If for a gas, `(R)/(C_V)=0.67`, the gas is

Options

A. monatomic
B. diatomic
C. polyatomic
D. mixture of aiatomic and polyatomic molecules

Explanation

To determine the type of gas given the ratio \(\frac{R}{C_V} = 0.67\), we can follow these steps:

### Step 1: Understand the relationship between \(R\) and \(C_V\)
The ratio \(\frac{R}{C_V}\) is related to the specific heat capacities of the gas. Here, \(R\) is the universal gas constant, and \(C_V\) is the molar heat capacity at constant volume.

### Step 2: Use the relation for ideal gases
For an ideal gas, the relationship between \(R\), \(C_V\), and \(C_P\) (molar heat capacity at constant pressure) is given by:
\[ C_P = C_V + R \]
Thus, we can express \(C_P\) in terms of \(C_V\):
\[ C_P = C_V + R \]

### Step 3: Use the degrees of freedom to find \(C_V\)
For a gas, the molar heat capacity at constant volume \(C_V\) can be expressed in terms of the degrees of freedom \(f\):
\[ C_V = \frac{f}{2} R \]
Where \(f\) is the degrees of freedom of the gas.

### Step 4: Relate \(R\) and \(C_V\) using the given ratio
Given that \(\frac{R}{C_V} = 0.67\), we can substitute \(C_V\) from the previous step:
\[ \frac{R}{\frac{f}{2} R} = 0.67 \]
This simplifies to:
\[ \frac{2}{f} = 0.67 \]

### Step 5: Solve for \(f\)
Rearranging the equation gives:
\[ f = \frac{2}{0.67} \approx 2.99 \]
Since \(f\) must be a whole number, we round \(f\) to 3.

### Step 6: Identify the type of gas
- For monatomic gases, \(f = 3\).
- For diatomic gases, \(f = 5\).
- For triatomic gases (non-linear), \(f = 6\).

Since we found \(f \approx 3\), this indicates that the gas is likely a **monatomic gas**.

### Conclusion
The gas is a **monatomic gas**.

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