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Question

An ideal gas having initial pressure P. volume V and temperature T is allowed to expand adiabatically until its volume becomes 27.914V, while its temperature falls to T3. Choose the correct option(s):

A
Gas molecules have 6 degree of freedom.
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B
Gas molecules are diatomic.
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C
Work done by the gas is 2.02PV.
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D
Workdone by the gas is 1.52PV.
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Solution

The correct option is C Work done by the gas is 2.02PV.
Initial position P,V,T
Final position P2,V2,T2
Given, V2=27.914V,T2=T3
For adiabatic expansion,
TVγ1=constant
T1Vγ11=T2Vγ12
TVγ1=T3(27.914V)γ1
(27.914)γ1=3
γ=1.33
Using γ=1+2F1.33=1+2F
F=6
No. of degrees of freedom =6. So, gas molecules are triatomic.
From Ideal gas equation, P1V1T1=P2V2T2
P2=P1V1T1×T2V2=PVT×T327.914V=P83.742
So, work done (W)=P1V1P2V2γ1=[PVP83.742×27.914V]11.331
=23×PV0.33=2.02PV
So work done by gas during expansion is 2.02PV.
Option (a) and (c) are correct.

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