0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Who said it is inversely proportional to $\sqrt T$? $\endgroup$ – Ivan Neretin Apr 11 '18 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. $\endgroup$ – AbhigyanC Apr 11 '18 at 11:27
  • 2
    $\begingroup$ @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? $\endgroup$ – Ivan Neretin Apr 11 '18 at 12:03
1
$\begingroup$

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.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.