One moles of a monatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is

Options

A. 1.625 R
B. 2.1R
C. 6.25 R
D. 3.625R

Answer

To find the molar specific heat of the mixture at constant volume for the given gases, we will follow these steps:

### Step 1: Identify the gases and their moles
We have:
- 1 mole of a monatomic gas (let's denote it as \( N_1 = 1 \))
- 3 moles of a diatomic gas (let's denote it as \( N_2 = 3 \))

### Step 2: Determine the molar specific heats of the individual gases
The molar specific heat at constant volume (\( C_v \)) for the gases are:
- For a monatomic gas: \( C_{v1} = \frac{3R}{2} \)
- For a diatomic gas: \( C_{v2} = \frac{5R}{2} \)

### Step 3: Use the formula for the molar specific heat of the mixture
The formula for the molar specific heat of the mixture at constant volume (\( C_{mixture} \)) is given by:

\[ C_{mixture} = \frac{N_1 C_{v1} + N_2 C_{v2}}{N_1 + N_2} \]

### Step 4: Substitute the values into the formula
Now substituting the values we have:

\[ C_{mixture} = \frac{(1 \cdot \frac{3R}{2}) + (3 \cdot \frac{5R}{2})}{1 + 3} \]

### Step 5: Simplify the expression
Calculating the numerator:

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

Combine the terms in the numerator:

\[ C_{mixture} = \frac{\frac{18R}{2}}{4} = \frac{9R}{4} \]

### Step 6: Final calculation
Thus, the molar specific heat of the mixture at constant volume is:

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

### Step 7: Convert to a numerical value
If we express this in terms of \( R \):

\[ C_{mixture} = 2.25R \]

### Conclusion
Thus, the final answer for the molar specific heat of the mixture at constant volume is:

\[ C_{mixture} = 2.25R \]