Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion,…

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=CPCv

  1. γ-12
  2. 3 γ + 5 6
  3. 3γ-56
  4. γ+12

Solution

For an adiabatic process TVγ1= constant.
We know that average time of collision between molecules

τ=1nπ2vrmsd2

where, n= number of molecules per unit volume Vrms = rms velocity of molecules

As n1V and vrmsT

τVT

Thus, we can write

n=K1V-1 and Vrms=K2T12.

where, K1 and K2 are constants.

For adiabatic process, TVγ-1= constant. Thus, we can write

τVT-12VV1-γ-12

or τVγ+12

Asked in: JEE Main 2015 (04 Apr)

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