Question

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

• A. $\frac{dInP}{dT}=\frac{-△H_v}{RT}$
• B. $\frac{dInP}{d{T}^2}=\frac{-△H_v}{T^2}$
• C. $\frac{dInP}{dT}=\frac{△H_v}{{RT}^2}$
• D. $\frac{dInG}{d{T}^2}=\frac{△H_v}{R{T}^2}$

Answer 1

The correct option is C $\frac{dInP}{dT}=\frac{△H_v}{{RT}^2}$

For liquid $\left(phase1\right)$ & vapour $\left(phase2\right)$ to be in equilibrium,

$g_1(P,T)=g_2(P,T)$

$g_1,g_2$ are the specific gibbs energy

$i.e.$ $d{g}_1=d{g}_2-\left(i\right)$

$dg={\left(\frac{\partial g}{\partial T}\right)}_{P}dT+{\left(\frac{\partial g}{\partial P}\right)}_{T}dp$

It in known that, ${\left(\frac{\partial g}{\partial T}\right)}_P=-s\&{\left(\frac{\partial g}{\partial P}\right)}_T=V$

Where $S-$specific entropy & $V\to$specific volume

$\therefore$ Equation $\left(i\right)$ becomes,

$-S_{1}dT+V_{1}dp=-S_{2}dT+V_{2}dp$

$i.e.\frac{dp}{dT}=\frac{S_2-S_1}{V_2-V_1}$

But the equilibrium state transition in a pure substance is an isothermal & isobaric process.

$\therefore$ Where $h_2$ &$h_1$ are the respective specific enthalpise of the phases.

$\therefore \frac{dp}{dT}=\frac{H_2-H_1}{T(V_G-V_L)}-\left(ii\right)$

$G$ & $L$ refers to vapour & liquid phase.

But $V_G=\frac{RT}{p}$ by ideal gas equation.

$\therefore$ Equation $\left(ii\right)$ becomes,

$\frac{dlnp}{dT}=\frac{△HV}{{RT}^2}$