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Unformatted text preview: 39 The Kinetic Theory of Gases 39—1 Properties of matter With this chapter we begin a new subject which will occupy us for some time.
It is the ﬁrst part of the analysis of the properties of matter from the physical point
of view, in which, recognizing that matter is made out of a great many atoms, or
elementary parts, which interact electrically and obey the laws of mechanics, we
try to understand why various aggregates of atoms behave the way they do. It is obvious that this is a difﬁcult subject, and we emphasize at the beginning
that it is in fact an extremely difficult subject, and that we have to deal with it
differently than we have dealt with the other subjects so far. In the case of me
chanics and in the case of light, we were able to begin with a precise statement of
some laws, like Newton’s laws, or the formula for the ﬁeld produced by an ac
celerating charge, from which a whole host of phenomena could be essentially
understood, and which would produce a basis for our understanding of mechanics
and of light from that time on. That is, we may learn more later, but we do not
learn different physics, we only learn better methods of mathematical analysis
to deal with the situation. We cannot use this approach effectively in studying the properties of matter.
We can discuss matter only in a most elementary way; it is much too complicated
a subject to analyze directly from its speciﬁc basic laws, which are none other than
the laws of mechanics and electricity. But these are a bit too far away from the
properties we wish to study; it takes too many steps to get from Newton’s laws to
the properties of matter, and these steps are, in themselves, fairly complicated.
We will now start to take some of these steps, but while many of our analyses
will be quite accurate, they will eventually get less and less accurate. We will
have only a rough understanding of the properties of matter. One of the reasons that we have to perform the analysis so imperfectly is that
the mathematics of it requires a deep understanding of the theory of probability;
we are not going to want to know where every atom is actually moving, but rather,
how many move here and there on the average, and what the odds are for different
effects. So this subject involves a knowledge of the theory of probability, and our
mathematics is not yet quite ready and we do not want to strain it too hard. Secondly, and more important from a physical standpoint, the actual behavior
of the atoms is not according to classical mechanics, but according to quantum
mechanics, and a correct understanding of the subject cannot be attained until we
understand quantum mechanics. Here, unlike the case of billiard balls and auto
mobiles, the difTerence between the classical mechanical laws and the quantum
mechanical laws is very important and very signiﬁcant, so that many things that
we will deduce by classical physics will be fundamentally incorrect. Therefore
there will be certain things to be partially unlearned; however, we shall indicate
in every case when a result is incorrect, so that we will know just where the “edges”
are. One of the reasons for discussing quantum mechanics in the preceding
chapters was to give an idea as to why, more or less, classical mechanics is incorrect
in the various directions. Why do we deal with the subject now at all? Why not wait a half a year, or
a year, until we know the mathematics of probability better, and we learn a little
quantum mechanics, and then we can do it in a more fundamental way? The
answer is that it is a difﬁcult subject, and the best way to learn is to do it slowly!
The ﬁrst thing to do is to get some idea, more or less, of what ought to happen in 391 39—1 Properties of matter 39—2 The pressure of a gas 39—3 Compressibility of radiation
39—4 Temperature and kinetic energy
39—5 The ideal gas law different circumstances, and then, later, when we know the laws better, we will
formulate them better. Anyone who wants to analyze the properties of matter in a real problem might
want to start by writing down the fundamental equations and then try to solve
them mathematically. Although there are people who try to use such an approach,
these people are the failures in this ﬁeld; the real successes come to those who start
from a physical point of view, people who have a rough idea where they are going
and then begin by making the right kind of approximations, knowing what is big
and what is small in a given complicated situation. These problems are so compli
cated that even an elementary understanding, although inaccurate and incomplete,
is worth while having, and so the subject will be one that we shall go over again
and again, each time with more and more accuracy, as we go through our course
in physics. Another reason for beginning the subject right now is that we have already
used many of these ideas in, for example, chemistry, and we have even heard of
some of them in high school. It is interesting to know the physical basis for
these things. As an interesting example, we all know that equal volumes of gases, at the
same pressure and temperature, contain the same number of molecules. The law
of multiple proportions, that when two gases combine in a chemical reaction the
volumes needed always stand in simple integral proportions, was understood ulti
mately by Avogadro to mean that equal volumes have equal numbers of atoms.
