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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?

Answer 1

Mateo Aguilar

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

Thus, we determine each mole fraction separately.

#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 significant depth, and if he or she had been breathing in nitrogen, such as from a typical air mix, they would probably have developed nitrogen narcosis.

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Answer 2

Kai Bowman

Partial pressure of oxygen: 0.167 atm.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.