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(C) The given graph is a $P-T$ diagram. The work done by an ideal gas in a cycle is given by the area enclosed by the cycle in a $P-V$ diagram. Howeve

Vedclass · Accepted Answer

$2400$

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(C) The given graph is a $P-T$ diagram. The work done by an ideal gas in a cycle is given by the area enclosed by the cycle in a $P-V$ diagram. However, for a $P-T$ diagram, we can use the relation $PV = nRT$. For an ideal gas, the work done in a process is $W = \int P dV$. From the ideal gas equation, $V = \frac{nRT}{P}$, so $dV = \frac{nR}{P} dT - \frac{nRT}{P^2} dP$. In the given $P-T$ diagram, the processes $AB$ and $CD$ are isochoric (since they lie on lines passing through the origin, $P \propto T \implies V = 	ext{constant}$), and processes $BC$ and $DA$ are isobaric (since $P$ is constant). Work done in isochoric processes ($AB$ and $CD$) is $0$. Work done in isobaric processes is $W = P \Delta V = nR \Delta T$. For process $BC$ (isobaric at $P_B$): $W_{BC} = nR(T_C - T_B) = 3R(2400 - 800) = 3R(1600) = 4800R$. For process $DA$ (isobaric at $P_A$): $W_{DA} = nR(T_A - T_D) = 3R(400 - 1200) = 3R(-800) = -2400R$. Total work done $W_{net} = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 4800R + 0 - 2400R = 2400R$.

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