# How should the rate of diffusion of helium compare to that of sulfur dioxide?

Jun 28, 2017

#### Explanation:

Graham's law of effusion holds that the relative rate of effusion/diffusion of 2 gases is inversely proportional to the square root of the molecular masses of the two diffusing gases.....

"rate 1"/"rate 2"=sqrt{"Molecular mass of Gas 1"/"Molecular mass of Gas 2"}

$\left(\text{rate" (He))/("rate} \left(S {O}_{2}\right)\right) = \sqrt{\frac{64 \cdot g \cdot m o {l}^{-} 1}{4 \cdot g \cdot m o {l}^{-} 1}}$

$\left(\text{rate" (He))/("rate} \left(S {O}_{2}\right)\right) = \left\{\frac{8}{2}\right\} = 4$

And so helium should diffuse/effuse FOUR TIMES as FAST as sulfur dioxide. What is special about the molecular mass of each gas; i.e. why were these 2 gases quoted for the problem?