A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ___________ .

Assuming  $Q=0,W=0$
We have      $\Delta U=0$ 

        $\Rightarrow n_1{C}_{V_1}\left(T_f-T_1\right)+n_2{C}_{V_2}\left(T_f-T_2\right)=0$

$\Rightarrow n_1{C}_{V_1}\left(T_f-T_1\right)=-n_2{C}_{V_2}\left(T_f-T_2\right)=0$

$\Rightarrow n_1\left(T_f-T_1\right)=-n_2\left(T_f-T_2\right)\left(\because For same gas{C}_{V_1}=C_{V_2} \right)$

$\Rightarrow n_1{T}_f-\hspace{0.17em}\hspace{0.17em}{n}_1{T}_1=-n_2{T}_f+n_2{T}_2$

$\Rightarrow n_1{T}_f+n_2{T}_f=\hspace{0.17em}\hspace{0.17em}{n}_1{T}_1+n_2{T}_2$

$\Rightarrow T_f=\frac{n_1{T}_1+n_2{T}_2}{n_1+n_2}=\frac{\left(0.1\right)\left(200\right)+\left(0.05\right)\left(400\right)}{0.1+0.05}$

$=\frac{20+20}{0.15}=266.6666 K$