# How does the rate of effusion of sulfur dioxide, SO_2, compare to that of helium?

Apr 14, 2016

Helium gas should effuse four times faster than sulfur dioxide.

#### Explanation:

"Rate of diffusion "prop" " 1/sqrt("Molecular mass")

Thus,

$\text{Rate of diffusion of helium"/" Rate of diffusion of sulfur dioxide}$

prop" " (sqrt("Molecular Mass " SO_2))/sqrt("Molecular mass "He)

$\propto \frac{\sqrt{64 \cdot g \cdot m o {l}^{-} 1}}{\sqrt{4 \cdot g \cdot m o {l}^{-} 1}}$

Conveniently, the molecular masses are perfect squares. And thus helium should effuse 4 times faster than sulfur dioxide.

This law is a manifestation of the ideal gas law. In fact, fissile (and volatile) ""^235UF_6 is separated from ""^238UF_6 with precisely this principle.