Kinetic Theory — Physics STD 11 Science — Question

$\lambda=\frac{\text{kT}}{\sqrt{2}\pi\text{d}^2\rho}=\frac{\text{m}}{\sqrt{2}\pi\text{d}^2\rho}=\frac{1}{\sqrt{2}\pi\text{n}\text{d}^2}$

$\therefore\lambda$ depends upon:

1. Diameter (d) of the molecule, smaller the 'd', larger is the mean free path $\lambda.$

2. $\lambda\propto\text{T}$ i.e., higher the temperature larger is the $\lambda.$

3. $\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the pressure larger is the $\lambda.$

4. $\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the density $(\rho)$ larger will be the $\lambda.$

5. $\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the number of molecules per unit volume of the gas, larger is the $\lambda.$