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## Main Question or Discussion Point

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

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