The Velocities of Gas Molecules ©
by
Flick Coleman
Department of Chemistry
Wellesley College
Wellesley MA 02181
© Copyright Flick Coleman 1996. All rights reserved. You are welcome to use this document in your
own classes but commercial use is not allowed without the permission of the author.
A knowledge of the velocities of gaseous molecules helps us understand a number of chemical
phenomena. As an example, the rates of chemical reactions involving gaseous species depend on
the kinetic energy of the molecules, which, in turn, depends on the velocity of the molecules.
When gas velocities are measured experimentally, a number of interesting features are observed.
Further down this page you will examine the effect of molecular mass on gas velocity, and later in
the document you can look at other factors. At the end of the first part of this document you will
be asked to write a paragraph about gas velocities based on your observations here. The
questions that you are to answer are shown in boldface.
Note that in the following graph, the molar mass of the three gases is given in kg/mol, so that N2
has a molar mass of 0.028 etc. In this first plot, change the masses of the three gases several
times, and change the temperature as you wish. Observe how the plots change. How will you
determine which curve corresponds to which gas? (You may also need to change the range for
u, the velocity, depending on your choice of masses and temperature. Consult your MathCad
guide if you do not remember how to change the range of a variable.)
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 1
fraction with given velocity
0.001
5 10
4
0
0
1000
2000
3000
4000
5000
velocity (m/sec)
trace 1
trace 2
trace 3
m1 .016 m2 .032 m3 .044
u 0 , 5 .. 5000
Molec. Wt. in kg/mol
T 2000
T in K
Velocity in m/sec
Do all molecules in a gas sample at a particular temperature have the same velocity?
Why do chemists speak of the distribution of gas velocities?
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 2
Before proceeding to the next page, predict the behavior of gas velocity as the
temperature changes. In the next graph, you choose a molecular weight and then plot the
distribution of velocities as the temperature changes (you can choose three different
temperatures at any one time).
fraction with given velocity
0.0015
0.001
5 10
4
0
0
1000
2000
3000
4000
5000
velocity (m/sec)
trace 1
trace 2
trace 3
m .032
Created: Fall 1995
Modified: June 1996
T1 1000
T2 2000
T3 3000
Molecular Velocities
Author: Flick Coleman
Page 3
Now you can begin to explore the question of the kinetic energy of gas molecules at a particular
temperature. Remember that the formula for kinetic energy is k.e. = (mv2/2). Since there is a
distribution of velocities in the gas sample, there is also a distribution of kinetic energies. You will
have noticed in the graphs above that one velocity is more likely than any others for a given
molecular weight at a given temperature. We call that the most probable velocity. We can use
that velocity to calculate a kinetic energy that is representative of the gas sample. Go back to the
first graph. Using the crosshair feature, determine the most probable velocity for each of the
three gases. Now calculate the kinetic energy for each gas, based on the most probable velocity
that you measure. What do you conclude about the kinetic energy of different gases at
the same temperature? Do your conclusions support the statement that temperature is
a measure of available energy?
Write a paragraph that discusses the velocity of gaseous molecules and the factors
which influence those velocities and kinetic energies. Alternatively, prepare a 5-10
minute talk on the same topic. In both cases you will find it useful to have some
graphics to support your words.
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 4
The Velocities of Gas Molecules - A Further Look
In this section we will examine the distribution of molecular velocities in more detail. Let's begin with a
mathematical form of the velocity distribution function. This equation was derived by the Scottish
physicist James Clerk Maxwell, and, independently, by the Austrian physicist Ludwig Boltzmann, in the
latter part of the 19th century. It is not unheard of for two researchers to reach the same result
independently of one another, although it will probably become less likely in the future as scientific results
are communicated more and more rapidly in print and electronic media. What is now known as the
Maxwell-Boltzmann distribution law is given by the expression:
3
p( u )
4. π.
m
2. π. R. T
2
. e
2
m. u
2 . R. T
. u2
Here u is the velocity in m/sec, R the gas constant in J/K-mol, T the temperature in K, and m the molar
mass in kg/mol.
