JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Match the LIST-I with LIST-II

$\begin{array}{|c|l|c|c|} \hline & \textbf{List-I (Thermodynamic Process)} & & \textbf{List-II} \\ & & & \textbf{(Magnitude in kJ)} \\ \hline A. & \begin{array}{l} \text{Work done in reversible,} \\ \text{isothermal expansion of }\\ \text{2 mol ideal gas from } 2\,\mathrm{dm}^3 \\ \text{ to } 20\,\mathrm{dm}^3 \text{ at } 300\,\mathrm{K} \end{array} & I. & 4 \\ \hline B. & \begin{array}{l} \text{Work done in irreversible} \\ \text{isothermal expansion of }\\ 1 \text{ mol ideal gas from } 1\,\mathrm{m}^3 \text{ to } 3\,\mathrm{m}^3 \\ \text{at } 300\,\mathrm{K} \text{ against} \\ \text{constant pressure } 3\,\mathrm{kPa} \end{array} & II. & 11.5 \\ \hline C. & \begin{array}{l} \text{Change in internal energy} \\ \text{for adiabatic expansion of }\\ 1 \text{ mol ideal gas, } \Delta T = 320\,\mathrm{K}, \; \\ \overline{C}_V=\dfrac{3}{2}R \end{array} & III. & 6 \\ \hline D. & \begin{array}{l} \text{Change in enthalpy at constant} \\ \text{pressure of }\\ 1 \text{ mol ideal gas, } \Delta T = 337\,\mathrm{K}, \; \\ \overline{C}_p=\dfrac{5}{2}R \end{array} & IV. & 7 \\ \hline \end{array}$

Choose the correct answer from the options given below:

Official previous-year question

Held on 22 Jan 2026 · Verified 6 Jul 2026.

Options

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

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Solution

Calculate each thermodynamic quantity:

A. Reversible isothermal expansion (2 mol, 2→20 dm³, 300 K):

$W = nRT\ln(V_2/V_1) = 2 \times 8.314 \times 300 \times \ln(10) = 2 \times 8.314 \times 300 \times 2.303 = 11,481$ J ≈ 11.5 kJ → II

B. Irreversible isothermal expansion (1 mol, 1→3 m³, 3 kPa constant pressure):

$W = P_{ext}\Delta V = 3000 \times (3-1) = 6000$ J = 6 kJ → III

C. Adiabatic expansion (1 mol, $\Delta T$ = -320 K, $C_V = \frac{3}{2}R$ ):

$\Delta U = nC_V|\Delta T| = 1 \times \frac{3}{2} \times 8.314 \times 320 = 3,991$ J ≈ 4 kJ → I

D. Enthalpy change (1 mol, $\Delta T$ = 337 K, $C_p = \frac{5}{2}R$ ):

$\Delta H = nC_p\Delta T = 1 \times \frac{5}{2} \times 8.314 \times 337 = 7,016$ J ≈ 7 kJ → IV

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

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