$\begin{array}{1 1} 1.9\;T_0 \\ 3.9\;T_0 \\ 3.8\;T_0 \\ 4.2\;T_0 \end{array}$

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Since the final internal energy will be the sum of the internal energies of individual gases, $u = u_1 + u_2$,

$\Rightarrow \large \frac{(n_1 + n_2) f R T}{2} $$ = \large \frac{n_1f_1 R 5T_0}{2} $$+ \large\frac{n_2f_2R 3T_0}{2}$, where $T$ is the temperature of the mixture and $f$ is the degrees of freedom of the mixture.

$\Rightarrow (2+3) T = 2 \times 5 T_0 + 3 \times 3 T_0$

$\Rightarrow T = \large\frac{19T_0}{5} $$=3.8 T_0$

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