A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture is

Options

A. `1.54`
B. `1.62`
C. `1.4`
D. `1.59`

Explanation

To find the ratio \( \frac{C_p}{C_v} \) of the gaseous mixture consisting of 16 g of helium and 16 g of oxygen, we will follow these steps:

### Step 1: Calculate the number of moles of each gas
- **For Helium (He)**:
- Molar mass of helium = 4 g/mol
- Number of moles of helium, \( n_1 = \frac{16 \text{ g}}{4 \text{ g/mol}} = 4 \text{ moles} \)

- **For Oxygen (O2)**:
- Molar mass of oxygen = 32 g/mol
- Number of moles of oxygen, \( n_2 = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \)

### Step 2: Determine the specific heat capacities
- **For Helium (monatomic gas)**:
- \( C_{v1} = \frac{3}{2} R \)

- **For Oxygen (diatomic gas)**:
- \( C_{v2} = \frac{5}{2} R \)

### Step 3: Calculate the average \( C_v \) of the mixture
Using the formula for the molar heat capacity at constant volume for a mixture:
\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Substituting the values:
\[ C_v = \frac{(4 \cdot \frac{3}{2} R) + (0.5 \cdot \frac{5}{2} R)}{4 + 0.5} \]
Calculating the numerator:
\[ = \frac{(6R) + (1.25R)}{4.5} = \frac{7.25R}{4.5} = \frac{29R}{18} \]

### Step 4: Calculate \( C_p \) using the relation \( C_p = C_v + R \)
\[ C_p = C_v + R = \frac{29R}{18} + R = \frac{29R}{18} + \frac{18R}{18} = \frac{47R}{18} \]

### Step 5: Calculate the ratio \( \frac{C_p}{C_v} \)
\[ \frac{C_p}{C_v} = \frac{\frac{47R}{18}}{\frac{29R}{18}} = \frac{47}{29} \]

### Final Answer
The ratio \( \frac{C_p}{C_v} \) of the mixture is \( \frac{47}{29} \) or approximately 1.62.

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