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
anor277
Dec 31, 2017

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^-3χO2=2.0⋅g⋅mol−132.0⋅g⋅mol−12.0⋅g32.0⋅g⋅mol−1+98.0⋅g4.0⋅g⋅mol−1=2.54×10−3

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)}=0.998χHe=98.0⋅g⋅mol−14.0⋅g⋅mol−12.0⋅g32.0⋅g⋅mol−1+98.0⋅g4.0⋅g⋅mol−1=0.998

And so p_(O_2)=2.54xx10^-3xx8.2*atm=0.0208*atmpO2=2.54×10−3×8.2⋅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....

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