1 mole of a gas with `gamma=7//5` is mixed with 1 mole of a gas with `gamma=5//3`, then the value of `gamma` for the resulting mixture is

Options

A. `7//5`
B. `2//5`
C. `24//16`
D. `12//7`

Explanation

To find the value of \( \gamma \) for the resulting mixture of gases, we can follow these steps:

### Step 1: Identify the given values of \( \gamma \)
We have two gases:
- Gas 1: \( \gamma_1 = \frac{7}{5} \) (diatomic gas)
- Gas 2: \( \gamma_2 = \frac{5}{3} \) (monatomic gas)

### Step 2: Calculate \( C_p \) and \( C_v \) for each gas
For the diatomic gas (Gas 1):
- \( C_{p1} = \frac{7}{2} R \)
- \( C_{v1} = \frac{5}{2} R \)

For the monatomic gas (Gas 2):
- \( C_{p2} = \frac{5}{2} R \)
- \( C_{v2} = \frac{3}{2} R \)

### Step 3: Calculate the total \( C_p \) and \( C_v \) for the mixture
Since we have 1 mole of each gas, the total \( C_p \) and \( C_v \) for the mixture can be calculated as follows:

Total \( C_p \):
\[ C_{p \text{ total}} = C_{p1} + C_{p2} = \frac{7}{2} R + \frac{5}{2} R = \frac{12}{2} R = 6R \]

Total \( C_v \):
\[ C_{v \text{ total}} = C_{v1} + C_{v2} = \frac{5}{2} R + \frac{3}{2} R = \frac{8}{2} R = 4R \]

### Step 4: Calculate the new \( \gamma \) for the mixture
Now, we can calculate the \( \gamma \) for the mixture using the formula:
\[ \gamma = \frac{C_p}{C_v} \]
Substituting the values we found:
\[ \gamma = \frac{C_{p \text{ total}}}{C_{v \text{ total}}} = \frac{6R}{4R} = \frac{6}{4} = \frac{3}{2} \]

### Conclusion
The value of \( \gamma \) for the resulting mixture is \( \frac{3}{2} \).