# Question #7157d

##### 1 Answer

#### Answer:

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

#"For methane: " M_M = "16 g mol"^(-1)#

#"For helium: " color(white)(a)M_M = "4 g mol"^(-1)#

This means that you can express the rate of effusion of methane, **relative** to the rate of effusion of helium,

#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 xx 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

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

#n_"effuse" = y + 2y = 3y" " " " color(orange)((1))#

Now, let's assume that the vessel **initially contained**

#n_"effuse" = 1/4 xx n_"total"#

Here

#n_"total" = x + x = 2x" " " "color(orange)((2))#

Combine equations

#3y = (2x)/4#

#3y = x/2 implies y = x/6#

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

#n_"methane" = (x - y)color(white)(a)"moles methane"#

#n_"helium" = (x - 2y)color(white)(a)"moles helium"#

This will be equivalent to

#n_"methane" = x - x/6 = (5/6x)color(white)(a)"moles methane"#

#n_"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)|)))#