Question

It was found that in a diving gas mixture for a certain altitude, the total predicted pressure is `"8.38 atm"`. If the partial pressure of oxygen gas must be `"0.21 atm"` underwater, what must the mol fraction be of oxygen gas?

Answer 1

This is asking you to use assume `"O"_2` is an ideal gas and use the definition of partial pressure:

`P_(O_2) = chi_(O_2(g))P_(t ot)`

where `chi_(O_2(g))` is the mol fraction of `"O"_2` in the diving gas mixture and `P_(O_2)` is its partial pressure in the diving gas mixture.

Since we want `P_(O_2) = "0.21 atm"`, we have:

`"0.21 atm" = chi_(O_2(g))("8.38 atm")`

`=> color(blue)(chi_(O_2(g)) = 0.025)`

or `2.5%`. This should make sense, since if we want only `"0.21 atm"` of `"O"_2` at a higher pressure, we want less than `21%` of the higher pressure to be `"O"_2`.