Two ideal gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy in this process. If n₁ and n₂ are the respective number of molecules of the gases, the temperature of the mixture will be 

Average kinetic energy per molecule of a perfect  gas =$\frac{3}{2}kT$.

$\therefore$ Average KE of molecules of the first gas $=\frac{3}{2}{n}_{1}k{T}_1$

Average KE of molecules of the second gas $=\frac{3}{2}{n}_{2}k{T}_2$

$\therefore$Total KE of the molecules of the two gases before they are mixed is

                        $$\begin{gathered} K=\frac{3}{2}{n}_{1}k{T}_1+\frac{3}{2}{n}_{2}k{T}_2 \\ =\frac{3}{2}\left(n_1{T}_1+n_2{T}_2\right)k \left(1\right) \end{gathered}$$                     

If T is the temperature of the mixture, the kinetic energy of the molecules (n₁ + n₂) in the mixture is       

                     $K\text{'}=\frac{3}{2}\left(n_1+n_2\right)kT \left(2\right)$

Since there is no loss of energy K = K'. Equating (1) and (2) we get 

                 $n_1{T}_1+n_2{T}_2=\left(n_1+n_2\right)T or T=\frac{n_1{T}_1+n_2{T}_2}{n_1+n_2}$