Consider a mixture of 1 moles of helium gas and 4 moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CP/CV value will be:

Options

A. 27/35
B. 24/23
C. 33/23
D. 40/27

Explanation

To find the ratio \( \frac{C_p}{C_v} \) for a mixture of helium and oxygen gases, we can follow these steps:

### Step 1: Identify the number of moles and types of gases
- We have 1 mole of helium (He) and 4 moles of oxygen (O₂).
- Helium is a monatomic gas, and oxygen is a diatomic gas.

### Step 2: Determine the degrees of freedom
- For helium (monatomic gas), the degrees of freedom \( F_1 = 3 \).
- For oxygen (diatomic gas), the degrees of freedom \( F_2 = 5 \).

### Step 3: Calculate \( C_p \) and \( C_v \) for each gas
- The formula for \( C_p \) is given by:
\[ C_p = \frac{F}{2} R + R = \left(\frac{F}{2} + 1\right) R \]

- For helium:
\[ C_{p1} = \left(\frac{3}{2} + 1\right) R = \frac{5}{2} R \]

- For oxygen:
\[ C_{p2} = \left(\frac{5}{2} + 1\right) R = \frac{7}{2} R \]

- The formula for \( C_v \) is:
\[ C_v = \frac{F}{2} R \]

- For helium:
\[ C_{v1} = \frac{3}{2} R \]

- For oxygen:
\[ C_{v2} = \frac{5}{2} R \]

### Step 4: Use the formula for the mixture
The ratio \( \frac{C_p}{C_v} \) for the mixture can be calculated using the formula:
\[ \frac{C_p}{C_v} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \]

Where:
- \( n_1 = 1 \) (moles of He)
- \( n_2 = 4 \) (moles of O₂)

### Step 5: Substitute the values into the formula
Substituting the values we calculated:
\[ \frac{C_p}{C_v} = \frac{1 \cdot \frac{5}{2} R + 4 \cdot \frac{7}{2} R}{1 \cdot \frac{3}{2} R + 4 \cdot \frac{5}{2} R} \]

Calculating the numerator:
\[ = \frac{\frac{5}{2} R + \frac{28}{2} R}{\frac{3}{2} R + \frac{20}{2} R} = \frac{\frac{33}{2} R}{\frac{23}{2} R} \]

### Step 6: Simplify the expression
\[ = \frac{33}{23} \]

### Final Answer
Thus, the value of \( \frac{C_p}{C_v} \) for the mixture is \( \frac{33}{23} \).