Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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 = C P / C V ]

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 ]

Grade:11

1 Answers

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.
 

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free