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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.

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  • $\begingroup$ I absolutely hated this topic. Anyway can you identify which equation you need to use? Once you've done that, it's just a matter of making sure your units are consistent. $\endgroup$ – orthocresol Oct 6 '15 at 21:50
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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

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

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  • $\begingroup$ Thanks alot, this made sense mainly, the only part from this I didnt understand is, i have the same equation as you provided, but I didnt know that it could be multipled by A (area). Any reason why? $\endgroup$ – Mathquestion-eer Oct 8 '15 at 1:36
  • $\begingroup$ @Mathquestion-eer I can only speculate on differences. Maybe, the result $Z_\mathrm w$ in your equation is defined as ‘rate per area’. You can always check your calculation with the help of the units. When you use coherent units, conversion factors between units are never required ($1\ \mathrm J=1\ \mathrm{kg\ m^2\ s^{-2}}$). If the units cancel out so that the final result has the correct unit, this is a nice indication, but not a proof, that your calculation is correct. Note that the number $N$ is a quantity of dimension one (a quantity of dimension one is often called dimensionless). $\endgroup$ – Loong Oct 8 '15 at 9:39

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