# Clausius-Clapeyron Equation

The Clausius-Clapeyron equation approximately relates a substance’s vapor pressure and its temperature. (The exact relationship is called the Clapeyron equation. A derivation of the Clausius-Clapeyron equation can be found in Section 4.5 of Atkins, Physical Chemistry 9th Ed.)

## Exponential Form

$P = Ae^{-\Delta H_{\mathrm{vap}}/RT}$

Variable Description
P vapor pressure
A pre-exponential factor (constant; substance dependent)
ΔH heat of vaporization (J mol–1)
R gas constant (8.315 J mol–1 K–1)
T temperature (in K)

## Linear Form

The Clausius-Clapeyron equation can be rewritten into a linear form (i.e. y = mx + b).

$\ln P = -\dfrac{\Delta H_{\mathrm{vap}}}{R}\left ( \dfrac{1}{T} \right ) + \ln A$

## Two-Point Form

It can also be written in a “two-point” form.

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

Below are the algebraic steps to solve for any variable in the Clausius-Clapeyron two-point form equation.

### ΔH

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\begin{align} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \\[2ex] R\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) &= \Delta H_{\mathrm{vap}} \\[2ex] \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] \mathrm{or}\\[2ex] \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln P_2 - \ln P_1}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \end{align}

### P1

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\begin{align} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \ln P_2 - \ln P_1 &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[2ex] - \ln P_1 &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) - \ln P_2\\[2ex] \ln P_1 &= -\dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) + \ln P_2\\[2ex] e^{(\ln P_1)} &= e^{\left(-\frac{\Delta H_{\mathrm{vap}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln P_2\right)}\\[2ex] P_1 &= e^{\left(-\frac{\Delta H_{\mathrm{vap}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln P_2\right)} \end{align}

### P2

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\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \ln P_2 - \ln P_1 &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[2ex] \ln P_2 &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) + \ln P_1\\[2ex] e^{(\ln P_2)} &= e^{\left(\frac{\Delta H_{\mathrm{vap}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln P_1\right)}\\[2ex] P_2 &= e^{\left(\frac{\Delta H_{\mathrm{vap}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln P_1\right)} \end{align*}

### T1

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\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} &= \dfrac{1}{T_1} - \dfrac{1}{T_2} \\[2ex] \dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_2} &= \dfrac{1}{T_1} \\[2ex] \dfrac{1}{\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_2}\right )} &= T_1 \\[2ex] T_1 &= \dfrac{1}{\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_2}\right ) } \end{align*}

### T2

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\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} &= \dfrac{1}{T_1} - \dfrac{1}{T_2} \\[2ex] \dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} - \dfrac{1}{T_1} &= -\dfrac{1}{T_2} \\[2ex] -\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_1} &= \dfrac{1}{T_2} \\[2ex] \dfrac{1}{\left ( -\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_1} \right )} &= T_2 \\[2ex] T_2 &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_1}\right )} \end{align*}

## Practice Problems

### Problem 1

How much heat (in kJ mol–1) is required to vaporize one mole of a substance that has a measured vapor pressure of 0.032 atm at 0 °C and 0.178 atm at 52 °C?

Solution

\begin{align} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[2ex] \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right )\\ \Delta H_{\mathrm{vap}} &= (8.315~\mathrm{J~mol^{-1}~K^{-1}}) \left (\dfrac{\ln \left ( \dfrac{0.178~\mathrm{atm}}{0.032~\mathrm{atm}} \right )} {\dfrac{1}{273.15~\mathrm{K}} - \dfrac{1}{325.15~\mathrm{K}}}\right ) \\ \Delta H_{\mathrm{vap}} &=24,370~\mathrm{J~mol^{-1}} \\ &= 24.37~\mathrm{kJ~mol^{-1}} \end{align}

### Problem 2

Jason Mick, et al. reported Clausius-Clapeyron plots for noble gases and is given below.

I have marked two points on this plot for Krypton and two for Argon with arrows. Below are the respective (approximate) values.

Krypton

• ln P1 = 1.347
• 1/T1 = 0.00718 K–1
• ln P2 = 0.71875
• 1/T2 = 0.007763 K–1

Argon

• ln P1 = 1.91667
• 1/T1 = 0.009085 K–1
• ln P2 = 1.15972
• 1/T2 = 0.01 K–1

Compute the heat of vaporization (ΔHvap in kJ mol–1) for Kr and Ar. Compare the computed value to the experimental value.

Solution for Kr

\begin{align*} \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln \left( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= R\left (\dfrac{\ln P_2 - \ln P_1}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= (8.315~\mathrm{J~mol^{-1}~K^{-1}}) \left ( \dfrac{0.71875 - 1.347}{0.00718~\mathrm{K^{-1}} - 0.007763~\mathrm{K^{-1}}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= 8.97~\mathrm{kJ~mol^{-1}} \end{align*}

The experimental heat of vaporization is $$\Delta H_{\mathrm{vap}} = 9.029~\mathrm{kJ~mol^{-1}}$$

Solution for Ar

\begin{align*} \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln \left( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= R\left (\dfrac{\ln P_2 - \ln P_1}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= (8.315~\mathrm{J~mol^{-1}~K^{-1}}) \left ( \dfrac{1.15972 - 1.91667}{0.009085~\mathrm{K^{-1}} - 0.01~\mathrm{K^{-1}}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= 6.88~\mathrm{kJ~mol^{-1}} \end{align*}

The experimental heat of vaporization is $$\Delta H_{\mathrm{vap}} = 6.45~\mathrm{kJ~mol^{-1}}$$

## Handout

Clausius-Clapeyron

VIDEO

1. 26.6 °C
2. –3.5 °C
3. 78.0 °C

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1. 23.3 °C
2. 46.0 °C
3. 13.9 °C

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1. 0.288 atm
2. 0.540 atm
3. 1.338 atm

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