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)`.

Options

A. `(T_(1)+T_(2))/(n_(1)+n_(2))`
B. `(T_(1))/(n_(1))+(T_(2))/(n_(2))`
C. `(n_(2)T_(1)+n_(1)T_(2))/(n_(1)+n_(2))`
D. `(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`

Answer

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 of a Gas**:
The internal energy \( U \) of a monoatomic gas at temperature \( T \) is given by the formula:
\[ U = n \cdot C_v \cdot T \]
where \( n \) is the number of moles and \( C_v \) is the specific heat capacity at constant volume.

2. **Specific Heat Capacity for Monoatomic Gases**:
For a monoatomic gas, the specific heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{3}{2} R \]
where \( R \) is the universal gas constant.

3. **Calculate Initial Energies**:
The initial energy of the first gas at temperature \( T_1 \) is:
\[ U_1 = n_1 \cdot C_v \cdot T_1 \]
The initial energy of the second gas at temperature \( T_2 \) is:
\[ U_2 = n_2 \cdot C_v \cdot T_2 \]

4. **Total Initial Energy**:
The total initial energy of the system is:
\[ U_{\text{initial}} = U_1 + U_2 = n_1 \cdot C_v \cdot T_1 + n_2 \cdot C_v \cdot T_2 \]

5. **Final Energy of the Mixture**:
After mixing, the total number of moles is \( n_1 + n_2 \), and the temperature of the mixture is \( T_{\text{mixture}} \). The final energy of the mixture is:
\[ U_{\text{final}} = (n_1 + n_2) \cdot C_v \cdot T_{\text{mixture}} \]

6. **Conservation of Energy**:
Since there is no loss of energy, we set the initial energy equal to the final energy:
\[ U_{\text{initial}} = U_{\text{final}} \]
This gives us:
\[ n_1 \cdot C_v \cdot T_1 + n_2 \cdot C_v \cdot T_2 = (n_1 + n_2) \cdot C_v \cdot T_{\text{mixture}} \]

7. **Cancel \( C_v \)**:
Since \( C_v \) is common in all terms, we can cancel it out:
\[ n_1 \cdot T_1 + n_2 \cdot T_2 = (n_1 + n_2) \cdot T_{\text{mixture}} \]

8. **Solve for \( T_{\text{mixture}} \)**:
Rearranging the equation to solve for \( T_{\text{mixture}} \):
\[ T_{\text{mixture}} = \frac{n_1 \cdot T_1 + n_2 \cdot T_2}{n_1 + n_2} \]

### Final Answer:
The temperature of the mixture is given by:
\[ T_{\text{mixture}} = \frac{n_1 \cdot T_1 + n_2 \cdot T_2}{n_1 + n_2} \]