Two moles of helium (He) are mixed with four moles of hydrogen `(H_2)`. Find
(a) `(C_(V)` of the mixture
(b) `(C_(P)` of the mixture and
( c) `(gamma) `of the mixture.

Explanation

To solve the problem of finding the specific heat capacities and the ratio of specific heats for a mixture of helium and hydrogen, we will follow these steps:

### Given:
- Moles of Helium (He), \( n_1 = 2 \)
- Moles of Hydrogen (H₂), \( n_2 = 4 \)

### Step 1: Determine the specific heat capacities for each gas.

**For Helium (He)**:
- Helium is a monatomic gas, so the degrees of freedom \( F = 3 \).
- The specific heat at constant volume \( C_{V1} \) is given by:
\[ C_{V1} = \frac{F}{2} R = \frac{3}{2} R \]
- The specific heat at constant pressure \( C_{P1} \) is given by:
\[ C_{P1} = C_{V1} + R = \frac{3}{2} R + R = \frac{5}{2} R \]

**For Hydrogen (H₂)**:
- Hydrogen is a diatomic gas, so the degrees of freedom \( F = 5 \).
- The specific heat at constant volume \( C_{V2} \) is given by:
\[ C_{V2} = \frac{F}{2} R = \frac{5}{2} R \]
- The specific heat at constant pressure \( C_{P2} \) is given by:
\[ C_{P2} = C_{V2} + R = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 2: Calculate the specific heat capacity at constant volume for the mixture \( C_{V \text{ mixture}} \).

Using the formula:
\[ C_{V \text{ mixture}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} \]
Substituting the values:
\[ C_{V \text{ mixture}} = \frac{2 \left(\frac{3}{2} R\right) + 4 \left(\frac{5}{2} R\right)}{2 + 4} \]
Calculating the numerator:
\[ = \frac{3R + 10R}{6} = \frac{13R}{6} \]
Thus,
\[ C_{V \text{ mixture}} = \frac{13}{6} R \]

### Step 3: Calculate the specific heat capacity at constant pressure for the mixture \( C_{P \text{ mixture}} \).

Using the relation:
\[ C_{P \text{ mixture}} = C_{V \text{ mixture}} + R \]
Substituting the value of \( C_{V \text{ mixture}} \):
\[ C_{P \text{ mixture}} = \frac{13}{6} R + R = \frac{13}{6} R + \frac{6}{6} R = \frac{19}{6} R \]

### Step 4: Calculate the ratio of specific heats \( \gamma \).

Using the formula:
\[ \gamma = \frac{C_{P \text{ mixture}}}{C_{V \text{ mixture}}} \]
Substituting the values:
\[ \gamma = \frac{\frac{19}{6} R}{\frac{13}{6} R} = \frac{19}{13} \]

### Final Results:
(a) \( C_{V \text{ mixture}} = \frac{13}{6} R \)
(b) \( C_{P \text{ mixture}} = \frac{19}{6} R \)
(c) \( \gamma = \frac{19}{13} \)