‘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^-6m³

V_C=700cc=700×10^-6m³

P_A=P_C=100kPa=100×10³ Pa

P_B=200kPa=200×10³Pa

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 ΔW₁=P_A(V_A-V_C)

=100×10³× (500-700)×10^-6

=-20J

Work done in path BC ΔW₂= P_avg(V_C-V_B)

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.