Oxygen Pressure Problem

  • #1
Pressure Problem!!

Homework Statement


Oxygen at pressures much greater
than 1 atm is toxic to lung cells. By weight, what ratio of
helium gas (He) to oxygen gas (O2 ) must be used by a
scuba diver who is to descend to an ocean depth of
50.0 m?

Homework Equations





The Attempt at a Solution


I first found the total pressure at that depth (note they didn't explicity give me rho_seawater, but I assumed):
P_50m = pgh = (1020kg/m^3)g(50m) = 4.93atm. Now here's where I get stuck. I assume that since O2 can only be 1atm, thus Helium must take up 4.93atm - 1atm = 3.93atm. Then I put these back into their own P = pgh, and solve for their 'p':
1atm =(p_O2)gh -> p_O2 = 206.7kg/m^3. Doing the same thing for p_He, I get 812.5kg/m^3. Then I simply take the ratio, and get phe/po2 = 3.93. The answer says 0.625. Any ideas??

Ari
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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Homework Statement


Oxygen at pressures much greater
than 1 atm is toxic to lung cells. By weight, what ratio of
helium gas (He) to oxygen gas (O2 ) must be used by a
scuba diver who is to descend to an ocean depth of
50.0 m?

Homework Equations





The Attempt at a Solution


I first found the total pressure at that depth (note they didn't explicity give me rho_seawater, but I assumed):
P_50m = pgh = (1020kg/m^3)g(50m) = 4.93atm. Now here's where I get stuck. I assume that since O2 can only be 1atm, thus Helium must take up 4.93atm - 1atm = 3.93atm. Then I put these back into their own P = pgh, and solve for their 'p':
1atm =(p_O2)gh -> p_O2 = 206.7kg/m^3. Doing the same thing for p_He, I get 812.5kg/m^3. Then I simply take the ratio, and get phe/po2 = 3.93. The answer says 0.625. Any ideas??

Ari
What is the pressure at 0 m depth? Add the additional pressure at 50m. to this to get the absolute pressure.

In order to determine the proportion of He to O2 by weight, you must determine first the molar proportions. It is the molar proportions that determine pressure: In the ideal gas law: PV=nRT so [itex]P \propto n[/itex]

Work that out and then determine what the relative weight would be.

I get a slightly different answer: assuming that water has a density of 1000 kg/m^3.

AM
 

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