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Numerical · Q21

Q.Two moles of a monatomic ideal gas (Cv=32RC_v = \tfrac32 R) are heated at constant volume so that their temperature rises by 50 K50\ \text{K}. Calculate the heat absorbed by the gas and the corresponding change in its internal energy. (Take R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}.)

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Given: n=2 moln=2\ \text{mol} (monatomic, Cv=32RC_v = \tfrac32 R), ΔT=50 K\Delta T = 50\ \text{K}, heated at constant volume.

Heat absorbed at constant volume:

Q=nCvΔT=2×32(8.314)×50=2×12.471×50≈1247 JQ = nC_v\Delta T = 2 \times \frac{3}{2}(8.314) \times 50 = 2 \times 12.471 \times 50 \approx 1247\ \text{J} …

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