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Diffusion Coefficient of O2 into N2 (Last Resort Please Help

  1. Mar 23, 2015 #1
    Hello everyone, I've been stuck on this problem for three days now I just can't seem to make it work out to a reasonable value

    1. The problem statement, all variables and given/known data
    Consider a horizontal glass tube with an inner diameter of 5 mm and a length, L, of 500 mm filled with pure nitrogen gas at a temperature of 25C, pressure of 101.3kPA. The tube is capped at both sides. At t=0 one cap is removed exposing one end to oxygen gas at same temperature and pressure.

    Calculate the Diffusion Coefficent of oxygen gas in a binary mixture of nitrogen gas with these conditions.

    2. Relevant equations

    Diffusion Equation
    [tex] D = \frac {3 \pi}{8} (\frac {k T}{2 \pi \mu})^{0.5} \frac {1}{\rho \sigma} [/tex]
    k is the boltzmann constant
    T is temperature
    [tex] \mu = \frac {m1 m2}{m1 + m2} [/tex] is reduce mass, m is mass
    [tex] \sigma N_2 = 0.43 nm^2 [/tex] is the collision cross section of Nitrogen (O2 = 0.4 nm2)
    [tex] \rho [/tex] is the density

    3. The attempt at a solution
    My attempt at a solution started with calculating the moles of Nitrogen Gas using the cylinder parameters
    [tex] n = \frac {PV}{RT} [/tex], where [tex] V = \pi r^2 L [/tex]
    [tex] n = \frac {101300 \times \pi \times (0.0025^2) 0.5}{8.31\times 298} [/tex]
    [tex] n = 4.00 \times 10^{-4} mol [/tex]

    Then solved for mass
    [tex] m = n\times Molar Mass Nitrogen/1000 [/tex]
    [tex] m = 4.00\times 10^{-4} \times 28.02/1000[/tex]
    [tex] m = 1.10\times 10^{-5} kg[/tex]

    Then solved for density
    [tex] \rho = \frac {1.10\times 10^{-5}}{ pi(0.0025^2)0.5} [/tex]
    [tex] \rho = 1.14 kg/m^3 [/tex]

    Heres where I get stuck: I originally DID NOT notice that [itex] \sigma [/itex] was in nm2 and continued using [itex] \sigma = 4.3\times 10^{-10} m [/itex] and solved for a diffusion coefficient of ~5 x 10-1 m2/s. I know this is wrong but the value makes sense as the two molecules are gasses meaning diffusion is fairly rapid. Using the correct [itex] \sigma = 1.89 \times 10^{-19} m [/itex] I get like 5000000 m2 /s which is incorrect obviously. However, I don't know where i went wrong I looked through my calculations several times and I really don't know what to do

    Please please help!

    Edit: did my best to correct the formatting, Borek.
    Last edited by a moderator: Mar 24, 2015
  2. jcsd
  3. Mar 29, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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