Thermodynamics Matching Question – Work Done, Internal Energy and Enthalpy | JEE Main 2026
List-I : Thermodynamic Process

A. Work done in reversible isothermal expansion of 2 mol of ideal gas from 2 dm³ to 20 dm³ at 300 K.

B. Work done in irreversible isothermal expansion of 1 mol ideal gas from 1 m³ to 3 m³ at 300 K against a constant pressure of 3 kPa.

C. Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and C̅v = 3/2 R.

D. Change in enthalpy at constant pressure of 1 mol ideal gas with change of temperature = 337 K and C̅p = 5/2 R.
A. A-I, B-II, C-III, D-IV
B. A-III, B-II, C-IV, D-I
C. A-II, B-I, C-III, D-IV
D. A-II, B-III, C-I, D-IV
Correct Answer: A-II, B-III, C-I, D-IV

Step-by-Step Solution

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.

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Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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