JEE Main 2026 — Thermodynamics (C) Question with 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, $C_V=\frac{3}{2}R$):
$\Delta U=n{C}_V|\Delta T|=1\times \frac{3}{2}\times 8.314\times 320=3,991$ J ≈ 4 kJ → I
D. Enthalpy change (1 mol, $\Delta T$ = 337 K, $C_p=\frac{5}{2}R$):
$\Delta H=n{C}_p\Delta T=1\times \frac{5}{2}\times 8.314\times 337=7,016$ J ≈ 7 kJ → IV
Matching: A-II, B-III, C-I, D-IV