n mole of He and 2n mole of `O_2` is mixed in a container. Then `(C_p/C_v)_(mix)` will be

Options

A. `19/13`
B. `40/27`
C. `1/3`
D. `1/4`

Explanation

To find the ratio \((C_p/C_v)_{mix}\) for a mixture of \(n\) moles of Helium (He) and \(2n\) moles of Oxygen (\(O_2\)), we can follow these steps:

### Step 1: Identify the specific heat capacities
1. **For Helium (He)**:
- Since Helium is a monatomic gas, its degrees of freedom \(f = 3\).
- The specific heat at constant volume \(C_{v1}\) is given by:
\[ C_{v1} = \frac{3}{2} R \]
- The specific heat at constant pressure \(C_{p1}\) is:
\[ C_{p1} = C_{v1} + R = \frac{3}{2} R + R = \frac{5}{2} R \]

2. **For Oxygen (\(O_2\))**:
- Oxygen is a diatomic gas, so its degrees of freedom \(f = 5\).
- The specific heat at constant volume \(C_{v2}\) is given by:
\[ C_{v2} = \frac{5}{2} R \]
- The specific heat at constant pressure \(C_{p2}\) is:
\[ C_{p2} = C_{v2} + R = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 2: Calculate the total heat capacities for the mixture
Using the formula for the mixture:
\[ \frac{C_p}{C_v} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \]
where:
- \(n_1 = n\) (moles of He)
- \(n_2 = 2n\) (moles of \(O_2\))

Substituting the values:
\[ C_p = n C_{p1} + 2n C_{p2} = n \left(\frac{5}{2} R\right) + 2n \left(\frac{7}{2} R\right) \]
\[ = n \left(\frac{5}{2} R + 7n R\right) = n \left(\frac{5 + 14}{2} R\right) = n \left(\frac{19}{2} R\right) \]

For \(C_v\):
\[ C_v = n C_{v1} + 2n C_{v2} = n \left(\frac{3}{2} R\right) + 2n \left(\frac{5}{2} R\right) \]
\[ = n \left(\frac{3}{2} R + 5n R\right) = n \left(\frac{3 + 10}{2} R\right) = n \left(\frac{13}{2} R\right) \]

### Step 3: Calculate the ratio \((C_p/C_v)_{mix}\)
Now substituting \(C_p\) and \(C_v\) into the ratio:
\[ \frac{C_p}{C_v} = \frac{n \left(\frac{19}{2} R\right)}{n \left(\frac{13}{2} R\right)} = \frac{\frac{19}{2} R}{\frac{13}{2} R} = \frac{19}{13} \]

### Final Answer:
\[ \frac{C_p}{C_v} = \frac{19}{13} \]