Usually the average speed of gas molecules is several hundred meters per second even at room temperature \( (27^{\circ}C) \). Odour from an open perfume bottle takes some time to reach us even if we are closer to the room. The time delay is because the odour of the molecules cannot travel straight to us as it undergoes a lot of collisions with the nearby air molecules and moves in a zigzag path. This average distance travelled by the molecule between two successive collisions is called mean free path \( (\lambda) \). We can calculate the mean free path based on kinetic theory.
Expression for mean free path
We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves with uniform velocity. The mean free path of a gas molecule is given by
\[ \lambda = \frac{1}{\sqrt{2} \pi n d^{2}} \]where \( n \) is the number density and \( d \) is the diameter of the molecule.
Using ideal gas law \( n = \frac{P}{kT} \), we can also write
\[ \lambda = \frac{kT}{\sqrt{2} \pi d^{2} P} \]EXAMPLE 9.6
An oxygen molecule is travelling in air at \( 300 \, \text{K} \) and 1 atm, and the diameter of oxygen molecule is \( 1.2 \times 10^{-10} \, \text{m} \). Calculate the mean free path of oxygen molecule.
Solution
\[ \lambda = \frac{1}{\sqrt{2} \pi n d^{2}} \]We have to find the number density \( n \). By using ideal gas law
\[ n = \frac{N}{V} = \frac{P}{kT} = \frac{101.3 \times 10^{3}}{1.381 \times 10^{-23} \times 300} = 2.449 \times 10^{25} \, \text{molecules/m}^{3} \]\[ \lambda = \frac{1}{\sqrt{2} \times \pi \times 2.449 \times 10^{25} \times (1.2 \times 10^{-10})^{2}} = \frac{1}{15.65 \times 10^{5}} = 0.63 \times 10^{-6} \, \text{m} \]