Match List-I with List-II Given V 1 and V 2 are initial and final volumes respectively: List-I (Isothermal process) List-II (Expression) --- --- A. Reversible expansion I. q=0 B. Free expansion II. q = nRT ln frac V 2 V 1 C. Irreversible Compression III. w =- p ext ( V 1 - V 2 ) D. Cyclic reversible IV. frac q rev T =0 Choose the correct answer from the options given below :

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 6 April, Evening Shift — Question 48

Match List-I with List-II Given $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ are initial and final volumes respectively:

List-I (Isothermal process)	List-II (Expression)
A. Reversible expansion	I. q=0
B. Free expansion	II. $\mathrm{q}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
C. Irreversible Compression	III. $\mathrm{w}=-\mathrm{p}_{\text {ext }}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$
D. Cyclic reversible	IV. $\frac{\mathrm{q}_{\text {rev }}}{\mathrm{T}}=0$

Choose the correct answer from the options given below :

(A) For isothermal process, $\Delta U=0$

$W=-nRT\ln\left(\frac{V_2}{V_1}\right)$

So,

$q=nRT\ln\left(\frac{V_2}{V_1}\right)$

(B) Free expansion which is isothermal must be adiabatic, so $q=0$

$W=-P_{\mathrm{ext}}\left[V_2-V_1\right]$

(D) Cyclic process, change in state function ( $S$ ) so change in entropy must be zero.

\begin{aligned} dS &= \frac{dq_{\mathrm{rev}}}{T} \\ \oint dS &= 0 \end{aligned}

Answer key and solution verified before publishing.

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