Three moles of an ideal gas undergo a cyclic | vedclass

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(A) The total work done in the cycle $ABCA$ is $W_{ABCA} = W_{AB} + W_{BC} + W_{CA}$.$1$. Process $AB$ is an isothermal process since $T_A = T_B = T_1

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$RT_1[3 \ln(3) - 2]$

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(A) The total work done in the cycle $ABCA$ is $W_{ABCA} = W_{AB} + W_{BC} + W_{CA}$.$1$. Process $AB$ is an isothermal process since $T_A = T_B = T_1$. The work done is $W_{AB} = nRT_1 \ln(V_B/V_A) = 3RT_1 \ln(3V_1/V_1) = 3RT_1 \ln(3)$.$2$. Process $BC$ is an isobaric process since $P_B = P_C = P_2$. The work done is $W_{BC} = P_2(V_C - V_B) = P_2(V_1 - 3V_1) = -2P_2V_1$. From the ideal gas law at point $B$, $P_2(3V_1) = nRT_1 = 3RT_1$, so $P_2V_1 = RT_1$. Thus, $W_{BC} = -2RT_1$.$3$. Process $CA$ is an isochoric process since $V_C = V_A = V_1$. The work done is $W_{CA} = 0$.$4$. Total work $W_{ABCA} = 3RT_1 \ln(3) - 2RT_1 + 0 = RT_1[3 \ln(3) - 2]$.

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