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Chem, 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 29, 2006 #2

    chemisttree

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    Try calculating the number of moles of ideal gas per breath and applying that answer to the number of moles of He/O2 available.

    I get 8.2 minutes... 8 minutes if significant figures are observed.
     
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