Calculating Gas Ratio for Safe Diving at 50m Depth

[AriAstronomer]

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

 

[Andrew Mason]

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 $P \propto n$

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