# A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm. What should be the mole fraction of #O_2# in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level?

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Note that the mole fraction of oxygen at sea level is 0.209.

Note that the mole fraction of oxygen at sea level is 0.209.

##### 1 Answer

#### Explanation:

The idea here is that the partial pressure of a gas that's part of a gaseous mixture depends on the **mole fraction** of said gas and on the **total pressure** of the mixture, as given by **Dalton's Law of Partial Pressures**.

In your case, you know that at a depth of

This means that you can write

#P_ ("O"_ 2|"355 m") = chi_ ("O"_ 2|"355 m") * P_ ("total"|"355 m"#

You also know that at *sea level*, the **total pressure** of the atmosphere is equal to

This means that at sea level, you have

#P_ ("O"_ 2|"sea level") = 0.209 * "1 atm"#

Since the partial pressure of oxygen gas at **equal** to the partial pressure of oxygen gas at sea level, you can say that you have

#0.209 * 1 color(red)(cancel(color(black)("atm"))) = chi_ ("O"_ 2|"355 m") * 36.5 color(red)(cancel(color(black)("atm")))#

which gets you

#chi_ ("O"_ 2|"355 m") = 0.209/36.5 = color(darkgreen)(ul(color(black)(0.00573)))#

The answer is rounded to three **sig figs**.