0.08 kg air is heated at constant volume through 5. The specific heat of air at constant volume is 0.17 kcal/kg and J = 4.18 joule/ cal. The change in its internal energy is approximately. (a) 318 J (b) 298 J (c) 284 J (d) 142 J

Answer: (c). At constant volume Δ U = Q = mcvΔ T. Δ U = 0.08 × 0.17 × 5 = 0.068 kcal = 68 cal 68 × 4.18 ≈ 284 J So the answer is (c) 284 J.

0.08 kg air is heated at constant volume through 5. The specific heat… | JEE Main 2024 (27 Jan, Shift 1)

Q114JEE Main 2024 · 27 JAN · SHIFT 1Medium

Thermodynamic Processes

$0.08\text{ kg}$ air is heated at constant volume through $5^\circ\text{C}$ . The specific heat of air at constant volume is $0.17\text{ kcal/kg}\,{}^\circ\text{C}$ and $J = 4.18$ joule/ $\text{cal}$ . The change in its internal energy is approximately.

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Answer: (c)

At constant volume $\Delta U = Q = mc_v\Delta T$ .
$\Delta U = 0.08 \times 0.17 \times 5 = 0.068\text{ kcal} = 68\text{ cal}$
$68 \times 4.18 \approx 284\text{ J}$
So the answer is (c) 284 J.

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