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Calculate N2/O2 ratio at 3255m

  1. Jan 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The ratio of nitrogen to oxygen at sea level is 4.0:1. Use Laplace's law of
    isothermal atmospheres to calculate the same ratio for 3255m above sea level
    Assume that T = 273 K

    2. Relevant equations
    [tex]P_fN2 = P_iN2 * e^-\Big({\frac{M_N2 g}{R T}}*h\Big)[/tex]
    [tex]P_fO2 = P_iO2 * e^-\Big({\frac{M_O2g}{R T}}*h\Big)[/tex]

    [tex]P_i = 1 atm[/tex]
    [tex]h = 3255 meters [/tex]
    [tex]M_N2 = 28 g/mol [/tex]
    [tex]M_O2 = 32 g/mol [/tex]
    T = 273K
    [tex]g = 9.8 m/s^2[/tex]
    R = 8.314 J/mol K



    3. The attempt at a solution
    I tried plugging in the numbers for both equations, but somehow that just makes my exponential go to 0 so I know that's not correct. I am wondering if I need to convert anything. I converted the initial pressure to Pa but that doesn't help me with an exponential of 0. For N2 I came up with approximately 393.52 for Mg/RT*h. e to -393.52 comes out zero on my calculator, I'm sure its not, because e to any power is never 0 but its too small for my calculator to read.
     
  2. jcsd
  3. Jan 15, 2015 #2

    SteamKing

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    Homework Helper

    Your molecular weights must be expressed in kg / mole, rather than grams / mole.

    Also, since you are given the ratio of nitrogen : oxygen at sea level, then the partial pressure of each element will not be 1.0 atm. That's the pressure of the whole atmosphere at sea level.
     
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