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 Vq, where V is the volume of the gas, the value of ‘q’ is
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a
γ+12
b
γ-12
c
1-γ2
d
γ+22
answer is A.
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Detailed Solution
for an adiabatic process temperature T, volume V relation is TVγ-1=constant ⇒time of collision t =1nπ2Vrmsd2----(1) n =number of molecules per unit volume Vrms=root mean square velocity d=diameter of molecule n = No. of moleculesvolume ⇒n α 1volume ----(2) ⇒Vrms =3RTM ⇒Vrms α Temperature ----(3) subtitute in equations (2) and (3) in (1) we get ⇒ t =1NVπ23RTMd2=VMNπ23RTd2 ⇒t α volumeTemperature ⇒t αVT-12 ⇒we know adiabatic relation TVγ-1=constant⇒T=V1-γ ;substitute in above equation of t ⇒τ α V(V1-γ)-12 τ α Vγ+12 ⇒q=γ+12