A thermodynamic process is shown in figure. The pressures and volumes corresponding to some points in the figure are:PA=3×104PaPB=8×104Pa and VA=2×10−3m3VD=5×10−3mIn process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be
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a
560 J
b
800 J
c
600 J
d
640 J
answer is A.
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Detailed Solution
By the graph, WAB=0 and WBC=8×104[5−2]×10−3=240J∴ WAC=WAB+WBC=0+240=240J Now, QAC=QAB+QBC=600+200=800JFrom the first law of thermodynamics, QAC=ΔUAC+WAC⇒ 800=ΔUAC+240 or ΔUAC=560J