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)`

Answer

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

### Step 1: Understand the given values
We have two gases:
- Gas 1: \( \gamma_1 = \frac{7}{5} \)
- Gas 2: \( \gamma_2 = \frac{4}{3} \)
- Number of moles of both gases: \( n_1 = n_2 = 1 \)

### Step 2: Write the formula for the specific heat of the mixture
The specific heat \( C_{mix} \) for the mixture can be expressed as:
\[ C_{mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Since \( n_1 = n_2 = 1 \), this simplifies to:
\[ C_{mix} = \frac{C_{v1} + C_{v2}}{2} \]

### Step 3: Relate specific heat to \( \gamma \)
The specific heat at constant volume \( C_v \) can be expressed in terms of the gas constant \( R \) and \( \gamma \):
\[ C_v = \frac{R}{\gamma - 1} \]
Thus, for our mixture:
\[ C_{mix} = \frac{\frac{R}{\gamma_1 - 1} + \frac{R}{\gamma_2 - 1}}{2} \]

### Step 4: Substitute the values of \( \gamma_1 \) and \( \gamma_2 \)
Substituting \( \gamma_1 = \frac{7}{5} \) and \( \gamma_2 = \frac{4}{3} \):
\[ C_{mix} = \frac{\frac{R}{\frac{7}{5} - 1} + \frac{R}{\frac{4}{3} - 1}}{2} \]

### Step 5: Calculate \( \frac{7}{5} - 1 \) and \( \frac{4}{3} - 1 \)
Calculating these:
\[ \frac{7}{5} - 1 = \frac{7}{5} - \frac{5}{5} = \frac{2}{5} \]
\[ \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]

### Step 6: Substitute back into the equation
Now substituting these values back:
\[ C_{mix} = \frac{\frac{R}{\frac{2}{5}} + \frac{R}{\frac{1}{3}}}{2} \]
This simplifies to:
\[ C_{mix} = \frac{R \cdot \frac{5}{2} + R \cdot 3}{2} \]
\[ C_{mix} = \frac{R \left(\frac{5}{2} + 3\right)}{2} = \frac{R \left(\frac{5}{2} + \frac{6}{2}\right)}{2} = \frac{R \cdot \frac{11}{2}}{2} = \frac{11R}{4} \]

### Step 7: Relate \( C_{mix} \) back to \( \gamma_{mix} \)
Now we can relate \( C_{mix} \) to \( \gamma_{mix} \):
\[ C_{mix} = \frac{R}{\gamma_{mix} - 1} \]
Setting these equal:
\[ \frac{11R}{4} = \frac{R}{\gamma_{mix} - 1} \]

### Step 8: Solve for \( \gamma_{mix} \)
Cross-multiplying gives:
\[ 11R(\gamma_{mix} - 1) = 4R \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 11(\gamma_{mix} - 1) = 4 \]
\[ 11\gamma_{mix} - 11 = 4 \]
\[ 11\gamma_{mix} = 15 \]
\[ \gamma_{mix} = \frac{15}{11} \]

### Final Answer
Thus, the value of \( \gamma \) for the mixture is:
\[ \gamma_{mix} = \frac{15}{11} \]