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Question

Physics Question on Thermodynamics

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as VqV^q , where VV is the volume of the gas. The value of qq is (γ=CpCv)\bigg(\gamma =\frac{C_p}{C_v}\bigg)

A

3γ+56\frac{3\gamma +5 }{6}

B

γ+12\frac{\gamma +1}{2}

C

3γ56\frac{3 \gamma-5}{6}

D

γ12\frac{\gamma-1}{2}

Answer

γ+12\frac{\gamma +1}{2}

Explanation

Solution

mean free path
λ=12πd2n\lambda=\frac{1}{\sqrt{2}\, \pi d ^{2} n}
n= no. of molecules  volume n =\frac{\text { no. of molecules }}{\text { volume }}
vavg. TT.Vγ1=Cv _{\text {avg. }} \propto \sqrt{ T }\,\,\,\,\,\,\,T .V ^{\gamma-1}= C
t=λVavg. VTvt =\frac{\lambda}{ V _{\text {avg. }}} \propto \frac{ V }{\sqrt{ T }} \,\,\,\,\,\,\,v \rightarrow is volume
VCvr1Vγ+12\frac{ V }{\sqrt{\frac{ C }{ v ^{ r -1}}}} \propto V ^{\frac{\gamma+1}{2}}
vqvγ+12v^{q} \propto v^{\frac{\gamma+1}{2}}
q=γ+12q =\frac{\gamma+1}{2}