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: Determine the degrees of freedom for each gas
- Helium is a monoatomic gas, so its degrees of freedom \( F_1 = 3 \).
- Oxygen is a diatomic gas, so its degrees of freedom \( F_2 = 5 \).

### Step 2: Calculate \( C_v \) for each gas
- For helium:
\[ C_{v1} = \frac{F_1}{2} R = \frac{3}{2} R \]
- For oxygen:
\[ C_{v2} = \frac{F_2}{2} R = \frac{5}{2} R \]

### Step 3: Calculate \( C_p \) for each gas
- 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 4: Use the mole fractions to calculate the mixture's \( C_p \) and \( C_v \)
- Given \( n_1 = n \) (moles of helium) and \( n_2 = 2n \) (moles of oxygen), we can calculate the total \( C_p \) and \( C_v \) for the mixture:
\[ C_{p, mixture} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{n \cdot \frac{5}{2} R + 2n \cdot \frac{7}{2} R}{n + 2n} \]
\[ = \frac{\frac{5}{2} n R + 7n R}{3n} = \frac{\frac{5}{2} + 7}{3} R = \frac{\frac{5 + 14}{2}}{3} R = \frac{19}{6} R \]

- Similarly, calculate \( C_v \):
\[ C_{v, mixture} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{n \cdot \frac{3}{2} R + 2n \cdot \frac{5}{2} R}{n + 2n} \]
\[ = \frac{\frac{3}{2} n R + 5n R}{3n} = \frac{\frac{3 + 10}{2}}{3} R = \frac{\frac{13}{2}}{3} R = \frac{13}{6} R \]

### Step 5: Calculate \( \frac{C_p}{C_v} \)
\[ \frac{C_p}{C_v} = \frac{\frac{19}{6} R}{\frac{13}{6} R} = \frac{19}{13} \]

Thus, the value of \( \frac{C_p}{C_v} \) for the mixture is \( \frac{19}{13} \).