Download the app

Questions  

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from τ1 to τ2 . If CPCV=γ for this gas then a good estimate for τ1τ2  is given by:

a
12
b
2
c
12γ
d
12γ+12

detailed solution

Correct option is D

Mean free time =λVrms=RT2πd2NAP3RTMt∝TP∝TRTV∝VTT∝VT∴t∝V1Vγ−1∝V1Vγ-12∝VVγ-12∝V1+γ-12∝V2+γ-12Mean Collision time , t∝VT                         ..........(1)For adiabatic process, TVγ−1=constant       .................(2)⇒t∝Vγ+12 ⇒t2t1=V2V1γ+12⇒t2t1=21γ+12 ⇒t1t2=12γ+12

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure P1=105Pa and volume Vi=103m3  changes to final state at Pf=132×105pa&Vf=8×103m3in an adiabatic quasi-static process, such that P3V5= constant. Consider another thermodynamic process that brings the system form the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (is volumetric) process at volumeVf. The amount of heat supplied to the system in the two-step process is approximately


phone icon
whats app icon