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

1 Answer
Apr 14, 2016

Answer:

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.