Q. 85.0( 1 Vote )

Consider two processes on a system as shown in the figure.



The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW1 and ΔW2 be the work done by the system in the processes A and B respectively.

A. ΔW1 > ΔW2

B. ΔW1 = ΔW2

C. ΔW1 < ΔW2

D. Nothing can be said about the relation between ΔW1 and ΔW2.

Answer :

Given V1=V2


We know that,


Work done = force ×displacement



Volume = area ×displacement


Therefore,


Work done=pressure ×volume


Let change in the volume of system = ΔV = V2-V1


Pressure =P


Thus, work done by the system W


ΔW=PΔV


For process A ΔW1=P1ΔV1


For process B ΔW2=P2ΔV2


Since ΔV1=ΔV2 we can write,




P1 < P2 (from graph)


Therefore ΔW1 < ΔW2.

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