## The Clapeyron Equation

The slopes of the lines on a one-component pressure-temperature phase diagram may be derived from the Clapeyron equation. For any two phases ( and ( we can write

where *V*^{(} is the molar volume of phase α and *S*^{(} is the molar entropy of phase ( and so on. We need not label the pressure and temperature since at equilibrium these properties are the same for all phases within the system.

At equilibrium,

so that

and

We can write the difference as , or more simply

But

since at equilibrium

and therefore

which is the basis of Trouton's Rule.

Substituting gives

This expression, the Clapeyron equation, is entirely general and applies to any phase change in a one-component system.

We can rearrange this expression to give

If we assume that the enthalpy and volume change are constant and do not vary with either temperature or pressure, then can integrate this expression to give

In the case of vaporisation or sublimation we can assume that since the volume of the gas formed is so much greater than that of the solid or liquid

or

Furthermore, if we assume that the vapour is ideal and work only with molar quantities, we can write

so that

and

which is the differential form of the Clausius-Clapeyron equation. Rearranging gives

which may be rewritten as

If we assume that (*H* does not vary with temperature and pressure this may be simplified to give

which, in its integrated form is

The equation relates the values of any pair of points (*p*_{1}, *T*_{1}) and (*p*_{2}, *T*_{2}) on the vaporisation or sublimation line. By measuring experimentally the gradient of a pressure-temperature line we may therefore determine an average value for the enthalpy of vaporisation or sublimation over that temperature range.

In reports of experimental measurements of vapour pressures above liquids or solids, it is common to fit the data to an equation of the form

which is often called the Antoine equation, and quote only the values of the parameters *A*, *B* and *C*. The use of the third fitting parameter, *C*, allows for deviations from the form of the Clausius-Clapeyron equation.

**Example 1**

The densities of ice and of liquid water vary little with temperature and pressure. We can therefore use these values to calculate the change in volume Δ_{fus}*V* on melting and, with a value for the enthalpy of fusion Δ_{fus}*H*^{o}, determine the melting temperature of ice at different pressures from a rearranged form the integrated form of the Clapeyron equation

Given the density of a substance ρ, we can calculate its molar volume *V*

where *M* is the molar mass. The densities of ice and liquid water are 0.917 g dm^{-3} and 1.000 g dm^{-3} and the molar mass, *M* = 18.02 g, so that the molar volumes of ice and liquid water are

*V*_{ice }= 18.02 g / 0.917 g dm^{-3} = 19.58 dm^{3}

and

*V*_{water }= 18.02 g / 1.000 g dm^{-3} = 18.02 cm^{3}

respectively. The volume change on melting is therefore

Δ_{fus}*V* = *V*_{water} - *V*_{ice} =18.02 - 19.58 = -1.56 cm^{3}.

We must remember to convert this value into units of m^{3}

Δ_{fus}*V* = -1.56 cm^{3} = -1.56 × 10^{-6} m^{3}

before substituting it into the Clapeyron equation.

We already know one point on the solid-liquid equilibrium line since we know that ice melts at a temperature of *T*_{1} = 273.15 K and pressure of *p*_{1} = 101325 Pa. The enthalpy of fusion of ice Δ_{fus}*H*^{o} = 6.030 kJ mol^{-1}.

Thus at a pressure of 70 bar ( 7093000 Pa, which is typical of the pressure exerted by an ice skater

and so

Thus at a pressure of 70 bar, the melting point of ice is *T*_{2} = 272.66 K, a decrease of 0.49 K.

The melting temperature of ice is therefore lowered by the effect of increased pressure. The application of pressure to a block of ice held at a constant temperature may therefore cause melting. This is unusual; for most substances, the melting point increases with pressure. The peculiar behaviour of water arises because the density of liquid water is greater than that of ice. The contraction on melting causes results in a negative value for Δ_{fus}*V* and so the solid-liquid line on the pressure-temperature phase diagram of water has a negative gradient.

**Example 2**

Given the normal boiling temperature and enthalpy of vaporisation of a substance, we can use the Clausius-Clapeyron equation to predict the vapour pressure at a range of different temperatures.

For example, the normal boiling temperature of benzene is 353.25 K, with a standard enthalpy of vaporisation Δ_{vap}*H*^{o} = 30.8 kJ mol^{-1}. If we assume that the enthalpy of vaporisation varies little with temperature and pressure, we may use the integrated form of the Clausius-Clapeyron equation

to determine the vapour pressure of benzene at another temperature, such as 298.15 K.

The normal boiling temperature is the temperature at which the vapour pressure of benzene is 101325 Pa. We may therefore take *T*_{1} = 353.25 K and *p*_{1} = 101325 Pa and substitute into the Clausius-Clapeyron equation. This will allow us to determine a value for the vapour pressure *p*_{2} at a temperature of *T*_{2} = 298.15.

giving *p*_{2} = 14600 Pa.

Note that we must assume that the enthalpy of vaporisation remains constant over the entire temperature and pressure range. In theory, the value quoted is only appropriate for a temperature of 353.25 K and a pressure of 10^{5} Pa. Significant variation in the enthalpy of vaporisation with temperature and pressure would be observed as a curve in the liquid-vapour equilibrium line on the benzene phase diagram.