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
B. 19
C. 20
D. 21

Answer

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

### Step 1: Identify the values
- **Number of moles of monoatomic gas (n1)** = 1 mole
- **Number of moles of diatomic gas (n2)** = 3 moles
- **Molar specific heat for monoatomic gas (Cv1)** = \( \frac{3}{2} R \)
- **Molar specific heat for diatomic gas (Cv2)** = \( \frac{5}{2} R \)
- **Universal gas constant (R)** = 8.3 J K\(^{-1}\) mol\(^{-1}\)

### Step 2: Write the formula for molar specific heat at constant volume
The formula for the molar specific heat at constant volume (Cv) for a mixture of gases is given by:

\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula, we have:

\[ C_v = \frac{1 \cdot \left(\frac{3}{2} R\right) + 3 \cdot \left(\frac{5}{2} R\right)}{1 + 3} \]

### Step 4: Simplify the equation
Calculating the numerator:

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

### Step 5: Substitute the value of R
Now, substituting the value of R (8.3 J K\(^{-1}\) mol\(^{-1}\)) into the equation:

\[ C_v = \frac{9 \cdot 8.3}{4} \]
\[ C_v = \frac{74.7}{4} \]
\[ C_v = 18.675 \text{ J K}^{-1} \text{ mol}^{-1} \]

### Step 6: Round to the nearest integer
Since the question asks for the molar specific heat in J K\(^{-1}\) mol\(^{-1}\), we can round 18.675 to 19.

### Final Answer
Therefore, the molar specific heat at constant volume is approximately **19 J K\(^{-1}\) mol\(^{-1}\)**.

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