Answer :

‘J’ is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to joule


From the graph we can write


VA=VB=500cc=500×10-6m3


VC=700cc=700×10-6m3


PA=PC=100kPa=100×103 Pa


PB=200kPa=200×103Pa


We know that work done by the gas is given as


ΔW=PΔV


Work done in path AB=0 as VA=VB.


Work done in path CA ΔW1=PA(VA-VC)


=100×103× (500-700)×10-6


=-20J


Work done in path BC ΔW2= Pavg(VC-VB)



ΔW2=150×103×(700-500)×10-6


=30J


Total work done in process ABCA=ΔW=ΔW1+ΔW2


=30-20=10J


We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.


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 ΔU=0, first law becomes


ΔQ=ΔW=10J


But it is given in question that ΔQ=2.4cal


So, 2.4×J=10Joule



value of ‘J’ is 4.17J/cal.


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