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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 22 January · Morning Shift · Chemistry · Thermodynamics (C)

Match the List-I with List-II ( List-I (Thermodynamic Process)...

Match the List-I with List-II
(
List-I (Thermodynamic Process)List-II
(Magnitude in kJ)
A.Work done in reversible,
isothermal expansion of
2 mol ideal gas from 2 dm^3
to 20 dm^3 at 300 K
I.4
B.Work done in irreversible
isothermal expansion of
1 mol ideal gas from 1 m³ to 3 m³
at 300 K against
constant pressure 3 kPa
II.11.5
C.Change in internal energy
for adiabatic expansion of
1 mol ideal gas, Δ\DeltaT = 320 K,
\overline{C}_V=32\dfrac{3}{2}R
III.6
D.Change in enthalpy at constant
pressure of
1 mol ideal gas, Δ\DeltaT = 337 K,
\overline{C}_p=52\dfrac{5}{2}R
IV.7
)
Choose the correct answer from the options given below:

Options

  1. AA-II, B-III, C-I, D-IV
  2. BA-III, B-II, C-IV, D-I
  3. CA-II, B-I, C-III, D-IV
  4. DA-I, B-II, C-III, D-IV

Correct 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=nRTln⁡(V2/V1)=2×8.314×300×ln⁡(10)=2×8.314×300×2.303=11,481W = 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=PextΔV=3000×(3−1)=6000W = P_{ext}\Delta V = 3000 \times (3-1) = 6000 J = 6 kJ → III
C. Adiabatic expansion (1 mol, ΔT\Delta T = -320 K, CV=32RC_V = \frac{3}{2}R):
ΔU=nCV∣ΔT∣=1×32×8.314×320=3,991\Delta U = nC_V|\Delta T| = 1 \times \frac{3}{2} \times 8.314 \times 320 = 3,991 J ≈ 4 kJ → I
D. Enthalpy change (1 mol, ΔT\Delta T = 337 K, Cp=52RC_p = \frac{5}{2}R):
ΔH=nCpΔT=1×52×8.314×337=7,016\Delta H = nC_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
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