Calculate the mean free path of gas molecules, if number of molecules per `cm^(3) is 3 xx 10^(19)` and diameter of each "molecule" is `2Å`.

Explanation

To calculate the mean free path of gas molecules, we will follow these steps:

### Step 1: Convert the number of molecules per cm³ to m³
Given:
- Number of molecules per cm³ = \(3 \times 10^{19}\)

To convert this to per m³, we use the conversion factor \(1 \text{ cm}^3 = 10^{-6} \text{ m}^3\):
\[ N = 3 \times 10^{19} \text{ molecules/cm}^3 = 3 \times 10^{19} \times 10^{6} \text{ molecules/m}^3 = 3 \times 10^{25} \text{ molecules/m}^3 \]

### Step 2: Convert the diameter of each molecule from Ångströms to meters
Given:
- Diameter of each molecule = \(2 \text{ Å} = 2 \times 10^{-10} \text{ m}\)

### Step 3: Use the formula for mean free path
The mean free path (\(\lambda\)) can be calculated using the formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi D^2 N} \]
Where:
- \(D\) = diameter of each molecule
- \(N\) = number density of molecules (in m³)

### Step 4: Substitute the values into the formula
Substituting \(D = 2 \times 10^{-10} \text{ m}\) and \(N = 3 \times 10^{25} \text{ m}^{-3}\):
\[ \lambda = \frac{1}{\sqrt{2} \pi (2 \times 10^{-10})^2 (3 \times 10^{25})} \]

### Step 5: Calculate \(D^2\)
Calculating \(D^2\):
\[ D^2 = (2 \times 10^{-10})^2 = 4 \times 10^{-20} \text{ m}^2 \]

### Step 6: Substitute \(D^2\) back into the formula
Now substituting \(D^2\) into the mean free path formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi (4 \times 10^{-20}) (3 \times 10^{25})} \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ \sqrt{2} \approx 1.414 \]
\[ \pi \approx 3.14 \]
So,
\[ \text{Denominator} = 1.414 \times 3.14 \times 4 \times 10^{-20} \times 3 \times 10^{25} \]
Calculating this gives:
\[ = 1.414 \times 3.14 \times 12 \times 10^{5} \approx 53.1 \times 10^{5} \approx 5.31 \times 10^{6} \]

### Step 8: Calculate the mean free path
Now we can find \(\lambda\):
\[ \lambda = \frac{1}{5.31 \times 10^{6}} \approx 1.88 \times 10^{-7} \text{ m} \]

### Final Answer
The mean free path of the gas molecules is approximately:
\[ \lambda \approx 1.876 \times 10^{-7} \text{ m} \]

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