Work done by a system under isothermal change from a volume $\mathbf{V}_1$ to $V_2$ for a gas which…

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Work done by a system under isothermal change from a volume $\mathbf{V}_1$ to $V_2$ for a gas which obeys Vander Waal's equation $(V-\boldsymbol{\beta n})\left(P+\frac{\alpha \boldsymbol{n}^{2}}{V}\right)=\boldsymbol{n}RT$
$(a) nRT \log_{e} \left(\frac{V_{2} - n \beta}{V_{1} - n \beta}\right) + \alpha n^{2} \left(\frac{V_{1} - V_{2}}{V_{1} V_{2}}\right) (b) nRT \log_{10} \left(\frac{V_{2} - \alpha \beta}{V_{1} - \alpha \beta}\right) + \alpha n^{2} \left(\frac{V_{1} - V_{2}}{V_{1} V_{2}}\right) (c) nRT \log_{e} \left(\frac{V_{2} - n \alpha}{V_{1} - n \alpha}\right) + \beta n^{2} \left(\frac{V_{1} - V_{2}}{V_{1} V_{2}}\right) (d) nRT \log_{e} \left(\frac{V_{1} - n \beta}{V_{2} - n \beta}\right) + \alpha n^{2} \left(\frac{V_{1} V_{2}}{V_{1} - V_{2}}\right)$

CGP EDU Academic Team · Accepted Answer

The correct answer is A. According to given Vander Waal’s equation
$P = \frac{nRT}{V - n\beta} - \frac{\alpha n^{2}}{V^{2}} \ Work \ done, \ W = \int_{V_{1}}^{V_{2}} PdV = nRT \int_{V_{1}}^{V_{2}} \frac{dV}{V - n\beta} - \alpha n^{2} \int_{V_{1}}^{V_{2}} \frac{dV}{V^{2}} \ = nRT [\log_{e} (V - n\beta)]_{V_{1}}^{V_{2}} + \alpha n^{2} \left[\frac{1}{V}ight]_{V_{1}}^{V_{2}} \ = nRT \log_{e} \frac{V_{2} - n\beta}{V_{1} - n\beta} + \alpha n^{2} \left(\frac{V_{1} - V_{2}}{V_{1} V_{2}}ight)$

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