Mean Free Path of Air Molecules

[cwill53]

Homework Statement

Calculate the mean free path of air molecules at $3.5\cdot 10^{-13}atm$ and 300K. Model the air molecules as spheres of radius $2.0\cdot 10^{-10}m$

Relevant Equations

$$\lambda =\frac{k_{b}T}{4\pi \sqrt{2}r^2p}$$

I used the form of the mean free path equation taking advantage of the fact that the Boltzmann constant is equal to the ideal gas constant R divided by Avogadro's number, because I didn't know if I could use the Boltzmann constant in the $1.381\cdot 10^{-23}J/(molecules\cdot K)$ form:

$$\lambda =\frac{RT}{4\pi \sqrt{2}r^2pN_{A}}=\frac{(008206(L\cdot atm)/(mol\cdot K))(300K)}{4\pi \sqrt{2}(2.0\cdot 10^{-10}m)^2(3.5\cdot 10^{-13}atm)(6.022\cdot 10^{23}mol^{-1})}=164308154.9m$$

There is no answer written in the back of the book for this particular question and when I looked the question up online I'm getting answers far from mine. I would have thought we would have to use the R constant with the units $\frac{L\cdot atm}{mol\cdot K}$.

 

[shjacks45]

10^-13 atm. A better vacuum than I've ever used, even in vacuum tubes. A long free path is reasonable.

 

[TSny]

Note that overall your units reduce to L/m². Did you convert L to m³? Also, you wrote the value of R as 008206 L⋅atm/(mol⋅K). Is there a decimal point somewhere?

 

[cwill53]

Note that overall your units reduce to L/m². Did you convert L to m³? Also, you wrote the value of R as 008206 L⋅atm. Is there a decimal point somewhere?

That was supposed to be 0.08206, I just forgot the decimal point only in the LaTeX.
I don’t understand which R value I should use if I have pressure in atm. Should I convert the pressure to pascals?

 

[TSny]

That was supposed to be 0.08206, I just forgot the decimal point only in the LaTeX.
I don’t understand which R value I should use if I have pressure in atm. Should I convert the pressure to pascals?

Your calculation is fine except that it gives the answer in units of L/m². You just need to reduce this to meters.

 

[cwill53]

Your calculation is fine except that it gives the answer in units of L/m². You just need to reduce this to meters.

I see. After converting units I got the correct answer of 164308 meters. I guess I got lazy and assumed that plug in chug would yield an answer in the correct units. Thanks.