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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, TVγ−1=constant ⇒T1Vγ−11=T2Vγ−12 ⇒TVγ−1=T3(27.914V)γ−1 ⇒(27.914)γ−1=3 ⇒γ=1.33 Using γ=1+2F⇒1.33=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)=P1V1−P2V2γ−1=[PV−P83.742×27.914V]11.33−1 =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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