From graph we can write

V_A =V_D = 200cc=200×10^-6m³

V_B=V_C = 400cc= 400×10^-6m³

P_B=P_D=155kPa= 155×10³Pa

P_A=P_C=50kPa=50×10³Pa

Given

Heat absorbed in process ABC ΔQ₁=50cal= 50×4.2 J=210J

Let heat absorbed into the system during process ADC =ΔQ₂

We know that work done by the gas is given as

ΔW=PΔV

Work done in path ACB W_ACB =ΔW₁= W_AC+W_BC

ΔW₁=P_A(V_C-V_A) + 0 (W_BC=0 because V_B=V_C)

=50×10³×(400-200)×10^-6

=10J

Work done in path ADB W_ADB=ΔW2=W_AD + W_DB

ΔW₂ = P_B (V_D-V_B) +0 (∵ W_AD=0 because V_A=V_D)

=155×10³×(400-200)×10^-6

=31 J

Now initial point A and final point C is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.

Therefore,

ΔU₁=ΔU₂=ΔU ……(i)

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

Using first law of thermodynamics for process ABC

ΔQ₁=ΔU₁+ΔW₁

ΔU₁=ΔQ₁-ΔW₁=210-10=200J

Using first law of thermodynamics for process adc

ΔQ₂=ΔU₂+ΔW₂

=ΔU₁+ΔW₂ (from (i))

=200+31=231J.

∴ heat supplied to the system during process ADC=231J.