Clausius-Clapeyron relation

The Clausius-Clapeyron relation, named after Rudolf Clausius and Émile Clapeyron, is a way of characterizing the phase transition between two phases of matter, such as solid and liquid. On a pressure-temperature (P-T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius-Clapeyron relation gives the slope of this curve. Mathematically,


 * $$\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{L}{T\,\Delta V} $$

where $$\mathrm{d}P/\mathrm{d}T$$ is the slope of the coexistence curve, $$L$$ is the latent heat, $$T$$ is the temperature, and $$\Delta V $$ is the volume change of the phase transition.

Disambiguation
The generalized equation given in the opening of this article is sometimes called the Clapeyron equation, while a less general form is sometimes called the Clausius-Clapeyron equation. The less general form neglects the magnitude of the specific volume of the liquid (or solid) state relative to that of the gas state and also approximates the specific volume of the gas state via the ideal gas law.

Derivation
Using the state postulate, take the specific entropy, $$s$$, for a homogeneous substance to be a function of specific volume, $$v$$, and temperature, $$T$$.


 * $$d s = \frac{\partial s}{\partial v} d v + \frac {\partial s}{\partial T} d T$$

During a phase change, the temperature is constant, so


 * $$d s = \frac{\partial s}{\partial v} d v$$.

Using the appropriate Maxwell relation gives


 * $$d s = \frac{\partial P}{\partial T} d v$$.

Since temperature and pressure are constant during a phase change, the derivative of pressure with respect to temperature is not a function of the specific volume. Thus the partial derivative may be changed into a total derivative and be factored out when taking an integral from one phase to another,


 * $$s_2 - s_1 = \frac{d P}{d T} (v_2 - v_1)$$,
 * $$\frac{d P}{d T} = \frac {s_2 - s_1}{v_2 - v_1} = \frac {\Delta s}{\Delta v}$$.
 * $$\Delta$$ is used as an operator to represent the change in the variable that follows it—final (2) minus initial (1)

For a closed system undergoing an internally reversible process, the first law is


 * $$d u = \delta q - \delta w = T d s - P d v\,$$.

Using the definition of specific enthalpy, $$h$$, and the fact that the temperature and pressure are constant, we have


 * $$d u + P d v = d h = T ds \Rightarrow ds = \frac {d h}{T} \Rightarrow \Delta s = \frac {\Delta h}{T}$$.

After substitution of this result into the derivative of the pressure, one finds


 * $$\frac{d P}{d T} = \frac {\Delta h}{T \Delta v} = \frac {\Delta H}{T \Delta V} = \frac {L}{T \Delta V}$$,

where the shift to capital letters indicates a shift to extensive variables. This last equation is called the Clausius-Clapeyron equation, though some thermodynamics texts just call it the Clapeyron equation, possibly to distinguish it from the approximation below.

When the transition is to a gas phase, the final specific volume can be many times the size of the initial specific volume. A natural approximation would be to replace $$\Delta v$$ with $$v_2$$. Furthermore, at low pressures, the gas phase may be approximated by the ideal gas law, so that $$v_2 = v_{gas} = R T / P$$, where R is the mass specific gas constant (forcing $$h$$ and $$v$$ to be mass specific). Thus,


 * $$\frac{d P}{d T} = \frac {P \Delta h}{T^2 R}$$.

This leads to a version of the Clausius-Clapeyron equation that is simpler to integrate:


 * $$\frac {d P}{P} = \frac {\Delta h}{R} \frac {dT}{T^2}$$,
 * $$\ln P = - \frac {\Delta h}{R} \frac {1}{T} + C$$, or
 * $$\ln \frac {P_2}{P_1} = \frac {\Delta h}{R} \left ( \frac {1}{T_1} - \frac {1}{T_2} \right )$$.
 * $$C$$ is a constant of integration

These last equations are useful because they relate saturation pressure and saturation temperature to the enthalpy of phase change, without requiring specific volume data. Note that in this last equation, the subscripts 1 and 2 correspond to different locations on the pressure versus temperature phase lines. In earlier equations, they corresponded to different specific volumes and entropies at the same saturation pressure and temperature.

