JEE Main 2026 Apr 6 Shift 2, Chemistry Q4: Basic Concepts of Thermodynamic

JEE Main2026Apr 6, Shift 2Chemistry

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. $q = n R T ln \frac{V _{2}}{V _{1}}$
C. Irreversible Compression	III. $w = - p_{e x t} (V_{1} - V_{2})$
D. Cyclic reversible	IV. $\oint \frac{q _{r e v}}{T} = 0$

Choose the correct answer from the options given below:

A
A-II, B-III, C-I, D-IV
B
A-II, B-I, C-IV, D-III
C
A-II, B-I, C-III, D-IV
D
A-I, B-II, C-III, D-IV

Solution

A. Reversible isothermal expansion: $Δ U = 0$ , so $q = - w = n R T ln (V_{2} / V_{1})$ . Match II.

B. Free expansion: against zero external pressure means $w = 0$ , and since the gas is thermally isolated in the standard treatment, $q = 0$ . Match I.

C. Irreversible compression at constant external pressure: $w = - p_{e x t} (V_{2} - V_{1}) = - p_{e x t} (V_{1} - V_{2}) \cdot (- 1)$ . Either sign convention matches III.

D. Cyclic reversible process: entropy is a state function, hence $\oint d q_{r e v} / T = 0$ . Match IV.

$∴$ A-II, B-I, C-III, D-IV.

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q4: Basic Concepts of Thermodynamic

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