Question

Consider the following liquid-vapour equilibrium ${Liquid <=> Vapour }$ Which of the following relations is correct ?

• A. ${ \frac{d \, In \, P}{dT} = \frac{ - \Delta H_v}{RT}}$
• B. ${ \frac{d \, In \, P}{dT^2} = \frac{ - \Delta H_v}{T^2}}$
• C. ${ \frac{d \, In \, P}{dT} = \frac{ \Delta H_v}{RT^2}}$
• D. ${ \frac{d \, In \, G}{dT^2} = \frac{ \Delta H_v}{RT^2}}$

Answer 1

The Correct Option is C

To solve this problem, we need to determine the correct relation for the liquid-vapour equilibrium. The given question involves the concept of vapour pressure and phase changes between liquid and vapour states, typically described by the Clausius-Clapeyron equation. This equation is used to relate the change in vapour pressure with temperature during a phase change.

The Clausius-Clapeyron equation is given by:

\[\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}\]

Where:

• P is the vapour pressure of the substance.
• \Delta H_v is the enthalpy change of vaporization.
• R is the universal gas constant.
• T is the temperature in Kelvin.

We need to determine which option corresponds to this equation:

• \[\frac{d \ln P}{dT} = \frac{-\Delta H_v}{RT}\] - This is not correct due to the incorrect placement of T.
• \[\frac{d \ln P}{dT^2} = \frac{-\Delta H_v}{T^2}\] - The approach is incorrect and does not match any established relation.
• \[\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}\] - This matches the Clausius-Clapeyron equation correctly.
• \[\frac{d \ln G}{dT^2} = \frac{\Delta H_v}{RT^2}\] - There is no accepted expression relating Gibbs free energy changes in this context.

By analyzing each option, we see that the third option \[\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}\] is indeed correct according to the Clausius-Clapeyron equation.

Conclusion: The correct relation according to the liquid-vapour equilibrium and the Clausius-Clapeyron equation is \[\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}\].