Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture is

Options

A. `(n_1T_1^2+n_2T_2^2+n_3T_3^2)/(n_1T_1+n_2T_2+n_3T_3)`
B. `(n_1^2T_1^2+n_2^2T_2^2+n_3^2T_3^2)/(n_1T_1+n_2T_2+n_3T_3)`
C. `(T_1+T_2+T_3)/3`
D. `(n_1T_1+n_2T_2+n_3T_3)/(n_1+n_2+n_3)`

Answer

D

Explanation

For a mixture of gases, sum of the heats given to each gas is equal to total energy supplied to the system.
Let T be the teperature of mixture, `T_1, T_2, T_3` be the teemperatures of individual gases before mixing.
`(n_1C_(V_(1))T_1 +n_2C_(V_(2))T_2+n_3C_(V_(3))T_3)=(n_1+n_2+n_3)C_vT`
As, `C_(V_1)=C_(V_2)=C_(V_3)` , So , `T = (n_1T_1+n_2T_2+n_3T_3)/(n_1+n_2+n_3)`