If 2 mole of an ideal monoatomic gas at temperature $$T$$, is mixed with 6 mole of another ideal monoatomic gas at temperature $$2T$$ then the temperature of mixture is :

use conservation of energy. When the two ideal gases are mixed, the total internal energy of the system remains constant (assuming no heat is lost to the surroundings).

$$U = nC_vT$$.

$$n_1 = 2$$

$$T_1 = T$$

$$C_{v1} = \frac{3}{2}R$$

$$n_2 = 6$$

$$T_2 = 2T$$

$$C_{v2} = \frac{3}{2}R$$

Since both are monoatomic gases, they have the same molar heat capacity. Let's call it $$C_v$$.

Let the final temperature of the mixture be $$T_{mix}$$.

$$ U_1 + U_2 = U_{mix} $$

$$ n_1 C_{v} T_1 + n_2 C_{v} T_2 = (n_1 + n_2) C_{v} T_{mix} $$

Since $$C_v$$ is common in all terms, it cancels out:

$$ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_{mix} $$

$$ 2(T) + 6(2T) = (2 + 6) T_{mix} $$

$$ 2T + 12T = 8 T_{mix} $$

$$ 14T = 8 T_{mix} $$

$$ T_{mix} = \frac{14T}{8} $$

$$ T_{mix} = \frac{7}{4}T $$