For a gas, `R=2/3C_(V)`. This suggests that the gas consists of molecules, which are

Options

A. Polyatomic
B. Diatomic
C. Monoatomic
D. A mixture of diatomic and polyatomic molecules

Explanation

To solve the problem, we need to analyze the relationship given in the question: \( R = \frac{2}{3} C_V \). We will derive the implications of this relationship regarding the type of gas molecules.

### Step-by-Step Solution:

1. **Understanding the Given Relationship**:
We start with the equation provided:
\[ R = \frac{2}{3} C_V \]
Rearranging this gives:
\[ C_V = \frac{3}{2} R \]

2. **Relating \( C_V \) to Degrees of Freedom**:
The specific heat at constant volume \( C_V \) can be expressed in terms of the degrees of freedom \( F \) of the gas molecules:
\[ C_V = \frac{F}{2} R \]
where \( F \) is the number of degrees of freedom.

3. **Equating the Two Expressions for \( C_V \)**:
Now, we have two expressions for \( C_V \):
\[ C_V = \frac{3}{2} R \quad \text{and} \quad C_V = \frac{F}{2} R \]
Setting these equal to each other gives:
\[ \frac{3}{2} R = \frac{F}{2} R \]

4. **Solving for Degrees of Freedom \( F \)**:
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ \frac{3}{2} = \frac{F}{2} \]
Multiplying both sides by 2:
\[ F = 3 \]

5. **Identifying the Type of Gas**:
The degrees of freedom \( F \) indicates the type of gas:
- For **monatomic gases**, \( F = 3 \) (only translational motion).
- For **diatomic gases**, \( F = 5 \) (3 translational and 2 rotational).
- For **polyatomic gases**, \( F \) is generally greater than 5.

Since we found \( F = 3 \), this indicates that the gas is **monatomic**.

### Conclusion:
Thus, the gas consists of molecules that are **monatomic**.

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