`C_p//C_v` ratio for a gas mixture consisting of 4g of helium and 32g of oxygen is N times 0.5. Then N is equal to

Answer

To find the ratio \( \frac{C_p}{C_v} \) for a gas mixture consisting of 4g of helium and 32g of oxygen, we will follow these steps:

### Step 1: Determine the number of moles of each gas

1. **Helium (He)**:
- Given mass = 4g
- Molar mass of helium = 4 g/mol
- Number of moles of helium (\( n_1 \)):
\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{4 \text{ g/mol}} = 1 \text{ mol} \]

2. **Oxygen (O\(_2\))**:
- Given mass = 32g
- Molar mass of oxygen = 32 g/mol
- Number of moles of oxygen (\( n_2 \)):
\[ n_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mol} \]

### Step 2: Calculate \( C_v \) for each gas

1. **For Helium** (monatomic gas):
- Degrees of freedom = 3
- Therefore, \( C_v \) for helium:
\[ C_{v1} = \frac{3}{2} R \]

2. **For Oxygen** (diatomic gas):
- Degrees of freedom = 5
- Therefore, \( C_v \) for oxygen:
\[ C_{v2} = \frac{5}{2} R \]

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

Using the formula for the specific heat capacity at constant volume for the mixture:
\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Substituting the values:
\[ C_v = \frac{1 \cdot \frac{3}{2} R + 1 \cdot \frac{5}{2} R}{1 + 1} \]
\[ C_v = \frac{\frac{3}{2} R + \frac{5}{2} R}{2} = \frac{\frac{8}{2} R}{2} = \frac{4R}{2} = 2R \]

### Step 4: Calculate \( C_p \) for the mixture

Using the relation \( C_p = C_v + R \):
\[ C_p = 2R + R = 3R \]

### Step 5: Calculate the ratio \( \frac{C_p}{C_v} \)

Now we can find the ratio:
\[ \frac{C_p}{C_v} = \frac{3R}{2R} = \frac{3}{2} \]

### Step 6: Relate it to \( N \)

According to the problem, this ratio is \( N \) times \( 0.5 \):
\[ \frac{3}{2} = N \times 0.5 \]
To find \( N \):
\[ N = \frac{3/2}{0.5} = \frac{3/2}{1/2} = 3 \]

### Final Answer
Thus, \( N \) is equal to **3**.

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