We were studying Graham's Law today and its dependence on Temperature, Pressure, Area, etc. However, our teacher said that, for diffusion: $$R \propto \sqrt T$$ and for effusion:$$R\propto1/\sqrt T$$ giving the reason that during effusion, transfer of gases takes place through a very small hole, and so, increasing temperature will cause increased vibration of those gases, thus making them harder to pass through.

However, for diffusion, he said, there is no hole, and so an increase in temperature will cause increased diffusion.

I've tried finding answers for this here, but all I got was that even in effusion, the formula is: $$r= \frac{PA}{\sqrt{2RTM\pi}}$$ and using formula $PV=nRT$, you can get that $r\propto \sqrt T$ even for effusion.

Can anybody clear this up for me?

Edit: I want to know whose explanation is correct and why, and also what is the (not too complicated) physical reason for this behaviour.

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    $\begingroup$ Does this answer your question? What is the relation between absolute temperature, and rates of diffusion and effusion of a gas? $\endgroup$
    – Tyberius
    Nov 25 at 17:03
  • $\begingroup$ Yeah, I'd seen that answer as well, but it didnt really answer my question. Primarily because whatever he said is way too advanced chemistry for me, and also I dont think what he said in his answer is for an ideal gas. $\endgroup$ Nov 25 at 17:42
  • $\begingroup$ Also, if $ R\propto \sqrt T$ for both diffusion and effusion, then how is that possible? I mean, whatever our teacher said to us made intuitive sense to me, so if someone could explain that as well, it would be helpful. $\endgroup$ Nov 25 at 17:46
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    $\begingroup$ What exactly is your question - are you looking for a discussion of whether your teacher's explanation is correct, or an entirely different physical rationalisation of why the dependence is of that form, or a mathematical proof of the dependence, or maybe a proof of the full formulae? As it stands, you wrote a number of statements, but it's not immediately obvious what about them you want "cleared up". $\endgroup$
    – orthocresol
    Nov 26 at 0:38
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    $\begingroup$ For both an explanation of my teacher's reasoning, and a (not too complicated) physical rationalisation of why the dependence is of that form, as you said. $\endgroup$ Nov 26 at 2:18

I think part of the confusion comes from assuming that these temperature dependences are fixed properties that remain true under any conditions. To explain what I mean, let's start from porphyrin's equations for rate of diffusion/effusion for ideal gases from the linked answer.

$$r_{\text{diff}}\propto\frac{1}{\rho}\sqrt{T}$$ $$r_{\text{eff}}\propto\rho\sqrt{T}$$

If we assume $\rho$ (the density) remains fixed while we change the temperature, then diffusion and effusion both have a $\sqrt{T}$ dependence. This makes sense, as various measures of the "typical" speed of the gas particles are proportional to $\sqrt{T}$ and an increase in speed should increase diffusion and effusion.

However, if we allow the density to change with the temperature, we will get a different temperature dependence for these two processes, since $\rho=\frac{P}{RT}$ for an ideal gas. So if we keep $P$ fixed, $r_{\text{diff}}$ would now have a $T^{3/2}$ dependence while $r_{\text{eff}}$ would have a $T^{-1/2}$ dependence.

Intuitively, the dependence of these properties on the density also makes sense. As we make the gas less dense by heating it, the particles have more space to move around before colliding, allowing them to diffuse faster. For effusion, we have essentially the opposite effect, where either increasing the volume or decreasing the number of gas particles decreases the opportunities to escape.

The overarching point is that the temperature dependence is related to the conditions you conduct your experiment with. The choice to keep density/pressure constant or not will change the overall temperature dependence. More generally, whether the system is a gas/liquid/solid will also affect the dependence, along with how sophisticated a model we use to describe these systems.

For your teachers explanation, I could see why they might arrive at the dependences they emphasized (admittedly I'm speculating a bit here). For diffusion, it may make sense to consider the constant density dependence in a situation where all the possible places the gas can go are included as your system (i.e. you aren't considering particles "escaping" via diffusion). Even if you started with a sample where the density was very concentrated, it should become more or less constant once it has spread to fill it's container. For effusion, it makes more sense to look at nonconstant density, as particles are escaping from the system through effusion.

However, I don't think the explanation in terms of vibrations makes sense. If the gas particles are single atoms, they wouldn't have any vibrational modes. Even if they were molecules, the ideal gas law treats them as point particles that don't occupy any volume, so at least within this model, we can't use vibrations to explain the temperature dependence (vibrational effects could play a role in real gas model of diffusion/effusion).


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