JEE Advanced 2022 Paper 1 · Q17 · First Law of Thermodynamics

Question

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I:
(I) 10⁻³ kg of water at 100°C is converted to steam at the same temperature, at a pressure of 10⁵ Pa. The volume of the system changes from 10⁻⁶ m³ to 10⁻³ m³ in the process. Latent heat of water = 2250 kJ/kg.
(II) 0.2 moles of a rigid diatomic ideal gas with volume V at temperature 500 K undergoes an isobaric expansion to volume 3V. Assume R = 8.0 J mol⁻¹ K⁻¹.
(III) One mole of a monatomic ideal gas is compressed adiabatically from volume V = (1/3) m³ and pressure 2 kPa to volume V/8.
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion.

List-II:
(P) 2 kJ
(Q) 7 kJ
(R) 4 kJ
(S) 5 kJ
(T) 3 kJ

Which one of the following options is correct?

SolveKar AI · Accepted Answer

(I) Q = mL = 10⁻³·2250×10³ = 2250 J. W = pΔV = 10⁵·(10⁻³ − 10⁻⁶) ≈ 100 J. ΔU = Q − W ≈ 2250 − 100 = 2150 J ≈ 2 kJ ⇒ (P). (II) Rigid diatomic: C_v = (5/2)R, C_p = (7/2)R. Isobaric expansion at p, V to 3V: T_2 = 3T_1 = 1500 K. ΔT = 1000 K. ΔU = nC_vΔT = 0.2·(5/2)·8·1000 = 4000 J = 4 kJ ⇒ (R). (III) Monatomic adiabatic, γ = 5/3. T_2/T_1 = (V_1/V_2)^(γ−1) = (8)^(2/3) = 4. Initial: pV = nRT ⇒ T_1 = pV/(nR) = (2000·1/3)/(1·8.31) ≈ 80.2 K. Using R ≈ 8.0: T_1 ≈ 250/3 K ≈ 83.3 K. ΔT = 3T_1 = 250 K. ΔU = nC_vΔT = 1·(3/2)·8·250 = 3000 J = 3 kJ ⇒ (T). (IV) Diatomic with vibration: degrees of freedom f = 7, C_v = (7/2)R, C_p = (9/2)R. Isobaric, Q = nC_pΔT ⇒ ΔT = 9000/(3·(9/2)·8) = 9000/108 ≈ 83.3 K. ΔU = nC_vΔT = 3·(7/2)·8·83.3 ≈ 7000 J = 7 kJ ⇒ (Q). Match: I → P, II → R, III → T, IV → Q = option (C). Units: all energies in J/kJ ✓.

JEE Advanced 2022 Paper 1 · Q17 · First Law of Thermodynamics

Solution