One mole of monoatomic gas and one mole of diatomic gas are mixed together. What is the molar specific heat at constant volume for the mixture ?

Options

A. 3/2R
B. 2R
C. 5/2R
D. 3R

Answer

To find the molar specific heat at constant volume for a mixture of one mole of monoatomic gas and one mole of diatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Gases**: We have one mole of a monoatomic gas and one mole of a diatomic gas.

2. **Determine Degrees of Freedom**:
- For a monoatomic gas, the degrees of freedom (F) is 3.
- For a diatomic gas, the degrees of freedom (F) is 5.

3. **Calculate Cv for Each Gas**:
- The formula for molar specific heat at constant volume (Cv) is given by:
\[ C_v = \frac{1}{2} F R \]
- For the monoatomic gas:
\[ C_{v1} = \frac{1}{2} \times 3 \times R = \frac{3}{2} R \]
- For the diatomic gas:
\[ C_{v2} = \frac{1}{2} \times 5 \times R = \frac{5}{2} R \]

4. **Use the Formula for Mixture**:
- The molar specific heat at constant volume for the mixture (C_v) can be calculated using the formula:
\[ C_v = \frac{N_1 C_{v1} + N_2 C_{v2}}{N_1 + N_2} \]
- Here, \(N_1\) and \(N_2\) are the number of moles of the monoatomic and diatomic gases, respectively. Since both are 1 mole:
\[ C_v = \frac{1 \cdot \frac{3}{2} R + 1 \cdot \frac{5}{2} R}{1 + 1} \]

5. **Calculate the Mixture's Cv**:
- Substitute the values:
\[ C_v = \frac{\frac{3}{2} R + \frac{5}{2} R}{2} = \frac{\frac{8}{2} R}{2} = \frac{8R}{4} = 2R \]

6. **Conclusion**:
- The molar specific heat at constant volume for the mixture of one mole of monoatomic gas and one mole of diatomic gas is:
\[ C_v = 2R \]