A thermodynamical process is shown in fig. With PA=3×104Pa;VA=2×10−3m3 ;PB=8×104N/m2;VD=5×10−3m3, In the processes AB and BC, 600J and 200J of heat is added to the system respectively . The change in internal energy of the system in process AC would be

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560J

800 J

600 J

640 J

answer is A.

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Detailed Solution

W=WA→B+WB→C=O+P(V2−V1)=8×104(5×10−3−2×10−3)=240JΔQ=ΔU+ΔW600+200=ΔU+240560J=ΔU

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