Question

3) Mean & Free. Calculate the mean free path in Earth's atmosphere. Use sea level conditions and the collision cross section section under what conditions would the mean free path equal 10km? Compare those conditions to your first calculation For the same cross-

Cross section:$\sigma =10^{-15}cm^{2}$

Answer 1

Ans:

This is we know:   λ = 1 / [(N/V) π r²]
where r is the radius of a molecule.

numerical estimates
Molecular cross-sectional area:   π r² ≈ 10^-19 m²

Mean free path (sea level):   &lambda ≈ 10000 m ≈ hundreds of molecular diameters

So,

Molecular density (sea level):   N/V = P/kT ≈ 1 x 10¹⁵ molecules/m³