Calculating Gas Ratio for Safe Diving at 50m Depth

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SUMMARY

The discussion focuses on calculating the gas ratio of helium (He) to oxygen (O2) for safe diving at a depth of 50 meters, where the total pressure is approximately 4.93 atm. The diver must maintain oxygen at 1 atm to avoid toxicity, leading to helium occupying the remaining pressure of 3.93 atm. The correct weight ratio of helium to oxygen is determined to be 0.625, contrary to the initial calculation of 3.93. Participants emphasize the importance of using molar proportions and the ideal gas law for accurate results.

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  • Understanding of gas laws, particularly the ideal gas law (PV=nRT).
  • Knowledge of pressure calculations in fluids (P = pgh).
  • Familiarity with the concept of molar proportions in gas mixtures.
  • Basic understanding of scuba diving principles and gas toxicity at depth.
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AriAstronomer
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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
 
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AriAstronomer said:

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
 

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