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What is the maximum work that can be obtained by the isothermal expansion of one mole of an ideal gas at 273 K from 2.24 dm3 to 22.4 dm3? |
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Answer» w = -2.303 n RT log \(\frac{V_2}{V_1}\) = -2.303 x 1 x 8.314 x 273 x log \(\frac{22.4}{2.24}\) = -2.303 x 1 x 8.314 x 273 x 1 = -5227.17 J |
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