One way to interpret p(u) is that it is the fraction of gas molecules that has a velocity between a given
value u and u+δu, where δu represents a small change in u. Go back to the graphs of the velocity
distributions. What do you notice about the area under each of the curves? You can use calculus to
evaluate the area under a curve since that area is equal to the integral of the distribution function over
the values of u. Below is an integral sign with MathCad placeholders for the integrand (the function to
be integrated), the variable of integration (u in this case) and the limits of integration. This is a numerical
integral and the result will be a number, not another function. Try changing the function (using p1
through p6 as possibilities) as evaluate the integral. Does the numerical value make sense given the
nature of the distribution? What do you observe if the upper limit for integration is too small?
5000
p1( u ) du = 1
0
Since there is a range of velocities at any temperature, how do we characterize the velocity? One way
would be to look at the peak of the distribution curve. The velocity at this peak is called the most
probable velocity (ump). How would you use calculus to find the value of ump? The maxima of a
function are values of the argument of the function (u in our case) for which the first derivative is zero.
(Yes, this is also true for the minima, and we will see how to distinguish one from the other for this
problem.) You can make use of MathCad's symbolic capabilities to differentiate p(u) with respect to u,
and derive an expression for ump. (You could also find dp(u)/du by hand but as you will see it is not a
trivial problem.) Load the MathCad symbolic processor (see handout). Under derivation format, click
on "show derivation comments" and "vertically, inserting lines" In the expression for p(u) below,
place the vertical cursor to the right of one of the u's (but before the superscript). Then choose
"differentiate on variable" in the symbolic menu.
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 5
3
4. π.
m
2. π. R . T
2
. e
2
m. u
2.R.T
. u2
Since this is a cubic equation in u (it is, isn't it), there are three roots, or three values of u for which the
derivative is zero. Place the cursor to the right of one of the u's in the derivative expression and choose
"solve for variable" from the symbolic menu. Which of the three roots corresponds to the most
probable velocity? Why?
Another measure of the velocity is the average velocity. Look at the shape of the velocity distribution
function. Is the average velocity going to lie to the left (lower velocity) or the right (higher velocity) of
the most probable velocity. For properties j, defined by distribution functions, F(j), the definition of the
average value of that property is given by:
∞
j. F( j) dj
j av
0
∞
F( j ) dj
0
Using this expression, evaluate the average velocity. Hint - substitute u for j and p(u) for F(j) and then
perform the appropriate integration. You already know the value of the denominator in the expression
for the average velocity (what is it).
The third commonly used measure of gas velocities is the root mean square velocity, urms. The text
contains a derivation of the pressure exerted by a gas based on the kinetic molecular theory. In this
model, the velocity appears as u2, and the average force exerted on the walls of a container by a gas
sample depends on the average of u2. The square root of the average of u2 is the root mean square
velocity. The expression for this velocity is:
u rms
Created: Fall 1995
Modified: June 1996
3. R. T
m
Molecular Velocities
Author: Flick Coleman
Page 6
Using the expression that you obtained for the most probable velocity and the
average velocity, compare these three approaches to characterizing the velocity of a
gas sample. If the most probable velocity is assigned a value of 1, what are the
relative values of uav and urms?
In the first part of this document you measured the most probable velocity from the
velocity distribution plot. How do the values that you measured compare with those
you would calculate using the formula above? You also calculated the most probable
kinetic energy for a gas based on your measured most probably velocity. Recalculate
the kinetic energies using the average and root mean squared velocities. Do you
reach the same conclusion that you did above about the kinetic energy of gas
molecules as a function of temperature? Most of the time when chemists speak of the
average kinetic energy of a sample of molecules they are speaking of a kinetic energy
based on the root mean squared velocity. Using the expression above for the root
mean squared velocity, derive a symbolic expression for the kinetic energy. Does this
expression make sense in light of what you have observed above regarding the kinetic
energy of different gases at the same temperature?
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 7
As a final example, how would you find the fraction of nitrogen molecules that have a kinetic
energy (based on the root mean squared velocity) greater than two times the average kinetic
energy?
One approach to this problem is to find the velocity that gives a kinetic energy twice the
average kinetic energy.
. .
2 2m
m. u
2
3. R . T
m
2
2
This is a mathematical statement of the
problem - the kinetic energy we want is
twice the average kinetic energy. Solve this
expression for u, the velocity which
produces this kinetic energy.
The solution we want is
This makes sense when you think about it. We
want a kinetic energy that is twice the average.
