# Question 7157d

Apr 29, 2016

Here's what I got.

#### Explanation:

The idea here is that you need to use Graham's Law of Effusion to determine how the rates of effusion of tho gases relate to each other.

This will allow you to estimate how many moles of one gas effused out of the vessel relative to the number of moles of the other gas that effused out of the vessel.

So, according to Graham's Law, the rate at which a gases effuses is inversely proportional to the square root of its molar mass.

color(blue)(|bar(ul(color(white)(a/a)"rate of effusion" prop 1/sqrt("molar mass")color(white)(a/a)|)))

You can approximate the molar masses of the two gases to

${\text{For methane: " M_M = "16 g mol}}^{- 1}$

${\text{For helium: " color(white)(a)M_M = "4 g mol}}^{- 1}$

This means that you can express the rate of effusion of methane, ${r}_{1}$, relative to the rate of effusion of helium, ${r}_{2}$, by writing

r_1/r_2 = (1/sqrt("16 g mol"^(-1)))/(1/sqrt("4 g mol"^(-1))) = sqrt( (4 color(red)(cancel(color(black)("g mol"^(-1)))))/(16color(red)(cancel(color(black)("g mol"^(-1)))))) = sqrt(1/4) = 1/2

Since you have

${r}_{2} = 2 \times {r}_{1}$

you can say that helium will effuse twice as fast as methane. This means that in the same period of time, you can expect twice as many molecules of helium than molecules of methane to effuse.

Let's assume that $y$ represents the number of moles of methane that effuse in a given period of time. You know that twice as many moles of helium, $2 y$, will effuse in the same period of time.

The total number of moles of gas that effuse will be equal to

${n}_{\text{effuse" = y + 2y = 3y" " " }} \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Now, let's assume that the vessel initially contained $x$ moles of methane and $x$ moles of helium. You know that $\frac{1}{4} \text{th}$ of the total number of moles present in the vessel effuse, which means that

${n}_{\text{effuse" = 1/4 xx n_"total}}$

Here ${n}_{\text{total}}$ will be equal to

${n}_{\text{total" = x + x = 2x" " " }} \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Combine equations $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ and $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to get

$3 y = \frac{2 x}{4}$

$3 y = \frac{x}{2} \implies y = \frac{x}{6}$

Now, after the effusion takes place, you know that the vessel will contain

${n}_{\text{methane" = (x - y)color(white)(a)"moles methane}}$

${n}_{\text{helium" = (x - 2y)color(white)(a)"moles helium}}$

This will be equivalent to

${n}_{\text{methane" = x - x/6 = (5/6x)color(white)(a)"moles methane}}$

${n}_{\text{helium" = x - 2 * x/6 = (2/3x)color(white)(a)"moles helium}}$

Assuming that you're interested in finding the ratio between the number of moles of helium and the number of moles of methane that remain in the vessel, you will have

n_"helium"/n_"methane" = ((2/3color(blue)(cancel(color(black)(x)))) color(red)(cancel(color(black)("moles"))))/((5/6color(blue)(cancel(color(black)(x))))color(red)(cancel(color(black)("moles")))) = 2/3 * 6/5 = color(green)(|bar(ul(color(white)(a/a)4/5color(white)(a/a)|)))#