Question

Consider the following liquid-vapour equilibrium
$\text{Liquid}\rightleftharpoons \text{Vapour}$
Which of the following relations is correct ?

• A. $\frac{d\left(\ln P\right)}{dT}=\frac{-\Delta H_V}{RT}$
• B. $\frac{d\left(\ln P\right)}{d{T}^2}=\frac{-\Delta H_V}{T^2}$
• C. $\frac{d\left(\ln P\right)}{dT}=\frac{\Delta H_V}{R{T}^2}$
• D. $\frac{d\left(\ln G\right)}{d{T}^2}=\frac{\Delta H_V}{R{T}^2}$

Answer 1

The correct option is C $\frac{d\left(\ln P\right)}{dT}=\frac{\Delta H_V}{R{T}^2}$

Acc. To Clausius Claperon equation:
$K=A{e}^{-\frac{\Delta H}{RT}}$
$\ln P=\ln A-\frac{\Delta H}{RT}$
$\Rightarrow \frac{d\left(\ln P\right)}{dT}=0-\frac{\Delta H}{R}\frac{d\left(T^{-1}\right)}{dT}$
$\frac{d}{dT}\left(\ln P\right)=\frac{\Delta H_V}{R{T}^2}$