1. Given data

• Initial gas in vessel: $n_1 = 0.1\ \text{mol}, \quad T_1 = 200\ \text{K}$
• Added gas: $n_2 = 0.05\ \text{mol}, \quad T_2 = 400\ \text{K}$
• Both are the same monoatomic ideal gas.

2. Physical idea

Since the vessel is closed and we are mixing the same ideal gas, the final equilibrium temperature is found by conservation of internal energy.

For a monoatomic ideal gas, $U = n C_V T$ with $C_V = \frac{3R}{2}$ which is the same for both samples.

So, $n_1 C_V T_1 + n_2 C_V T_2 = (n_1+n_2) C_V T_f$

Cancelling $C_V$: $n_1 T_1 + n_2 T_2 = (n_1+n_2) T_f$

3. Substitute values

$T_f = \frac{n_1 T_1 + n_2 T_2}{n_1+n_2}$

$T_f = \frac{(0.1)(200) + (0.05)(400)}{0.1+0.05}$

$T_f = \frac{20 + 20}{0.15}$

$T_f = \frac{40}{0.15} = 266.67\ \text{K}$

4. Final integer value

The temperature is close to $\boxed{267\ \text{K}}$

5. Comparison with stored answer

Stored correct answer: $266$

Our calculated value is $266.67\,\text{K}$, which is closer to $267$ than $266$. So I disagree with the stored answer if the question expects the nearest integer.

If the exam intended truncation or simply “close to”, then $266$ may have been accepted, but mathematically the nearest integer is $267$.