Clausius–Clapeyron relation

Derivations

Derivation from Gibbs–Duhem relation

Suppose two phases, ${\displaystyle \alpha }$ and ${\displaystyle \beta }$, are in contact and at equilibrium with each other. Their chemical potentials are related by $${\displaystyle {\mu }_{\alpha }={\mu }_{\beta }.}$$

Furthermore, along the coexistence curve, $${\displaystyle \mathrm{d}{\mu }_{\alpha }=\mathrm{d}{\mu }_{\beta }.}$$

One may therefore use the Gibbs–Duhem relation $${\displaystyle \mathrm{d}\mu =M(-s\mathrm{d}T+v\mathrm{d}P)}$$ (where ${\displaystyle s}$ is the specific entropy, ${\displaystyle v}$ is the specific volume, and ${\displaystyle M}$ is the molar mass) to obtain $${\displaystyle -(s_{\beta }-s_{\alpha })\mathrm{d}T+(v_{\beta }-v_{\alpha })\mathrm{d}P=0.}$$

Rearrangement gives $${\displaystyle \frac{\mathrm{d}P}{\mathrm{d}T}=\frac{s_{\beta }-s_{\alpha }}{v_{\beta }-v_{\alpha }}=\frac{\mathrm{\Delta }s}{\mathrm{\Delta }v},}$$

from which the derivation of the Clapeyron equation continues as in the previous section.

Ideal gas approximation at low temperatures

When the phase transition of a substance is between a gas phase and a condensed phase (liquid or solid), and occurs at temperatures much lower than the critical temperature of that substance, the specific volume of the gas phase ${\displaystyle v_{\text{g}}}$ greatly exceeds that of the condensed phase ${\displaystyle v_{\text{c}}}$. Therefore, one may approximate $${\displaystyle \mathrm{\Delta }v=v_{\text{g}}\left(1-\frac{v_{\text{c}}}{v_{\text{g}}}\right)\approx v_{\text{g}}}$$ at low temperatures. If pressure is also low, the gas may be approximated by the ideal gas law, so that $${\displaystyle v_{\text{g}}=\frac{RT}{P},}$$

where ${\displaystyle P}$ is the pressure, ${\displaystyle R}$ is the specific gas constant, and ${\displaystyle T}$ is the temperature. Substituting into the Clapeyron equation $${\displaystyle \frac{\mathrm{d}P}{\mathrm{d}T}=\frac{L}{T\mathrm{\Delta }v},}$$ we can obtain the Clausius–Clapeyron equation $${\displaystyle \frac{\mathrm{d}P}{\mathrm{d}T}=\frac{PL}{T^{2}R}}$$ for low temperatures and pressures, where ${\displaystyle L}$ is the specific latent heat of the substance. Instead of the specific, corresponding molar values (i.e. ${\displaystyle L}$ in kJ/mol and R = 8.31 J/(mol⋅K)) may also be used.

Let ${\displaystyle (P_1,T_1)}$ and ${\displaystyle (P_2,T_2)}$ be any two points along the coexistence curve between two phases ${\displaystyle \alpha }$ and ${\displaystyle \beta }$. In general, ${\displaystyle L}$ varies between any two such points, as a function of temperature. But if ${\displaystyle L}$ is approximated as constant, $${\displaystyle \frac{\mathrm{d}P}{P}\cong \frac{L}{R}\frac{\mathrm{d}T}{T^2},}$$ $${\displaystyle {\int }_{P_1}^{P_2}\frac{\mathrm{d}P}{P}\cong \frac{L}{R}{\int }_{T_1}^{T_2}\frac{\mathrm{d}T}{T^2},}$$ $${\displaystyle \ln P{|}_{P=P_1}^{P_2}\cong -\frac{L}{R}\cdot {\frac{1}{T}|}_{T=T_1}^{T_2},}$$ or $${\displaystyle \ln \frac{P_2}{P_1}\cong -\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right).}$$

These last equations are useful because they relate equilibrium or saturation vapor pressure and temperature to the latent heat of the phase change without requiring specific-volume data. For instance, for water near its normal boiling point, with a molar enthalpy of vaporization of 40.7 kJ/mol and R = 8.31 J/(mol⋅K), $${\displaystyle P_{\text{vap}}(T)\cong 1\text{bar}\cdot \exp \left[-\frac{40700\text{K}}{8.31}\left(\frac{1}{T}-\frac{1}{373\text{K}}\right)\right].}$$

Clapeyron's derivation

In the original work by Clapeyron, the following argument is advanced. Clapeyron considered a Carnot process of saturated water vapor with horizontal isobars. As the pressure is a function of temperature alone, the isobars are also isotherms. If the process involves an infinitesimal amount of water, ${\displaystyle \mathrm{d}x}$, and an infinitesimal difference in temperature ${\displaystyle \mathrm{d}T}$, the heat absorbed is $${\displaystyle Q=L\mathrm{d}x,}$$ and the corresponding work is $${\displaystyle W=\frac{\mathrm{d}p}{\mathrm{d}T}\mathrm{d}T(V^{″}-V^{\prime }),}$$ where ${\displaystyle V^{″}-V^{\prime }}$ is the difference between the volumes of ${\displaystyle \mathrm{d}x}$ in the liquid phase and vapor phases. The ratio ${\displaystyle W/Q}$ is the efficiency of the Carnot engine, ${\displaystyle \mathrm{d}T/T}$. Substituting and rearranging gives $${\displaystyle \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{L}{T(v^{″}-v^{\prime })},}$$ where lowercase ${\displaystyle v^{″}-v^{\prime }}$ denotes the change in specific volume during the transition.

