# 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?

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

Aug 2, 2017

$0.00573$

#### 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 $\text{355 m}$, the total pressure is equal to $\text{36.5 atm}$.

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 $\text{1 atm}$ and the mole fraction of oxygen gas is $0.209$.

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 $\text{355 m}$ must be equal to the partial pressure of oxygen gas at sea level, you can say that you have

$0.209 \cdot 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) = chi_ ("O"_ 2|"355 m") * 36.5 color(red)(cancel(color(black)("atm}}}}$

which gets you

${\chi}_{\text{O"_ 2|"355 m}} = \frac{0.209}{36.5} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.00573}}}$

The answer is rounded to three sig figs.