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 of finding the ratio of the number of moles of monoatomic gas \( n_1 \) to diatomic gas \( n_2 \) in an ideal gas mixture undergoing an adiabatic process, we follow these steps:

### Step 1: Understand the given information
We know that the adiabatic process for the mixture is described by the equation:
\[ PV^\gamma = \text{constant} \]
where \( \gamma = \frac{11}{7} \).

### Step 2: Recall the definition of \( \gamma \)
For an ideal gas, \( \gamma \) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
where \( C_p \) is the specific heat at constant pressure and \( C_v \) is the specific heat at constant volume.

### Step 3: Determine \( C_p \) and \( C_v \) for monoatomic and diatomic gases
For a monoatomic gas:
- \( C_v = \frac{3}{2} R \)
- \( C_p = \frac{5}{2} R \)

For a diatomic gas:
- \( C_v = \frac{5}{2} R \)
- \( C_p = \frac{7}{2} R \)

### Step 4: Write the expressions for \( C_p \) and \( C_v \) of the mixture
The specific heats for the mixture can be expressed as:
\[ C_{p,\text{mixture}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} \]
\[ C_{v,\text{mixture}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]

### Step 5: Substitute the values of \( C_p \) and \( C_v \)
Substituting the values into the equations:
\[ C_{p,\text{mixture}} = \frac{n_1 \left(\frac{5}{2} R\right) + n_2 \left(\frac{7}{2} R\right)}{n_1 + n_2} \]
\[ C_{v,\text{mixture}} = \frac{n_1 \left(\frac{3}{2} R\right) + n_2 \left(\frac{5}{2} R\right)}{n_1 + n_2} \]

### Step 6: Write the expression for \( \gamma \) of the mixture
Now, substituting these into the expression for \( \gamma \):
\[ \gamma = \frac{C_{p,\text{mixture}}}{C_{v,\text{mixture}}} = \frac{\frac{n_1 \left(\frac{5}{2} R\right) + n_2 \left(\frac{7}{2} R\right)}{n_1 + n_2}}{\frac{n_1 \left(\frac{3}{2} R\right) + n_2 \left(\frac{5}{2} R\right)}{n_1 + n_2}} \]

### Step 7: Simplify the expression
This simplifies to:
\[ \gamma = \frac{n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right)}{n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)} \]

### Step 8: Substitute the value of \( \gamma \)
Substituting \( \gamma = \frac{11}{7} \):
\[ \frac{n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right)}{n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)} = \frac{11}{7} \]

### Step 9: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 7\left(n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right)\right) = 11\left(n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)\right) \]

### Step 10: Expand and simplify
Expanding both sides:
\[ \frac{35}{2} n_1 + \frac{49}{2} n_2 = \frac{33}{2} n_1 + \frac{55}{2} n_2 \]

### Step 11: Rearranging the equation
Rearranging gives:
\[ 35 n_1 + 49 n_2 = 33 n_1 + 55 n_2 \]
\[ 2 n_1 = 6 n_2 \]

### Step 12: Solve for the ratio \( \frac{n_1}{n_2} \)
Thus, we find:
\[ \frac{n_1}{n_2} = \frac{6}{2} = 3 \]

### Final Answer
The ratio of the number of moles of monoatomic gas to diatomic gas is:
\[ \frac{n_1}{n_2} = 3 \]