Calculate the diameter of one molecule of an ideal gas having number density `2xx10^(8) cm^(-3)` and mean free path of the molecule is `10^(-8)` cm.

Answer

To calculate the diameter of one molecule of an ideal gas, we can use the formula for the mean free path (λ):

\[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \]

Where:
- \( \lambda \) is the mean free path,
- \( d \) is the diameter of the molecule,
- \( n \) is the number density of the gas.

We can rearrange this formula to solve for the diameter \( d \):

\[ d = \sqrt{\frac{1}{\sqrt{2} \pi n \lambda}} \]

### Step 1: Substitute the given values
We are given:
- Number density \( n = 2 \times 10^8 \, \text{cm}^{-3} \)
- Mean free path \( \lambda = 10^{-8} \, \text{cm} \)

Substituting these values into the rearranged formula:

\[ d = \sqrt{\frac{1}{\sqrt{2} \pi (2 \times 10^8) (10^{-8})}} \]

### Step 2: Calculate the denominator
First, we calculate the product in the denominator:

\[ \sqrt{2} \approx 1.414 \]
\[ \pi \approx 3.14 \]

Now, calculate:

\[ \sqrt{2} \pi n \lambda = 1.414 \times 3.14 \times (2 \times 10^8) \times (10^{-8}) \]

This simplifies to:

\[ = 1.414 \times 3.14 \times 2 = 8.88 \quad \text{(since } 10^8 \times 10^{-8} = 1\text{)} \]

### Step 3: Substitute back into the diameter formula
Now substituting back into the formula for \( d \):

\[ d = \sqrt{\frac{1}{8.88}} \]

### Step 4: Calculate the square root
Now we calculate:

\[ \frac{1}{8.88} \approx 0.1126 \]

Taking the square root:

\[ d \approx \sqrt{0.1126} \approx 0.335 \, \text{cm} \]

### Step 5: Final result
Thus, the diameter of one molecule of the ideal gas is approximately:

\[ d \approx 0.33 \, \text{cm} \]