# How fast is the rate of effusion of hydrogen compared to that of oxygen?

Dec 3, 2016

"Rate"_text(H₂)/"Rate"_text(O₂) = 3.99

#### Explanation:

Graham's Law of Effusion is:

color(blue)(bar(ul(|color(white)(a/a) "Rate"_1/"Rate"_2 = sqrt(M_2/M_1)color(white)(a/a)|)))" "

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can re-write this to get

$P V = \left(\frac{m}{M}\right) R T = \frac{m R T}{M}$

P = (m/V)((RT)/M) = (ρRT)/M

M = (ρRT)/P

"Rate"_1/"Rate"_2 = sqrt(M_2/M_1) = sqrt((ρ_2(color(red)(cancel(color(black)((RT)/P)))))/(ρ_1color(red)(cancel(color(black)(((RT)/P))))) )=sqrt(ρ_2/ρ_1)

Let component 1 be hydrogen and component 2 be oxygen.

"Rate"_2/"Rate"_1 = sqrt((1.43 color(red)(cancel(color(black)("g/L"))))/(0.0899 color(red)(cancel(color(black)("g/L"))))) = sqrt15.9 = 3.99