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# connsider an ideal gas confinned in an isolated closed chamber. As the gas undergoes an adiabatic expansion . the average time of collison between molecules increases as V^q , where  V is the volume of the gas. the value of q ?[ γ = CP / CV ]

Piyush Kumar Behera
436 Points
3 years ago

@subodh this is a previous year JEE main  question.See the solution below

We know the relation that for an adiabatic process the following relation is true :-

$\boxed{P V^{\gamma}=constant}$

$\boxed{P \propto V^{-\gamma}}$

where,

P = Pressure of gas

V = Volume occupied by the gas

$\gamma =$ Specific heat ratio of the gas

Now the mean free path of the gas $\left( \lambda \right)$ is given as :-

$\boxed{\lambda \propto V}$

The average time between the collisions of the gas molecules is nothing but the mean free path divided by the root mean square speed of the gas molecules.

So,

$\implies \boxed{Time = t =\frac{ \lambda }{ v_{RMS}}}$

Now we also know that :-

$\boxed{v_{RMS} = \sqrt{\frac{3RT}{M}}}$

$\boxed{v_{RMS}\propto \sqrt{{RT}}}$

$\boxed{v_{RMS} \propto \sqrt{PV}}$

Using the above we get :-

$\implies t \propto V \times \sqrt{\dfrac{1}{PV}}$

$\implies t \propto \sqrt{\dfrac{V^2}{PV}}$

$\implies t \propto \sqrt{\dfrac{V}{P}}$

Now, using the adiabatic relationship :-

$\implies t \propto \sqrt{\dfrac{V}{V^{-\gamma}}}$

$\implies t \propto V^{\frac{\gamma + 1}{2}}$

and hence,

$\implies \boxed{q = \dfrac{\gamma + 1}{2}}$

Hope it is clear.