If the molar specific heat of a gas at constant pressure is `7/2R`, then the atomicity of gas is

Options

A. one
B. two
C. three
D. four

Explanation

To solve the problem, we need to determine the atomicity of a gas given that its molar specific heat at constant pressure (Cp) is \( \frac{7}{2}R \).

### Step-by-step Solution:

1. **Understanding Molar Specific Heat at Constant Pressure**:
The molar specific heat at constant pressure (\( C_p \)) for a gas can be expressed in terms of its degrees of freedom (\( F \)) as:
\[ C_p = \left( 1 + \frac{F}{2} \right) R \]
where \( R \) is the universal gas constant.

2. **Setting Up the Equation**:
We are given that:
\[ C_p = \frac{7}{2} R \]
Therefore, we can set up the equation:
\[ \left( 1 + \frac{F}{2} \right) R = \frac{7}{2} R \]

3. **Canceling \( R \)**:
Since \( R \) is a common factor on both sides of the equation, we can cancel it out:
\[ 1 + \frac{F}{2} = \frac{7}{2} \]

4. **Isolating \( F \)**:
Next, we isolate \( \frac{F}{2} \):
\[ \frac{F}{2} = \frac{7}{2} - 1 \]
Simplifying the right side:
\[ \frac{F}{2} = \frac{7}{2} - \frac{2}{2} = \frac{5}{2} \]

5. **Finding \( F \)**:
Now, we multiply both sides by 2 to find \( F \):
\[ F = 5 \]

6. **Determining Atomicity**:
The degrees of freedom (\( F \)) are related to the atomicity of the gas. For different types of gases:
- Monatomic gases have \( F = 3 \)
- Diatomic gases have \( F = 5 \)
- Triatomic gases have \( F = 6 \) or more depending on their structure.

Since we found \( F = 5 \), this corresponds to a diatomic gas.

7. **Conclusion**:
Therefore, the atomicity of the gas is \( 2 \).

### Final Answer:
The atomicity of the gas is \( 2 \).

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