Answer :

Given


Constant pressure p=1.0×105 Pa


Change in volume ΔV=(200-100)×10-6m3=10-4m3


Internal energy U= 1.5pV


So, change in internal energy ΔU=1.5pΔV


=1.5×1.0×105×10-4


=15J


We know that work done by the gas is given as


ΔW=pΔV=1.0×105×10-4=10J


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


Therefore,


ΔQ=15+10=25J


heat absorbed by the system=25J.


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