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JEE Main 2021 (27 Jul, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2021 (27 Jul, Shift 2)

Q190JEE Main 2021 · 27 JUL · SHIFT 2Medium

Thermodynamic Processes

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27∘^\circ C to 37∘^\circ C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol−1{}^{-1} K−1{}^{-1}]

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Answer: (b)

  1. Degrees of freedom: 3 translational + 3 rotational + 2×42\times4 vibrational =14= 14, so Cv=142R=7RC_v = \frac{14}{2}R = 7R.
  2. Adiabatic: Q=0Q = 0, so Won=ΔU=nCvΔTW_{on} = \Delta U = nC_v\Delta T.
  3. ΔU=1×7×8.314×10≈582\Delta U = 1 \times 7 \times 8.314 \times 10 \approx 582 J (positive, temperature rises, so work is done on the gas).
  4. So the answer is (b) work done on the gas is close to 582 J.
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