What is the mean free path of oxygen molecules at an altitude of 300 km, where the pressure is only 10−11 atmospheres? (Assume the temperature is 300 K). r= 14.5 nm

The mean free path, \( \lambda \), is the average distance a molecule travels before colliding with another molecule. For gases, the mean free path can be calculated using the following formula:

\[
\lambda = \frac{k_B \cdot T}{\sqrt{2} \cdot \pi \cdot r^2 \cdot p}
\]

where:
- \( k_B \) is the Boltzmann constant, approximately \( 1.38 \times 10^{-23} \, \text{J/K} \),
- \( T \) is the temperature in Kelvin (given as \( 300 \, \text{K} \)),
- \( r \) is the effective radius of the molecule (given as \( 14.5 \, \text{nm} = 14.5 \times 10^{-9} \, \text{m} \)),
- \( p \) is the pressure (given as \( 10^{-11} \, \text{atm} \)).

### Step 1: Convert Pressure to SI Units

We need the pressure in Pascals (Pa), not in atmospheres. The conversion factor is:
\[
1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}
\]

So,
\[
p = 10^{-11} \, \text{atm} \times 1.013 \times 10^5 \, \text{Pa/atm} = 1.013 \times 10^{-6} \, \text{Pa}
\]

### Step 2: Substitute Values and Solve

Now substitute the known values:

\[
\lambda = \frac{(1.38 \times 10^{-23} \, \text{J/K}) \cdot 300 \, \text{K}}{\sqrt{2} \cdot \pi \cdot (14.5 \times 10^{-9} \, \text{m})^2 \cdot 1.013 \times 10^{-6} \, \text{Pa}}
\]

Calculating this will give the mean free path of the oxygen molecules at 300 km altitude.

Let's perform the calculation.The mean free path \( \lambda \) of oxygen molecules at an altitude of 300 km is approximately \( 4.38 \, \text{meters} \).

Would you like more details on this solution or have further questions?

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Here are related questions to explore further:

1. How would the mean free path change if the pressure decreased by another factor of 10?
2. What is the impact of temperature on the mean free path in similar conditions?
3. How would the mean free path differ for nitrogen molecules at the same altitude?
4. What are the implications of such a long mean free path for molecular collisions?
5. How does mean free path affect the behavior of gases in different layers of the atmosphere?

**Tip:** At very high altitudes, the low density of air makes the mean free path long, causing gases to act more like particles in free motion than interacting molecules.

What is the mean free path of oxygen molecules at an altitude of 300 km, where the pressure is only 10^-11 atmospheres? (Assume the temperature is 300 K). r= 14.5 nm

Explore the mean free path calculation for oxygen molecules at 300 km altitude with low pressure (10^-11 atm) and temperature of 300 K. This solution includes the application of kinetic theory, unit conversions, and the mean free path formula, providing insights suitable for advanced high school and college-level physics students.

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