JEE Main 2023 — Thermodynamics (C) Question with Solution

The Gibbs free energy can be written as a function of enthalpy and entropy as follows,

(B) $\mathrm{G}=\mathrm{H}-\mathrm{TS}$

At constant $\mathrm{T}$ 

$\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}$

(A) According to first law of thermodynamics, $\mathrm{\Delta U}=\mathrm{Q}+\mathrm{W}$

If we apply constant $\mathrm{P}$ and reversible work.

$\mathrm{W}=-\mathrm{P}∆\mathrm{V}$

$\mathrm{\Delta U}=\mathrm{Q}-\mathrm{P\Delta V}$

(C) By definition of entropy change $\mathrm{dS}=\frac{{\mathrm{dq}}_{\mathrm{rev}}}{\mathrm{T}}$

At constant $\mathrm{T}$

$\mathrm{\Delta S}=\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$

Enthalpy can be written as a function of internal energy, pressure and volume as follows,

(D) $\mathrm{H}=\mathrm{U}+\mathrm{PV}$

For ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$

$\mathrm{H}=\mathrm{U}+\mathrm{nRT}$

At constant $\mathrm{T}$

$\mathrm{\Delta H}=\mathrm{\Delta U}+\mathrm{\Delta nRT}$