If oxygen, O_2, effuses (spreads out) at a rate of 622.0 ms^-1 under certain conditions of temperature and pressure, how fast will methane, CH_4, effuse under the same conditions?

Feb 17, 2016

We know the mass of an ${O}_{2}$ molecule is $32$ $g m o {l}^{-} 1$ and the mass of a $C {H}_{4}$ molecule is $16$ $g m o {l}^{-} 1$, so Graham's Law of Effusion shows that the rate of effusion will be $879.6$ $m {s}^{-} 1$.

Explanation:

This question requires Graham's Law of Effusion, which states:

${\text{Rate"_1/"Rate}}_{2} = \sqrt{{M}_{2} / {M}_{1}}$ where ${M}_{1}$ and ${M}_{2}$ are the molar masses of the effusing gases.

To make the math easier, let's call the $C {H}_{4}$ rate ${\text{Rate}}_{1}$. (Which we call which doesn't matter, as long as we're consistent: the math will sort itself out!)

"Rate"_1=?
${M}_{1} = 16$ $g m o {l}^{-} 1$
${\text{Rate}}_{2} = 622.0$ $m {s}^{-} 1$
${M}_{2} = 32$ $g m o {l}^{-} 1$

Rearranging:

${\text{Rate"_1 = sqrt(M_2/M_1)*"Rate}}_{2}$
$= \sqrt{\frac{32}{16}} \cdot 622.0 = 879.6$ $m {s}^{-} 1$