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Grade 11Thermal Physics

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 ]

Profile image of subodh
8 Years agoGrade 11
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1 Answer

Profile image of Piyush Kumar Behera
8 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.