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Graham's law states that rate of effusion of a gas is inversely proportional to its molar mass which is mathematically written as $\frac{R_A}{R_B}=\sqrt{\frac{M_B}{M_A}}$.

Here we see that the rate depends only on the molar mass. But let us consider two gases $\ce{H2}$ and $\ce{O2}$ in a container. $\ce{O2}$ has a molar mass of $32~\mathrm{g}$ and $\ce{H2}$ is of $2~\mathrm{g}$. So from the above formula, it gives that rate of $\ce{H2}$ will be more,but here is my doubt. Suppose we have only 1 molecule of $\ce{O2}$ in the container and billions of molecules of $\ce{H2}$ in the container. In that case, $\ce{O2}$ will take less time to escape than $\ce{H2}$, but Graham's law says it only depends on molar mass. This is something i don't understand.

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  • $\begingroup$ Could you give the definition of $R_A$ for a given type of moelcule? I'm thinking it will have some involvement of the number density of molecules. $\endgroup$
    – Dan
    Feb 2 at 19:50

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The reason why the actual density of the gasses is not included in Graham's law (leading to the counter-intuitive conclusion in the question) is, that one implicitly assumes, that the effusion experiment confirming Graham's law is conducted with each of the gasses separately, and at the same temperature and pressure.

According to these course notes, the number current density hitting the container walls due to pressure (also called the "impingement current") can be calculated from kinetic gas theory as $$j=\frac{1}{4}\rho_n \bar{v}$$ where $\bar{v}$ is the average speed of the molecules in the gas and $\rho_n$ is the number density of the molecules. This is kind of intuitive because if all the gas molecules moved at the same speed $v$ in the same direction perpendicular to the wall, the current density would be most basically $j=\rho_n v$. The additional factor $1/4$ occuring in the impingement current simply accounts for the fact that the Maxwell-Boltzmann distribution also contains molecules moving at different than average speeds and different than perpendicular directions.

The rate of effusion then can be simply calculated from the cross-sectional area $S$ of the pinhole and the impingement current density by $$R=j S=\frac{1}{4}\rho_n v S$$ Hence, for different gasses with possibly different number densities effusing through the same pinhole area $S$ out of the same container, the ratio of the effusion rates is actually $$\frac{R_A}{R_B}=\frac{\rho_{n,A} v_A}{\rho_{n,B} v_B}$$ Now, if temperature and pressure are assumed the same in both cases, the number densities of the two gasses are also the same, at least for ideal gasses: $$\rho_{n,A}=\frac{p}{k_B T}=\rho_{n,B}$$ Therefore, under the same pressure, same temperature assumption, the ratio of effusion currents simply reduces to the ratio of average molecule velocity: $$\frac{R_A}{R_B}=\frac{v_A}{v_B}$$ Using the fact, that at the same temperature velocity is related to the kinetic energy of the molecules and their molecular mass, leads to Graham's law in the known form: $$\frac{R_A}{R_B}=\sqrt{\frac{M_B}{M_A}}$$ If pressure is not the same (due to different number densities) or if a mixture of gasses with different partial pressures is considered, the more general relation is $$\frac{R_A}{R_B}=\frac{\rho_{n,A}}{\rho_{n,B}}\sqrt{\frac{M_B}{M_A}}$$ If in the same container there are only a few hydrogen molecules in bulk oxygen ($\rho_{n,A}\ll\rho_{n,B}$), then the effusion rate of hydrogen is also much smaller than the effusion rate of oxygen ($R_A\ll R_B$), even though the molecular mass of oxygen is greater than that of hydrogen (and thus, the effusion rate of hydrogen would be greater at the same pressure).

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Actually, $$\frac{R_A}{R_B}=\frac{C_{rms_A}}{C_{rms_B}},$$ where $C_{rms}$ is the root mean square speed. Now, $$C_{rms}=\sqrt{\frac{3RT}{M}}$$ So, the ratio of the rates of dissociation becomes $$\frac{R_A}{R_B}=\sqrt{\frac{T_AM_B}{T_BM_A}}$$ So, the rates don't just depend on the molecular mass. They also depend on temperature. Back to your question. Let's not think about something as small as 1 particle. Say we have 1 mole of $O_2$ in and 100 moles of $H_2$ in a container.

Sure, there may be more moles of Hydrogen than there are of Oxygen. But the important thing is that, at the same temperature, the partial pressure due to the Hydrogen molecules will also be greater than that of the Oxygen molecules, simply because there are many more molecules of $H_2$ than there are of $O_2$, making the former move faster than the latter.

So, the ratio of the rate of diffusion ultimately depends on the molecular mass only(if the temperatures are equal).

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  • $\begingroup$ I don't think that this answers the question. Oxygen is 16 times heavier than hydrogen, resulting in hydrogen molecules being 4 times faster (as per rms velocity) than oxgen molecules at the same temperature. So far, so good. But if there is only a single hydrogen molecule (I think the OP confused the roles) in a container of 22.4 liters otherwise full of oxygen, then why should a lame 4 times as many hydrogen molecules escape the container per second while one would naively expect around a 6*10^23 higher likelihood of an oxygen molecule being close to the pinhole of the container at any time? $\endgroup$
    – oliver
    Feb 2 at 16:38
  • $\begingroup$ In the case of one particle, $C_{rms}$ of $H_2$ will be 4 times greater than that of $O_2$, not the rate of diffusion. This is exactly why I said - "Let's not think about something as small as 1 particle. Say we have 1 mole of O2 and 100 moles of H2 in a container." $\endgroup$ Feb 3 at 5:06

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