Now why do they have equal numbers of atoms? Can we deduce from Newton’s
laws that the number of atoms should be equal? We shall address ourselves to
that speciﬁc matter in this chapter. In succeeding chapters, we shall discuss various
other phenomena involving pressures, volumes, temperature, and heat. We shall also ﬁnd that the subject can be attacked from a nonatomic point of
View, and that there are many interrelationships of the properties of substances.
For instance, when we compress something, it heats; if we heat it, it expands.
There is a relationship between these two facts which can be deduced independently
of the machinery underneath. This subject is called thermodynamics. The deepest
understanding of thermodynamics comes, of course, from understanding the actual
machinery underneath, and that is what we shall do: we shall take the atomic
viewpoint from the beginning and use it to understand the various properties of
matter and the laws of thermodynamics. Let us, then, discuss the properties of gases from the standpoint of Newton’s
laws of mechanics. 39—2 The pressure of a gas First, we know that a gas exerts a pressure, and we must clearly understand
what this is due to. If our ears were a few times more sensitive, we would hear a
perpetual rushing noise. Evolution has not developed the ear to that point,
because it would be useless if it were so much more sensitive—we would hear a
perpetual racket. The reason is that the eardrum is in contact with the air, and air
is a lot of molecules in perpetual motion and these bang against the eardrums.
In banging against the eardrums they make an irregular tattoo—boom, boom,
boom—which we do not hear because the atoms are so small, and the sensitivity
of the ear is not quite enough to notice it. The result of this perpetual bombardment
is to push the drum away, but of course there is an equal perpetual bombardment
of atoms on the other side of the eardrum, so the net force on it is zero. If we were
to take the air away from one side, or change the relative amounts of air on the
two sides, the eardrum would then be pushed one way or the other, because the
amount of bombardment on one side would be greater than on the other. We
sometimes feel this uncomfortable eﬂect when we go up too fast in an elevator or
an airplane, especially if we also have a bad cold (when we have a cold, inﬂammation
closes the tube which connects the air on the inside of the eardrum with the outside
air through the throat, so that the two pressures cannot readily equalize). 392 In considering how to analyze the situation quantitatively, we imagine that
we have a volume of gas in a box, at one end of which is a piston which can be
moved (Fig. 39—1). We would like to ﬁnd out what force on the piston results from
the fact that there are atoms in this box. The volume of the box is V, and as the
atoms move around inside the box with various velocities they bang against the
piston. Suppose there is nothing, a vacuum, on the outside of the piston. What of
it? If the piston were left alone, and nobody held onto it, each time it got banged
it would pick up a little momentum and it would gradually get pushed out of the
box. So in order to keep it from being pushed out of the box, we have to hold it
with a force F. The problem is, how much force? One way of expressing the force
is to talk about the force per unit area: if A is the area of the piston, then the force
on the piston will be written as a number times the area. We deﬁne the pressure,
then, as equal to the force that we have to apply on a piston, divided by the area
of the piston: P = F /A. (39.1)
To make sure we understand the idea (we have to derive it for another purpose
anyway), the differential work dW done on the gas in compressing it by moving
the piston in a dilferential amount —dx would be the force times the distance
that we compress it, which, according to (39.1), would be the pressure times the
area, times the distance, which is equal to minus the pressure times the change
in the volume: dW = F(—dx) = —PA dx = —P dV. (39.2)
(The area A times the distance dx is the volume change.) The minus sign is there
because, as we compress it, we decrease the volume; if we think about it we can
see that if a gas is compressed, work is done on it. How much force do we have to apply to balance the banging of the molecules?
The piston receives from each collision a certain amount of momentum. A certain
amount of momentum per second will pour into the piston, and it will start to move.
