If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture is

Explanation

To find the value of gamma for the mixture of one mole of a monatomic gas and one mole of a diatomic gas, we can follow these steps:

### Step 1: Identify the values of gamma for each gas
- For the monatomic gas, \( \gamma_1 = \frac{5}{3} \).
- For the diatomic gas, \( \gamma_2 = \frac{7}{5} \).

### Step 2: Calculate the specific heats for each gas
Using the relationships between \( C_p \), \( C_v \), and \( \gamma \):
- For a monatomic gas:
\[ C_p = \frac{5}{2} R, \quad C_v = \frac{3}{2} R \]
- For a diatomic gas:
\[ C_p = \frac{7}{2} R, \quad C_v = \frac{5}{2} R \]

### Step 3: Calculate the total specific heats for the mixture
The total specific heats for the mixture can be calculated as follows:
- Total \( C_p \) for the mixture:
\[ C_{p, \text{net}} = C_{p1} + C_{p2} = \frac{5}{2} R + \frac{7}{2} R = \frac{12}{2} R = 6R \]
- Total \( C_v \) for the mixture:
\[ C_{v, \text{net}} = C_{v1} + C_{v2} = \frac{3}{2} R + \frac{5}{2} R = \frac{8}{2} R = 4R \]

### Step 4: Calculate the gamma for the mixture
Using the formula for gamma:
\[ \gamma_{\text{mixture}} = \frac{C_{p, \text{net}}}{C_{v, \text{net}}} \]
Substituting the values we calculated:
\[ \gamma_{\text{mixture}} = \frac{6R}{4R} = \frac{6}{4} = \frac{3}{2} = 1.5 \]

### Final Answer
The value of gamma for the mixture is \( \frac{3}{2} \) or \( 1.5 \).

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