Question
Consider the following liquid-vapour equilibrium
$${\text{Liquid }} \rightleftharpoons {\text{Vapour}}$$
Which of the following relations is correct?
A.
$$\frac{{d\ln P}}{{dT}} = \frac{{ - \Delta {H_v}}}{{RT}}$$
B.
$$\frac{{d\ln P}}{{d{T^2}}} = \frac{{ - \Delta {H_v}}}{{{T^2}}}$$
C.
$$\frac{{d\ln P}}{{dT}} = \frac{{ - \Delta {H_v}}}{{R{T^2}}}$$
D.
$$\frac{{d\ln G}}{{d{T^2}}} = \frac{{ - \Delta {H_v}}}{{R{T^2}}}$$
Answer :
$$\frac{{d\ln P}}{{dT}} = \frac{{ - \Delta {H_v}}}{{R{T^2}}}$$
Solution :
The given phase equilibria is $${\text{Liquid}} \rightleftharpoons {\text{Vapour}}.$$ This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius Clapeyron and the relationship is written as,
$$\frac{{d\ln P}}{{dT}} = - \frac{{\Delta {H_v}}}{{R{T^2}}}$$
where, $$\Delta {H_v} = $$ Heat of vaporisation