To keep it from moving, we must pour back into it the same amount of momentum
per second from our force. Of course, the force is the amount of momentum per
second that we must pour in. There is another way to put it: if we let go of the
piston it will pick up speed because of the bombardments; with each collision we
get a little more speed, and the speed thus accelerates. The rate at which the piston
picks up speed, or accelerates, is proportional to the force on it. So we see that
the force, which we already have said is the pressure times the area, is equal to the
momentum per second delivered to the piston by the colliding molecules. To calculate the momentum per second is easy—~we can do it in two parts:
ﬁrst, we ﬁnd the momentum delivered to the piston by one particular atom in a
collision with the piston, then we have to multiply by the number of collisions per
second that the atoms have with the wall. The force will be the product of these
two factors. Now let us see what the two factors are: In the ﬁrst place, we shall
suppose that the piston is a perfect “reﬂector” for the atoms. If it is not, the whole
theory is wrong, and the piston will start to heat up and things will change, but
eventually, when equilibrium has set in, the net result is that the collisions are
effectively perfectly elastic. On the average, every particle that comes in leaves
with the same energy. So we shall imagine that the gas is in a steady condition, and
we lose no energy to the piston because the piston is standing still. In those cir
cumstances, if a particle comes in with a certain speed, it comes out with the same
speed and, we will say, with the same mass. If v is the velocity of an atom, and 1),, is the xcomponent of v, then m1), is the
xcomponent of momentum “in”; but we also have an equal component of mo
mentum “out,” and so the total momentum delivered to the piston by the particle,
in one collision, is 2mm, because it is “reﬂected.” Now, we need the number of collisions made by the atoms in a second, or in
a certain amount of time dt; then we divide by dt. How many atoms are hitting?
Let us suppose that there are N atoms in the volume V, or n = N / V in each unit
volume. To ﬁnd how many atoms hit the piston, we note that, given a certain 39—3 4— x ———»!dx¢— Fig. 39—1. Atoms of a gas in a box
with a frictionless piston. amount of time t, if a particle has a certain velocity toward the piston it will hit
during the time t, provided it is close enough. If it is too far away, it goes only
part way toward the piston in the time I, but does not reach the piston. Therefore
it is clear that only those molecules which are within a distance vxt from the piston
are going to hit the piston in the time t. Thus the number of collisions in a time t is
equal to the number of atoms which are in the region within a distance 1231, and
since the area of the piston is A, the volume occupied by the atoms which are going
to hit the piston is vxtA. But the number of atoms that are going to hit the piston
is that volume times the number of atoms per unit volume, nvItA. Of course we
do not want the number that hit in a time t, we want the number that hit per second,
so we divide by the time t, to get nva. (This time t could be made very short; if
we feel we want to be more elegant, we call it dt, then differentiate, but it is the same
thing.)
So we ﬁnd that the force is F = nvIA  2mvz. (39.3) See, the force is proportional to the area, if we keep the particle density ﬁxed as
we change the area! The pressure is then P = 2nmvi. (39.4) Now we notice a little trouble with this analysis: First, all the molecules do
not have the same velocity, and they do not move in the same direction. So, all
the v,’s are different! So what we must do, of course, is to take an average of the
03’s, since each one makes its own contribution. What we want is the square of
2);, averaged over all the molecules: P = moi). (39.5) Did we forget to include the factor 2? No; of all the atoms, only half are
headed toward the piston. The other half are headed the other way, and if we take
(v3), we are averaging the negative vx’s squared, as well as the positive vx’s. So
when we just take (v3), without looking, we are getting twice as much as we want.
The average of vi, for positive 1),, is equal to the average of v2 for all 2),, times one
half. Now as the atoms bounce around, it is clear that there is nothing special about
the “xdirection”; the atoms may also be moving up and down, back and forth,
in and out. Therefore it is going to be true that (vi), the average motion of the
atoms in one direction, and the average in the other two directions, are all going
to be equal: (vi) = (vi) = (03>. (39.6) It is only a matter of rather tricky mathematics to notice, therefore, that they are
each equal to onethird of their sum, which is of course the square of the magnitude
of the velocity: <2)?» = av: + vi + v3) = <v2>/3. (39.7) This has the advantage that we do not have to worry about any particular direction,
and so we write our pressure formula again in this form: P = (%)n(mv2/2). (39.8) The reason we wrote the last factor as (mvz/Z) is that this is the kinetic energy of
the centerofmass motion of the molecule. We ﬁnd, therefore, that PV = N(%)<mv2/2). (39.9) With this equation we can calculate how much the pressure is, if we know the speeds. As a very simple example let us take helium gas, or any other gas, like mercury
vapor, or potassium vapor of high enough temperature, or argon, in which all the
molecules are single atoms, for which we may suppose that there is no internal 39—4 motion in the atom. If we had a complex molecule, there might be some internal
motion, mutual vibrations, or something. We suppose that we may disregard that;
this is actually a serious matter that we will have to come back to, but it turns out
to be all right. We suppose that the internal motion of the atoms can be disre
garded, and therefore, for this purpose, that the kinetic energy of the centerofmass
motion is all the energy there is. So for a monatomic gas, the kinetic energy is the
total energy. In general, we are going to call U the total energy (it is sometimes
called the total internal energy—we may wonder why, since there is no external
energy to a gas), i.e., all the energy of all the molecules in the gas, or the object,
whatever it is. For a monatomic gas we will suppose that the total energy U is equal to a num
ber of atoms times the average kinetic energy of each, because we are disregarding
any possibility of excitation or motion inside the atoms themselves. Then, in these
circumstances, we would have PV = gU. (39.10) Incidentally, we can stop here and ﬁnd the answer to the following question:
Suppose that we take a can of gas and compress the gas slowly, how much pressure
do we need to squeeze the volume down? It is easy to ﬁnd out, since the pressure
is % the energy divided by V. As we squeeze it down, we do work on the gas and we
thereby increase the energy U. So we are going to have some kind of a diﬂerential
equation: If we start out in a given circumstance with a certain energy and a certain
volume, we then know the pressure. Now we start to squeeze, but the moment we
do, the energy U increases and the volume V decreases, so the pressure goes up. So, we have to solve a differential equation, and we will solve it in a moment.
