How Accurate Is the Pressure Calculation for Argon in a Cubic Container?

[v_pino]

Can I please check if I've converted these units right? I'm doing a question and got the answer a few order of magnitudes out.

Question: Argon has a collision cross section of 0.36nm^2. At what pressure is the mean free path of argon atoms equal to the length of the side of a 1cm^3 cubic container in which they are kept at a temperature of 10K?

Given info
==========
cross section = 0.36 nm^2
volume of cube = 1 cm^3

Conversions
==========
cross section = 0.36 * 10^-18 m^2
volume of cube = 10^-6 m^3
length of one side of cube = 0.01 m

Attempt
=======

mean free path = 1 / ((2^0.5) * cross section area * number of molecules )

number of molecules =1 / ((2^0.5) * (0.36*10^-18) *0.01) = 1.96*10^20 molecules

N = number of molecules

pV = NkT
p = (NkT)/V = ((1.96*10^20)*(1.38*10^-23)*10) / 10^-6
= 2.71*10^4 Pa


Given Solution
============

0.027 Pa

 

[AndyW]

After reviewing your conversions and calculations, it appears that you have made a small error in your conversion of the cross section. The correct conversion for 0.36 nm^2 is 3.6 * 10^-20 m^2, not 0.36 * 10^-18 m^2. This small error has resulted in your final answer being off by a few orders of magnitude.

With the correct conversion, your final answer should be 2.71 * 10^-10 Pa, which is equivalent to 0.027 Pa. This is in line with the given solution of 0.027 Pa.

I hope this helps and please double check your conversions in the future to ensure accurate results.