A cyclic process for 1 , text{mole} of an ide | vedclass

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Question

(A) For process $AB$: The volume $V$ is constant $(V = V_1)$. Therefore, the work done $W_{AB} = \int P dV = 0$.For process $BC$: The temperature $T$

Vedclass · Accepted Answer

$0; R T_2 \ln \left( \frac{V_2}{V_1} \right); R(T_1 - T_2)$

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(A) For process $AB$: The volume $V$ is constant $(V = V_1)$. Therefore, the work done $W_{AB} = \int P dV = 0$.For process $BC$: The temperature $T$ is constant $(T = T_2)$. The work done for an isothermal process is $W_{BC} = nRT_2 \ln \left( \frac{V_f}{V_i} \right)$. Here, $n = 1$, $V_i = V_1$, and $V_f = V_2$. So, $W_{BC} = R T_2 \ln \left( \frac{V_2}{V_1} \right)$.For process $CA$: This is a process where the graph is a straight line passing through the origin in the $V-T$ diagram, which implies $V \propto T$, meaning pressure $P$ is constant (isobaric process). The work done is $W_{CA} = nR \Delta T = R(T_1 - T_2)$.

$A$ cyclic process for $1 \, \text{mole}$ of an ideal gas is shown in the $V-T$ diagram. Find the work done in processes $AB$, $BC$, and $CA$ respectively.

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