# A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.2 atm?

Dec 31, 2017

The partial pressure is proportional to the mole fraction....

#### Explanation:

And so we calculate the individual mole fractions....

${\chi}_{{O}_{2}} = \frac{\frac{2.0 \cdot g \cdot m o {l}^{-} 1}{32.0 \cdot g \cdot m o {l}^{-} 1}}{\frac{2.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} + \frac{98.0 \cdot g}{4.0 \cdot g \cdot m o {l}^{-} 1}} = 2.54 \times {10}^{-} 3$

${\chi}_{H e} = \frac{\frac{98.0 \cdot g \cdot m o {l}^{-} 1}{4.0 \cdot g \cdot m o {l}^{-} 1}}{\frac{2.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} + \frac{98.0 \cdot g}{4.0 \cdot g \cdot m o {l}^{-} 1}} = 0.998$

And so ${p}_{{O}_{2}} = 2.54 \times {10}^{-} 3 \times 8.2 \cdot a t m = 0.0208 \cdot a t m$

The diver is at a serious depth....and had he or she been inhaling dinitrogen, i.e. from a normal air mix, they would likely suffer nitrogen narcosis....