The value of `gamma=C_p/C_v ` for a gaseous mixture consisting of 2.0 moles of oxygen and 3.0 moles of helium. The gases are assumed to be ideal.

Options

A. `7/5`
B. `29/19`
C. `23/19`
D. `25/19`

Explanation

To find the value of \( \gamma \) (gamma) for a gaseous mixture consisting of 2.0 moles of oxygen and 3.0 moles of helium, we will use the formula:

\[ \gamma = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \]

### Step 1: Identify the properties of the gases
- For **Oxygen (O₂)**:
- It is a diatomic gas, so:
- \( C_{p1} = \frac{7R}{2} \)
- \( C_{v1} = \frac{5R}{2} \)
- Number of moles, \( n_1 = 2.0 \)

- For **Helium (He)**:
- It is a monatomic gas, so:
- \( C_{p2} = \frac{5R}{2} \)
- \( C_{v2} = \frac{3R}{2} \)
- Number of moles, \( n_2 = 3.0 \)

### Step 2: Substitute values into the formula
Substituting the values into the gamma formula:

\[ \gamma = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} = \frac{(2)(\frac{7R}{2}) + (3)(\frac{5R}{2})}{(2)(\frac{5R}{2}) + (3)(\frac{3R}{2})} \]

### Step 3: Simplify the numerator
Calculating the numerator:

\[ = \frac{(2 \cdot \frac{7R}{2}) + (3 \cdot \frac{5R}{2})} = \frac{7R + 15R}{2} = \frac{22R}{2} = 11R \]

### Step 4: Simplify the denominator
Calculating the denominator:

\[ = \frac{(2 \cdot \frac{5R}{2}) + (3 \cdot \frac{3R}{2})} = \frac{5R + 9R}{2} = \frac{14R}{2} = 7R \]

### Step 5: Calculate gamma
Now substituting back into the gamma formula:

\[ \gamma = \frac{11R}{7R} = \frac{11}{7} \]

### Final Result
Thus, the value of \( \gamma \) for the gaseous mixture is:

\[ \gamma = \frac{11}{7} \]