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. `3/2`
D. `(12)/(7)`

Answer

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

### Step 1: Identify the specific heat capacities
For the first gas (Gas A) with \( \gamma = \frac{7}{5} \):
- The specific heat at constant pressure \( C_{P1} \) is given by:
\[ C_{P1} = \frac{7}{2} R \]
- The specific heat at constant volume \( C_{V1} \) can be calculated using the relation:
\[ C_{V1} = C_{P1} - R = \frac{7}{2} R - R = \frac{5}{2} R \]

For the second gas (Gas B) with \( \gamma = \frac{5}{3} \):
- The specific heat at constant pressure \( C_{P2} \) is given by:
\[ C_{P2} = \frac{5}{2} R \]
- The specific heat at constant volume \( C_{V2} \) can be calculated using the relation:
\[ C_{V2} = C_{P2} - R = \frac{5}{2} R - R = \frac{3}{2} R \]

### Step 2: Calculate the effective specific heats for the mixture
Since we are mixing equal moles of both gases, we can find the effective specific heats.

**Effective \( C_P \)**:
\[ C_{P_{\text{effective}}} = \frac{C_{P1} + C_{P2}}{2} = \frac{\frac{7}{2} R + \frac{5}{2} R}{2} = \frac{12}{2} R \cdot \frac{1}{2} = \frac{6}{2} R = 3R \]

**Effective \( C_V \)**:
\[ C_{V_{\text{effective}}} = \frac{C_{V1} + C_{V2}}{2} = \frac{\frac{5}{2} R + \frac{3}{2} R}{2} = \frac{8}{2} R \cdot \frac{1}{2} = \frac{4}{2} R = 2R \]

### Step 3: Calculate the effective gamma for the mixture
Now we can find the effective \( \gamma \) for the mixture using the formula:
\[ \gamma_{\text{effective}} = \frac{C_{P_{\text{effective}}}}{C_{V_{\text{effective}}}} = \frac{3R}{2R} = \frac{3}{2} \]

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

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