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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 28 January · Morning Shift · Chemistry · Thermodynamics (C)

20.0 dm^ 3 of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes...

20.0dm320.0 \mathrm{dm}^{3} of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct?
(Given: log⁡2=0.3010\log 2=0.3010 and log⁡5=0.6989\log 5=0.6989)

Options

  1. Aw=−9.1 kJ,ΔU=0,ΔH=0,q=9.1 kJ\mathrm{w}=-9.1 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0, \mathrm{q}=9.1 \mathrm{~kJ}
  2. Bw=+4.1 kJ,ΔU=0,ΔH=0;q=−4.1 kJ\mathrm{w}=+4.1 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0 ; \mathrm{q}=-4.1 \mathrm{~kJ}
  3. Cw=9.1 J,ΔU=9.1 J,ΔH=0;q=0\mathrm{w}=9.1 \mathrm{~J}, \Delta \mathrm{U}=9.1 \mathrm{~J}, \Delta \mathrm{H}=0 ; \mathrm{q}=0
  4. Dw=−3.9 kJ,ΔU=0,ΔH=0;q=3.9 kJ\mathrm{w}=-3.9 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0 ; \mathrm{q}=3.9 \mathrm{~kJ}

Correct answer

A. w=−9.1 kJ,ΔU=0,ΔH=0,q=9.1 kJ\mathrm{w}=-9.1 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0, \mathrm{q}=9.1 \mathrm{~kJ}

Step-by-step solution

For isothermal reversible process ΔU=ΔH=0\Delta U = \Delta H = 0
wiso=−p1v1ln⁡P1P2w_{iso} = -p_1 v_1 \ln \dfrac{P_1}{P_2}
wiso=−0.5×106×20×10−3ln⁡0.50.2w_{iso} = -0.5 \times 10^6 \times 20 \times 10^{-3} \ln \dfrac{0.5}{0.2}
wiso=−0.5×106×20×10−3×2.303×(.6989−.3010)w_{iso} = -0.5 \times 10^6 \times 20 \times 10^{-3} \times 2.303 \times (.6989 - .3010)
w≃−9.1 kJw \simeq -9.1 \text{ kJ}
q=−w=9.1 kJq = -w = 9.1 \text{ kJ}
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