At what pressure does the mean free path of argon at 20 degrees celsius become comparable to the diameter of a 100 cm3vessel that contains it? note sigma= 0.36 nm2. sigma= collision cross section of argon.

The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa

Explanation:

Step 1: Calculate the volume of a spherical container V

V = (4π*r³)/3

r = (3V/4π)^1/3

2r = d = 2*(3V/4π)^1/3

with r= radius

with d= diameter

The diameter is:

d= 2*(3V/4π)^1/3

d= 2*(3*100cm³/4π)^1/3

d= 5.76 cm

Step2 : Define the free path lambda  λ of argon

with  λ =k*T/ σp

with p = kT/σλ

with T= temperature = 20°C = 293.15 Kelvin

with k = Boltzmann's constant = 1.381 * 10^-23 J/K

with p = the atmospheric pressure

with σ = 0.36 nm²

p = kT/σλ

p = (1.38 * 10^-23 J*K^-1 * 1Pa *m³/1J)*(293,15K) /(0.36 nm²*(10^-9/ 1nm)² *(5.76cm* 10^-2m/1cm)

p = 0.195 Pa

The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa

d.) circle motion requires a force to give it a centripetal acceleration.

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