Question

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

Answer 1

Helium gas should effuse four times faster than sulfur dioxide.

Explanation:

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

Thus,

`"Rate of diffusion of helium"/" Rate of diffusion of sulfur dioxide"`

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

`prop (sqrt(64*g*mol^-1))/sqrt(4*g*mol^-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.