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Kinetic theory of ideal gases

Question

An adiabatic vessel contains  n1 moles of diatomic gas. Moment of inertia of each  molecule is I=2.76×1046kgm2 and root mean square angular velocity is ω0=5×1012rads1 .  Another adiabatic vessel contains   n2= 5 moles of a monoatomic gas at a  temperature 470K. Assume gases to be ideal, calculate root-mean square  angular velocity of diatomic molecules (in multiple of 1012rads1) when the two  vessels are connected by a thin tube of negligible volume.

Very difficult
Solution

By energy conservation,
f1n1+f2n2Tf=f1n1T1+f2n2T2      (1)
Here, f1=5,  n1=3,T1=250K  and   
f2=3,  n2=5,T2=470K 
Substituting in (1),
Tf=5×3×250+3×5×4705×3+3×5=360K 
If RMS angular velocity of diatomic gas molecules is  ωRMSf, then according to law  of equipartition of energy,
12IωRMSf2=kT 
Or,
ωRMSf=2kTI=2×1.38×1023×3602.76×1046=6×1012rad/s.
 



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