What is the mean free path for oxygen molecules at 300 K and atmospheric pressure (105 Pa)?
Calculate the average frequency of collisions for a particular molecule. (The diameter of an oxygen molecule is 0.29 nm).

To calculate the mean free path (\( \lambda \)) and average frequency of collisions (\( f \)) for oxygen molecules, we use the following formulas:

### Step 1: Mean Free Path Calculation

The mean free path \( \lambda \) for gas molecules is given by:
\[
\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}
\]
where:
- \( k_B \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \( T \) is the temperature in Kelvin (300 K in this case),
- \( d \) is the diameter of an oxygen molecule (\(0.29 \, \text{nm} = 0.29 \times 10^{-9} \, \text{m}\)),
- \( P \) is the pressure (given as \(10^5 \, \text{Pa}\)).

#### Calculation:
Substituting the values:
\[
\lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi (0.29 \times 10^{-9})^2 \times 10^5}
\]
Let's calculate this:

1. \( \sqrt{2} \pi \approx 4.44288 \)
2. \( d^2 = (0.29 \times 10^{-9})^2 = 8.41 \times 10^{-20} \, \text{m}^2 \)
3. \( \lambda = \frac{1.38 \times 10^{-23} \times 300}{4.44288 \times 8.41 \times 10^{-20} \times 10^5} \approx 9.71 \times 10^{-8} \, \text{m} \)

So, the mean free path \( \lambda \approx 9.71 \times 10^{-8} \, \text{m} \).

### Step 2: Collision Frequency Calculation

The average frequency of collisions \( f \) for a molecule is given by:
\[
f = \frac{\bar{v}}{\lambda}
\]
where:
- \( \bar{v} \) is the average speed of an oxygen molecule, given by:
  \[
  \bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}
  \]
- \( m \) is the mass of an oxygen molecule. For \( \text{O}_2 \), \( m = \frac{32 \times 10^{-3}}{6.022 \times 10^{23}} \approx 5.32 \times 10^{-26} \, \text{kg} \).

#### Calculate \( \bar{v} \):
\[
\bar{v} = \sqrt{\frac{8 \times 1.38 \times 10^{-23} \times 300}{\pi \times 5.32 \times 10^{-26}}}
\]
\[
\bar{v} \approx 394 \, \text{m/s}
\]

#### Calculate \( f \):
\[
f = \frac{394}{9.71 \times 10^{-8}} \approx 4.06 \times 10^9 \, \text{collisions per second}
\]

### Summary
- **Mean free path** (\( \lambda \)) ≈ \( 9.71 \times 10^{-8} \, \text{m} \)
- **Collision frequency** (\( f \)) ≈ \( 4.06 \times 10^9 \, \text{Hz} \)

Would you like further details or have any questions?

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**Related Questions for Exploration:**
1. How would increasing temperature affect the mean free path of oxygen molecules?
2. What impact does reducing the pressure have on collision frequency?
3. How is mean free path related to gas density?
4. Can you derive the formula for mean free path from kinetic theory principles?
5. What are real-world applications of understanding mean free path in gases?

**Tip:** Mean free path decreases with increased pressure, as molecules are packed closer together, increasing collision frequency.

What is the mean free path for oxygen molecules at 300 K and atmospheric pressure (105 Pa)? Calculate the average frequency of collisions for a particular molecule. (The diameter of an oxygen molecule is 0.29 nm).

Explore how to calculate the mean free path and collision frequency for oxygen molecules at 300 K and atmospheric pressure (105 Pa) using principles from the kinetic theory of gases. This solution includes key formulas, such as mean free path and collision frequency calculations, suitable for undergraduate physics students.

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