Q. 135.0( 1 Vote )

# A gas is taken along the path AB as shown in the figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

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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