For a gas `R/C_V` = 0.4, where R is the universal gas constant and C, is molar specific heat at constant volume. The gas is made up of molecules which are

Options

A. rigid diatomic
B. monoatomic
C. non - rigid diatomic
D. polyatomic

Answer

To solve the problem, we need to analyze the given information about the gas, specifically the ratio \( \frac{R}{C_V} = 0.4 \), where \( R \) is the universal gas constant and \( C_V \) is the molar specific heat at constant volume.

### Step-by-Step Solution:

1. **Understanding the Given Ratio**:
We start with the equation:
\[ \frac{R}{C_V} = 0.4 \]
From this, we can express \( C_V \) in terms of \( R \):
\[ C_V = \frac{R}{0.4} \]

2. **Finding \( C_P \)**:
We know the relationship between \( C_P \) (molar specific heat at constant pressure) and \( C_V \):
\[ C_P = C_V + R \]
Substituting \( C_V \) from the previous step:
\[ C_P = \frac{R}{0.4} + R = \frac{R}{0.4} + \frac{0.4R}{0.4} = \frac{R + 0.4R}{0.4} = \frac{1.4R}{0.4} = \frac{7R}{2} \]

3. **Calculating the Ratio \( \gamma \)**:
The ratio \( \gamma \) (gamma) is defined as:
\[ \gamma = \frac{C_P}{C_V} \]
Substituting the values we found:
\[ \gamma = \frac{\frac{7R}{2}}{\frac{R}{0.4}} = \frac{7R}{2} \times \frac{0.4}{R} = \frac{7 \times 0.4}{2} = \frac{2.8}{2} = 1.4 \]

4. **Identifying the Type of Gas**:
The value of \( \gamma = 1.4 \) indicates the type of gas. For diatomic gases, \( \gamma \) typically ranges around 1.4, which suggests that the gas is made up of diatomic molecules.

### Conclusion:
The gas must be made up of diatomic molecules, as indicated by the calculated value of \( \gamma \).