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JEE Mains · Physics · STD 11 - 11. thermodynamics

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, \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

  1. A \(72.9 \mathrm{~J}\)
  2. B \(24.9 \mathrm{~J}\)
  3. C \(48 \mathrm{~J}\)
  4. D \(23.1 \mathrm{~J}\)
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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 \(\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}\) \(\Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W}\) \(\Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W}\)…
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