InterviewSolution
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InterviewSolution
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Find the entropy increment of one mole of a Van der Waals gas due to the isothermal variation of volume from `V_1` to `V_2`. The Van der Walls corrections are assumed to be known. |
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Answer» For a Vander Waal gas `(p + (a)/(V^2))(V - b) = RT` The entropy change along an isotherm can be calculated from `Delta S = int_(V_1)^(V_2) ((del S)/(del V))_T dV` If follows from (2.129) that `((del S)/(del V))_T = ((del p)/(del T))_V = (R)/(V - b)` Assuming `a,b` to be known constants. Thus `Delta S = R 1n (V_2 - b)/(V_1 - b)`. |
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