The pressure of gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc.

(a) Calculate the work done by the gas.


(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?


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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