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Q.

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
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