Question

At what pressure does the mean free path of argon at 25 degree Celsius become comparable to the diameter of a spherical vessel of volume 1dm^3 that contains it? Take Sigma = 0.36nm^2. PLEASE SHOW YOUR WORK

***I'm confused about what to take as the mean free path when calculating pressure. Do i take the diameter of the vessel or the diameter from
(sigma=pi×d^2 which is the cross-sectional collision area)

Answer 1

Given-

T = 25 degree Celsius = 25 +273 = 298K

Let λ = 1dm = 0.1m

δ = 0.36nm^2

δ = 0.36 x 10^-18 m^2

λ = Kt/(2)^0.5δP

Where

λ = the mean free path = 0.1mh22

K = Boltzmann constant 1.38e-23 m^2 kg s^-2 K^-1

Sigma = mean free path= 0.36nm^2 = 3.6 x 10^-17m^2

P = Kt/√2δ λ

Plugging all the values in equation-

P = 1.38e-23 m^2 kg s^-2 K^-1 x 298K /1.414 x 3.6 x 10^-17m^2 x 0.1 m

P = 8.07 x 10^-4 KPa

P = 0.0807Pa = 0.081Pa

You should take (sigma=pi×d^2 which is the cross-sectional collision area), which is already given.