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, \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)
- A \(72.9 \mathrm{~J}\)
- B \(24.9 \mathrm{~J}\)
- C \(48 \mathrm{~J}\)
- D \(23.1 \mathrm{~J}\)
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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