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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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Most popular questions from this chapter

A flat square of side \(s_{0}\) at temperature \(T_{0}\) expands by \(\Delta s\) in both length and width when the temperature increases by \(\Delta T\). The original area is \(s_{0}^{2}=A_{0}\) and the final area is $\left(s_{0}+\Delta s\right)^{2}=A .\( Show that if \)\Delta s \ll s_{0}$$$\frac{\Delta A}{A_{0}}=2 \alpha \Delta T$$(Although we derive this relation for a square plate, it applies to a flat area of any shape.)
The volume of air taken in by a warm-blooded vertebrate in the Andes mountains is \(210 \mathrm{L} /\) day at standard temperature and pressure (i.e., \(\left.0^{\circ} \mathrm{C} \text { and } 1 \text { atm }\right) .\) If the air in the lungs is at \(39^{\circ} \mathrm{C},\) under a pressure of $450 \mathrm{mm} \mathrm{Hg}$ and we assume that the vertebrate takes in an average volume of \(100 \mathrm{cm}^{3}\) per breath at the temperature and pressure of its lungs, how many breaths does this vertebrate take per day?
An emphysema patient is breathing pure \(\mathrm{O}_{2}\) through a face mask. The cylinder of \(\mathrm{O}_{2}\) contains \(0.60 \mathrm{ft}^{3}\) of \(\mathrm{O}_{2}\) gas at a pressure of \(2200 \mathrm{lb} / \mathrm{in}^{2} .\) (a) What volume would the oxygen occupy at atmospheric pressure (and the same temperature)? (b) If the patient takes in \(8 \mathrm{L} / \mathrm{min}\) of \(\mathrm{O}_{2}\) at atmospheric pressure, how long will the cylinder last?
A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)
Consider the expansion of an ideal gas at constant pressure. The initial temperature is \(T_{0}\) and the initial volume is \(V_{0} .\) (a) Show that \(\Delta V / V_{0}=\beta \Delta T,\) where \(\beta=1 / T_{0}\) (b) Compare the coefficient of volume expansion \(\beta\) for an ideal gas at \(20^{\circ} \mathrm{C}\) to the values for liquids and gases listed in Table 13.3
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