If one mole of an ideal gas at (P_{1}, V_{1}) | vedclass

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(A) For the isothermal process $A-B$:$W_{AB} = nRT \ln \left(\frac{V_B}{V_A}\right) = RT \ln \left(\frac{2V_1}{V_1}\right) = RT \ln 2$.For the isochor

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$RT \left(\ln 2 - \frac{1}{2(\gamma-1)}\right)$

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(A) For the isothermal process $A-B$:$W_{AB} = nRT \ln \left(\frac{V_B}{V_A}\right) = RT \ln \left(\frac{2V_1}{V_1}\right) = RT \ln 2$.For the isochoric process $B-C$:Since the volume is constant,$W_{BC} = 0$.For the adiabatic process $C-A$:The work done in an adiabatic process is $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.Here,$P_C = \frac{P_1}{4}$,$V_C = 2V_1$,$P_A = P_1$,$V_A = V_1$.$W_{CA} = \frac{P_C V_C - P_A V_A}{\gamma - 1} = \frac{(\frac{P_1}{4})(2V_1) - P_1 V_1}{\gamma - 1} = \frac{\frac{P_1 V_1}{2} - P_1 V_1}{\gamma - 1} = \frac{-\frac{P_1 V_1}{2}}{\gamma - 1} = -\frac{RT}{2(\gamma - 1)}$.The net work done is $W_{net} = W_{AB} + W_{BC} + W_{CA} = RT \ln 2 + 0 - \frac{RT}{2(\gamma - 1)} = RT \left(\ln 2 - \frac{1}{2(\gamma - 1)}\right)$.

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