Two perfect gases at absolute temperature `T_(1) and T_(2)` are mixed. There is no loss of energy. The masses of the molecules are `m_(1) and m_(2)`. The number of molecules in the gases are `n_(1) and n_(2)`. The temperature of the mixture is

Explanation

According to the kinetic theory of gases, the kinetic energy of an ideal gas molecule at temperature is given by `E=(3//2)KT`.
Now if T is the temperature of the mixture, conservation of energy `n_(1)E_(1)+_n_(2)E_(2)=(n_(1)+n_(2))E`
`therefore n_(1)((3)/(2)k_(B_T_(1))+n_(2)((3)/(2)k_(B)T_(2))=(n_(1)+n_(2))((3)/(2)k_(B)T)`
i.e. `T=((n_(1)T_(1)+n_(2)T_(2)))/((n_(1)+n_(2)))`