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 :

Using the principle of conservation of internal energy for the mixture of ideal gases:

\(U_{mix} = U_1 + U_2\)

\((n_1 + n_2) C_{V} T_{mix} = n_1 C_{V} T_1 + n_2 C_{V} T_2\)

Since both gases are monoatomic, their molar heat capacities at constant volume are equal (\(C_{V} = \dfrac{3}{2}R\)).

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

Substituting the given values \(n_1 = 2\), \(T_1 = T\), \(n_2 = 6\), and \(T_2 = 2T\):

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

\(8 T_{mix} = 2T + 12T\)

\(8 T_{mix} = 14T\)

\(T_{mix} = \dfrac{14}{8}T = \dfrac{7}{4}T\)

Answer: \(\dfrac{7}{4}T\)