Question

The two gases, H2 and O2, were allowed to effuse under the same conditions through a pinhole. If the rate of effusion of O2 was 4.80 x 10-2 ms-1, what is the rate of effusion of H2?​

Answer 1

The rate of effusion of H₂ : 7.2 x 10⁻² m/s

Further explanation

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

`\rm (r_1)/(r_2)=\sqrt{(M_2)/(M_1) }`

or

`\rm M_1* r_1^2=M_2* r_2^2`

MW₁ O₂ = 32 g/mol

MW₂ H₂ = 2 g/mol

`\tt (1.8* 10^(-2))/(r_2)=\sqrt{(2)/(32) }\\\\((1.8* 10^(-2))^2)/(r_2^2)=(2)/(32)\\\\r_2=0.072=7.2* 10^(-2)~m/s`