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?

1 Answer
Dec 31, 2017

Answer:

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

Explanation:

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

#chi_(O_2)=((2.0*g*mol^-1)/(32.0*g*mol^-1))/{(2.0*g)/(32.0*g*mol^-1)+(98.0*g)/(4.0*g*mol^-1)}=2.54xx10^-3#

#chi_(He)=((98.0*g*mol^-1)/(4.0*g*mol^-1))/{(2.0*g)/(32.0*g*mol^-1)+(98.0*g)/(4.0*g*mol^-1)}=0.998#

And so #p_(O_2)=2.54xx10^-3xx8.2*atm=0.0208*atm#

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....