The Clapeyron equation provides a general description of the slope of a phase boundary:
\[ \frac{dp}{dT} = \frac{\Delta H} {T\Delta V} \]While this equation is completely general, it is not always convenient to use because the volume change \(\Delta V\) may be difficult to determine. Fortunately, for phase transitions involving a gas phase, several reasonable approximations can be made.
Consider the liquid-vapor equilibrium of a pure substance. During vaporization, the molar volume of the vapor is much larger than that of the liquid:
\[ \bar V_{\text{vapor}} \gg \bar V_{\text{liquid}} \]Consequently,
\[ \Delta V = \bar V_{\text{vapor}} - \bar V_{\text{liquid}} \approx \bar V_{\text{vapor}} \]If the vapor behaves approximately as an ideal gas,
\[ \bar V_{\text{vapor}} = \frac{RT}{p} \]Substituting this expression into the Clapeyron equation gives
\[ \frac{dp}{dT} = \frac{\Delta H_{\text{vap}}\,p} {RT^2} \]This result contains only measurable quantities and is much easier to work with than the original Clapeyron equation.
Rearranging the equation gives
\[ \frac{dp}{p} = \frac{\Delta H_{\text{vap}}} {R} \frac{dT}{T^2} \]If the enthalpy of vaporization is approximately constant over the temperature range of interest, both sides may be integrated:
\[ \int_{p_1}^{p_2} \frac{dp}{p} = \frac{\Delta H_{\text{vap}}}{R} \int_{T_1}^{T_2} \frac{dT}{T^2} \]Evaluating the integrals yields
\[ \ln\!\left( \frac{p_2}{p_1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]This expression is known as the Clausius–Clapeyron equation.
It relates changes in vapor pressure directly to the enthalpy of vaporization.
The Clausius–Clapeyron equation can be rearranged into a linear form:
\[ \ln p = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T} \right) + C \]This equation has the same form as the equation of a straight line:
\[ y=mx+b \]Therefore, a plot of
\[ \ln p \]versus
\[ \frac{1}{T} \]should be approximately linear.
The slope of the line is
\[ -\frac{\Delta H_{\text{vap}}}{R} \]allowing the enthalpy of vaporization to be determined directly from experimental vapor-pressure measurements.
Adjust the enthalpy of vaporization and the vapor pressure at 25 °C to see how the vapor-pressure curve changes. The applet calculates the normal boiling point, where the vapor pressure reaches 760 Torr.
| \(\Delta H_{\text{vap}}\) | kJ mol-1 |
| Vapor pressure at 25 °C | Torr |
The Clausius–Clapeyron equation converts vapor-pressure measurements into thermodynamic information. By measuring how vapor pressure changes with temperature, we can determine the enthalpy required to convert a liquid into a vapor.
More importantly, the derivation illustrates a common strategy in thermodynamics: begin with a completely general relationship, then introduce physically reasonable approximations to obtain a model that is simple enough to use while still retaining the essential physics.
Big picture: The Clausius–Clapeyron equation is a simplified form of the Clapeyron equation that applies to phase equilibria involving a gas phase. It provides a direct connection between vapor pressure and the enthalpy of vaporization and forms the basis for many experimental methods used to characterize liquids and solids.
A volatile liquid has a vapor pressure of 457 Torr at 35 °C and an enthalpy of vaporization of
\[ \Delta H_{\text{vap}} = 37.8\ \text{kJ mol}^{-1} \]Estimate the normal boiling point of the liquid.
The normal boiling point is the temperature at which the vapor pressure reaches
\[ 760\ \text{Torr} \]We therefore know
\[ p_1=457\ \text{Torr} \] \[ T_1=35+273.15=308.15\ \text{K} \] \[ p_2=760\ \text{Torr} \]and wish to determine \(T_2\).
Begin with the integrated Clausius-Clapeyron equation:
\[ \ln\!\left(\frac{p_2}{p_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Rearranging to solve for \(1/T_2\):
\[ \frac{1}{T_2} = \frac{1}{T_1} - \frac{R}{\Delta H_{\text{vap}}} \ln\!\left(\frac{p_2}{p_1}\right) \]Substituting the known values:
\[ \frac{1}{T_2} = \frac{1}{308.15\ \text{K}} - \frac{8.314\ \text{J mol}^{-1}\text{K}^{-1}} {3.78\times10^4\ \text{J mol}^{-1}} \ln\!\left(\frac{760}{457}\right) \] \[ \frac{1}{T_2} = 0.003245 - 0.000112 \] \[ \frac{1}{T_2} = 0.003133\ \text{K}^{-1} \]Therefore,
\[ T_2 = \frac{1}{0.003133} = 319.2\ \text{K} \]Converting back to degrees Celsius:
\[ T_2 = 319.2-273.15 = 46.1^\circ\text{C} \]Thus, the estimated normal boiling point is
\[ \boxed{46.1^\circ\text{C}} \]Physical interpretation: The liquid already has a relatively high vapor pressure at 35 °C, so only a modest increase in temperature is required for the vapor pressure to reach 760 Torr. Once that pressure is reached, the liquid boils under atmospheric conditions.
The integrated Clausius-Clapeyron equation is
\[ \ln\!\left(\frac{p_2}{p_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Four of the five quantities are given below. Calculate the missing quantity.
| Quantity | Value |
|---|---|
| \(p_1\) | |
| \(p_2\) | |
| \(T_1\) | |
| \(T_2\) | |
| \(\Delta H_{\text{vap}}\) | |
| Your answer | |
Big picture: The Clausius-Clapeyron equation is one of the most useful results in thermodynamics because it connects measurable vapor-pressure data directly to the enthalpy of a phase transition. It demonstrates how a general thermodynamic relationship can be simplified through physically reasonable approximations to produce a powerful and experimentally useful model.