# Question #24c45

##### 1 Answer

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.

#"rate" prop 1/sqrt(M_M)#

In your case, you have

#M_ "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

#"rate He"/"rate O"_ 2 = sqrt(32.0/4.0) = sqrt(8) = 2sqrt(2)#

This means that you have

#"rate O"_ 2= "rate He"/(2sqrt(2))#

#"rate O"_ 2 = sqrt(2)/4 * "rate He " -># matches what you know from#color(orange)("(*)")#

Now, if you take *amount* of helium that effuses from the container in **seconds**, you can say that

#"rate He" = (x color(white)(.)"dm"^3)/"5 s" = (x/5)# #"dm"^3# #"s"^(-1)#

Plug this into the equation to find

#"rate O"_ 2 = sqrt(2)/4 * (x/5)# #"dm"^3# #"s"^(-1)#

#"rate O"_2 = (sqrt(2)/20 * x)color(white)(.)"dm"^3 color(white)(.)"s"^(-1)#

This, of course, means that

#color(blue)(cancel(color(black)(x))) color(red)(cancel(color(black)("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 **seconds** for **all the helium** present in the container to effuse, i.e. for

#"rate He" = "10 dm"^3/"5 s" = "2 dm"^3# #"s"^(-1)#

This will get you

#"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 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"#