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(A) In a $P-T$ diagram,the work done $W$ is given by the area under the process curve. For an ideal gas,$PV = nRT$,so $V = \frac{nRT}{P}$.Processes $1

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$19.94$

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(A) In a $P-T$ diagram,the work done $W$ is given by the area under the process curve. For an ideal gas,$PV = nRT$,so $V = \frac{nRT}{P}$.Processes $1-2$ and $3-4$ are isochoric (constant volume) because they lie on lines passing through the origin $(P \propto T)$,so $W_{12} = 0$ and $W_{34} = 0$.Processes $2-3$ and $4-1$ are isobaric (constant pressure),so $W = P\Delta V = nR\Delta T$.Total work done $W = W_{12} + W_{23} + W_{34} + W_{41} = 0 + nR(T_3 - T_2) + 0 + nR(T_1 - T_4)$.$W = nR(T_3 - T_2 + T_1 - T_4) = 3 	imes 8.31 	imes (2400 - 800 + 400 - 1200) \, J$.$W = 3 	imes 8.31 	imes (800) \, J = 19944 \, J = 19.944 \, kJ$.

Three moles of an ideal monatomic gas perform a cyclic process as shown in the figure. The temperatures in different states are $T_1 = 400\,K, T_2 = 800\,K, T_3 = 2400\,K$ and $T_4 = 1200\,K$. Determine the work done by the gas during the cycle in $kJ$. (Given $R = 8.31\, J\, mol^{-1} K^{-1}$)

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