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JEE Main 2024 (9 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2024 (9 Apr, Shift 1)

Q119JEE Main 2024 · 9 APR · SHIFT 1Medium

Thermodynamic Processes

A sample of 1 mole gas at temperature TT is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32\gamma = \frac{3}{2}, then the work done by the gas in the process is:

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Answer: (b)

  1. Adiabatic: T2=T(12)γ−1=T2T_2 = T\left(\frac{1}{2}\right)^{\gamma-1} = \frac{T}{\sqrt{2}}.
  2. W=R(T−T2)γ−1=2RT(1−12)=RT(2−2)W = \frac{R(T - T_2)}{\gamma - 1} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = RT(2 - \sqrt{2})
  3. So the answer is (b) RT[2−2]RT[2 - \sqrt{2}].
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