If the number density of molecule of an ideal gas becomes 4 time then its mean free path becomes

Options

A. Half of its initial value
B. Double of its initial value
C. Four time of its initial value
D. One fourth of its initial value

Answer

To solve the problem, we need to analyze how the mean free path of an ideal gas changes when the number density of its molecules increases. Here are the steps to arrive at the solution:

### Step 1: Understand the Mean Free Path Formula
The mean free path (λ) of gas molecules is given by the formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \]
where:
- \( \lambda \) = mean free path
- \( d \) = diameter of the gas molecules
- \( n \) = number density of the gas molecules (number of molecules per unit volume)

### Step 2: Define Initial Conditions
Let the initial number density be \( n_1 \). Therefore, the initial mean free path \( \lambda_1 \) can be expressed as:
\[ \lambda_1 = \frac{1}{\sqrt{2} \pi d^2 n_1} \]

### Step 3: Define New Conditions
According to the problem, the number density becomes 4 times the initial value:
\[ n_2 = 4n_1 \]
Now, we can express the new mean free path \( \lambda_2 \) as:
\[ \lambda_2 = \frac{1}{\sqrt{2} \pi d^2 n_2} = \frac{1}{\sqrt{2} \pi d^2 (4n_1)} \]

### Step 4: Substitute the New Number Density
Substituting \( n_2 \) into the equation for \( \lambda_2 \):
\[ \lambda_2 = \frac{1}{\sqrt{2} \pi d^2 (4n_1)} = \frac{1}{4} \cdot \frac{1}{\sqrt{2} \pi d^2 n_1} \]
This can be rewritten as:
\[ \lambda_2 = \frac{1}{4} \lambda_1 \]

### Step 5: Conclusion
From the calculations, we find that when the number density of the gas increases to four times its initial value, the mean free path becomes one fourth of its initial value:
\[ \lambda_2 = \frac{1}{4} \lambda_1 \]

### Final Answer
Thus, the mean free path becomes one fourth of its initial value.

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