If the mean free path of a molecule of an ideal gas having diameter `10^(-8)` cm is `10^(-4)` m, calculate the number density of the gas.

Answer

To solve the problem of calculating the number density of an ideal gas given the mean free path and the diameter of its molecules, we can follow these steps:

### Step 1: Convert Units
First, we need to ensure that all measurements are in consistent units. The diameter \( D \) is given in centimeters, and the mean free path \( \lambda \) is given in meters.

- Given:
- Diameter \( D = 10^{-8} \) cm
- Mean free path \( \lambda = 10^{-4} \) m

Convert \( \lambda \) from meters to centimeters:
\[ \lambda = 10^{-4} \text{ m} \times 100 \text{ cm/m} = 10^{-2} \text{ cm} \]

### Step 2: Use the Mean Free Path Formula
The mean free path \( \lambda \) is related to the number density \( n \) by the formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi D^2 n} \]
Rearranging this formula to solve for the number density \( n \):
\[ n = \frac{1}{\sqrt{2} \pi D^2 \lambda} \]

### Step 3: Substitute Values
Now, substitute the values of \( D \) and \( \lambda \) into the formula.

- \( D = 10^{-8} \) cm
- \( \lambda = 10^{-2} \) cm

Calculating \( D^2 \):
\[ D^2 = (10^{-8})^2 = 10^{-16} \text{ cm}^2 \]

Now substitute into the equation for \( n \):
\[ n = \frac{1}{\sqrt{2} \pi (10^{-16}) (10^{-2})} \]
\[ n = \frac{1}{\sqrt{2} \pi (10^{-18})} \]

### Step 4: Calculate the Numerical Value
Now we can calculate \( n \):
\[ n = \frac{1}{\sqrt{2} \times 3.14 \times 10^{-18}} \]
Calculating \( \sqrt{2} \approx 1.414 \):
\[ n \approx \frac{1}{1.414 \times 3.14 \times 10^{-18}} \approx \frac{1}{4.442 \times 10^{-18}} \approx 2.25 \times 10^{16} \text{ molecules/cm}^3 \]

### Final Answer
Thus, the number density \( n \) of the gas is approximately:
\[ n \approx 2.25 \times 10^{16} \text{ molecules/cm}^3 \]

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