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Given

Heat extracted from the system ΔQ=-70cal=-70×4.2=-294J

The negative sign is because heat is extracted from the system.

From graph

V_A=250cc=250×10^-6m³

V_B=100cc=100×10^-6m³

P_A=200kPa=200×10³Pa

P_B=500kPa=500×10³Pa

We know that work done by the gas is given as

ΔW=PΔV

Here since we have two values of pressure we will take average pressure.

ΔW=P_avg(V_A-V_B)

= 350×10³× (250-100) ×10^-6

=-52.5J

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

Therefore,

ΔU=-294-(-52.5) =-241.5J

∴ change in internal energy is -241.5J.