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Solving the ideal gas law for moles

  1. Nov 12, 2007 #1
    Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:

    1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
    2. The temperature of the air is constant (body temperature).
    3. The air acts as an ideal gas.
    4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.

    If the temperature of air in Gabor's lungs is 37 Celsius and the volume is 6 L, how many moles of air n must be released by the time he reaches the surface? Let the molar gas constant be given by R = 8.31 J/ (mol*K)
    I know that the number of moles of air in 6L at the underwater pressure of is 0.006(?)
    Now I know that the difference between this and the moles of air in the lungs that I calculate at the surface pressure gives me my answer but how do I know what the conditions are at the surface? am I just approaching this problem in an incorrect way?? Thanks in advance..
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    his lungs don't change is size, and neither should his body temp. also, check your units carefully
  4. Nov 12, 2007 #3
    ah ok i get it
    Last edited: Nov 12, 2007
  5. Nov 12, 2007 #4
    no i don't get it. i get n=0.006 at the depth of 15 m but my units are atm*L*mol/J when I use the formula n=PV/(RT). what do i do?????? so at the depth of 15m, i got p=2.5 atm, V=6L, R=8.31 J/(mol*K), T=37+273 K, and the surface, p=1 atm, V=6L, R=8.31 J/(mol*K), T=37+273 K. i can't use the 1 mol=22.4 L rule because conditions are not "standard pressure and temp"
    Last edited: Nov 12, 2007
  6. Nov 12, 2007 #5
    the equation you used was n = PV/RT

    T is in kelvin, P is in either atmospheres or mmHg, and volume is in litres. make sure that your values have these units.
  7. Nov 12, 2007 #6
    that's what i had. but i'm NOT getting the right answer. i will spell what i did out
    @ 15 m --> n=PV/(RT)= (2.5)(6)/(8.31*(37+273)) = 0.005823
    @surface--> n=PV/(RT) = 1(6)/(8.31*(37+273)) = 0.002329
    you said that temp stays the same and volume stays the same, correct? then why am i not getting the right ans? I get 0.00349 atm*L*mol/J when I should be getting my ans in moles, am i not right?
  8. Nov 12, 2007 #7
    i believe that you have the incorrect units for the pressure. in addition your final units should be in moles
  9. Nov 12, 2007 #8
    no, you said that p should be in either atm or mmHg, I DO have it in atm. 2.5atm at 15 m, got from (15m(1atm/10m)+1atm) and 1atm at the surface. help anyone? maybe you meant pascuals...
    Last edited: Nov 12, 2007
  10. Nov 12, 2007 #9
  11. Nov 12, 2007 #10
    oh... you meant the universal gas constant! ok, thanks
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