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:

\[ \gamma = \frac{C_p}{C_v} \]

where \( C_p \) is the heat capacity at constant pressure and \( C_v \) is the heat capacity at constant volume.

### Step 1: Identify the components and their properties
- For oxygen (O₂):
- Moles (\( N_1 \)) = 2.0
- \( C_{p1} = \frac{7}{2} R \)
- \( C_{v1} = \frac{5}{2} R \)

- For helium (He):
- Moles (\( N_2 \)) = 3.0
- \( C_{p2} = \frac{5}{2} R \)
- \( C_{v2} = \frac{3}{2} R \)

### Step 2: Calculate \( C_p \) for the mixture
Using the formula for \( C_p \) of the mixture:

\[ C_p = \frac{N_1 C_{p1} + N_2 C_{p2}}{N_1 + N_2} \]

Substituting the values:

\[ C_p = \frac{2 \cdot \frac{7}{2} R + 3 \cdot \frac{5}{2} R}{2 + 3} \]

Calculating the numerator:

\[ = \frac{(7R + \frac{15}{2} R)}{5} = \frac{(7R + 7.5R)}{5} = \frac{14.5R}{5} = \frac{29R}{10} \]

### Step 3: Calculate \( C_v \) for the mixture
Using the formula for \( C_v \) of the mixture:

\[ C_v = \frac{N_1 C_{v1} + N_2 C_{v2}}{N_1 + N_2} \]

Substituting the values:

\[ C_v = \frac{2 \cdot \frac{5}{2} R + 3 \cdot \frac{3}{2} R}{2 + 3} \]

Calculating the numerator:

\[ = \frac{(5R + \frac{9}{2} R)}{5} = \frac{(5R + 4.5R)}{5} = \frac{9.5R}{5} = \frac{19R}{10} \]

### Step 4: Calculate \( \gamma \)
Now we can calculate \( \gamma \):

\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{29R}{10}}{\frac{19R}{10}} \]

The \( R \) and \( 10 \) cancel out:

\[ \gamma = \frac{29}{19} \]

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

\[ \gamma = \frac{29}{19} \]

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