A thermodynamical process is shown in with PA= 3 X 104 Pa; VA = 2 X 10-3 m3; PB = 8 x 104 N/m2; VD = 5 x 10-3 m3.In the processes AB and BC, 600 J and 200 J 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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560 J

800 J

600 J

640 J

answer is A.

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Detailed Solution

PBFor process A→B→C ;Q=W+∆U ⇒ ∆U=Q-W=600+200-(8×104)(5-2)×10-3 =800-240=560 J

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