Discussion Forum

 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 V^q, where V  is the Volume of the gas. The value of q is $\left(\gamma=\frac{C_{p}}{C_{V}}\right)$

Answer:

Explanation:

Central Idea  For an adiabtic  process $TV^{\gamma-1}$  = constant

 We know that average time of collision between molecules

  $\tau=\frac{1}{n\pi \sqrt{2}V_{rms}d^{2}}$

 where  , n= number of molecules per unit volume

     $V_{rms}$  = rms velocity of molecules

   As    $n\propto\frac{1}{V}$    and    $V_{rms}\propto \sqrt{T}$

     $\tau \propto \frac{V}{\sqrt{T}}$

  Thus, we can write  $n=K_{1}V^{-1}$      and   $V_{rms}=K_{2}T^{1/2}$

 where K₁    and K₂    are constants

   For adiabatic process,    $TV^{\gamma-1}$  = constant

 Thus , we can write

       $\tau \propto  VT^{-1/2}\propto V(V^{1-\gamma})^{-1/2}$

        or   $\tau \propto V^{\frac{\gamma+1}{2}}$