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Air in the Lungs thermodynamics help!

  1. Aug 24, 2013 #1
    A submarine has run into trouble and is stuck at the bottom of the ocean. Several people are on board and must make their way to the surface without any diving gear. The air pressure aboard the submarine is 3.000 atm. The air temperature inside the submarine is 18.02 °C and you can take body temperature (inside the lungs) to be 36.04 °C.

    1. The second person to leave also takes a breath as deep as possible by exhaling as far as possible (leaving a volume of 1.190 L in their lungs), and then slowly inhaling to increase their lung volume by 4.080 L. His body temperature is also 36.04 °C. This person breathes out all the way to the surface in order to maintain a constant lung volume. How many moles of gas remain in the lungs?


    Relevant equations:
    PV = nRT


    The attempt at a solution:
    i found the total volume of the lungs to be
    Vtotal = 1.190L + 4.080L = 5.270L = 0.005270m^3

    and using PV = nRT i found the number of moles at full capacity lungs
    (3.00 x 101300)(0.005270) = n(8.314)(36.04+273.15)
    n = 0.6230 moles

    then i found the number of moles when the person exhales using PV = nRT
    (3.00 x 101300)(0.004080 - 0.001190) = n(8.314)(36.04+273.15)
    n = 0.365

    ∴ the change in moles is n = 0.6230 - 0.341 = 0.281mols

    BUT the actual answer is 0.208 mols and i don't know what i'm doing wrong HELP!!
     
  2. jcsd
  3. Aug 24, 2013 #2

    TSny

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    Hello, Nigzzzz.

    What is volume of the lungs when the person reaches the surface?

    What is the pressure of the air in the lungs at the surface?

    What is the temperature of the air in the lungs at the surface?
     
  4. Aug 24, 2013 #3
    have to assume that lung pressure is the same as air pressure as no other pressure is given so 3.00atm, and it's provided that lung temperature is at 36.04 °C, since the person exhaled all the way i would assume the remaining volume would be 1.190L
     
  5. Aug 24, 2013 #4

    TSny

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    I think you can assume that the pressure inside the lungs will be the same as the pressure of the surroundings at the surface.

    Yes, that sounds right.

    Note that the problem stated that the volume of the lungs remains constant while going to the surface.
     
  6. Aug 24, 2013 #5
    would that mean it remains at the inhaled capacity - 4.080L? or the difference between 2 volumes?
     
  7. Aug 24, 2013 #6

    TSny

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    What did you find for the total volume of the lungs after the inhale? That would be the volume of the lungs at the the start of the ascent to the surface.
     
  8. Aug 24, 2013 #7
    it says that he slowly inhaled by 4.080L with a residual volume of 1.190L so after inhalation he would have 5.270L, would this be the volume constant in the ascend?
     
  9. Aug 24, 2013 #8

    TSny

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    That's how I interpret it.
     
  10. Aug 24, 2013 #9
    but that would only be useful in giving me the number of moles in the lungs after inhalation? and it says that he exhaled to the surface to find the number of moles after exhalation?
     
  11. Aug 24, 2013 #10

    TSny

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    But if you know the three quantities that I asked about in post #2, can you find the number of moles left in the lungs at the surface?
     
  12. Aug 24, 2013 #11
    OHHH

    so you use PV = nRT

    101300 x 0.005270 = n(8.314)(36.04 + 273.15)

    n = 0.208 moles

    OK i think i get it! hahaha
     
    Last edited: Aug 24, 2013
  13. Aug 24, 2013 #12
    thanks so much!
     
  14. Aug 24, 2013 #13

    TSny

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    OK, good. (But it's a little strange the way you wrote your last equation. The amount on the far left of the equation does not equal the amount in the middle or the amount on the right. :smile:)
     
  15. Aug 24, 2013 #14
    woops forgot the n thats why
     
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