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Ratio of oxygen in atmosphere

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Air at sealevel is composed of 80% nitrogen and 20% oxygen, but at the top of Mount Everest the composition of air is different. Give an expression for the ratio ([itex]f_O[/itex]) of oxygen in the air as a function of height z above sealevel, assuming the temperature T is constant and the air is an ideal gas. The gravitational acceleration is g and the mass of a molecule of nitrogen is [itex]m_N[/itex] and the mass of a molecule of oxygen is [itex]m_O[/itex].

    3. The attempt at a solution

    The potential energy of a molecule of oxygen is [itex]m_O g z[/itex], so the probability of finding it at height z is: [itex]P_{mO}(z) \propto e^{-\frac{m_O g z}{k_B T}}[/itex]. Similarly for nitrogen: [itex]P_{mN}(z) \propto e^{-\frac{m_N g z}{k_B T}}[/itex]. So the ratio is simply [itex]f_O = \frac{e^{-\frac{m_O g z}{k_B T}}}{e^{-\frac{m_N g z}{k_B T}}} = e^{-\frac{(m_O - m_N)gz}{k_B T}}?[/itex]
  2. jcsd
  3. Mar 3, 2012 #2

    D H

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    No, it isn't. The troposphere is well mixed. That's why it's called the troposphere. Is this a problem from a physics text, or one that your teacher made up?

    I'll ignore this minor problem.

    That gives a ratio of one at sea level.

    Hint: Assume each gas is individually in hydrostatic equilibrium.
  4. Mar 3, 2012 #3
    Just a question my teacher came up with.

    The hydrostatic equation is [itex]dp = -\rho g dz[/itex] with [itex]dp = k_B T dn[/itex] and [itex]\rho = nm[/itex]:

    [tex]\frac{dn}{n} = -\frac{m_O g}{k_B T}dz, \Leftrightarrow \\
    \ln{\frac{n}{n_0}} = -\frac{m_O g }{k_B T}z \Leftrightarrow \\
    n = n(0) e^{-\frac{m_O g }{k_B T}z} [/tex]

    If I do the same thing for nitrogen and divide again I get the same answer, so I'm not sure what to do...
  5. Mar 6, 2012 #4
  6. Mar 6, 2012 #5

    D H

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    You do not get the same answer. Oxygen and nitrogen have different molar masses.

    Try computing nO(z)/ nN(z), the ratio of the oxygen density to nitrogen density. What does that tell you?
  7. Mar 6, 2012 #6
    Since [itex]n_n = n(0) e^{-\frac{m_N g }{k_B T}z}[/itex] I get [itex]\frac{n_O}{n_N} = \frac{n_0 e^{-\frac{m_O g }{k_B T}z}}{n_0 e^{-\frac{m_N g }{k_B T}z}} = e^{-\frac{(m_O - m_N) g }{k_B T}z}[/itex]...or am I missing something? Because that's the same thing I got in my original post.
  8. Mar 7, 2012 #7

    D H

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    You are assuming that oxygen and nitrogen have the same density at sea level (i.e., the same n0. They obviously don't since the the atmosphere at sea level is 80% nitrogen, 20% oxygen.
  9. Mar 7, 2012 #8
    You're right, so I should get [itex]\frac{0.8}{0.2}e^{-\frac{(m_O - m_N)g}{k_B T}z} = 4 e^{-\frac{(m_O - m_N)g}{k_B T}z}[/itex].
  10. Mar 7, 2012 #9

    D H

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    Correct. So now it's just a matter of plugging in the numbers with z=height of Everest and then converting the resulting nitrogen:eek:xygen ratio to the percentage of the air at the top of Everest that is oxygen.

    The result will be wrong (wrong in reality that is; the result will be correct for this homework problem). The makeup is still more or less 80%/20% ratio at the top of Everest. The troposphere is just too well-mixed for any disparities to show up. The upper reaches of the atmosphere (above the stratosphere) do exhibit this differentiation.
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