Clausius-Clapeyron Equation

It is possible to characterize the phase-equilibrium line in terms of a differential equation. Referring to Figure 9.1, consider a point, such as $A$ , that lies on the phase-equilibrium line, and corresponds to the temperature $T$ and the pressure $p$ . The condition (9.63) implies that

$$g_1(T,p)= g_2(T,p).$$ (9.64)

Now, consider a neighboring point, such as $B$ , that also lies on the phase-equilibrium line, and corresponds to the temperature $T+dT$ and the pressure $p+dp$ . The condition (9.63) yields

$$g_1(T+dT,p+dp)= g_2(T+dT,p+dp).$$ (9.65)

Taking the difference between the previous two equations, we obtain

$$dg_1=dg_2,$$ (9.66)

where

$$dg_i =\left(\frac{\partial g_i}{\partial T}\right)_p dT + \left(\frac{\partial g_i}{\partial p}\right)_T dp$$ (9.67)

is the change in Gibbs free energy per mole of phase $i$ in going from point $A$ to point $B$ .

The change, $dg$ , for each phase can also be obtained from the fundamental thermodynamic relation

$$d\epsilon = T ds -p dv.$$ (9.68)

Here, $\epsilon$ refers to molar energy (i.e., energy per mole), $s$ to molar entropy, and $v$ to molar volume. Thus,

$$dg \equiv d(\epsilon -T s+p v) = -s dT+v dp.$$ (9.69)

Hence, Equation (9.66) implies that

$$-s_1 dT+v_1 dp = -s_2 dT+v_2 dp,$$ (9.70)

or

$$(s_2-s_1) dT=(v_2-v_2) dp,$$ (9.71)

which reduces to

$$\frac{dp}{dT} = \frac{{\mit\Delta}s}{{\mit\Delta} v},$$ (9.72)

where ${\mit\Delta}s\equiv s_2-s_1$ and ${\mit\Delta v}\equiv v_2-v_1$ . This result is known as the Clausius-Clapeyron equation.

Consider any point on the phase-equilibrium line at temperature $T$ and pressure $p$ . The Clausius-Clapeyron equation then relates the local slope of the line to the molar entropy change, ${\mit\Delta}s$ , and the molar volume change, ${\mit\Delta}v$ , of the substance in crossing the line at this point. Note, incidentally, that the quantities on the right-hand side of the Clausius-Clapeyron equation do not necessarily need to refer to one mole of the substance. In fact, both numerator and denominator can be multiplied by the same number of moles, leaving $dp/dT$ unchanged.

Because there is an entropy change associated with a phase transformation, heat must also be absorbed during such a process. The latent heat of transformation, $L_{12}$ , is defined as the heat absorbed when a given amount of phase 1 is transformed to phase 2. Because this process takes place at the constant temperature $T$ , the corresponding entropy change is

$${\mit\Delta} S = S_2-S_1 = \frac{L_{12}}{T},$$ (9.73)

where $L_{12}$ is the latent heat at this temperature. Thus, the Clausius-Clapeyron equation, (9.72), can also be written

$$\frac{dp}{dT} = \frac{{\mit\Delta}S}{{\mit\Delta} V}= \frac{L_{12}}{T {\mit\Delta}V}.$$ (9.74)