Calculating Gas Consumption for Scuba Diving: A Homework Problem Solution

[teleport]

Hi, I have been struggling a little with this question.

Scuba divers breathe a mixture of O2(g) and He(g) to avoid "the bends, a condition caused by nitrogen in the blood. If 65.0g O2(g) and 2.00g He(g) are placed in a 5.0L tank at 25oC, calculate:

If the average human takes 15 breaths per minute, and breathes in 0.50L at 1.00 atm, calculate how long the gas in the tank will last?

This is what I've done:

Pressure in the tank:

n(He) = (2.00 g He)/(4.00 g/mol) = 0.500 mol He

n(O2) = (65.0 g)/(32 .00 g/mol) = 2.03125 mol O2

n(total) = n(He) + n(O2) = 2.53125 mol

P(total) = (n(total)RT)/V = (2.53125)(0.082057)(298)/5.0
P(total) = 12.379 atm

time to empty:

P1V1 = P2V2
(1.00 atm)(0.50 L) = (12 atm)x,

where x is the volume breathed in one breath

x = 0.0416667 L

in one min: Vbreathed = 15x = 0.625 L

(1 min)/(0.625 L) = t/(5.0 L)

Therefore t = 8.0 min.

Is all that right or instead of using the total pressure in the tank I should use the partial pressure of O2? The answer should be 7.8 min which I'm not getting. But I have also tried it with O2 partial volume and I don't get the answer. Am I missing something? Any help is appreciated. Thanks.

 

[Gokul43201]

Hi, I have been struggling a little with this question.

Scuba divers breathe a mixture of O2(g) and He(g) to avoid "the bends, a condition caused by nitrogen in the blood. If 65.0g O2(g) and 2.00g He(g) are placed in a 5.0L tank at 25oC, calculate:

If the average human takes 15 breaths per minute, and breathes in 0.50L at 1.00 atm, calculate how long the gas in the tank will last?

This is what I've done:

Pressure in the tank:

n(He) = (2.00 g He)/(4.00 g/mol) = 0.500 mol He

n(O2) = (65.0 g)/(32 .00 g/mol) = 2.03125 mol O2

n(total) = n(He) + n(O2) = 2.53125 mol

P(total) = (n(total)RT)/V = (2.53125)(0.082057)(298)/5.0
P(total) = 12.379 atm

time to empty:

P1V1 = P2V2
(1.00 atm)(0.50 L) = (12 atm)x,

Does the pressure in the tank always stay at 12 atm? Doesn't it keep decreasing with time?

You've counted the total number of moles of gas in the tank. Why don't you simply count the number of moles of gas in a single breath as well?

Also, be careful with the numbers - what was 12.379 at one step became 12 in the next.

Finally, I think 7.8 min is wrong.

PS : This question belongs in the Homework & Coursework section.

 

[Blueshift5]

I would say that your calculations and approach to solving this problem are correct. However, there are a few points to consider:

1. The ideal gas law assumes that the gases in the tank behave ideally, which may not be the case in real life. Factors such as temperature, pressure, and gas composition can affect the behavior of gases.

2. The calculation of the total pressure in the tank is correct, as you have correctly accounted for the moles of both gases present. However, in real life, the partial pressure of each gas would be important to consider, as it is the partial pressure that determines the amount of gas that can dissolve in the blood.

3. The answer of 7.8 minutes may be an approximation or an ideal scenario. In real life, factors such as breathing rate, depth of the dive, and individual differences in physiology can affect the duration of the gas in the tank.

Overall, your calculations are correct, but it is important to keep in mind the limitations and assumptions of the ideal gas law and how they may affect the real-life scenario of scuba diving. I would recommend reaching out to your instructor or a more experienced scuba diver for further clarification and understanding.