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JEE Main 2026 (2 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2026 (2 Apr, Shift 1)

Q12JEE Main 2026 · 2 APR · SHIFT 1Medium

Thermodynamic Processes

A vessel contains 0.15 m3 of a gas at pressure 8 bar and temperature 140 °C with cp=3Rc_p = 3R and cv=2Rc_v = 2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is _________ kJ. (R is gas constant)

Numerical answer type. Work it out, then open the answer.

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Answer: 120

  1. γ=cpcv=3R2R=1.5\gamma = \dfrac{c_p}{c_v} = \dfrac{3R}{2R} = 1.5
  2. V2=V1(P1P2)1/γ=0.15×82/3=0.15×4=0.6 m3V_2 = V_1\left(\dfrac{P_1}{P_2}\right)^{1/\gamma} = 0.15 \times 8^{2/3} = 0.15 \times 4 = 0.6\text{ m}^3
  3. P1V1=8×105×0.15=1.2×105 JP_1V_1 = 8\times10^5 \times 0.15 = 1.2\times10^5\text{ J}, P2V2=105×0.6=0.6×105 JP_2V_2 = 10^5 \times 0.6 = 0.6\times10^5\text{ J}
  4. W=P1V1−P2V2γ−1=0.6×1050.5=1.2×105 JW = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1} = \dfrac{0.6\times10^5}{0.5} = 1.2\times10^5\text{ J}
  5. Work done =120 kJ= 120\text{ kJ}.
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