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Numerical · Q24

Q.One mole of a monatomic ideal gas (γ=5/3\gamma = 5/3) expands adiabatically and reversibly, its temperature falling from 500 K500\ \text{K} to 350 K350\ \text{K}. Calculate the work done by the gas during this expansion. (Take R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}.)

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Given: n=1 moln=1\ \text{mol}, γ=5/3\gamma = 5/3, T1=500 KT_1 = 500\ \text{K}, T2=350 KT_2 = 350\ \text{K} (so T1−T2=150 KT_1 - T_2 = 150\ \text{K}).

Work done by the gas in an adiabatic process:

W=nR(T1−T2)γ−1=1×8.314×1505/3−1=1247.12/3W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{1 \times 8.314 \times 150}{5/3 - 1} = \frac{1247.1}{2/3}

W=1247.1×32≈1870.6 J≈1871 JW = 1247.1 \times \frac{3}{2} \approx 1870.6\ \text{J} \approx 1871\ \text{J} …

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