Applications

Chemistry and chemical engineering

For transitions between a gas and a condensed phase with the approximations described above, the expression may be rewritten as $${\displaystyle \ln \left(\frac{P_1}{P_0}\right)=\frac{L}{R}\left(\frac{1}{T_0}-\frac{1}{T_1}\right)}$$ where ${\displaystyle P_0,P_1}$ are the pressures at temperatures ${\displaystyle T_0,T_1}$ respectively and ${\displaystyle R}$ is the ideal gas constant. For a liquid–gas transition, ${\displaystyle L}$ is the molar latent heat (or molar enthalpy) of vaporization; for a solid–gas transition, ${\displaystyle L}$ is the molar latent heat of sublimation. If the latent heat is known, then knowledge of one point on the coexistence curve, for instance (1 bar, 373 K) for water, determines the rest of the curve. Conversely, the relationship between ${\displaystyle \ln P}$ and ${\displaystyle 1/T}$ is linear, and so linear regression is used to estimate the latent heat.

Meteorology and climatology

Atmospheric water vapor drives many important meteorologic phenomena (notably, precipitation), motivating interest in its dynamics. The Clausius–Clapeyron equation for water vapor under typical atmospheric conditions (near standard temperature and pressure) is$${\displaystyle \frac{\mathrm{d}{e}_{\text{s}}}{\mathrm{d}T}=\frac{L_{\text{v}}(T)e_{\text{s}}}{R_{\text{v}}{T}^2},}$$where

• ${\displaystyle e_{\text{s}}}$ is saturation vapor pressure,
• ${\displaystyle T}$ is temperature,
• ${\displaystyle L_{\text{v}}}$ is the specific latent heat of evaporation of water, and
• ${\displaystyle R_{\text{v}}}$ is the gas constant of water vapor.

The temperature dependence of the latent heat $L_{\text{v}}(T)$ can be neglected in this application. The August–Roche–Magnus formula provides a solution under that approximation: $${\displaystyle e_{\text{s}}(T)=6.1094\exp \left(\frac{17.625T}{T+243.04}\right),}$$ where $e_{\text{s}}$ is in hPa and ${\displaystyle T}$ is in degrees Celsius (whereas everywhere else on this page, ${\displaystyle T}$ is an absolute temperature in units of kelvin).

The above formula is also sometimes called the Magnus or Magnus–Tetens approximation, though this attribution is historically inaccurate; see also the discussion of the accuracy of different approximating formulae for saturation vapour pressure of water.

Under typical atmospheric conditions, the denominator in the exponent $T+243.04$ depends weakly on $T$. Therefore, the August–Roche–Magnus formula implies that saturation vapor pressure changes approximately exponentially with temperature under typical atmospheric conditions, and hence the water-holding capacity of the atmosphere increases by about 7% for every 1 °C rise in temperature.

Example

One of the uses of this equation is to determine if a phase transition will occur in a given situation. Consider the question of how much pressure is needed to melt ice at a temperature ${\displaystyle \mathrm{\Delta }T}$ below 0 °C. Note that water is unusual in that its change in volume upon melting is negative. We can assume $${\displaystyle \mathrm{\Delta }P=\frac{L}{T\mathrm{\Delta }v}\mathrm{\Delta }T,}$$ and substituting in

• ${\displaystyle L=3.34\times {10}^5\mathrm{J}/\mathrm{k}\mathrm{g}}$ (latent heat of fusion for water),
• ${\displaystyle T=273\mathrm{K}}$ (absolute temperature in kelvins),
• ${\displaystyle \mathrm{\Delta }v=-9.05\times {10}^{-5}{\mathrm{m}}^3/\mathrm{k}\mathrm{g}}$ (change in specific volume from solid to liquid),

we obtain $${\displaystyle \frac{\mathrm{\Delta }P}{\mathrm{\Delta }T}=-13.5\text{MPa}/\text{K}.}$$

To provide a rough example of how much pressure this is, to melt ice at −7 °C (the temperature many ice skating rinks are set at) would require balancing a small car on an area around 1cm². This shows that ice skating cannot be simply explained by pressure-caused melting point depression, and in fact the mechanism is quite complex.

Second derivative

While the Clausius–Clapeyron relation gives the slope of the coexistence curve, it does not provide any information about its curvature or second derivative. The second derivative of the coexistence curve of phases 1 and 2 is given by $${\displaystyle \begin{array}{cc}\frac{{\mathrm{d}}^{2}P}{\mathrm{d}{T}^2}& =\frac{1}{v_2-v_1}\left[\frac{c_{p2}-c_{p1}}{T}-2(v_2{\alpha }_2-v_1{\alpha }_1)\frac{\mathrm{d}P}{\mathrm{d}T}\right]\\ & +\frac{1}{v_2-v_1}\left[(v_2{\kappa }_{T2}-v_1{\kappa }_{T1}){\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}^2\right],\end{array}}$$ where subscripts 1 and 2 denote the different phases, ${\displaystyle c_p}$ is the specific heat capacity at constant pressure, ${\displaystyle \alpha =(1/v)(\mathrm{d}v/\mathrm{d}T{)}_P}$ is the thermal expansion coefficient, and ${\displaystyle {\kappa }_T=-(1/v)(\mathrm{d}v/\mathrm{d}P{)}_T}$ is the isothermal compressibility.