Oxygen partial pressure in deep-sea diving air at 9.1 atm

[jmandas]

In deep-sea dives the pressure of the air that a diver breathes may be quite large. The mixture of gases breathed is adjusted so that the partial pressure of oxygen is the same as on the surface of the earth. If a person is breathing air with a pressure of 9.1 atm, what % of the molecules in that air should be oxygen in order for the partial pressure(pO2) to be 150 mmHg.

I think I use F=(p1-p2)/R
I know atomspheric pressure is 1.013810E5N/m^2
blood has pO2 of 40 mmHg

Now of this seems to be going together though.
Please help...! Answer should be 2.17%

 

[dextercioby]

Well,it's not difficult.Think of Dalton's law.Consider 100 molecules of air which are at the pressure of 9.1 atm.Compute how many O_{2} molecules from that air would creat a pressure of 150 mmHg.

Pay attention to the units.

Daniel.

P.S.1 atm=101,325 Pa (u don't need it,though).

 

[thepatientmental]

Hi there, it seems like you are trying to calculate the percentage of oxygen molecules in a deep-sea dive where the partial pressure of oxygen needs to be 150 mmHg. To do this, we can use the formula for partial pressure: pO2 = (F * p) / (F + p), where p is the total pressure and F is the fraction of oxygen. We can rearrange this formula to solve for F: F = (pO2 * p) / (p - pO2).

Plugging in the given values, we have p = 9.1 atm, pO2 = 150 mmHg, and converting mmHg to atm, we get pO2 = 0.197 atm. Solving for F, we get F = (0.197 * 9.1) / (9.1 - 0.197) = 2.17%. This means that 2.17% of the molecules in the air should be oxygen for the partial pressure of oxygen to be 150 mmHg in a deep-sea dive.

I hope this helps clarify things for you. Remember, when solving problems involving gases, it's important to use the correct units and formulas. Keep practicing and you'll get the hang of it!