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.

Clausius-Clapeyron plots of some noble gases.


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 \(\Delta H_{\mathrm{vap}} = 6.45~\mathrm{kJ~mol^{-1}}\)


Timed Assessments


Solve for T1

VIDEO

Answers
  1. 26.6 °C
  2. –3.5 °C
  3. 78.0 °C


Solve for T2

VIDEO

Answers
  1. 23.3 °C
  2. 46.0 °C
  3. 13.9 °C


Solve for P1

VIDEO

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


Solve for P2

VIDEO

Answers
  1. 1.960 atm
  2. 3.583 atm
  3. 4.075 atm


Solve for ΔH

VIDEO

Answers
  1. 24.4 kJ mol–1
  2. 92.0 kJ mol–1
  3. 4.61 kJ mol–1


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