Deduce Avogadro’s law from the kinetic theory of gases.

Avogadro law states that under similar physical conditions of temperature and pressure, equal volume of all the gases contains equal number of molecules. 

Let P be the pressure, V be the volume and T be the absolute temperature. 

Let us consider two gases having same P, V and T. Let one gas contains n₁ molecules, each of mass m₁ and second gas contains n₂ molecules each of mass m₂. Let v₁, and v₂ be the r.m.s. velocities of two gases.

Now, According to kinetic theory of gases, pressure of an ideal gas is given by 

$$\begin{gathered} \mathrm{P} = \frac{1}{3}{\mathrm{\rho v}}^2 = \frac{1}{3}\frac{\mathrm{M}}{\mathrm{V}}{\mathrm{v}}^2 = \frac{1}{3} \frac{\mathrm{nm}}{\mathrm{V}}{\mathrm{v}}^2 \\ \mathrm{For} \mathrm{gas} \mathrm{having} \mathrm{mass} {\mathrm{m}}_1, \\ \mathrm{P}\hspace{0.17em}= \frac{1}{3} \frac{{\mathrm{n}}_1{\mathrm{m}}_1}{\mathrm{V}} {{\mathrm{v}}_1}^2 ... \left(1\right) \\ \mathrm{For} \mathrm{gas} \mathrm{having} \mathrm{mass} {\mathrm{m}}_{2,} \\ \mathrm{P}\hspace{0.17em}= \frac{1}{3} \frac{{\mathrm{n}}_2{\mathrm{m}}_2}{\mathrm{V}} {{\mathrm{v}}_2}^2 ... \left(2\right) \end{gathered}$$

Now, from equations (1) and (2), we have

${\mathrm{m}}_1{\mathrm{n}}_1{{\mathrm{v}}_1}^2 = {\mathrm{m}}_2{\mathrm{n}}_2 {{\mathrm{v}}_2}^2 ... \left(3\right)$

Now, since temperature of both the gases is the same, the average K.E of both the gases will be same.

That is, $\frac{1}{2}{m}_1{v_1}^2 = \frac{1}{2}{m}_2{v_2}^2$       ... (4) 

So, from equation (3) and (4), we have

                      n₁ = n₂

That, is the number of molecules for both gases is the same. 

Hence, Avogadro's law is proved.