JEE Main
Chemistry
Thermodynamics (C)
2026
JEE Main 2026 (22 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Question
Match the LIST-I with LIST-II \( array |c|l|c|c| & List-I (Thermodynamic Process) & & List-II \\ & & & (Magnitude in kJ) \\ A. & array l Work done in reversible, \\ isothermal expansion of \\ 2 mol ideal gas from 2\, dm ^3 \\ to 20\, dm ^3 at 300\, K array & I. & 4 \\ B. & array l Work done in irreversible \\ isothermal expansion of \\ 1 mol ideal gas from 1\, m ^3 to 3\, m ^3 \\ at 300\, K against \\ constant pressure 3\, kPa array & II. & 11.5 \\ C. & array l Change in internal energy \\ for adiabatic expansion of \\ 1 mol ideal gas, T = 320\, K , \; \\ C _V= 3 2 R array & III. & 6 \\ D. & array l Change in enthalpy at constant \\ pressure of \\ 1 mol ideal gas, T = 337\, K , \; \\ C _p= 5 2 R array & IV. & 7 \\ array \) Choose the correct answer from the options given below:
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
Answer
A. A-II, B-III, C-I, D-IV
Step-by-step solution
Calculate each thermodynamic quantity: A. Reversible isothermal expansion (2 mol, 2→20 dm³, 300 K): W = nRT (V_2/V_1) = 2 8.314 300 (10) = 2 8.314 300 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 V = 3000 (3-1) = 6000 J = 6 kJ → III C. Adiabatic expansion (1 mol, T = -320 K, C_V = 3 2 R): U = nC_V| T| = 1 3 2 8.314 320 = 3,991 J ≈ 4 kJ → I D. Enthalpy change (1 mol, T = 337 K, C_p = 5 2 R): H = nC_p T = 1 5 2 8.314 337 = 7,016 J ≈ 7 kJ → IV Matching: A-II, B-III, C-I, D-IV
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