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

Explanation

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

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

2. **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 \( C_p \) and \( C_v \) for each gas
- For helium (monatomic gas):
- \( C_{p, He} = \frac{5}{2} R \)
- \( C_{v, He} = \frac{3}{2} R \)

- For oxygen (diatomic gas):
- \( C_{p, O_2} = \frac{7}{2} R \)
- \( C_{v, O_2} = \frac{5}{2} R \)

### Step 3: Calculate \( C_p \) and \( C_v \) for the mixture
1. **Total \( C_p \)**:
\[ C_p = n_1 C_{p, He} + n_2 C_{p, O_2} \]
\[ C_p = 4 \left(\frac{5}{2} R\right) + 0.5 \left(\frac{7}{2} R\right) = 10R + 1.75R = 11.75R \]

2. **Total \( C_v \)**:
\[ C_v = n_1 C_{v, He} + n_2 C_{v, O_2} \]
\[ C_v = 4 \left(\frac{3}{2} R\right) + 0.5 \left(\frac{5}{2} R\right) = 6R + 1.25R = 7.25R \]

### Step 4: Calculate the ratio \( \frac{C_p}{C_v} \)
\[ \frac{C_p}{C_v} = \frac{11.75R}{7.25R} = \frac{11.75}{7.25} \approx 1.62 \]

### Conclusion
The ratio \( \frac{C_p}{C_v} \) for the gaseous mixture is approximately **1.62**.

---