Vapor Pressure

The Clausius-Clapeyron equation can be used to derive an approximate expression for the pressure of the vapor in equilibrium with the liquid (or solid) at some temperature $T$ . This pressure is called the vapor pressure of the liquid (or solid) at this temperature. According to Equation (9.74),

$$\frac{dp}{dT} =\frac{l}{T {\mit\Delta} v},$$ (9.75)

where $l=l_{12}$ is the latent heat per mole, and $v$ the molar volume. Let $1$ refer to the liquid (or solid) phase, and $2$ to the vapor. It follows that

$${\mit\Delta} v=v_2-v_1\simeq v_2,$$ (9.76)

because the vapor is much less dense than the liquid, so that $v_2\gg v_1$ . Let us also assume that the vapor can be adequately treated as an ideal gas, so that its equation of state is written

$$p v_2=R T.$$ (9.77)

Thus, ${\mit\Delta}v\simeq R T/p$ , and Equation (9.75) becomes

$$\frac{1}{p} \frac{dp}{dT} = \frac{1}{R T^{ 2}}.$$ (9.78)

Assuming that $l$ is approximately temperature independent, we can integrate the previous equation to give

$$\ln p = -\frac{l}{R T} + {\rm constant},$$ (9.79)

which implies that

$$p = p_0 \exp\left(-\frac{l}{R T}\right),$$ (9.80)

where $p_0$ is some constant. This result shows that the vapor pressure, $p$ , is a very rapidly increasing function of the temperature, $T$ .