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(D) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta U = q + W = 0$,which impli

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$0.5$

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(D) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta U = q + W = 0$,which implies $q = -W$. Total work done $W_{total} = W_{AB} + W_{BC} + W_{CA}$. Process $A ightarrow B$: Isochoric (constant volume),so $W_{AB} = 0$. Process $B ightarrow C$: Given $W_{BC} = 1 \, L \, atm$. Process $C ightarrow A$: Isobaric (constant pressure at $2 \, atm$),so $W_{CA} = -p \Delta V = -2 \, atm 	imes (1 \, L - 1.5 \, L) = -2 	imes (-0.5) = 1 \, L \, atm$. Total work $W_{total} = 0 + 1 + 1 = 2 \, L \, atm$. Since $q = -W_{total}$,the heat exchanged is $q = -2 \, L \, atm$. Re-evaluating the provided options and the cyclic nature,the net heat exchanged in a cycle is equal to the negative of the total work done. Based on the graph,the area enclosed is the work. The area of the triangle $ABC$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2-1) \, atm 	imes (1.5-1) \, L = 0.25 \, L \, atm$. Given the specific path $A$ $ightarrow B$ $ightarrow C$ $ightarrow A$,the net work is $W_{AB} + W_{BC} + W_{CA} = 0 + 1 + 1 = 2 \, L \, atm$. There appears to be a discrepancy in the provided options. Based on standard thermodynamic cycles,the correct magnitude is $2.0 \, L \, atm$.

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