Two polyatomic ideal gases are mixed together temperatures `T_1` and `T_2`, number of molecules `N_1` and `N_2`, mass of particles `m_1` and `m_2`, degrees of freedom `f_1` and `f_2` find final temperatures of mixtures.

Options

A. `(N_1 T_1 + N_2 T_2)/(N_1 + N_2)`
B. `(N_1 f_1 T_1 + N_2 f_2 T_2)/(N_1 f_1 + N_2 f_2)`
C. `(f_1 T_1 + f_2 T_2)/(f_1 + f_2)`
D. `(T_1 + T_2)/2`

Answer

To find the final temperature of the mixture of two polyatomic ideal gases, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Variables:**
- Let \( T_1 \) and \( T_2 \) be the initial temperatures of the two gases.
- Let \( N_1 \) and \( N_2 \) be the number of molecules of the two gases.
- Let \( m_1 \) and \( m_2 \) be the masses of the particles of the two gases.
- Let \( f_1 \) and \( f_2 \) be the degrees of freedom of the two gases.

2. **Calculate the Equivalent Degrees of Freedom:**
- The equivalent degrees of freedom \( f_{eq} \) for the mixture can be calculated using:
\[ f_{eq} = \frac{N_1 f_1 + N_2 f_2}{N_1 + N_2} \]

3. **Calculate the Internal Energy of Each Gas:**
- The internal energy \( U \) of an ideal gas can be expressed as:
\[ U = \frac{N}{N_A} \cdot \frac{f}{2} R T \]
- For gas 1, the internal energy \( U_1 \) is:
\[ U_1 = \frac{N_1}{N_A} \cdot \frac{f_1}{2} R T_1 \]
- For gas 2, the internal energy \( U_2 \) is:
\[ U_2 = \frac{N_2}{N_A} \cdot \frac{f_2}{2} R T_2 \]

4. **Total Internal Energy of the Mixture:**
- The total internal energy \( U_{total} \) of the mixture is:
\[ U_{total} = U_1 + U_2 = \frac{N_1 f_1 R T_1}{2 N_A} + \frac{N_2 f_2 R T_2}{2 N_A} \]

5. **Internal Energy of the Mixture:**
- The internal energy of the mixture can also be expressed as:
\[ U_{mix} = \frac{N_1 + N_2}{N_A} \cdot \frac{f_{eq}}{2} R T \]
- Where \( T \) is the final temperature of the mixture.

6. **Set the Total Internal Energy Equal to the Internal Energy of the Mixture:**
- Equating the two expressions for internal energy:
\[ \frac{N_1 f_1 R T_1}{2 N_A} + \frac{N_2 f_2 R T_2}{2 N_A} = \frac{N_1 + N_2}{N_A} \cdot \frac{f_{eq}}{2} R T \]

7. **Cancel Common Terms:**
- Cancel \( R \) and \( N_A \) from both sides:
\[ \frac{N_1 f_1 T_1}{2} + \frac{N_2 f_2 T_2}{2} = \frac{(N_1 + N_2) f_{eq}}{2} T \]

8. **Solve for Final Temperature \( T \):**
- Rearranging gives:
\[ T = \frac{N_1 f_1 T_1 + N_2 f_2 T_2}{N_1 f_1 + N_2 f_2} \]

### Final Result:
The final temperature \( T \) of the mixture is given by:
\[ T = \frac{N_1 f_1 T_1 + N_2 f_2 T_2}{N_1 f_1 + N_2 f_2} \]