Consider a mixture of 3 moles of helium gas and 2 moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its `C_P//C_V` value will be:

Options

A. 67/45
B. 19/13
C. 29/19
D. 40/27

Explanation

To find the value of \( \frac{C_P}{C_V} \) for a mixture of 3 moles of helium gas and 2 moles of oxygen gas, we can follow these steps:

### Step 1: Identify the number of moles
- We have:
- \( n_1 = 3 \) moles of helium (He)
- \( n_2 = 2 \) moles of oxygen (O₂)

### Step 2: Determine the degrees of freedom
- Helium is a monatomic gas, which has 3 degrees of freedom.
- Oxygen is a diatomic gas, which has 5 degrees of freedom.

### Step 3: Calculate \( C_V \) for each gas
- For helium (monatomic):
\[ C_{V1} = \frac{f}{2} R = \frac{3}{2} R \]
- For oxygen (diatomic):
\[ C_{V2} = \frac{f}{2} R = \frac{5}{2} R \]

### Step 4: Calculate \( C_P \) for each gas
- For helium:
\[ C_{P1} = C_{V1} + R = \frac{3}{2} R + R = \frac{5}{2} R \]
- For oxygen:
\[ C_{P2} = C_{V2} + R = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 5: Use the formula for \( \frac{C_P}{C_V} \) of the mixture
The formula for the ratio \( \frac{C_P}{C_V} \) for a mixture is given by:
\[ \frac{C_P}{C_V} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}} \]

### Step 6: Substitute the values into the formula
Substituting the known values:
\[ \frac{C_P}{C_V} = \frac{3 \cdot \frac{5}{2} R + 2 \cdot \frac{7}{2} R}{3 \cdot \frac{3}{2} R + 2 \cdot \frac{5}{2} R} \]

### Step 7: Simplify the numerator and denominator
- Numerator:
\[ = \frac{15R}{2} + \frac{14R}{2} = \frac{29R}{2} \]
- Denominator:
\[ = \frac{9R}{2} + \frac{10R}{2} = \frac{19R}{2} \]

### Step 8: Final calculation
Now, substituting back into the ratio:
\[ \frac{C_P}{C_V} = \frac{\frac{29R}{2}}{\frac{19R}{2}} = \frac{29}{19} \]

### Conclusion
Thus, the value of \( \frac{C_P}{C_V} \) for the mixture is:
\[ \frac{C_P}{C_V} = \frac{29}{19} \]