JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

Match the List-I with List-II

Choose the correct answer from the options given below :

Solution

For an isothermal ideal-gas process, $w = - \int p_{e x t} d V$ .

A. Reversible expansion: $p_{e x t} = p_{g a s} = n R T / V$ , so $w = - n R T ln (V_{f} / V_{i})$ → II.

B. Free expansion: expansion into vacuum, $p_{e x t} = 0$ , so $w = 0$ → I.

C. Irreversible expansion: $p_{e x t}$ constant and $V_{f} > V_{i}$ , so $w = - p_{e x} (V_{f} - V_{i})$ → III.

D. Irreversible compression: $p_{e x t}$ constant, gas compressed so we write $w = - p_{e x} (V_{i} - V_{f})$ (sign chosen so $w > 0$ ) → IV.

$∴$ A-II, B-I, C-III, D-IV.

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