We must ﬁrst emphasize, however, that as we are compressing this gas, we are
supposing that all the work goes into increasing the energy of the atoms inside.
We may ask, “Isn’t that necessary? Where else could it go?” It turns out that it
can go another place. There are what we call “heat leaks” through the walls:
the hot (i.e., fastmoving) atoms that bombard the walls, heat the walls, and energy
goes away. We shall suppose for the present that this is not the case. For somewhat wider generality, although we are still making some very special
assumptions about our gas, we shall write, not PV = %U, but PV = (7 — 1)U. (39.11) It is written (‘7 — 1) times U for conventional reasons, because we will deal with
a few other cases later where the number in front of U will not be g, but will be a
diﬂerent number. So, in order to do the thing in general, we call it 7 — 1, because
people have been calling it that for almost one hundred years. This 7, then, is g,
because g — 1 is g for a monatomic gas like helium. We have already noticed that when we compress a gas the work done is
—P dV. A compression in which there is no heat energy added or removed is
called an adiabatic compression, from the Greek a (not) + dia (through) + bainein
(to go). (The word adiabatic is used in physics in several ways, and it is sometimes
hard to see what is common about them.) That is, for an adiabatic compression
all the work done goes into changing the internal energy. That is the key—that there are no other losses of energy—for then we have PdV = —dU. But since
U = PV/(‘r — 1), we may write
dU = (PdV + VdP)/('Y — 1). (39.12) So we have PdV = —(P dV+ VdP)/(7 —— l), or, rearranging the terms,
'YPdV = —VdP, or
(7 dV/V) + (dP/P) = 0. (39.13) Fortunately, assuming that 7 is constant, as it is for a monatomic gas, we can inte
grate this: it gives 7 1n V + In P = ln C, where In C is the constant of integration.
If we take the exponential of both sides, we get the law PV" = C (a constant). (39.14)
395 In other words, under adiabatic conditions, where the temperature rises as we
compress because no heat is being lost, the pressure times the volume to the g
power is a constant for a monatomic gas! Although we derived it theoretically,
this is, in fact, the way monatomic gases behave experimentally. 393 Compressibility of radiation We may give one other example of the kinetic theory of a gas, one which is
not used in chemistry so much, but is used in astronomy. We have a large number
of photons in a box in which the temperature is very high. (The box is, of course,
the gas in a very hot star. The sun is not hot enough; there are still too many
atoms, but at still higher temperatures in certain very hot stars, we may neglect the
atoms and suppose that the only objects that we have in the box are photons.)
