Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between…
JEE Main 2015 — Physics Thermodynamics
2015mcqhard
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:
(γ=CvCP)
Official previous-year question
Held on 4 Apr 2015 · Verified 6 Jul 2026.
Options
A
2γ−1
B
63γ+5
C
63γ−5
D
2γ+1
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Solution
For an adiabatic process TVγ−1= constant.
We know that average time of collision between molecules
τ=nπ2vrmsd21
where, n= number of molecules per unit volume Vrms= rms velocity of molecules
As n∝V1 and vrms∝T
τ∝TV
Thus, we can write
n=K1V−1 and Vrms=K2T21.
where, K1 and K2 are constants.
For adiabatic process, TVγ−1= constant. Thus, we can write
τ∝VT−21∝V(V1−γ)2−1
or τ∝V2γ+1
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