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Partial pressure ratio of gases at different altitudes

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data

    On a planet not entirely unlike earth, the ratio of the partial pressure of N2 to that of O2 equals 1 at an altitude of 1 km: [itex]\frac{p_{\text{N}_2}}{p_{\text{O}_2}} = 1[/itex]. Assuming that T = 200 K, and the gravitational constant is 5 m/s2, what is the ratio [itex]\frac{p_{\text{N}_2}}{p_{\text{O}_2}}[/itex] at an altitude of 0 km? (Hint: You can use the Boltzmann distribution to calculate the law of atmospheres.)

    2. Relevant equations

    The Boltzmann distribution (it's a probability equation): [itex] P_n = \frac{d_n e^{-\frac{E_n}{kT}}}{Z}[/itex] where:
    [itex] Z = \sum \limits_{i=1}^N d_i e^{-\frac{E_i}{kT}}[/itex] for N energy levels
    di is the degeneracy for energy level i
    Ei is the energy of energy level i
    k is the Boltzmann constant 1.38 ×10−23 J/K
    T is the temperature

    Here, the energy of an energy level is the potential energy of a particle, E = mgh, where h is the height of the particle, g = 5 m/s2, and m is its mass.

    Note that I was NOT given the mass of nitrogen or oxygen, and I was not given a periodic table, so I am assuming that I either do not need the masses of these molecules, or I can deduce them from what I am given.

    3. The attempt at a solution

    I didn't know what the law of atmospheres was so I cheated by looking it up. I at least know what I am working toward but I am still having a lot of issues.

    Using the Boltzmann distribution, the probability of a particle at height h is [itex]P(h) = \frac{d_h e^{-\frac{mgh}{kT}}}{Z}[/itex] and the probability of a particle at height 0 is [itex]P(0) = \frac{d_h e^{-\frac{mg\cdot 0}{kT}}}{Z} = \frac{d_h e^0}{Z} = \frac{d_h}{Z}[/itex]. So [itex]P(h) = P(0) \cdot e^{-\frac{mgh}{kT}}[/itex], which looks almost the same as the "law of atmospheres" except instead of probability, it is pressure. I know that probability and pressure are probably directly related (more probability means more particles in a certain place which implies more pressure; also note the p and N on opposing sides of the ideal gas pV = NkT equation), but I can't completely figure out the connection.

    Using this, for probabilities, you get:
    [tex]
    \frac{P_{\text{N}_2}(h)}{P_{\text{O}_2}(h)} = \frac{P_{\text{N}_2}(0)\cdot e^{-\frac{m_{\text{N}_2}gh}{kT}}}{P_{\text{O}_2}(0)\cdot e^{-\frac{m_{\text{O}_2}gh}{kT}}}
    [/tex]
    where mC is the mass of the particle with chemical formula C.

    Now I'm stuck. I know the left side of the equals sign is 1 if I set h to 1 km, since it was given. The right side has the value I am trying to calculate, multiplied by something that requires knowledge of the masses of N2 and O2, which I am assuming I do not know. Also, this is still probability, and needs to be put in terms of pressure.

    Could someone please point me in the right direction? Thank you! (And if the right direction is that I should just use the masses of the two molecules, I will. That would be negligence on the question-writer to include that, since it usually is included, but it wouldn't surprise me. I once spent over 2 hours on a physics problem wondering how I was supposed to solve it without knowing the mass ("the mass is m" was never in the problem statement) and when I finally caved and peeked at the solution, the first thing they do is give the object the mass m and use it in the final result >.>)
     
  2. jcsd
  3. Jul 26, 2014 #2

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    You are very close to solving this, closer than you probably realize.

    The problem is with that very assumption. It is generally assumed that you do know whatever you have been taught in previous coursework. This includes the ability to look things up in a periodic table. Also, without checking a periodic table, you may even know the masses of common elements like nitrogen and oxygen to the nearest integer value in atomic mass units -- which could be potentially useful on future exams.

    The probability would be proportional to the total number of molecules per unit volume, N/V. And you know how N/V is related to the pressure.

    Hope that helps.
     
  4. Jul 27, 2014 #3
    Yeah, this was an old exam that they gave us, so I was pretending I only had the resources I'll be provided, which doesn't seem like it includes a periodic table. We are given the mass of a proton though, so I'll just memorize the four most common that we seem to get (1 proton mass for hydrogen atoms, 4 for helium, 14 for nitrogen, and 16 for oxygen atoms).

    I did use the masses and got the answer, and also, thanks for explaining the relation between probability and N/V! That makes more sense than what I was trying.
     
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