Q. 5

# <span lang="EN-US

Given

Initial pressure P1=10kPa=10×103Pa

Final pressure P2=50kPa=50×103Pa

Initial volume V1=200cc=200×10-6m3

Final volume V2=50cc= 50×10-6m3

a) We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V2-V1

Pressure =P

Thus, work done by the gas

ΔW=PΔV

Here we have given two values of pressure. So, we will take the average value of pressure

Average pressure P

Therefore, ΔW = 30×103× (50-200)×10-6

ΔW= -4.5J

Work done by the gas is -4.5J.

b) Given that no heat is supplied or extracted from the gas.

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

Since ΔQ=0

Therefore ΔU = -ΔW = -(-4.5) J=4.5J

the change in internal energy of the gas is 4.5J.

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