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How does the rate of effusion of sulfur dioxide, #SO_2#, compare to that of helium?

Answer 1

Caleb Adams

Helium gas should effuse four times faster than sulfur dioxide.

#"Rate of diffusion "prop" " 1/sqrt("Molecular mass")#

Consequently,

#"Rate of diffusion of helium"/" Rate of diffusion of sulfur dioxide"#

#prop" " (sqrt("Molecular Mass " SO_2))/sqrt("Molecular mass "He)#

#prop (sqrt(64*g*mol^-1))/sqrt(4*g*mol^-1)#

Since the molecular masses are conveniently perfect squares, helium should erupt four times more quickly than sulfur dioxide.

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

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Answer 2

Owen Ambrose

The rate of effusion of sulfur dioxide (SO2) is slower than that of helium.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.