One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in `JK​^(-1)"​ mol​ "^(-1)`​) at constant volume is (R = 8.3 `J K​^(-1​) " mol"​^(-1​)` )

Options

A. 18.7
B. 18.9
C. 19.2
D. 19.7

Explanation

To find the molar specific heat at constant volume (Cv) for the combination of one mole of monoatomic gas and three moles of diatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Moles and Specific Heats:**
- Let \( N_1 = 1 \) mole (monoatomic gas).
- Let \( N_2 = 3 \) moles (diatomic gas).
- The molar specific heat at constant volume for monoatomic gas \( Cv_1 = \frac{3}{2} R \).
- The molar specific heat at constant volume for diatomic gas \( Cv_2 = \frac{5}{2} R \).

2. **Calculate the Total Molar Specific Heat:**
- The formula for the molar specific heat of the mixture is given by:
\[ Cv = \frac{N_1 Cv_1 + N_2 Cv_2}{N_1 + N_2} \]

3. **Substituting the Values:**
- Substitute \( N_1, N_2, Cv_1, \) and \( Cv_2 \) into the equation:
\[ Cv = \frac{1 \cdot \left(\frac{3}{2} R\right) + 3 \cdot \left(\frac{5}{2} R\right)}{1 + 3} \]

4. **Simplifying the Equation:**
- Calculate the numerator:
\[ Cv = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} \]
\[ = \frac{\frac{18}{2} R}{4} = \frac{9R}{4} \]

5. **Substituting the Value of R:**
- Given \( R = 8.3 \, J \, K^{-1} \, mol^{-1} \):
\[ Cv = \frac{9 \cdot 8.3}{4} \]
\[ = \frac{74.7}{4} = 18.675 \, J \, K^{-1} \, mol^{-1} \]

6. **Final Answer:**
- Rounding to one decimal place, the molar specific heat at constant volume is:
\[ Cv \approx 18.7 \, J \, K^{-1} \, mol^{-1} \]