Two 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 two moles of a monatomic gas mixed with three moles of a diatomic gas, we can follow these steps:

### Step 1: Identify the number of moles and specific heats
- Let \( N_1 = 2 \) moles (monatomic gas)
- Let \( N_2 = 3 \) moles (diatomic gas)

### Step 2: Determine the specific heat capacities at constant volume
- For a monatomic gas, the molar specific heat at constant volume \( C_{V1} \) is given by:
\[ C_{V1} = \frac{3}{2} R \]
- For a diatomic gas, the molar specific heat at constant volume \( C_{V2} \) is given by:
\[ C_{V2} = \frac{5}{2} R \]

### Step 3: Use the formula for the specific heat of the mixture
The molar specific heat of the mixture at constant volume \( C_{V\text{mix}} \) can be calculated using the formula:
\[ C_{V\text{mix}} = \frac{N_1 C_{V1} + N_2 C_{V2}}{N_1 + N_2} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ C_{V\text{mix}} = \frac{2 \left(\frac{3}{2} R\right) + 3 \left(\frac{5}{2} R\right)}{2 + 3} \]

### Step 5: Simplify the equation
Calculating the numerator:
\[ = \frac{2 \cdot \frac{3}{2} R + 3 \cdot \frac{5}{2} R}{5} \]
\[ = \frac{3R + \frac{15}{2} R}{5} \]
\[ = \frac{\frac{6}{2} R + \frac{15}{2} R}{5} \]
\[ = \frac{\frac{21}{2} R}{5} \]
\[ = \frac{21R}{10} \]

### Step 6: Final result
Thus, the molar specific heat of the mixture at constant volume is:
\[ C_{V\text{mix}} = 2.1 R \]