If 2 moles of diatomic gas and 1 mole of monatomic gas are mixed, then the ratio of specific heats for the mixture is

Options

A. `(7)/(3)`
B. `(5)/(4)`
C. `(19)/(13)`
D. `(15)/(19)`

Answer

To find the ratio of specific heats for a mixture of 2 moles of diatomic gas and 1 mole of monatomic gas, we will follow these steps:

### Step 1: Identify the specific heat capacities of the gases
- For a diatomic gas:
- The molar specific heat at constant volume, \( C_{V1} = \frac{5}{2} R \)
- The molar specific heat at constant pressure, \( C_{P1} = \frac{7}{2} R \)

- For a monatomic gas:
- The molar specific heat at constant volume, \( C_{V2} = \frac{3}{2} R \)
- The molar specific heat at constant pressure, \( C_{P2} = \frac{5}{2} R \)

### Step 2: Calculate the total heat capacities for the mixture
- The total heat capacity at constant pressure for the mixture:
\[ C_{P_{mix}} = n_1 C_{P1} + n_2 C_{P2} \]
where \( n_1 = 2 \) moles (diatomic gas) and \( n_2 = 1 \) mole (monatomic gas).

Substituting the values:
\[ C_{P_{mix}} = 2 \times \frac{7}{2} R + 1 \times \frac{5}{2} R = 7R + \frac{5}{2} R = \frac{14}{2} R + \frac{5}{2} R = \frac{19}{2} R \]

- The total heat capacity at constant volume for the mixture:
\[ C_{V_{mix}} = n_1 C_{V1} + n_2 C_{V2} \]

Substituting the values:
\[ C_{V_{mix}} = 2 \times \frac{5}{2} R + 1 \times \frac{3}{2} R = 5R + \frac{3}{2} R = \frac{10}{2} R + \frac{3}{2} R = \frac{13}{2} R \]

### Step 3: Calculate the ratio of specific heats for the mixture
The ratio of specific heats \( \gamma_{mix} \) is given by:
\[ \gamma_{mix} = \frac{C_{P_{mix}}}{C_{V_{mix}}} \]

Substituting the values we calculated:
\[ \gamma_{mix} = \frac{\frac{19}{2} R}{\frac{13}{2} R} = \frac{19}{13} \]

### Conclusion
The ratio of specific heats for the mixture of 2 moles of diatomic gas and 1 mole of monatomic gas is \( \frac{19}{13} \).