Question

Calculate the mean free path of air molecules at 3.50 * 10-13 atm and 300 k. (this pressure is readily attainable in the laboratory; see exercise 18.23.) as in example 1

Answer 1

The mean free path can be defined as the average distance between collisions for gas molecules. It can be calculated using the formula:

∧ = $\frac{kT}{\sqrt{2}\pi d^{2}N_{A}P    }$---------(1)

Given :

T = 300 K

P = 3.50 * 10-13 atm = 3.55 * 10^-8 pascals

d = diameter of air molecules = 2* radius of air molecules

                                                 = 2* 2*10^-10 m = 4*10^-10 m

k = Boltzmann constant = 1.38 * 10^-23 J/K

Na = avogadro's number = 6.023 * 10^23

Substituting in equation(1) we get:

∧ = 1.38 *10^-23* 300/√2*π*(4*10^-10)²*3.55*10^-8

  = 1.641*10^5 m