Thermodynamics

JEE Main level, single correct, medium

31 of 190

Question 31 of 190

0.08kg0.08kg air is heated at constant volume through 5∘C5^{\circ}C. The specific heat of air at constant volume is 0.17kcal{kg}^{-1}^{\circ}{C}^{-1} and 1J=4.18joulecal−11J=4.18joule{cal}^{-1}. The change in its internal energy is approximately.

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Answer: (C) 284J284J

The amount of work done for the given process should be zero, as the process is an isochoric process.
Thus, the change in internal energy will be equal to the heat exchanged in the process. Mathematically,
Q=∆UQ=∆U
Now,
∆U=ms∆T...(1)∆U=ms∆T...\left(1\right)
From equation (1), it follows that
∆U=0.08×(170×4.18)×5J≈284J\begin{matrix}∆U & = & 0.08\times (170\times 4.18)\times 5J \\ & \approx & 284J\end{matrix}

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