Since the kinetic energy is proportional to the
velocity squared, the velocity to produce a kinetic
energy that is twice the average must be the square
root of two times the velocity to produce the
average kinetic energy.
6. R. T
m
The solution to our problem is then one minus the fraction which have a velocity less than the
desired velocity (why is this the case):
6. R. T
m
3
4. π.
1
m
2. π. R . T
2
. e
2
m. u
2.R.T
. u2 du = 0.112
0
What percentage of gas molecules satisfy our criterion at any temperature.?
Additional exercises:
A
fast ball
Created:
Fallthrown
1995 by
Modified: June 1996
professional baseballMolecular
pitcher may
approach a velocity of 100Author:
miles Flick
per Coleman
Velocities
Page 8
A fast ball thrown by professional baseball pitcher may approach a velocity of 100 miles per
hour. What fraction of oxygen molecules have a velocity greater than this at room temperature?
The velocity that a body must have in order to completely escape the earth's gravitational field
is known as the escape velocity. The value of the escape velocity is about 7 miles per second.
Using calculations similar to those above, what do you conclude about the number of gas
molecules which escape the earth's gravitational field. Try a variety of reasonable molecular
weights and choose reasonable temperatures. Explain how your results are consistent or
inconsistent with your understanding of the earth's atmosphere.
Two additional concepts which are useful when describing the behavior of gas molecules are
the collision number, Z, and the mean free path, λ. The collision number is the number of
collisions a particular gas molecule would experience with other molecules in a second. These
collisions are very important in determining the rates of chemical reactions. The number of such
collisions must depend on the number of gas molecules in a given volume, N/V, the size of the
molecules, d, and the average velocity of the molecules. The expression that is derived for Z is:
Z 4.
N . 2. π. R. T
d
V
m
You should recognize elements of the average velocity in this expression.
Closely related to the collision number is the mean free path, the distance that a molecule
travels, on average, between collisions. Since Z is the number of collisions per second, 1/Z is
the time between collisions and 1/Z times the average velocity is the distance between
collisions. Thus the expression for the mean free path is:
λ
uav
Z
Substitute the appropriate expressions for uav and Z and obtain an expression for the
mean free path. Write a brief paragraph explaining why the formula you have obtained
seems reasonable.
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 9
The value of d for O2 molecules is 3.61* 10-8 m. Calculate the collision number and mean
free path for O2 molecules at room temperature and one atmosphere pressure.
Suppose you were devising an experiment in which it was important for molecules, on
average, to go at least one microsecond without experiencing a collision. Using O2 as the
sample, and room temperature conditions, what pressure would be necessary to achieve this
desired time between collisions? (To begin with, what will the relationship between the
number of collisions per second and the average time between collisions be?)
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 10
A digression: One way toconsider the M-B distribution, is that it represents the balance
between the number of ways of achieving a particular velocity u, a number which goes as
u2, and the probability of having a particular velocity, and hence a particular kinetic energy,
at any given temperature. This latter is given by the term exp(-mu2/2RT). We will
frequently encounter this comparison between the kinetic energy of the molecules and the
thermal energy RT (or kT if we are discussing single molecules). We can evaluate the
constant terms by realizing that we wish the result to be a distribution function which
integrates to unity. Do this, and see if you get the same form of the distribution seen earlier.
(The solution below would not appear in the document students see.)
∞
N.
e
2
m. u
2 . R. T
. u2 du 1
0
has solution(s)
1
3
2
limit exp
1. . u
. u . R. T
m
.
2
(R T) m
1
3
. R 2 . T 2 . π. 2. erf 1 . 2.
3
2
2
2. m
m
. u , u ∞ , left
R. T
1
3
1
3
2. m
3
3
. R 2 . T 2 . π. 2
p( u )
4. π.
2
R
( 2. π)
2
3
2
. T 2 . π
Created: Fall 1995
Modified: June 1996
. u2
3
3
3
. e
2
m. u
2 . R. T
4. π
simplifies to
m
m
2. π. R. T
2
. 2
2
simplifies to
1.
2
π
Molecular Velocities
Author: Flick Coleman
Page 11
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 12
Created: Fall 1995
Modified: June 1996
Molecular Velocities
Author: Flick Coleman
Page 13