Q. 134.5( 2 Votes )
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Answer :
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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