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Homework Problem. Gases

  1. Nov 14, 2006 #1
    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.
  2. jcsd
  3. Nov 14, 2006 #2


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    Staff Emeritus
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    Gold Member

    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.
    Last edited: Nov 14, 2006
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