Q70 / 90 · JEE Main 2015 (04 Apr) · Physics

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

  • A
    γ-12
  • B
    3 γ + 5 6
  • C
    3γ-56
  • Correct answer
    D
    γ+12

Solution

For an adiabatic process TVγ−1= 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 n∝1V and vrms∝T

τ∝VT

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

τ∝VT-12∝VV1-γ-12

or τ∝Vγ+12

Hence, the correct answer is option: 'D'.

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