Q. 135.0( 1 Vote )

A gas is taken along the path AB as shown in the figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.



Answer :

Given


Heat extracted from the system ΔQ=-70cal=-70×4.2=-294J


The negative sign is because heat is extracted from the system.


From graph


VA=250cc=250×10-6m3


VB=100cc=100×10-6m3


PA=200kPa=200×103Pa


PB=500kPa=500×103Pa


We know that work done by the gas is given as


ΔW=PΔV


Here since we have two values of pressure we will take average pressure.



ΔW=Pavg(VA-VB)


= 350×103× (250-100) ×10-6


=-52.5J


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,


ΔU=-294-(-52.5) =-241.5J


change in internal energy is -241.5J.


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