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JEE Main 2023 (1 Feb, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2023 (1 Feb, Shift 1)

Q154JEE Main 2023 · 1 FEB · SHIFT 1Medium

Thermodynamic Processes

A sample of gas at temperature TT is adiabatically expanded to double its volume. The work done by the gas in the process is (given, γ=32)\left(\text{given, } \gamma = \frac{3}{2}\right):

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Answer: (d)

  1. Adiabatic: TVγ−1=constTV^{\gamma-1} = \text{const}, so T2=T2T_2 = \frac{T}{\sqrt{2}}.
  2. Work by gas (taking 1 mole): W=R(T−T/2)1/2=RT(2−2)W = \frac{R(T - T/\sqrt{2})}{1/2} = RT(2 - \sqrt{2})
  3. So the answer is (d) W=RT[2−2]W = RT[2 - \sqrt{2}].
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