Question

What is the partial pressure of `"O"_2` in air at `"1 atm"` given that `21%` of air molecules are `"O"_2` ?

Answer 1

Here's what I got.

Explanation:

For starters, you know that the partial pressure of a gas that's part of a gaseous mixture can be calculated using the following equation--think Dalton's Law of Partial Pressures here!

`color(blue)(ul(color(black)(P_ i =chi_ i * P_"total")))`

Here

• `P_i` is the partial pressure of gas `i`
• `chi_i` is the mole fraction of gas `i` in the mixture
• `P_"total"` is the total pressure of the mixture

Now, you know that you have a sample of air at a total pressure of `"1 atm"` and that `21%` of all the molecules of gas that make up this sample are molecules of oxygen gas.

In order to be able to calculate the partial pressure of oxygen gas, you need to figure out the mole fraction of oxygen gas in the sample. The mole fraction of oxygen gas is calculated by dividing the number of moles of oxygen gas by the total number of moles of gas present in the sample.

`chi_ ("O"_ 2) = "moles of O"_2/"total moles of gas" " "" "color(darkorange)("(*)")`

As you know, Avogadro's constant allows you to convert the number of molecules to moles.

`color(white)(overbrace(color(blue)(ul(color(black)("1 mole gas" = 6.022 * 10^(23) quad "molecules gas"))))^color(red)(ul("Avogadro's constant")))`

So, let's assume that this sample contains `N` molecules of gas. Since `21%` of these molecules are molecules of oxygen gas, you can say that this sample contains

`N color(red)(cancel(color(black)("molecules gas"))) * "21 molecules O"_2/(100color(red)(cancel(color(black)("molecules gas")))) = ((21 * N)/100) quad "molecules O"_2`

This means that the number of moles of oxygen gas present in the sample is equal to

`((21 * N)/100) color(red)(cancel(color(black)("molecules O"_2))) * "1 mole O"_2/(6.022 * 10^(23) color(red)(cancel(color(black)("molecules O"_2)))) = ((0.21 * N)/(6.022 * 10^(23))) quad "moles O"_2`

Similarly, the total number of moles of gas present in the sample will be

`N color(red)(cancel(color(black)("molecules gas"))) * "1 mole gas"/(6.022 * 10^(23) color(red)(cancel(color(black)("molecules gas")))) = (N/(6.022 * 10^(23))) quad "moles gas"`

Plug this into equation `color(darkorange)("(*)")` to find the mole fraction of oxygen gas in the sample.

`chi_ ("O"_ 2) = ( (0.21 * N)/(6.022 * 10^(23)) quad color(red)(cancel(color(black)("moles"))))/(N/(6.022 * 10^(23)) quad quad color(red)(cancel(color(black)("moles"))))`

This gets you

`chi_ ("O"_ 2) = ( 0.21 * color(blue)(cancel(color(black)(N))))/(color(red)(cancel(color(black)(6.022 * 10^(23))))) * color(red)(cancel(color(black)(6.022 * 10^(23))))/color(blue)(cancel(color(black)(N)))`

`chi_ ("O"_ 2) = 0.21`

As you can see, the mole fraction of oxygen gas does not depend on the number of molecules of gas present in the sample, it only depends on the concentration of the sample!

You can thus say that the partial pressure of oxygen gas in air at `"1 atm"` is

`color(darkgeen)(ul(color(black)(P_( "O"_ 2)))) = 0.21 * "1 atm" = color(darkgreen)(ul(color(black)("0.21 atm")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the total pressure of the sample, which means that you should report the answer as

`P_ ("O"_ 2) = "0.2 atm"`