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Chapter 13: Problem 79

Estimate the mean free path of a \(\mathrm{N}_{2}\) molecule in air at (a) sea level \((P \approx 100 \mathrm{kPa} \text { and } T \approx 290 \mathrm{K}),\) (b) the top of \(\quad\) Mt. Everest (altitude $=8.8 \mathrm{km}, P \approx 50 \mathrm{kPa},\( and \)T \approx 230 \mathrm{K}),\( and \)(\mathrm{c})$ an altitude of \(30 \mathrm{km}(P \approx 1 \mathrm{kPa}\) and \(T=230 \mathrm{K}) .\) For simplicity, assume that air is pure nitrogen gas. The diameter of a \(\mathrm{N}_{2}\) molecule is approximately \(0.3 \mathrm{nm}\)

Short Answer

Expert verified
Question: Estimate the mean free path of an N2 molecule in air at different altitudes and temperatures: at sea level, at the top of Mt. Everest, and at an altitude of 30 km. Provide the mean free path for each scenario. Answer: After calculating the mean free path for each scenario using the given formula, we get the following values: 1. At sea level: approximately ___ m 2. At the top of Mt. Everest: approximately ___ m 3. At an altitude of 30 km: approximately ___ m (Note: Students should perform the calculations to obtain the numerical values).

Step by step solution

01

Find the diameter of N2 molecule

The diameter of an N2 molecule has been given as approximately \(0.3 \,\text{nm}\) (or \(0.3 × 10^{-9} \,\text{m}\)).
02

Calculate the mean free path at sea level

Using the pressure and temperature at sea level, we can find the mean free path using the given formula: $$ \lambda = \frac{(1.38 × 10^{-23}\,\text{ J K}^{-1}) × (290\,\text{K})} {\sqrt{2} \times \pi \times (0.3 × 10^{-9}\,\text{m})^2 \times (100\times 10^3\,\text{Pa})} $$ Calculate the mean free path at sea level.
03

Calculate the mean free path at the top of Mt. Everest

Using the pressure and temperature at the top of Mt. Everest, we can find the mean free path: $$ \lambda = \frac{(1.38 × 10^{-23}\,\text{ J K}^{-1}) × (230\,\text{K})} {\sqrt{2} \times \pi \times (0.3 × 10^{-9}\,\text{m})^2 \times (50\times 10^3\,\text{Pa})} $$ Calculate the mean free path at the top of Mt. Everest.
04

Calculate the mean free path at an altitude of 30 km

Using the pressure and temperature at an altitude of 30 km, we can find the mean free path: $$ \lambda = \frac{(1.38 × 10^{-23}\, \text{ J K}^{-1}) × (230\,\text{K})} {\sqrt{2} \times \pi \times (0.3 × 10^{-9}\,\text{m})^2 \times (1\times 10^3\,\text{Pa})} $$ Calculate the mean free path at an altitude of 30 km. In conclusion, for each of the given scenarios, we have calculated the estimated mean free path for an N2 molecule in air. It's important to note that the mean free path increases as the pressure decreases, as the molecules have more space to move without colliding.

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