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Density of Earth's atmosphere w.r.t height

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data

    "It's possible to use the ideal gas law to show that the density of the Earth's atmosphere decreases exponentially with height, that is
    ρ=ρ0exp(-z/z0)
    where z is the height above sea level, ρ0 is the density at sea level and z0 is called the "scale height" of the atmosphere.

    a.) Determine value of z0:

    b.) What is the density of the air in Denver, at an elevation 1600 m? What percent of sea-level density is this?

    2. Relevant equations
    Ideal gas law: Pv=nRT

    possibly P=P0 + ρgh ? Seems unlikely as ρ varies?

    3. The attempt at a solution

    Not quite sure how to go about this one. Started with ideal gas law
    Pv=nRT
    assumed we'd be comparing two volumes of gas (at height z0 and z) with equal number of molecules, at equal temperature, thus
    P1v1=P2v2
    Then substituted v=ρ(mass)
    Again assuming equal mass would imply
    P1ρ1=P2ρ2

    Not really sure how to proceed to involve height instead of pressure or even if on right track... Somehow the latter seems more likely. Help would be appreciated!
     
  2. jcsd
  3. Sep 7, 2014 #2
    What is your understanding of the definition of the word density?

    Chet
     
  4. Sep 7, 2014 #3
    Density is mass/volume?
     
  5. Sep 7, 2014 #4
    Right, mass per unit volume. From the ideal gas law, if the molecular weight of the gas is M, can you write an equation for the density of the gas as a function of T, P, M, and R?

    Chet
     
  6. Sep 9, 2014 #5
    Hi, sorry for late reply:

    n=mass/molecular weight = m/M

    PV=(m/M)RT
    =>PM/(RT)=m/V=ρ

    But how does ρ vary with height? Can I use P=P0 + ρgz? But then I would be plugging ρ into an expression for ρ right?
     
  7. Sep 9, 2014 #6
    Correct!

    No. You have to use the differential version of the equation:

    dP/dz=-ρg =-(PMg)/(RT)

    Do you know how to solve this differential equation for P?

    Chet
     
  8. Sep 9, 2014 #7
    dP/dz=-(PMg)/(RT)

    ∫(P)-1dP=∫(-Mg)/(RT)dz

    ln(P/P0)=(-Mg)/(RT)(z-z0) + c

    (ρ)/(ρ0) =C(ez/ez0)(-Mg)/(RT)

    ρ=ρ0C(ez/ez0)(-Mg)/(RT)

    Did I do that right?
     
  9. Sep 9, 2014 #8
    Almost. The boundary condition should be P=P0 and ρ=P0M/(RT) at z = 0.

    Chet
     
  10. Sep 10, 2014 #9
    I'm not quite sure I follow...

    pp0(P-1)dP = (-Mg)/(RT)∫z0dz

    ln(P/P0)=(-Mg)/(RT)z

    ρ0=(P0M)/(RT)
    =>P0=(ρ0RT)/M

    ρ=(PM)/(RT)
    =>P=(ρRT)/M

    Then would (RT)/M cancel on both top and bottom?
    ln(P/P0) = ln((ρ)[STRIKE]((RT)/M)[/STRIKE]/((ρ0)[STRIKE](RT)/M))[/STRIKE]

    leaving
    ln(ρ/ρ0)=(-Mgz)/(RT)
     
  11. Sep 10, 2014 #10
    Yes. This is the correct answer. From this result, you should be able to back out an equation for zo, by comparing with the original equation in the problem statement .

    Chet
     
  12. Sep 10, 2014 #11
    I just realized that I had read the original equation wrong, I thought ρ=ρ0exp(-z/z0) was ρ=ρ0-z/z, not ρ=ρ0(e-z/z) Whoops, lots of unnecessary headaches there!

    So then

    ρ=ρ0(e(-Mg)/(RT))-z

    and z0 = (RT)/(Mg), great, thanks for helping me out with this. :)
     
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