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# A gas is taken through a cyclic process ABCA as shown in the figure. If 2.4 cal of heat is given in the process, what is the value of J? ‘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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