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Step-by-Step Solution
Step 1: Write the expression for total initial internal energy
Each ideal polyatomic gas with degrees of freedom $F$ has an internal energy given by
$U = \tfrac{F}{2} \, n \, R \, T$.
Therefore, the total initial internal energy (before mixing) is the sum of the internal energies of both gases:
$U_\text{initial}
= \frac{F_1}{2} \, n_1 \, R \, T_1 \;+\; \frac{F_2}{2} \, n_2 \, R \, T_2.$
Step 2: Write the expression for total final internal energy
When the two gases are mixed without any loss of energy, the final internal energy is the sum of their individual internal energies at the common final temperature $T$:
$U_\text{final}
= \frac{F_1}{2} \, n_1 \, R \, T \;+\; \frac{F_2}{2} \, n_2 \, R \, T.$
Step 3: Equate initial and final internal energies
Since energy is conserved (no energy loss):
$\frac{F_1}{2} \, n_1 \, R \, T_1 \;+\; \frac{F_2}{2} \, n_2 \, R \, T_2
= \frac{F_1}{2} \, n_1 \, R \, T \;+\; \frac{F_2}{2} \, n_2 \, R \, T.$
We can factor out $\tfrac{R}{2}$ from both sides:
$F_1 \, n_1 \, T_1 \;+\; F_2 \, n_2 \, T_2
= F_1 \, n_1 \, T \;+\; F_2 \, n_2 \, T.
Step 4: Solve for the final temperature $T$
Rearrange to isolate $T$:
$F_1 \, n_1 \, T_1 + F_2 \, n_2 \, T_2
= \bigl(F_1 \, n_1 + F_2 \, n_2 \bigr)\, T.
$\implies T = \frac{F_1 \, n_1 \, T_1 + F_2 \, n_2 \, T_2}{F_1 \, n_1 + F_2 \, n_2}.
Final Answer
The temperature of the mixture is
$\displaystyle T \;=\; \frac{\,n_1 \, F_1 \, T_1 \;+\; n_2 \, F_2 \, T_2\,}{\,n_1 \, F_1 \;+\; n_2 \, F_2\,}.$
