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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?

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

V_{A}=V_{B}=500cc=500×10^{-6}m^{3}

V_{C}=700cc=700×10^{-6}m^{3}

P_{A}=P_{C}=100kPa=100×10^{3} Pa

P_{B}=200kPa=200×10^{3}Pa

We know that work done by the gas is given as

ΔW=PΔV

Work done in path AB=0 as V_{A}=V_{B}.

Work done in path CA ΔW_{1}=P_{A}(V_{A}-V_{C})

=100×10^{3}× (500-700)×10^{-6}

=-20J

Work done in path BC ΔW_{2}= P_{avg}(V_{C}-V_{B})

ΔW_{2}=150×10^{3}×(700-500)×10^{-6}

=30J

Total work done in process ABCA=ΔW=ΔW_{1}+ΔW_{2}

=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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