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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.