Two perfect monoatomic gases at absolute temperature `T_(1)` and `T_(2)` are mixed. There is no loss of energy. Find the temperature of the mixture if the number of moles in the gases are` n_(1)` and `n_(2)`.

Explanation

To find the temperature of the mixture of two perfect monoatomic gases at absolute temperatures \( T_1 \) and \( T_2 \) with number of moles \( n_1 \) and \( n_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Conservation**: Since there is no loss of energy when the two gases are mixed, we can use the principle of conservation of energy. The total energy before mixing should equal the total energy after mixing.

2. **Calculate Initial Energies**: For a monoatomic gas, the energy \( E \) can be expressed as:
\[ E = \frac{3}{2} nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. Therefore, the initial energies of the two gases are:
- For gas 1:
\[ E_1 = \frac{3}{2} n_1 R T_1 \]
- For gas 2:
\[ E_2 = \frac{3}{2} n_2 R T_2 \]

3. **Total Initial Energy**: The total initial energy \( E_{\text{initial}} \) of the system is:
\[ E_{\text{initial}} = E_1 + E_2 = \frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 \]

4. **Final Energy of the Mixture**: Let \( T \) be the final temperature of the mixture. The total energy of the mixture can be expressed as:
\[ E_{\text{final}} = \frac{3}{2} (n_1 + n_2) R T \]

5. **Set Initial Energy Equal to Final Energy**: According to the conservation of energy:
\[ E_{\text{initial}} = E_{\text{final}} \]
Thus, we have:
\[ \frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 = \frac{3}{2} (n_1 + n_2) R T \]

6. **Cancel Common Terms**: We can cancel \( \frac{3}{2} R \) from both sides of the equation:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

7. **Solve for Final Temperature \( T \)**: Rearranging the equation gives:
\[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]

### Final Answer:
The temperature of the mixture is:
\[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]