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Homework Help: Finding density of air in air column from just pressure?

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data

    This is for a lab I finished but cannot get this question, I spoke with the TA today and she made it make sense but I am still not getting the question correct.

    (2) The difference in pressure between the lowest and the highest measurement point is related to the weight of air between these points. Use your observations to calculate the density of air between these points? How does this number compare with the textbook density at 24oC? (Hint – convert your measurements to pounds per square inch and figure out how many pounds of air you observed in a column of air that is 1 inch square by the height of the stairs.)

    For the lab we build a pressure sensor and measured pressure at 5 heights in a stairwell. The pressures increased from 692.5 torr to 745.12 torr, so a difference in pressure of 52.6 torr.

    No temperature readings, just pressure was taken.
    The height from the bottom to the highest meaurement was 17.094 meters.

    2. Relevant equations

    Not sure what equations would be relevant. F=mg or conversion factors.

    3. The attempt at a solution

    I first converted the 52.6 torr to Pa, 1torr=133.332Pa.

    7012.7 Pa or N/m^2, but got stuck here.

    then I tried converting the torr to psi, got 1.017 psi. Multiplied that by the height of the column of air I want to find the mass of and got 684.4 pounds. But now what? divide that by the volume, then i just get the same number since the volume is 673 in^3.

    I'm sure I'm just missing something here.

    Ok, I think I got it.

    converted 52.6 torr to 7012.7 N/m^2.
    P=F/area(m^2) --> 7012.7 N
    7012.7 N = m x g
    715.7 N/(m/s^2) --> 715.7 kg / 17.094 m^3 = 41.87 kg/m^3

    But that density seems really high, shouldn't atmospheric density be around 1.3kg/m^3

    Am I using the wrong volume? I am pretty confident in the mass, I think?

    to get a density around 1.3 I would need a volume of around 550m^3.
    Last edited: May 13, 2010
  2. jcsd
  3. May 13, 2010 #2

    Andrew Mason

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    There is something wrong with your measurements. There is no way that the pressure will increase that much due to the weight of a 17 m column of air.
  4. May 13, 2010 #3
    thats what I am thinking now, that my calculations are fine, but the original measurements are wrong. Which is completely possible considering we just connected a data logger to a pressure sensor with random wire and such.

    I'll note it in my lab report as an error.
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