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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2024
  • Hard
  • MCQ

A sample of gas at temperature \(\mathrm{T}\) 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 _______. \((\mu=1 \mathrm{~mole})\)

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Answer & Solution

Correct Answer

(D) \(\mathrm{RT}[2-\sqrt{2}]\)

Step-by-step Solution

Detailed explanation

\(\mathrm{W}=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}\)

\(\mathrm{TV}^{\gamma-1}=\text { constant }=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\gamma-1}\)

\(\mathrm{~T}_{\mathrm{f}}=\mathrm{T}\left(\frac{1}{2}\right)^{1 / 2}=\frac{\mathrm{T}}{\sqrt{2}}\)

\(\mathrm{~W}=\frac{\mathrm{R}\left(\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right)}{1-\frac{3}{2}}=2 \mathrm{RT} \frac{(\sqrt{2}-1)}{\sqrt{2}}\)

\(=\mathrm{RT}(2-\sqrt{2})\)

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