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Exercise · Q27

Q.2 mol2\ \text{mol} of an ideal gas expands isothermally and reversibly from 10 L10\ \text{L} to 20 L20\ \text{L} at a constant temperature. Calculate the entropy change of the gas. (R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}.)

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For the isothermal, reversible expansion of nn moles of an ideal gas from volume V1V_1 to V2V_2, the entropy change is given by ΔS=nRln⁡ ⁣(V2V1)\Delta S = nR\ln\!\left(\dfrac{V_2}{V_1}\right). Here n=2 moln=2\ \text{mol}, R=8.314 J K−1mol−1R=8.314\ \text{J K}^{-1}\text{mol}^{-1}, V1=10 LV_1=10\ \text{L} and V2=20 LV_2=20\ \text{L}, so V2V1=2\dfrac{V_2}{V_1}=2. Computing ln⁡2≈0.6931\ln 2 \approx 0.6931: ΔS=2×8.314×0.6931≈11.53 J K−1\Delta S = 2 \times 8.314 \times 0.6931 \approx 11.53\ \text{J K}^{-1}. The positive entropy change reflects the gas molecules having access to a larger volu …

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