and 101 kPa) and the heat of vaporization (45,051 J/mol at 0 °C).

Exercise $$\PageIndex{2}$$

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 $$\PageIndex{3}$$: 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)

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c \nonumber$

Substituting into the above equation twice produces:

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

and

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

Subtract these two equations, to produce:

\begin{align} \ln 150 -\ln 760 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \nonumber \\[5pt] -1.623 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ 0.0032 - 0.0028 \right] \nonumber \end{align} \nonumber

Solving for $$\Delta{H_{vap}}$$ :

\begin{align} \Delta{H_{vap}} &= 3.90 \times 10^4 \text{ joule/mole} \nonumber \\[5pt] &= 39.0 \text{ kJ/mole} \nonumber \end{align} \nonumber

$\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}}{T \Delta \bar{V}} \label{14.8} \nonumber$
where $$\Delta \bar{H}$$ and $$\Delta \bar{V}$$ is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition.