# Question 24c45

Jul 14, 2017

Here's what I got.

#### Explanation:

The idea here is that oxygen gas will actually take longer to effuse from your container, i.e. oxygen will have a lower rate of effusion, because it has a bigger molar mass than helium.

So right from the start, you know that you must have

"rate O"_2 < "rate He"" "color(orange)("(*)")

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

$\text{rate} \propto \frac{1}{\sqrt{{M}_{M}}}$

${M}_{\text{M He" ~~ "4.0 g mol}}^{- 1}$

M_ ("M O"_2) ~~ "32.0 g mol"^(-1)

Now, both gases are kept under the same conditions for pressure and temperature and they must effuse through the same hole, so you can say that their respective rates of effusion will satisfy the equation

"rate He"/"rate O"_ 2 = (1/sqrt(M_ "M He"))/(1/sqrt(M_ ("M O"_ 2))) = sqrt(M_ ("M O"_ 2))/sqrt(M_ "M He")

This will be equivalent to

"rate He"/"rate O"_ 2 = sqrt(32.0 color(red)(cancel(color(black)("g mol"^(-1)))))/sqrt(4.0color(red)(cancel(color(black)("g mol"^(-1)))))

which gets you

${\text{rate He"/"rate O}}_{2} = \sqrt{\frac{32.0}{4.0}} = \sqrt{8} = 2 \sqrt{2}$

This means that you have

$\frac{\text{rate O"_ 2= "rate He}}{2 \sqrt{2}}$

$\text{rate O"_ 2 = sqrt(2)/4 * "rate He } \to$ matches what you know from $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$

Now, if you take $x$ ${\text{dm}}^{3}$ to be the amount of helium that effuses from the container in $5$ seconds, you can say that

$\text{rate He" = (x color(white)(.)"dm"^3)/"5 s} = \left(\frac{x}{5}\right)$ ${\text{dm}}^{3}$ ${\text{s}}^{- 1}$

Plug this into the equation to find

${\text{rate O}}_{2} = \frac{\sqrt{2}}{4} \cdot \left(\frac{x}{5}\right)$ ${\text{dm}}^{3}$ ${\text{s}}^{- 1}$

${\text{rate O"_2 = (sqrt(2)/20 * x)color(white)(.)"dm"^3 color(white)(.)"s}}^{- 1}$

This, of course, means that $x$ ${\text{dm}}^{3}$ of oxygen gas will effuse from the container in

$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{x}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{dm"^3))) * "1 s"/((sqrt(2)/20 * color(blue)(cancel(color(black)(x)))) * color(red)(cancel(color(black)("dm"^3)))) = 20/sqrt(2)color(white)(.)"s" = 10sqrt(2) color(white)(.)"s" ~~ color(darkgreen)(ul(color(black)("14 s}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for your values.

If you assume that it takes $5$ seconds for all the helium present in the container to effuse, i.e. for $10$ ${\text{dm}}^{3}$ of helium, you will have

${\text{rate He" = "10 dm"^3/"5 s" = "2 dm}}^{3}$ ${\text{s}}^{- 1}$

This will get you

${\text{rate O"_2 = sqrt(2)/4 * "2 dm"^3 color(white)(.)"s"^(-1) = sqrt(2)/2color(white)(.)"dm"^3color(white)(.)"s}}^{- 1}$

Once again, the time needed for $10$ ${\text{dm}}^{3}$ of oxygen gas to effuse will be

 10 color(red)(cancel(color(black)("dm"^3))) * "1 s"/(sqrt(2)/2color(red)(cancel(color(black)("dm"^3)))) = 20/sqrt(2)color(white)(.)"s" = 10sqrt(2) color(white)(.)"s" ~~ "14 s"#