Calculating Gas Consumption for Scuba Diving: A Homework Problem Solution

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SUMMARY

The discussion focuses on calculating gas consumption for scuba diving using a mixture of 65.0g O2 and 2.00g He in a 5.0L tank at 25°C. The total pressure calculated in the tank is 12.379 atm, leading to an estimated duration of 8.0 minutes for the gas supply based on an average breathing rate of 15 breaths per minute. Participants debated whether to use total pressure or partial pressure of O2 for accurate calculations, with some suggesting that the pressure in the tank decreases over time. The correct answer for the duration is contested, with some claiming it should be 7.8 minutes.

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  • Understanding of gas laws, specifically the Ideal Gas Law
  • Basic knowledge of scuba diving physiology and the effects of gas mixtures
  • Familiarity with unit conversions in chemistry (grams to moles)
  • Ability to perform calculations involving pressure, volume, and moles of gas
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  • Study the Ideal Gas Law and its applications in real-world scenarios
  • Learn about the physiological effects of different gas mixtures in scuba diving
  • Explore calculations involving partial pressures and their significance in gas mixtures
  • Investigate the impact of pressure changes over time in closed systems
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teleport
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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.
 
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teleport said:
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.
 
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