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

1 Answer
Dec 3, 2016

Answer:

#"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

#color(blue)(bar(ul(|color(white)(a/a) PV = nRT color(white)(a/a)|)))" "#

We can re-write this to get

#PV = (m/M)RT= (mRT)/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#