A system undergoes three quasi-static processes sequentially as indicated in the given figure. 1-2 is an isobaric process, 2-3 is a polytropic process with γ = 4/3 and 3-1 is a process in which PV= constant. P2 = P1 = 4 x 105 N/m2, P3 =1 x 105 N/m2 and V1 = 1 m3.The heat transfer for the cycle is ΔQ, the change in internal energy is ΔU and the work done is ΔW. Then
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a
ΔW=0
b
ΔQ=1.08×105J
c
ΔU=0
d
ΔQ>ΔW
answer is B.
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Detailed Solution
For the process 3→1P1V1=P3V3⇒V3=P1V1P3=4×105×11×105=4m3For the process 2→3P2V2γ=P3V3γ⇒V2=P3P21γV3=41434=414m3=2m3For the process 1→2,W12=P1V2−V1=(2−1)4×105JFor the process 2→3,W23=P2V2−P3V3(γ−1)=(42−1×4)×10543−1=12(2−1)×105JFor the process 3→1,W31=−P1V1lnV3V1 =−4×105×1×ln41ΔU=0 ΔQ=ΔU+ΔW1→2→3→1=4×105(2−1)+12×105(2−1)−4×105×1.386≃1.08×105JΔQ=ΔW≃1.08×105J