Two ideal monoatomic and diatomic gases are mixed with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is `PV^(gamma)=` constant, where `gamma=(11)/(7)` If `n_(1)` and `n_(2)` are the number of moles of the monoatomic and diatomic gases in the mixture respectively, find the ratio `(n_(1))/(n_(2))`

Answer

To solve the problem, we need to find the ratio of the number of moles of the monoatomic gas \( n_1 \) to the number of moles of the diatomic gas \( n_2 \) given that the adiabatic process of the mixture follows the equation \( PV^{\gamma} = \text{constant} \) with \( \gamma = \frac{11}{7} \).

### Step-by-step Solution:

1. **Identify the specific heat capacities:**
- For a monoatomic gas, the specific heat at constant volume \( C_{V1} = \frac{3}{2} R \) and the specific heat at constant pressure \( C_{P1} = \frac{5}{2} R \).
- For a diatomic gas, the specific heat at constant volume \( C_{V2} = \frac{5}{2} R \) and the specific heat at constant pressure \( C_{P2} = \frac{7}{2} R \).

2. **Calculate the specific heat capacities of the mixture:**
- The specific heat at constant pressure for the mixture \( C_{P} \) is given by:
\[ C_{P} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2} \]
- The specific heat at constant volume for the mixture \( C_{V} \) is given by:
\[ C_{V} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} \]

3. **Substituting the values:**
- Substitute \( C_{P1}, C_{P2}, C_{V1}, C_{V2} \):
\[ C_{P} = \frac{n_1 \left(\frac{5}{2} R\right) + n_2 \left(\frac{7}{2} R\right)}{n_1 + n_2} \]
\[ C_{V} = \frac{n_1 \left(\frac{3}{2} R\right) + n_2 \left(\frac{5}{2} R\right)}{n_1 + n_2} \]

4. **Finding the ratio \( \gamma \):**
- The ratio \( \gamma \) for the mixture is given by:
\[ \gamma = \frac{C_{P}}{C_{V}} \]
- Substitute the expressions for \( C_{P} \) and \( C_{V} \):
\[ \gamma = \frac{\frac{n_1 \left(\frac{5}{2} R\right) + n_2 \left(\frac{7}{2} R\right)}{n_1 + n_2}}{\frac{n_1 \left(\frac{3}{2} R\right) + n_2 \left(\frac{5}{2} R\right)}{n_1 + n_2}} \]
- Simplifying this gives:
\[ \gamma = \frac{n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right)}{n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)} \]

5. **Setting up the equation:**
- Given \( \gamma = \frac{11}{7} \), we can set up the equation:
\[ \frac{5n_1 + 7n_2}{3n_1 + 5n_2} = \frac{11}{7} \]

6. **Cross-multiplying:**
- Cross-multiply to eliminate the fraction:
\[ 7(5n_1 + 7n_2) = 11(3n_1 + 5n_2) \]
- Expanding both sides:
\[ 35n_1 + 49n_2 = 33n_1 + 55n_2 \]

7. **Rearranging the equation:**
- Rearranging gives:
\[ 35n_1 - 33n_1 = 55n_2 - 49n_2 \]
\[ 2n_1 = 6n_2 \]

8. **Finding the ratio:**
- Dividing both sides by \( n_2 \):
\[ \frac{n_1}{n_2} = \frac{6}{2} = 3 \]

### Final Answer:
The ratio \( \frac{n_1}{n_2} = 3 \).