Q. 5

# The pressure of gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc.

(a) Calculate the work done by the gas.

(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Answer :

Given

Initial pressure P_{1}=10kPa=10×10^{3}Pa

Final pressure P_{2}=50kPa=50×10^{3}Pa

Initial volume V_{1}=200cc=200×10^{-6}m^{3}

Final volume V_{2}=50cc= 50×10^{-6}m^{3}

a) We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V_{2}-V_{1}

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 = 30×10^{3}× (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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