One mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the molar specific heat of the mixture at constant volume?

Options

A. `(5)/(4)R`
B. `(9)/(4)R`
C. `(3)/(4)R`
D. `R`

Explanation

To find the molar specific heat of the mixture at constant volume (\(C_V\)), we can use the formula for the specific heat of a mixture of gases:

\[ C_{V,\text{mixture}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} \]

Where:
- \(n_1\) = number of moles of the first gas (monatomic)
- \(C_{V1}\) = molar specific heat at constant volume of the first gas
- \(n_2\) = number of moles of the second gas (diatomic)
- \(C_{V2}\) = molar specific heat at constant volume of the second gas

### Step 1: Identify the values for \(n_1\), \(n_2\), \(C_{V1}\), and \(C_{V2}\)

From the question:
- \(n_1 = 1\) mole (monatomic gas)
- \(n_2 = 3\) moles (diatomic gas)

For the specific heats:
- For a monatomic gas, \(C_{V1} = \frac{3}{2} R\)
- For a diatomic gas, \(C_{V2} = \frac{5}{2} R\)

### Step 2: Substitute the values into the formula

Now, we can substitute these values into the formula:

\[ C_{V,\text{mixture}} = \frac{1 \cdot \frac{3}{2} R + 3 \cdot \frac{5}{2} R}{1 + 3} \]

### Step 3: Calculate the numerator

Calculating the numerator:

\[ = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} \]

Combine the terms in the numerator:

\[ = \frac{\frac{3 + 15}{2} R}{4} = \frac{\frac{18}{2} R}{4} = \frac{9R}{4} \]

### Step 4: Calculate the final value of \(C_{V,\text{mixture}}\)

Now, we divide by the total number of moles:

\[ C_{V,\text{mixture}} = \frac{9R/4}{4} = \frac{9R}{16} \]

### Final Result

Thus, the molar specific heat of the mixture at constant volume is:

\[ C_{V,\text{mixture}} = \frac{9}{4} R \]