Ideal Gas in Piston (Constant Pressure vs Rate of collision)

[Stunner]

A syringe contains ideal gas. The piston is frictionless and no gas escapes. Once heated slowly, the piston moves outwards. The piston stops moving when the temperature is steady. The pressure of the gas after the piston stops moving remains the same (no change in pressure). What can we say about the rate of collision between the gas molecules and piston?
1) Ok we know that the average speed of the molecules increases (hence increased KE).
2) It is said that the rate of collision “decreases”.
Reason: KE increases. If frequency of collision remains the same (or increases), the pressure would increase. But pressure remains constant, so the frequency of collision has to be lower.

This is a very logical approach to answering the problem. But is there a mathematical derivation to show how this is so? The closest I can find is P=(Nmv^2)/(3V) where v is the average speed per molecule and V is the volume. N is the number of molecules and m is the mass.
Is there a formula showing how rate of collision is linked to an increased temperature (given increased volume and constant pressure).

Any help would be greatly appreciated. Also let me know if there’s anyone I can ask. Thank you so much in advance!

 

[Philip Wood]

We can show that the number, f, of collisions per unit wall area, per unit time is given by $$f\:=\:\frac{1}{4}\frac{N}{V}\overline{v}$$
in which $\overline{v}$ is the mean speed of the molecules.

It can be shown from the Maxwell distribution of molecular speeds that $\overline{v}=0.921v_{rms}$.

So $f\:=\:\frac{1}{4}0.921\frac{N}{V}v_{rms}$.

But $pV=\frac{1}{3}Nmv_{rms}^2$

Eliminating $v_{rms}$ we obtain
$$f\:=\:\frac{1}{4}0.921\sqrt{\frac{3Np}{mV}}$$
and, finally, eliminating V using pV = NkT

We have $f\:=\:\frac{1}{4}0.921\sqrt{\frac{3p^2}{mRT}}$

So we have established that for constant p, f is proportional to $\frac{1}{\sqrt T}$

 

[Philip Wood]

Stunner: Is this what you wanted?