Answer:
The mean free path = 2.16*10^-6 m
Explanation:
Given:
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
To determine:
The mean free path, λ
Calculation:
The mean free path is related to the collision cross section by the following equation:
[tex]\lambda =\frac{1}{n\sigma }------(1)[/tex]
where n = number density
[tex]n = \frac{P}{kT}-----(2)[/tex]
Substituting for P, k and T in equation (2) gives:
[tex]n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}[/tex]
Next, substituting for n and σ in equation (1) gives:
[tex]\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m[/tex]
