A closed vessel contains 0.1 mole of a monoatomic 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 __________

We have a rigid, perfectly insulated (adiabatic) vessel, so there is no work done (because the volume is constant) and no heat exchanged with the surroundings. Under these conditions the total internal energy of the gas in the vessel remains conserved during the mixing process.

For an ideal gas the molar internal energy depends only on temperature. Specifically, for a mono-atomic ideal gas the internal energy per mole is given by the well-known formula

$$U_{\text{molar}}=\frac{3}{2}RT,$$

where $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

If a sample contains $$n$$ moles, its total internal energy is therefore

$$U=\frac{3}{2}nRT.$$

Initially the vessel holds

$$n_1 = 0.1\ \text{mol}, \qquad T_1 = 200\ \text{K}.$$

So its internal energy is

$$U_1=\frac{3}{2} n_1 R T_1 = \frac{3}{2} R (0.1)(200).$$

An additional sample is injected containing

$$n_2 = 0.05\ \text{mol}, \qquad T_2 = 400\ \text{K},$$

whose internal energy before mixing is

$$U_2=\frac{3}{2} n_2 R T_2 = \frac{3}{2} R (0.05)(400).$$

After the two portions mix and thermal equilibrium is reached, the total number of moles inside the vessel becomes

$$n_f = n_1 + n_2 = 0.1 + 0.05 = 0.15\ \text{mol},$$

and let the common final temperature be $$T_f$$. Because energy is conserved, we set the sum of the initial internal energies equal to the final internal energy:

$$U_1 + U_2 = U_f.$$

Substituting the expressions for each internal energy, we have

$$\frac{3}{2} R n_1 T_1 + \frac{3}{2} R n_2 T_2 = \frac{3}{2} R n_f T_f.$$

Every term contains the common factor $$\dfrac{3}{2}R$$, so it cancels out, yielding the simple relation

$$n_1 T_1 + n_2 T_2 = n_f T_f.$$

Now we substitute the numerical values step by step:

$$\bigl(0.1\bigr)(200) + \bigl(0.05\bigr)(400) = \bigl(0.15\bigr) T_f.$$

First, calculate each product:

$$0.1 \times 200 = 20,$$

$$0.05 \times 400 = 20.$$

Adding them gives

$$20 + 20 = 40.$$

Hence,

$$40 = 0.15\, T_f.$$

Solving for $$T_f$$, we divide both sides by $$0.15$$:

$$T_f = \frac{40}{0.15}.$$

Carrying out the division,

$$\frac{40}{0.15} = 266.\overline{6}\ \text{K},$$

which is very close to $$267\ \text{K}$$ when rounded to the nearest whole number.

So, the answer is $$267\ \text{K}$$.