How many atoms of helium are in the chamber?

Question

A $22.7 \,\mathrm{L}$ helium-filled chamber at $298\,\mathrm{K}$ has a tiny hole of area $1.63 \times 10^{-5}\,\mathrm{mm^2}$ punched in it through which is leaking helium at a rate of $2.25 \times 10^{15}$ atoms per second. How many atoms of helium are in the chamber?

I'm personally not too sure how to answer this, I'm thinking that it probably uses the root-mean-square speed or collisions with the wall equation ($Z_\mathrm w$). But I'm sort of lost when it comes to substituting the values and so on. (This is also a question on last years Gas Laws test, and I was planning on solving it for practice questions)

What I tried doing was using the Collisions with the wall equation: https://en.wikipedia.org/wiki/Kinetic_theory#Collisions_with_container then solved for $N$, but the answer is supposed to be $N = 9.98 \times 10^{21}$, and the closest I've gotten was getting $3.38 \times 10^{22}$, so I'm sort of stuck on the procedure of solving this question.

Answer 1

You already mentioned the equation for the rate of collisions of gas molecules with a section of wall:

$$Z_\mathrm w=\frac14\frac NV\sqrt{\frac{8RT}{\pi M}}A$$

We know the values of
rate $Z_\mathrm w=2.25\times10^{15}\ \mathrm{s^{-1}}$,
volume $V=22.7\ \mathrm l=0.0227\ \mathrm{m^3}$,
molar gas constant $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$,
temperature $T=298\ \mathrm K$,
molar mass of helium $M=4.003\ \mathrm{g\ mol^{-1}}=0.004003\ \mathrm{kg\ mol^{-1}}$, and
area $A=1.63\times10^{-5}\ \mathrm{mm^2}=1.63\times10^{-11}\ \mathrm{m^2}$.

Thus, the number $N$ of helium atoms is

$$\begin{align} N&=\frac{4Z_\mathrm wV}{\sqrt{\dfrac{8RT}{\pi M}}A}\\[6pt] &=\frac{4\times2.25\times10^{15}\ \mathrm{s^{-1}}\times 0.0227\ \mathrm{m^3}}{\sqrt{\dfrac{8\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm K}{\pi \times 0.004003\ \mathrm{kg\ mol^{-1}}}}\times1.63\times10^{-11}\ \mathrm{m^2}}\\[6pt] &=9.98\times10^{21} \end{align}$$