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(C) Given,$n = 6$ moles.The process is shown on a $P-T$ diagram.$AB$ and $CD$ are isochoric processes (lines passing through the origin in a $P-T$ dia

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$40$

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(C) Given,$n = 6$ moles.The process is shown on a $P-T$ diagram.$AB$ and $CD$ are isochoric processes (lines passing through the origin in a $P-T$ diagram represent constant volume,but here they are lines of constant pressure/volume ratios; specifically,in a $P-T$ diagram,$P \propto T$ represents an isochoric process). Since $A, B$ and $C, D$ lie on lines passing through the origin,$AB$ and $CD$ are isochoric processes.Work done in isochoric processes $AB$ and $CD$ is $\Delta W_{AB} = 0$ and $\Delta W_{CD} = 0$.$BC$ and $DA$ are isobaric processes (constant pressure lines).Work done in an isobaric process is $\Delta W = nR\Delta T$.For process $BC$: $\Delta W_{BC} = nR(T_C - T_B) = 6 	imes R 	imes (2200 - 800) = 6R 	imes 1400 = 8400R$.For process $DA$: $\Delta W_{DA} = nR(T_A - T_D) = 6 	imes R 	imes (600 - 1200) = 6R 	imes (-600) = -3600R$.Total work done per cycle: $\Delta W_{cyclic} = \Delta W_{AB} + \Delta W_{BC} + \Delta W_{CD} + \Delta W_{DA} = 0 + 8400R + 0 - 3600R = 4800R$.Using $R \approx 8.314\,J/(mol\cdot K)$:$\Delta W_{cyclic} = 4800 	imes 8.314 = 39907.2\,J \approx 40\,kJ$.

Six moles of an ideal gas perform a cycle shown in the figure. If the temperatures are $T_A = 600\,K, T_B = 800\,K, T_C = 2200\,K$ and $T_D = 1200\,K,$ the work done per cycle is...... $kJ$

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