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The Velocities of Gas Molecules ©

by

Flick ColemanDepartment 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 1995Modified: June 1996

Molecular Velocities Author: Flick ColemanPage 1

0 1000 2000 3000 4000 50000

5 10 4

0.001

trace 1trace 2trace 3

velocity (m/sec)

frac

tion

with

giv

en v

eloc

ity

m1 .016 m2 .032 m3 .044 Molec. Wt. in kg/mol T 2000 T in K

u ..,0 5 5000 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?



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).

0 1000 2000 3000 4000 50000

5 10 4

0.001

0.0015

trace 1trace 2trace 3

velocity (m/sec)

frac

tion

with

giv

en v

eloc

ity

m .032 T1 1000 T2 2000 T3 3000



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 thefirst 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.



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:

p( )u ....4 πm

...2 π R T

3

2

e

.m u2

..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?

=d0

5000up1( )u 1

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.



....4 πm

...2 π R T

3

2

e

.m u2

..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 av

d0

∞j.j F( )j

d0

∞jF( )j

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..3 R T

m



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?



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.

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.

.m u2

2

..2 m..3 R T

m

2

2

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):

=1 d

0

..6 R T

m

u....4 πm

...2 π R T

3

2

e

.m u2

..2 R T u2 0.112

What percentage of gas molecules satisfy our criterion at any temperature.?

Additional exercises:

A fast ball thrown by professional baseball pitcher may approach a velocity of 100 miles per Created: Fall 1995Modified: June 1996


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 ...4N

Vd2 ..π R T

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:

λuav

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.



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?)



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 d0

∞

u.e

.m u2

..2 R T u2 1

has solution(s)

1

limit ,,...exp ..1

2m

u2

( ).R T

u

mR T .....1

.2 m

3

2

R

3

2 T

3

2 π 2 erf ...1

22

m

.R Tu u ∞ left

1

....1

.2 m

3

2

R

3

2 T

3

2 π 2p( )u ....4 π

m...2 π R T

3

2

e

.m u2

..2 R T u2

simplifies to.4 π

( ).2 π

3

2

.m

3

2

.R

3

2 .T

3

2 π

2 simplifies to

.1

π2







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