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smoez

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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

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.

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 (O

[tex] \rho [/tex] is the density

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 nm

Please please help!

## Homework Statement

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.

## Homework 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 (O

_{2}= 0.4 nm^{2})[tex] \rho [/tex] is the density

## 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 nm

^{2}and continued using [itex] \sigma = 4.3\times 10^{-10} m [/itex] and solved for a diffusion coefficient of ~5 x 10^{-1}m^{2}/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 m^{2}/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 doPlease please help!

*Edit: did my best to correct the formatting, Borek.*
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