Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is : (γ=CpCv)
A
3γ+56
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B
3γ−56
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C
γ+12
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D
γ−12
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Solution
The correct option is Cγ+12
Fro an adiabatic process Tvr−1=constant, we know that average time of collision between molecules τ=1nπ√2Vrmsd2 where n= no, of molecules per unit volume Vrms=rms velocity of molecules As n∝1V and Vrms∝√t τ∝V√T Thus we can write: n=k1v−1 and Vrms=K2T1/2 where k1 and k−2 are constant for adiabatic process TVr−1= constant Thus we can write τ∝VT−1/2∝V(v1−r)−1/2 or τ∝Vr+12 Average time between collision =meansfreepathVrms t=1πd2N/V√3RTM;t=CV√T where C=√Mπd2B√3R ⇒T∝V2t2 For adiabatic TVγ−1=k V2t2Vγ−1=k Vγ−1t2=k,t∝γ+12 so q=γ+12