Answer:
The rate of effusion of helium is always about twice that of oxygen.
Explanation:
Recall Graham's law of effusion:
[tex]\displaystyle \frac{r_1}{r_2} = \sqrt{\frac{\mathcal{M}_2}{\mathcal{M}_1}}[/tex]
Where r are the rates of effusion and M are the respective molecular weights.
Let r₁ be the effusion rate of helium and r₂ be the effusion rate of oxygen. Hence, M₁ = 8.00 g/mol and M₂ = 32.00 g/mol.
Substitute:
[tex]\displaystyle \begin{aligned}\frac{r_1}{r_2} & = \sqrt{\frac{(32.00\text{ g/mol})}{(8.00\text{ g/mol})}} \\ \\ & = \sqrt{4.00} \\ \\ & = 2.00 \end{aligned}[/tex]
Solving for r₁ yields:
[tex]\displaystyle \begin{aligned} \frac{r_1}{r_2} & = 2.00\\ \\ r_1 & = 2.00r_2\end{aligned}[/tex]
Hence, the rate of effusion of helium is always about twice that of oxygen. This is expected, as helium is smaller than oxygen.
