Process A: Reversible isothermal expansion
Formula: W = nRT ln(V₂/V₁)
n = 2 mol, R = 8.314 J mol⁻¹ K⁻¹, T = 300 K
V₂/V₁ = 20/2 = 10
W = 2 × 8.314 × 300 × ln(10)
W ≈ 11490 J ≈ 11.5 kJ
So, A → II
Process B: Irreversible isothermal expansion at constant pressure
Formula: W = Pext(V₂ − V₁)
P = 3 kPa = 3000 Pa
ΔV = 3 − 1 = 2 m³
W = 3000 × 2 = 6000 J = 6 kJ
So, B → III
Process C: Change in internal energy
Formula: ΔU = n C̅v ΔT
n = 1 mol, C̅v = 3/2 R, ΔT = 320 K
ΔU = 1 × (3/2 × 8.314) × 320
ΔU ≈ 3990 J ≈ 4 kJ
So, C → I
Process D: Change in enthalpy at constant pressure
Formula: ΔH = n C̅p ΔT
n = 1 mol, C̅p = 5/2 R, ΔT = 337 K
ΔH = (5/2 × 8.314) × 337
ΔH ≈ 7000 J = 7 kJ
So, D → IV
Hence final matching is: A-II, B-III, C-I, D-IV
Very important matching-type question for JEE Main, JEE Advanced and IIT JEE.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.