Q73 / 90 · JEE Main 2024 (04 Apr Shift 2) · Physics

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $\gamma=3 / 2$. The work done by the gas in the process is: $(\mu=1 \mathrm{~mole})$

  • A
    $R T[1-2 \sqrt{2}]$
  • B
    $R T[\sqrt{2}-2]$
  • Correct answer
    C
    $R T[2-\sqrt{2}]$
  • D
    $R T[2 \sqrt{2}-1]$

Solution

$\begin{aligned} & \mathrm{W}=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma} \\ & \mathrm{TV}^{\gamma-1}=\text { constan } \mathrm{t}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\gamma-1} \\ & \mathrm{~T}_{\mathrm{f}}=\mathrm{T}\left(\frac{1}{2}\right)^{1 / 2}=\frac{\mathrm{T}}{\sqrt{2}} \\ & \mathrm{~W}=\frac{\mathrm{R}\left(\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right)}{1-\frac{3}{2}}=2 \mathrm{RT} \frac{(\sqrt{2}-1)}{\sqrt{2}} \\ & =\mathrm{RT}(2-\sqrt{2})\end{aligned}$

Hence, the correct answer is option: 'C'.

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