Diffusion Coefficient of O2 into N2 (Last Resort

[smoez]

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

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
$$D = \frac {3 \pi}{8} (\frac {k T}{2 \pi \mu})^{0.5} \frac {1}{\rho \sigma}$$
k is the Boltzmann constant
T is temperature
$$\mu = \frac {m1 m2}{m1 + m2}$$ is reduce mass, m is mass
$$\sigma N_2 = 0.43 nm^2$$ is the collision cross section of Nitrogen (O₂ = 0.4 nm²)
$$\rho$$ 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
$$n = \frac {PV}{RT}$$, where $$V = \pi r^2 L$$
$$n = \frac {101300 \times \pi \times (0.0025^2) 0.5}{8.31\times 298}$$
$$n = 4.00 \times 10^{-4} mol$$

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

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

Heres where I get stuck: I originally DID NOT notice that $\sigma$ was in nm² and continued using $\sigma = 4.3\times 10^{-10} m$ 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 $\sigma = 1.89 \times 10^{-19} m$ 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.

 

[Nino MMP]

The issue here is that you have used the wrong collision cross section for nitrogen. While the collision cross section of oxygen is 0.4 nm2, the collision cross section of nitrogen is 0.43 nm2. So, when calculating the diffusion coefficient using the equation mentioned in the homework statement, use the correct cross section value (0.43 nm2). Once you have done that, you need to convert the cross section into m2 by multiplying it with 10^-19. Then, you can plug in the other values in the equation and calculate the diffusion coefficient.