Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 21, Problem 21.62AP

A vessel contains 1.00 × 104 oxygen molecules at 500 K. (a) Make an accurate graph of the Maxwell speed distribution function versus speed with points at speed intervals of 100 m/s. (b) Determine the most probable speed from this graph. (c) Calculate the average and rms speeds for the molecules and label these points on your graph. (d) From the graph, estimate the fraction of molecules with speeds in the range 300 m/s to 600 m/s.

(a)

Expert Solution
Check Mark
To determine

The graph of the Maxwell speed distribution function versus speed with points at speed intervals of 100m/s.

Answer to Problem 21.62AP

The of the Maxwell speed distribution function versus speed with points at speed intervals of 100m/s is shown below;

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  1

Explanation of Solution

The Maxwell distribution curve is the graph between the distribution of speed and the change in speed or speed interval.

The number of molecules of oxygen in vessel is 1.00×104 and the temperature is 500K. The interval of speed is 100m/s. The range of speed is 300m/s to 600m/s.

Write the expression of Maxwell’s speed distribution function.

    Nv=4πN(m02πkBT)v2em0v2/2kBT                                             (1)

Here, N is the total number of molecules of oxygen in the vessel, m0 is the mass of molecule of oxygen, kB is the Boltzmann’s constant, T is the absolute temperature and em0v2/2kBT is the Boltzmann factor.

The mass of the molecules of oxygen

    m0=MNA

Here, M is the molecular mass of the oxygen in kg and NA is the Avogadro number.

The molecular mass of the oxygen molecules in kg is 0.032.

Substitute MNA for m0 in equation (1).

    Nv=4πN((MNA)2πkBT)v2e(MNA)v2/2kBT

Substitute 1.00×104 for N, 0.032kg for M, 6.02×1023 for NA, 500K for T and 1.38×1023J/moleculesK for kB in above equation.

    Nv=4π(1.00×104)[((0.032kg6.02×1023)2π(1.38×1023J/moleculesK)(500K))v2e(0.032kg6.02×1023)v2/2(1.38×1023J/moleculesK)(500K)]=(1.71×104)v2e(3.85×106)v2

Substitute the values of v and form a table.

v(m/s)NV
00
1001.64
2005.86
30010.88
40014.78
50016.33
60015.39
70012.7
8009.31
9006.13
10003.64
11001.961
12000.96
13000.43
14000.18
15000.07

On the basis of the table, a graph is plotted below;

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  2

(b)

Expert Solution
Check Mark
To determine

The most probable speed from the graph.

Answer to Problem 21.62AP

The most probable speed is 510m/s.

Explanation of Solution

The most probable speed occurs where NV is maximum. From the above graph the value of speed when Nv is maximum is 510m/s.

Conclusion:

Therefore, the most probable speed is 510m/s.

(c)

Expert Solution
Check Mark
To determine

The average and rms speeds for the molecules and label these points on the graph.

Answer to Problem 21.62AP

The average and rms speeds for the molecules is 575m/s and 624m/s respectively and the graph is shown below,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  3

Explanation of Solution

Write the expression of average velocity.

    vavg=8kBTπm0

The mass of the molecules of oxygen

    m0=MNA

Substitute MNA for m0 in above equation.

    vavg=8kBTπ(MNA)

The molecular mass of the oxygen molecules in kg is 0.032.

Substitute 1.38×1023J/moleculeK for kB, 500K for T, 0.032kg for M and 6.02×1023 for NA in above equation.

    vavg=8(1.38×1023J/moleculeK)500Kπ(0.032kg6.02×1023)=575m/s

Thus, the average speed is 575m/s.

Write the expression of rms velocity.

    vrms=3kBTm0

Substitute MNA for m0 in above equation.

    vrm=3kBT(MNA)

Substitute 1.38×1023J/moleculeK for kB, 500K for T, 0.032kg for M and 6.02×1023 for NA in above equation.

    vrms=3(1.38×1023J/moleculeK)500K(0.032kg6.02×1023)=624m/s

Thus, the rms velocity of the oxygen molecules is 624m/s.

The graph of Maxwell’s curve is shown below;

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  4

The point A on the graph is the average velocity of the oxygen molecules and the point B is the average velocity of the oxygen molecules.

Conclusion:

Therefore, the average and rms speeds for the molecules is 575m/s and 624m/s respectively and the graph is shown below,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  5

(d)

Expert Solution
Check Mark
To determine

The fraction of molecules with the speed in the range of 300m/s to 600m/s.

Answer to Problem 21.62AP

The fraction of molecules with the speed in the range of 300m/s to 600m/s is 44%.

Explanation of Solution

The figure given below shows the Maxwell’s curve,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 21, Problem 21.62AP , additional homework tip  6

The area under the distribution curve in the range 300m/s to 600m/s gives the fraction of molecules with speed in the range 300m/s to 600m/s.

Conclusion:

Write the area under the curve in the range 300m/s to 600m/s

    F11×300+(100×5.5)+12(200×5.5)=4400=4400×1100%=44%

Therefore, the fraction of molecules with the speed in the range of 300m/s to 600m/s is 44%.

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Students have asked these similar questions
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Chapter 21 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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