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JEE PHYSICS MCQ: THERMODYNAMICS

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An ideal gas has molar heat capacity cvc_v at constant volume. The gas undergoes a process as T=T0(1+αV2)T=T_0\left(1+\alpha V^2\right) where TT and VV are temperature and volume respectively, T0T_0 and α\alpha are positive constants. The molar heat capacity CC of the gas is given as C=CV+Rf(V)C=C_V+R f(V), where f(V)f(V) is a function of volume. The expression for f(V)f(V) is
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Correct option: B
1+αV22αV2\frac{1+\alpha V^2}{2 \alpha V^2}
Option A (αV21+αV2\frac{{\alpha V^2}}{{1+\alpha V^2}}) is incorrect. The correct answer is B (1+αV22αV2\frac{{1+\alpha V^2}}{{2 \alpha V^2}}) because it is derived from the first law of thermodynamics and the given temperature-volume relationship.

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