The internal energy of a gas is given by U = 1.5 pV. It expands from 100 cm3 to 200 cm3 against a constant pressure of 1.0 × 105 Pa. Calculate the heat absorbed by the gas in the process.

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.


1