Other Derivation
Suppose two phases, I and II, are in contact and at equilibrium with each other. Then the chemical potentials are related by $$\mu_{I} = \mu_{II}$$. Along the coexistence curve, we also have $$\mathrm{d}\mu_{I} = \mathrm{d}\mu_{II}$$. We now use the Gibbs-Duhem relation $$\mathrm{d}\mu = -s\mathrm{d}T + v\mathrm{d}P$$, where $$s$$ and $$v$$ are, respectively, the entropy and volume per particle, to obtain


 * $$-(s_I-s_{II}) \mathrm{d}T + (v_I-v_{II}) \mathrm{d}P = 0. \,$$

Hence, rearranging, we have


 * $$\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{s_I-s_{II}}{v_I-v_{II}}.$$

From the relation between heat and change of entropy in a reversible process δQ = T dS, we have that the quantity of heat added in the transformation is


 * $$L= T (s_I-s_{II}). \,$$

Combining the last two equations we obtain the standard relation.

Chemistry and chemical engineering
The Clausius-Clapeyron equation for the liquid-vapor boundary may be used in either of two equivalent forms.
 * $$\ln \left(\frac{P_2}{P_1} \right) = \frac{\Delta H_\mathrm{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right)$$

where
 * $$T_1$$ and $$P_1$$ are a corresponding temperature (in kelvin or other absolute temperature units) and vapor pressure
 * $$T_2$$ and $$P_2$$ are the corresponding temperature and pressure at another point
 * $$\Delta H_\mathrm{vap}$$ is the molar enthalpy of vaporization
 * $$R$$ is the gas constant (8.314 J mol-1K-1)

This can be used to predict the temperature at a certain pressure, given the temperature at another pressure, or vice versa. Alternatively, if the corresponding temperature and pressure is known at two points, the enthalpy of vaporization can be determined.

The equivalent formulation, in which the values associated with one P,T point are combined into a constant (the constant of integration as above), is
 * $$\ln P = -\frac{\Delta H_\mathrm{vap}}{RT}+C$$

For instance, if the p,T values are known for a series of data points along the phase boundary, then the enthalpy of vaporization may be determined from a plot of $$\ln P$$ against $$1/T$$.

Notes:
 * As in the derivation above, the enthalpy of vaporization is assumed to be constant over the pressure/temperature range considered
 * Equivalent expressions for the solid-vapor boundary are found by replacing the molar enthalpy of vaporization by the molar enthalpy of sublimation, $$\Delta H_\mathrm{sub}$$

Meteorology
In meteorology, a specific derivation of the Clausius-Clapeyron equation is used to describe dependence of saturated water vapor pressure on temperature. This is similar to its use in chemistry and chemical engineering.

It plays a crucial role in the current debate on climate change because its solution predicts exponential behavior of saturation water vapor pressure (and, therefore water vapor concentration) as a function of temperature. In turn, because water vapor is a greenhouse gas, it might lead to further increase in the sea surface temperature leading to runaway greenhouse effect. Debate on iris hypothesis and intensity of tropical cyclones dependence on temperature depends in part on “Clausius-Clapeyron” solution.

Clausius-Clapeyron equations is given for typical atmospheric conditions as


 * $$ \frac{\mathrm{d}e_s}{\mathrm{d}T} = \frac{L_v e_s}{R_v T^2} $$

where: One can solve this equation to give
 * $$ e_s $$ is saturation water vapor pressure
 * $$T$$ is a temperature
 * $$ L_v $$ is latent heat of evaporation
 * $$R_v $$ is water vapor gas constant.


 * $$ e_s(T)= 6.112 \exp \left( \frac{17.67 T}{T+243.5} \right) $$

where: Thus, neglecting the weak variation of (T+243.5) at normal temperatures, one observes that saturation water vapor pressure changes exponentially with $$T$$.
 * $$ e_s(T) $$ is in hPa (mbar)
 * $$ T $$ is in degrees Celsius.

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 $${\Delta T}$$ below 0 °C. We can assume
 * $$ {\Delta P} = \frac{L}{T\,\Delta V} {\Delta T} $$

and substituting in
 * $$L$$ = 3.34 J/kg (latent heat of water),
 * $$T$$ = 273 K (absolute temperature), and
 * $$\Delta V $$ = -9.05 m³/kg (change in volume from solid to liquid),

we obtain
 * $$\frac{\Delta P}{\Delta T} $$ = -13.1 MPa/°C.

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 (mass = 1000 kg ) on a thimble (area = 1 cm²).