The Clausius–Clapeyron relation, named after Rudolf Clausius and Émile Clapeyron, who defined it sometime after 1834, is a way of characterizing a discontinuous phase transition between two phases of matter. 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,
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.:509
During a phase change, the temperature is constant, so:508
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.:57, 62 & 671 Thus the partial derivative may be changed into a total derivative and be factored out when taking an integral from one phase to another,:508
For a closed system undergoing an internally reversible process, the first law is
Using the definition of specific enthalpy, h, and the fact that the temperature and pressure are constant, we have:508
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 Δv with v2. Furthermore, at low pressures, the gas phase may be approximated by the ideal gas law, so that v2 = vgas = RT / P, where R is the mass specific gas constant (forcing h and v to be mass specific). Thus,:509
This leads to a version of the Clausius–Clapeyron equation that is simpler to integrate::509
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.
Suppose two phases, I and II, are in contact and at equilibrium with each other. Then the chemical potentials are related by μI = μII. Along the coexistence curve, we also have dμI = dμII. We now use the Gibbs–Duhem relation dμ = − sdT + vdP, where s and v are, respectively, the entropy and volume per particle, to obtain
Hence, rearranging, we have
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
Combining the last two equations we obtain the standard relation.
The Clausius–Clapeyron equation for the liquid–vapor boundary may be used in either of two equivalent forms.
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
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 lnP against 1 / T.
Clausius–Clapeyron equations is given for typical atmospheric conditions as
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 ΔT below 0°C. Note that water is unusual in that its change in volume upon melting is negative. We can assume
and substituting in
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).