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

A total of \(48 \mathrm{~J}\) heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by \(2^{\circ} \mathrm{C}\). The work done by the gas is _______. (Given, \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

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Answer & Solution

Correct Answer

(D) \(23.1 \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

\(1^{\text {st }}\) law of thermodynamics

\(\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}\)

\(\Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W}\)

\(\Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W}\)

\(\Rightarrow \mathrm{W}=48-3 \times \mathrm{R}\)

\(\Rightarrow \mathrm{W}=48-3 \times(8.3)\)

\(\Rightarrow \mathrm{W}=23.1 \text { Joule }\)

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