# 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?

##### 1 Answer
May 18, 2017

This is asking you to use assume ${\text{O}}_{2}$ is an ideal gas and use the definition of partial pressure:

${P}_{{O}_{2}} = {\chi}_{{O}_{2} \left(g\right)} {P}_{t o t}$

where ${\chi}_{{O}_{2} \left(g\right)}$ is the mol fraction of ${\text{O}}_{2}$ in the diving gas mixture and ${P}_{{O}_{2}}$ is its partial pressure in the diving gas mixture.

Since we want ${P}_{{O}_{2}} = \text{0.21 atm}$, we have:

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

$\implies \textcolor{b l u e}{{\chi}_{{O}_{2} \left(g\right)} = 0.025}$

or 2.5%. This should make sense, since if we want only $\text{0.21 atm}$ of ${\text{O}}_{2}$ at a higher pressure, we want less than 21% of the higher pressure to be ${\text{O}}_{2}$.