Answer :

Given


Initial pressure P1=10kPa=10×103Pa


Final pressure P2=50kPa=50×103Pa


Initial volume V1=200cc=200×10-6m3


Final volume V2=50cc= 50×10-6m3


a) 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 gas


ΔW=PΔV


Here we have given two values of pressure. So, we will take the average value of pressure


Average pressure P




Therefore, ΔW = 30×103× (50-200)×10-6


ΔW= -4.5J


Work done by the gas is -4.5J.


b) Given that no heat is supplied or extracted from the gas.


From first law of thermodynamics, we know that,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


Since ΔQ=0


Therefore ΔU = -ΔW = -(-4.5) J=4.5J


the change in internal energy of the gas is 4.5J.


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