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Air In Lungs - Thermal Physics

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data
    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 3.36 atm. the air temperature inside the submarine is 18.51 C and you can take the body temp (inside the lungs) to be 36.38 C.

    2. Relevant equations
    A) 1)The first person to leave takes a breath as deep as possible by exhaling as far as possible leaving 1.03 L in their lungs then slowly inhaling to increase their lung volume by 4.98L
    If the air temp in submarine was 18.51 C what volume of air is inhaled from surrounding?

    2)The person unwisely holds their breath all the way up tp the surface. What would the volume of their lungs be?

    B) The 2nd person to leave also takes a breath as deep as possible by exhaling to 1.06L of air left in the lungs and then slowly inhaling to increase their lung volume by 4.92 L, same body temp as above 36.38 C

    How many particle of air do their lungs contain after inhaling?

    The person breaths out all the way to the surface in order to maintain constant lung volume.How many moles of gas remain in lungs?

    C)Given that the molar mass of C02 is 44.0 g/mol what is the root mean square velocity of a C02 molecule in the lungs?

    3. The attempt at a solution

    1) I used V=Beta * temp change * initial volume

    V = 0.0034 * (273+(36.38-18.51))*1.03
    V=1.0186 L
    The answer is suppose to be 4.69 L i have tried other formulas like the ideal gas law but cannot come to the same answer

    Havent attempted the other questions as i dont know where to start
    Would appretaite some help please on what to look at, have looked in a couple of text books and searched the net with no luck
    Thank you in advance for any help that can be given
  2. jcsd
  3. Aug 3, 2009 #2
    Hey ducksmad. Sounds a little like physics 160 there..

    but anyway.

    for A-1). just use PV = nRT. the question suggests constant pressure. So Charles law applies?

    [tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]

    Um... so .

    Let [tex]{V_{1}}[/tex] = volume of the lung (4.98L)
    Let [tex]{T_{1}}[/tex] = temperature of the body (309.38K)
    Let [tex]{T_{2}}[/tex] = temperature of the submarine (291.51K)
    Let [tex]{V_{2}}[/tex] = be your unknown

    substitute, rearrange and calculate as necessary, and you should get 4.69 to 3 s.f.

    The others, Im not so sure.

    For A-2) I assumed that temperature was constant, and that the surface pressure was 1 atm.

    Hence, [tex]{P_{1}}{V_{1}} = {P_{2}}{V_{2}}[/tex]

    For this one, [tex]{V_{1}}[/tex] is the sum of the initial and the change in volume.

    Other questions.. still being worked out.

    Meanwhile, I'd also appreciate an explanation as to why the total lung volume was used rather than the change in volume. //edit: volume used is the change, misread question, which says increased "by" xx L. not "to" xx L


    PS. love how the forum supports LaTeX
    Last edited: Aug 3, 2009
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