1. Use internal energy of an ideal gas

For an ideal gas with degrees of freedom $F$, number of molecules $n$, and temperature $T$, the total internal energy is

$U = \frac{F}{2} n k T$

where $k$ is Boltzmann constant.

2. Write initial total energy of the two gases

For the first gas,

$U_1 = \frac{F_1}{2} n_1 k T_1$

For the second gas,

$U_2 = \frac{F_2}{2} n_2 k T_2$

So initial total energy is

$U_{\text{initial}} = \frac{k}{2}(n_1F_1T_1 + n_2F_2T_2)$

3. After mixing

Since there is no loss of energy, total internal energy remains conserved.

Let final temperature of mixture be $T$.

Then total internal energy of the mixture is

$U_{\text{final}} = \frac{F_1}{2} n_1 k T + \frac{F_2}{2} n_2 k T$

$U_{\text{final}} = \frac{k}{2}(n_1F_1 + n_2F_2)T$

4. Apply conservation of energy

$U_{\text{initial}} = U_{\text{final}}$

$\frac{k}{2}(n_1F_1T_1 + n_2F_2T_2) = \frac{k}{2}(n_1F_1 + n_2F_2)T$

Cancel $\frac{k}{2}$:

$n_1F_1T_1 + n_2F_2T_2 = (n_1F_1 + n_2F_2)T$

Hence,

$T = \frac{n_1F_1T_1 + n_2F_2T_2}{n_1F_1 + n_2F_2}$

5. Match with options

This matches Option B.

6. Check note about masses

The masses $m_1, m_2$ are irrelevant here because internal energy of an ideal gas depends on temperature and degrees of freedom, not directly on molecular mass.