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According to Graham's diffusion law, rate of diffusion is directly proportional to square root of temperature but rate of effusion is inversely proportional to square root of temperature.

Why is this so? I would assume that rate of effusion would also increase with increase in temperature due to increased speed of the gas.

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  • $\begingroup$ Who said it is inversely proportional to $\sqrt T$? $\endgroup$ Apr 11, 2018 at 10:19
  • $\begingroup$ @IvanNeretin Actually, it is inversely proportional to $\sqrt{T}$. See here: en.m.wikipedia.org/wiki/Effusion under measures of flow rate. $\endgroup$
    – Abhigyan
    Apr 11, 2018 at 11:27
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    $\begingroup$ @AbhigyanC That depends on your point of view. Say, you measured it in a system of two connected constant-volume vessels. Then you increase T. Pressure in both vessels will increase proportionally to T, and so will $\Delta P$. Also, the denominator will grow as $\sqrt T$. Guess what will happen to the overall expression? $\endgroup$ Apr 11, 2018 at 12:03

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The self diffusion coefficient in a gas is given by $\displaystyle D=\frac{\lambda}{2}\bar c$ and the rate of effusion by $\displaystyle \frac{\rho}{4}\bar c$ where $\bar c$ is the average velocity of the gas molecules of mass $m$, and $\displaystyle \bar c= \sqrt{\frac{8k_BT}{\pi m}} $, $\lambda$ the mean free path and $\rho$ the gas density. The diffusion constant is $\displaystyle D=\frac{1}{\pi \sigma^2 n}\sqrt{\frac{k_BT}{\pi m}}$ where $\sigma$ is the molecule's diameter, and the rate of effusion $\displaystyle \rho \sqrt{\frac{k_BT}{2\pi m}}$. The diffusion constant can also be written as $\displaystyle D=\frac{m}{\pi \sigma^2 \rho }\sqrt{\frac{k_BT}{\pi m}}$ thus the dependence on temperature is the same for diffusion and effusion if the gas density is held constant.

[ The rate of effusion is calculated by considering the gas bouncing around in a container and then suddenly opening a tiny trap door of area $dA$ in the wall so that some minute proportion of the gas streams out in time $dt$. The trap door diameter should be no more than 1/10 of the mean free path. If each molecule travels with average speed $\bar c$ the number leaving would be $n\bar c dA dt$ if there are $n$ molecules /unit volume. The efflux in terms of the mass/unit area/unit time is then $mn\bar c =\rho \bar c$.

However, half the molecules are moving away from the trap door at this instant so the rate is divided by 2. Additionally molecules come out at different angles so it is necessary to integrate over the volume element $\sin(\theta)d\theta d\phi$ and this leads to a further factor of 1/2.]

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I just have a simple answer for this. Rate of effusion is the rate at which a gas is allowed to escape from a given cross sectional area(orifice) due to pressure difference.

While diffusion is the tendency of gases to mix with each other spontaneously and form a homogeneous mixture.

In case of diffusion, increasing temperature increases the rate of diffusion because it increases the kinetic energy of molecules which increases the mixing ability of the gases and thus, diffusion.

In effusion the same thing happens, kinetic energy of molecules increases, which increases the speed of the molecules. But Increase in speed of the molecules decreases the probability of the molecules to escape through the given orifice/hole in a given time.

Therefore, rate of effusion decreases with increase in temperature.

Thanks:)

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