Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its `C_(p)//C_(y)` value will be:

Options

A. `40//27`
B. `67//45`
C. `19//13`
D. `23//15`

Answer

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

### Step 1: Identify the properties of the gases
- Helium is a monatomic gas.
- Oxygen is a diatomic gas.

### Step 2: Determine the degrees of freedom
- For a monatomic gas (helium), the degrees of freedom \( f_1 = 3 \).
- For a diatomic gas (oxygen), the degrees of freedom \( f_2 = 5 \).

### Step 3: Calculate \( C_v \) for each gas
- The formula for \( C_v \) is given by:
\[ C_v = \frac{f}{2} R \]
- For helium:
\[ C_{v1} = \frac{3}{2} R \]
- For oxygen:
\[ C_{v2} = \frac{5}{2} R \]

### Step 4: Calculate \( C_p \) for each gas
- The relationship between \( C_p \) and \( C_v \) is:
\[ C_p = C_v + R \]
- 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: Set up the mixture
- Let \( n_1 = n \) (moles of helium) and \( n_2 = 2n \) (moles of oxygen).

### Step 6: Calculate the total \( C_p \) and \( C_v \) for the mixture
- Total \( C_p \) for the mixture:
\[ C_p = n_1 C_{p1} + n_2 C_{p2} = n \left(\frac{5}{2} R\right) + 2n \left(\frac{7}{2} R\right) \]
\[ = \frac{5nR}{2} + \frac{14nR}{2} = \frac{19nR}{2} \]

- Total \( C_v \) for the mixture:
\[ C_v = n_1 C_{v1} + n_2 C_{v2} = n \left(\frac{3}{2} R\right) + 2n \left(\frac{5}{2} R\right) \]
\[ = \frac{3nR}{2} + \frac{10nR}{2} = \frac{13nR}{2} \]

### Step 7: Calculate \( \frac{C_p}{C_v} \)
- Now we can find \( \frac{C_p}{C_v} \):
\[ \frac{C_p}{C_v} = \frac{\frac{19nR}{2}}{\frac{13nR}{2}} = \frac{19}{13} \]

### Final Answer
The value of \( \frac{C_p}{C_v} \) for the mixture of helium and oxygen is \( \frac{19}{13} \).

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