What is the relation between absolute temperature, and rates of diffusion and effusion of a gas?
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.]