One mole of gas having `gamma = 7//5` is mixed with 1 mole of a gas having `gamma = 4//3`. What will be `gamma` for the mixture ?

Options

A. `(15)/(11)`
B. `(5)/(13)`
C. `(5)/(11)`
D. `(15)/(13)`

Explanation

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

### Step 1: Identify the values of gamma for each gas
- For gas 1, \( \gamma_1 = \frac{7}{5} \)
- For gas 2, \( \gamma_2 = \frac{4}{3} \)

### Step 2: Calculate \( C_v \) for each gas
- For gas 1:
\[ C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2} \]

- For gas 2:
\[ C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{4}{3} - 1} = \frac{R}{\frac{1}{3}} = 3R \]

### Step 3: Calculate \( C_p \) for each gas
- For gas 1:
\[ C_{p1} = \gamma_1 R \div (\gamma_1 - 1) = \frac{7}{5} R \div \left(\frac{2}{5}\right) = \frac{7R}{2} \]

- For gas 2:
\[ C_{p2} = \gamma_2 R \div (\gamma_2 - 1) = \frac{4}{3} R \div \left(\frac{1}{3}\right) = 4R \]

### Step 4: Calculate total \( C_p \) and \( C_v \) for the mixture
- Total \( C_p \) for the mixture:
\[ C_p = C_{p1} + C_{p2} = \frac{7R}{2} + 4R = \frac{7R}{2} + \frac{8R}{2} = \frac{15R}{2} \]

- Total \( C_v \) for the mixture:
\[ C_v = C_{v1} + C_{v2} = \frac{5R}{2} + 3R = \frac{5R}{2} + \frac{6R}{2} = \frac{11R}{2} \]

### Step 5: Calculate \( \gamma \) for the mixture
\[ \gamma_{mixture} = \frac{C_p}{C_v} = \frac{\frac{15R}{2}}{\frac{11R}{2}} = \frac{15}{11} \]

### Final Answer
The value of \( \gamma \) for the mixture is \( \frac{15}{11} \).

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