If typical size of a gas molecule is `2Å`, average distance between the molecules is

Options

A. `1 Å`
B. `2 Å`
C. `lt20 Å`
D. `ge 20 Å`

Explanation

To find the average distance between gas molecules given that the typical size of a gas molecule is \(2 \, \text{Å}\) (angstrom), we can use the formula for the mean free path, which is a measure of the average distance a molecule travels between collisions. The formula for the mean free path (\(\lambda\)) is given by:

\[ \lambda = \frac{k_B T}{\sqrt{2} \pi D^2 P} \]

Where:
- \(k_B\) is the Boltzmann constant, approximately \(1.38 \times 10^{-23} \, \text{J/K}\)
- \(T\) is the temperature in Kelvin
- \(D\) is the diameter of the gas molecule
- \(P\) is the pressure in Pascals

### Step-by-Step Solution:

1. **Identify the Values**:
- Diameter of the gas molecule \(D = 2 \, \text{Å} = 2 \times 10^{-10} \, \text{m}\)
- Temperature \(T = 273 \, \text{K}\) (standard temperature)
- Pressure \(P = 1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa}\)

2. **Substitute the Values into the Formula**:
\[ \lambda = \frac{(1.38 \times 10^{-23} \, \text{J/K}) \times (273 \, \text{K})}{\sqrt{2} \pi (2 \times 10^{-10} \, \text{m})^2 (1.01 \times 10^5 \, \text{Pa})} \]

3. **Calculate the Denominator**:
- First, calculate \(D^2\):
\[ D^2 = (2 \times 10^{-10})^2 = 4 \times 10^{-20} \, \text{m}^2 \]
- Now calculate \(\sqrt{2} \pi D^2 P\):
\[ \sqrt{2} \approx 1.414, \quad \pi \approx 3.14 \]
\[ \sqrt{2} \pi D^2 P = 1.414 \times 3.14 \times (4 \times 10^{-20}) \times (1.01 \times 10^5) \]
\[ \approx 1.414 \times 3.14 \times 4.04 \times 10^{-15} \approx 1.78 \times 10^{-14} \]

4. **Calculate the Numerator**:
\[ k_B T = (1.38 \times 10^{-23}) \times (273) \approx 3.77 \times 10^{-21} \, \text{J} \]

5. **Calculate the Mean Free Path**:
\[ \lambda = \frac{3.77 \times 10^{-21}}{1.78 \times 10^{-14}} \approx 2.11 \times 10^{-7} \, \text{m} \]

6. **Convert to Angstroms**:
\[ 2.11 \times 10^{-7} \, \text{m} = 2110 \, \text{Å} \]

### Final Answer:
The average distance between the molecules is approximately \(2110 \, \text{Å}\).