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Let's start with a question:

"One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm takes 57 s to diffuse through the same hole. Calculate the molecular formula of the compound."

It is a question where pressures of the gases are not similar. Could be solved quite easily, if we calculate the rates of diffusion ($r_{N_2}$ and $r_{un}$) of each gas separately as no. of moles of the respective gas diffused per unit time, find the ratio of the rates and then equate this ratio, with another ratio found using pressures and molar masses of the given gases, i.e, with $$\frac{r_{N_2}}{r_{un}}= \frac{p_1}{p_2}\sqrt{\frac{M_{un}}{M_{N_2}}}$$, ($M_{un}$ and $M_{N_2}$ are molar masses of the unknown gas and nitrogen gas respectively) and finally find the no. of fluorine atoms attached with xenon from the value of molar mass of the unknown gas.

It's fine that way. But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures. Consequently, the first ratio (the ratio involving volume and time) would have been different compared to the ratio involving moles and time. And that would have affected the final value of molar mass and thus messed up our solution.
So my question is, what would be the actual cause of such a situation? And how to fare in those situations?

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  • $\begingroup$ If you write the rate of diffusion as number of moles divided by time then multiply both sides by p2 and divide by p1. Now on the left side of equation, multiply divide by RT. Now use gas laws. $\endgroup$ – jatin Mar 20 '16 at 8:10
  • $\begingroup$ You finally get that volume rate of diffusion is not proportional to pressure. I still don't know why :/ $\endgroup$ – jatin Mar 20 '16 at 8:14
  • $\begingroup$ @jatin- Could you write the steps for me? i am getting an equation with squares on the volume variables... $\endgroup$ – Phill2 Mar 20 '16 at 14:44
  • $\begingroup$ Just write the no. of moles you've written in the equation as P1*V1/RT. $\endgroup$ – jatin Mar 20 '16 at 16:23
  • $\begingroup$ Yeah, P*V is more important than just V. I get it.. $\endgroup$ – Phill2 Mar 20 '16 at 16:33
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"But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures."

If you wanted to measure the effused gas in terms of volume, you had to somehow calculate this volume from the number of moles effused. This volume has nothing to do with the volume or the pressure inside the vessel from which the gas effused.

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From the number of particles that escaped to the right, you could calculate a volume, given an arbitrary pressure (e.g. standard pressure). With this you could express the rate of diffusion in units of volume per time unit.

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  • $\begingroup$ Are you saying no. of moles of a gas effused or diffused could be treated as volumes themselves and in any case, equal moles of two gases correspond to equal volumes irrespective of pressures? Please elaborate and state reasons. $\endgroup$ – Phill2 Mar 20 '16 at 15:17
  • $\begingroup$ After your answer I have begun to doubt my conceptual depth on 'pressures'. $\endgroup$ – Phill2 Mar 20 '16 at 15:21
  • $\begingroup$ The rate of effusion is the number of particles that escape through the orifice per time unit. If you wanted to measure the escaped quantity as a volume you had to calculate this volume from the number of particles escaped. Hereby the pressure inside the vessel from which the gas effuses through the hole is considered to be constant, and the volume of the vessel is considered large enough to keep pressure constant. $\endgroup$ – aventurin Mar 20 '16 at 15:37
  • $\begingroup$ So an answer to how i would fare in such situations would be that i would multiply the volumes with their respective pressures because that would give us P*V which remains same at constant temperatures and is only dependent on no. of moles? $\endgroup$ – Phill2 Mar 20 '16 at 16:08
  • $\begingroup$ Isn't that what you mean by "calculate volume from the number of moles"? $\endgroup$ – Phill2 Mar 20 '16 at 16:14

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