for a Maxwellian gas, show that
,
1 4
V
V
where
V
is the average
speed.
Answers
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LMS
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In the context of the Kinetic Molecular Theory of Gases, a gas contains a large number of particles in rapid motions. Each particle has a different speed, and each collision between particles changes the speeds of the particles. An understanding of the properties of the gas requires an understanding of the distribution of particle speeds.
Many molecules, many velocities
At temperatures above absolute zero, all molecules are in motion. In the case of a gas, this motion consists of straight-line jumps whose lengths are quite great compared to the dimensions of the molecule. Although we can never predict the velocity of a particular individual molecule, the fact that we are usually dealing with a huge number of them allows us to know what fraction of the molecules have kinetic energies (and hence velocities) that lie within any given range.
M
Figure 27.3.4: The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures.Image used with permission from OpenStax. The speed at the top of the curve is called the most probable speed because the largest number of molecules have that speed.
The average speed is the sum of the speeds of all the molecules divided by the number of molecules.
vavg=
ˉ
v
=∫
∞
0
vf(v)dv=
√
8RT
πM
The root-mean-square speed is square root of the average speed-squared.
vrms=
¯
v2
=
√
3RT
M
where
R is the gas constant,
T is the absolute temperature and
M is the molar mass of the gas.
It always follows that for gases that follow the Maxwell-Boltzmann distribution:
vmp<vavg<vrms
Problems
Using the Maxwell-Boltzman function, calculate the fraction of argon gas molecules with a speed of 305 m/s at 500 K.
If the system in problem 1 has 0.46 moles of argon gas, how many molecules have the speed of 305 m/s?
Calculate the values of Cmp, Cavg, and Crms for xenon gas at 298 K.
From the values calculated above, label the Boltzmann distribution plot (Figure 1) with the approximate locations of (C_{mp}\), Cavg, and Crms.
What will have a larger speed distribution, helium at 500 K or argon at 300 K? Helium at 300 K or argon at 500 K? Argon at 400 K or argon at 1000 K?
Answers
0.00141
3.92×1020 argon molecules
cmp = 194.27 m/s, cavg = 219.21 m/s, crms = 237.93 m/s
As stated above, Cmp is the most probable speed, thus it will be at the top of the distribution curve. To the right of the most probable speed will be the average speed, followed by the root-mean-square speed.
Hint: Use the related speed expressions to determine the distribution of the gas molecules: helium at 500 K. helium at at 300 K. argon at 1000 K.
References
Dunbar, R.C. Deriving the Maxwell Distribution J. Chem. Ed. 1982, 59, 22-23.
Peckham, G.D.; McNaught, I.J.; Applications of the Maxwell-Boltzmann Distribution J. Chem. Ed. 1992, 69, 554-558.
Chang, R. Physical Chemistry for the Biosciences, 25-27.
Contributors
Prof. David Blauch (Davidson College)
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
Adam Maley (Hope College)
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27.2: The Distribution of the Components of Molecular Speeds are Described by a Gaussian Distribution 27.4: The Frequency of Collisions
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