# <span lang="EN-US

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

VA=250cc=250×10-6m3

VB=100cc=100×10-6m3

PA=200kPa=200×103Pa

PB=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=Pavg(VA-VB)

= 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.

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