Two ideal polyatomic gases at temperatures $$T_1$$ and $$T_2$$ are mixed so that there is no loss of energy. If $$F_1$$ and $$F_2$$, $$m_1$$ and $$m_2$$, $$n_1$$ and $$n_2$$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:

When two ideal gases are mixed with no loss of energy, the total internal energy is conserved. The internal energy of an ideal gas with $$n$$ molecules, each having $$F$$ degrees of freedom, at temperature $$T$$ is $$U = n \cdot \frac{F}{2} k_B T$$, where $$k_B$$ is the Boltzmann constant.

Before mixing: $$U_{\text{total}} = n_1 \cdot \frac{F_1}{2} k_B T_1 + n_2 \cdot \frac{F_2}{2} k_B T_2$$. After mixing at temperature $$T$$: $$U_{\text{total}} = n_1 \cdot \frac{F_1}{2} k_B T + n_2 \cdot \frac{F_2}{2} k_B T$$.

Equating these and cancelling $$\frac{k_B}{2}$$ from both sides: $$n_1 F_1 T_1 + n_2 F_2 T_2 = (n_1 F_1 + n_2 F_2) T$$.

Solving for the mixture temperature gives $$T = \frac{n_1 F_1 T_1 + n_2 F_2 T_2}{n_1 F_1 + n_2 F_2}$$.

The correct answer is option 2.