Work done by a system under isothermal change | vedclass

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Question

(A) According to the given Van der Waals' equation:$(V - n\beta) \left( P + \frac{\alpha n^2}{V^2} \right) = nRT$Rearranging for $P$:$P = \frac{nRT}{V

Vedclass · Accepted Answer

$nRT \log_e \left( \frac{V_2 - n\beta}{V_1 - n\beta} \right) + \alpha n^2 \left( \frac{V_1 - V_2}{V_1 V_2} \right)$

Vedclass · Answer

(A) According to the given Van der Waals' equation:$(V - n\beta) \left( P + \frac{\alpha n^2}{V^2} \right) = nRT$Rearranging for $P$:$P = \frac{nRT}{V - n\beta} - \frac{\alpha n^2}{V^2}$Work done $W$ is given by the integral:$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \left( \frac{nRT}{V - n\beta} - \frac{\alpha n^2}{V^2} \right) dV$$W = nRT \int_{V_1}^{V_2} \frac{dV}{V - n\beta} - \alpha n^2 \int_{V_1}^{V_2} V^{-2} dV$Integrating the terms:$W = nRT [\log_e(V - n\beta)]_{V_1}^{V_2} - \alpha n^2 \left[ -\frac{1}{V} \right]_{V_1}^{V_2}$$W = nRT \log_e \left( \frac{V_2 - n\beta}{V_1 - n\beta} \right) + \alpha n^2 \left( \frac{1}{V_2} - \frac{1}{V_1} \right)$$W = nRT \log_e \left( \frac{V_2 - n\beta}{V_1 - n\beta} \right) + \alpha n^2 \left( \frac{V_1 - V_2}{V_1 V_2} \right)$

Work done by a system under isothermal change from a volume $V_1$ to $V_2$ for a gas which obeys Van der Waals' equation $(V - n\beta) \left( P + \frac{\alpha n^2}{V^2} \right) = nRT$.

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