How does the rate of effusion of sulfur dioxide, $S O_{2}$ $SO_2$ , compare to that of helium?

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How does the rate of effusion of sulfur dioxide, $S O_{2}$ $SO_2$ , compare to that of helium?

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Answer 1 · 2016-04-14T21:30:36

Helium gas should effuse four times faster than sulfur dioxide.

Explanation:

$Rate of diffusion \propto \frac{1}{\sqrt{Molecular mass}}$ $"Rate of diffusion "prop" " 1/sqrt("Molecular mass")$

Thus,

$\frac{Rate of diffusion of helium}{Rate of diffusion of sulfur dioxide}$ $"Rate of diffusion of helium"/" Rate of diffusion of sulfur dioxide"$

$\propto \frac{\sqrt{Molecular Mass S O_{2}}}{\sqrt{Molecular mass H e}}$ $prop" " (sqrt("Molecular Mass " SO_2))/sqrt("Molecular mass "He)$

$\propto \frac{\sqrt{64 \cdot g \cdot m o l^{- 1}}}{\sqrt{4 \cdot g \cdot m o l^{- 1}}}$ $prop (sqrt(64*g*mol^-1))/sqrt(4*g*mol^-1)$

Conveniently, the molecular masses are perfect squares. And thus helium should effuse 4 times faster than sulfur dioxide.

This law is a manifestation of the ideal gas law. In fact, fissile (and volatile) $^{235} U F_{6}$ $""^235UF_6$ is separated from $^{238} U F_{6}$ $""^238UF_6$ with precisely this principle.

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How does the rate of effusion of sulfur dioxide, $S O_{2}$ $SO_2$ , compare to that of helium?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How does the rate of effusion of sulfur dioxide, SO2SO_2, compare to that of helium?

1 Answer

Explanation:

Related questions

Impact of this question

How does the rate of effusion of sulfur dioxide, $S O_{2}$ $SO_2$ , compare to that of helium?