/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Estimate the mean free path of a... [FREE SOLUTION] | 91影视

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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 \(0.3 脳 10^{-9} \,\text{m}\)).
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: $$ \lambda = \frac{(1.38 脳 10^{-23}\,\text{ J K}^{-1}) 脳 (290\,\text{K})} {\sqrt{2} \times \pi \times (0.3 脳 10^{-9}\,\text{m})^2 \times (100\times 10^3\,\text{Pa})} $$ 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: $$ \lambda = \frac{(1.38 脳 10^{-23}\,\text{ J K}^{-1}) 脳 (230\,\text{K})} {\sqrt{2} \times \pi \times (0.3 脳 10^{-9}\,\text{m})^2 \times (50\times 10^3\,\text{Pa})} $$ 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: $$ \lambda = \frac{(1.38 脳 10^{-23}\, \text{ J K}^{-1}) 脳 (230\,\text{K})} {\sqrt{2} \times \pi \times (0.3 脳 10^{-9}\,\text{m})^2 \times (1\times 10^3\,\text{Pa})} $$ 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

The driver from Practice Problem 13.3 fills his \(18.9-\mathrm{L}\) steel gasoline can in the morning when the temperature of the can and the gasoline is \(15.0^{\circ} \mathrm{C}\) and the pressure is 1.0 atm, but this time he remembers to replace the tightly fitting cap after filling the can. Assume that the can is completely full of gasoline (no air space) and that the cap does not leak. The temperature climbs to \(30.0^{\circ} \mathrm{C}\) Ignoring the expansion of the steel can, what would be the pressure of the heated gasoline? The bulk modulus for gasoline is $1.00 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$
(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 40.0 psi. If the pump is a cylinder of length 18.0 in. with a cross- sectional area of 3.00 in. \(^{2},\) how far down must the piston be pushed before air will flow into the tire?
A cylinder with an interior cross-sectional area of \(70.0 \mathrm{cm}^{2}\) has a moveable piston of mass \(5.40 \mathrm{kg}\) at the top that can move up and down without friction. The cylinder contains $2.25 \times 10^{-3} \mathrm{mol}\( of an ideal gas at \)23.0^{\circ} \mathrm{C} .$ (a) What is the volume of the gas when the piston is in equilibrium? Assume the air pressure outside the cylinder is 1.00 atm. (b) By what factor does the volume change if the gas temperature is raised to \(223.0^{\circ} \mathrm{C}\) and the piston moves until it is again in equilibrium?

Verify that the SI units of \(P V\) (pressure times volume) are joules.
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