If the mean free path of atoms is doubled then the pressure of gas will become

Options

A. `P//4`
B. `P//2`
C. `P//8`
D. `P`

Explanation

To solve the problem, we need to analyze the relationship between the mean free path of gas molecules and the pressure of the gas. The mean free path (λ) is the average distance a molecule travels between collisions, and it is related to pressure (P) as follows:

### Step-by-Step Solution:

1. **Understand the Mean Free Path Formula**:
The mean free path (λ) can be expressed using the formula:
\[ \lambda = \frac{k_B T}{\sqrt{2} \pi D^2 P} \]
where:
- \(k_B\) is the Boltzmann constant,
- \(T\) is the absolute temperature,
- \(D\) is the diameter of the gas molecules,
- \(P\) is the pressure of the gas.

2. **Identify the Relationship**:
From the formula, we can see that the mean free path (λ) is inversely proportional to the pressure (P):
\[ \lambda \propto \frac{1}{P} \]
This means if the mean free path increases, the pressure must decrease, and vice versa.

3. **Doubling the Mean Free Path**:
According to the problem, the mean free path is doubled:
\[ \lambda' = 2\lambda \]
If we denote the new pressure as \(P'\), we can set up the relationship:
\[ \frac{\lambda}{\lambda'} = \frac{P'}{P} \]

4. **Substituting the Values**:
Substituting \(\lambda' = 2\lambda\) into the equation gives:
\[ \frac{\lambda}{2\lambda} = \frac{P'}{P} \]
Simplifying this, we find:
\[ \frac{1}{2} = \frac{P'}{P} \]

5. **Solving for the New Pressure**:
Rearranging the equation gives:
\[ P' = \frac{P}{2} \]
This indicates that the new pressure \(P'\) is half of the original pressure \(P\).

6. **Conclusion**:
Thus, if the mean free path of atoms is doubled, the pressure of the gas will become half of its initial value.

### Final Answer:
The pressure of the gas will become \( \frac{P}{2} \).

---