If oxygen, $O_{2}$ $O_2$ , effuses (spreads out) at a rate of $622.0$ $622.0$ $m s^{- 1}$ $ms^-1$ under certain conditions of temperature and pressure, how fast will methane, $C H_{4}$ $CH_4$ , effuse under the same conditions?

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If oxygen, $O_{2}$ $O_2$ , effuses (spreads out) at a rate of $622.0$ $622.0$ $m s^{- 1}$ $ms^-1$ under certain conditions of temperature and pressure, how fast will methane, $C H_{4}$ $CH_4$ , effuse under the same conditions?

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Answer 1 · 2016-02-17T05:43:25

We know the mass of an $O_{2}$ $O_2$ molecule is $32$ $32$ $g m o l^{- 1}$ $gmol^-1$ and the mass of a $C H_{4}$ $CH_4$ molecule is $16$ $16$ $g m o l^{- 1}$ $gmol^-1$ , so Graham's Law of Effusion shows that the rate of effusion will be $879.6$ $879.6$ $m s^{- 1}$ $ms^-1$ .

Explanation:

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

$\frac{{Rate}_{1}}{{Rate}_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ $"Rate"_1/"Rate"_2 = sqrt(M_2/M_1)$ where $M_{1}$ $M_1$ and $M_{2}$ $M_2$ are the molar masses of the effusing gases.

To make the math easier, let's call the $C H_{4}$ $CH_4$ rate ${Rate}_{1}$ $"Rate"_1$ . (Which we call which doesn't matter, as long as we're consistent: the math will sort itself out!)

${Rate}_{1} = ?$ $"Rate"_1=?$
$M_{1} = 16$ $M_1=16$ $g m o l^{- 1}$ $gmol^-1$
${Rate}_{2} = 622.0$ $"Rate"_2=622.0$ $m s^{- 1}$ $ms^-1$
$M_{2} = 32$ $M_2=32$ $g m o l^{- 1}$ $gmol^-1$

Rearranging:

${Rate}_{1} = \sqrt{\frac{M_{2}}{M_{1}}} \cdot {Rate}_{2}$ $"Rate"_1 = sqrt(M_2/M_1)*"Rate"_2$
$= \sqrt{\frac{32}{16}} \cdot 622.0 = 879.6$ $=sqrt(32/16)*622.0=879.6$ $m s^{- 1}$ $ms^-1$

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If oxygen, $O_{2}$ $O_2$ , effuses (spreads out) at a rate of $622.0$ $622.0$ $m s^{- 1}$ $ms^-1$ under certain conditions of temperature and pressure, how fast will methane, $C H_{4}$ $CH_4$ , effuse under the same conditions?

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Explanation:

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If oxygen, O2O_2, effuses (spreads out) at a rate of 622.0622.0 ms−1ms^-1 under certain conditions of temperature and pressure, how fast will methane, CH4CH_4, effuse under the same conditions?

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Explanation:

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If oxygen, $O_{2}$ $O_2$ , effuses (spreads out) at a rate of $622.0$ $622.0$ $m s^{- 1}$ $ms^-1$ under certain conditions of temperature and pressure, how fast will methane, $C H_{4}$ $CH_4$ , effuse under the same conditions?