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JEE PHYSICS MCQ: THERMODYNAMICS

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A total of 48 J48 \mathrm{~J} heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2∘C2^{\circ} \mathrm{C}. The work done by the gas is : (Given, R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}.)
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Correct option: D
23.1 J23.1 \mathrm{~J}
Option D (23.1 J23.1 \mathrm{~J}) is correct because it correctly applies the first law of thermodynamics to find the work done by the gas.

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