Now then, a photon has a certain momentum p. (We always ﬁnd that we are in
terrible trouble when we do kinetic theory: p is the pressure, but p is the momentum;
v is the volume, but 2; is the velocity; T is the temperature, but T is the kinetic
energy or the time or the torque; one must keep one’s wits about one!) This
p is momentum, it is a vector. Going through the same analysis as before, it is
the xcomponent of the vector p which generates the “kick,” and twice the xcom
ponent of the vector p is the momentum which is given in the kick. Thus 2p, re
places 2mvz, and in evaluating the number of collisions, v, is still 1),, so when we
get all the way through, we ﬁnd that the pressure in Eq. (39.4) is, instead, P = 2np$vx. (39.15) Then, in the averaging, it becomes n times the average of plug, (the same factor of
2) and, ﬁnally, putting in the other two directions, we ﬁnd PV = N(p  v)/3. (39.16) This checks with the formula (39.9), because the momentum is mv; it is a little
more general, that is all. The pressure times the volume is the total number of
atoms times §(p  v), averaged. Now, for photons, what is p  v? The momentum and the velocity are in the
same direction, and the velocity is the speed of light, so this is the momentum of
each of the objects, times the speed of light. The momentum times the speed of
light of every photon is its energy: E = pc, so these terms are the energies of
each of the photons, and we should, of course, take an average energy, times the
number of photons. So we have % of the energy inside the gas: PV = U/3 (photon gas). (39.17) For photons, then, since we have % in front, (7 — 1) in (39.11) is %, or ‘Y = %, and
we have discovered that radiation in a box obeys the law PV‘“3 = C. (39.18) So we know the compressibility of radiation! That is what is used in an analysis
of the contribution of radiation pressure in a star, that is how we calculate it, and
how it changes when we compress it. What wonderful things are already within
our power! 39—4 Temperature and kinetic energy So far we have not dealt with temperature; we have purposely been avoiding
the temperature. As we compress a gas, we know that the energy of the molecules
increases, and we are used to saying that the gas gets hotter; we would like to
understand what this has to do with the temperature. If we try to do the experi
ment, not adiabatically but at what we call constant temperature, what are we doing?
We know that if we take two boxes of gas and let them sit next to each other long
enough, even if at the start they were at what we call diﬁerent temperatures, they 39—6 will in the end come to the same temperature. Now what does that mean? That
means that they get to a condition that they would get to if we left them alone
long enough! What we mean by equal temperature is just that—the ﬁnal condition
when things have been sitting around interacting with each other long enough. Let us consider, now, what happens if we have two gases in containers sepa
rated by a movable piston as in Fig. 39—2 (just for simplicity we shall take two
monatomic gases, say helium and neon). In container (1) the atoms have mass
m1, velocity v1, and there are n1 per unit volume, and in the other container the
atoms have mass m2, velocity 02, there are n2 atoms per unit volume. What are
the conditions for equilibrium? Obviously, the bombardment from the left side must be such that it moves
the piston to the right and compresses the other gas until its pressure builds up,
and the thing will thus slosh back and forth, and will gradually come to rest at
a place where the pressures are equal on both sides. So we can arrange that the
pressures are equal; that just means that the internal energies per unit volume are
equal, or that the numbers n times the average kinetic energies on each side are
equal. What we have to try to prove, eventually, is that the numbers themselves
are equal. So far, all we know is that the numbers times the kinetic energies
are equal, 7110711032) = "2(m203/2X from (39.8), because the pressures are equal. We must realize that this is not the
only condition over the long run, but something else must happen more slowly
as the true complete equilibrium corresponding to equal temperatures sets in. To see the idea, suppose that the pressure on the left side were developed by
having a very high density but a low velocity. By having a large n and a small 2),
we can get the same pressure as by having a small n and a large 2). The atoms may
be moving slowly but be packed nearly solidly, or there may be fewer but they are
hitting harder. Will it stay like that forever? At ﬁrst we might think so, but then
we think again and ﬁnd we have forgotten one important point. That is, that the
intermediate piston does not receive a steady pressure; it wiggles, just like the
eardrum that we were ﬁrst talking about, because the bangings are not absolutely
uniform. There is not a perpetual, steady pressure, but a tattoo—the pressure
varies, and so the thing jiggles. Suppose that the atoms on the right side are not
jiggling much, but those on the left are few and far between and very energetic.
The piston will, now and then, get a big impulse from the left, and will be driven
against the slow atoms on the right, giving them more speed. (As each atom
collides with the piston, it either gains or loses energy, depending upon whether
the piston is moving one way or the other when the atom strikes it.) So, as a result
of the collisions, the piston ﬁnds itself jiggling, jiggling, jiggling, and this shakes
the other gas—it gives energy to the other atoms, and they build up faster motions,
until they balance the jiggling that the piston is giving to them. The system comes
to some equilibrium where the piston is moving at such a mean square speed that
it picks up energy from the atoms at about the same rate as it puts energy back
into them. So the piston picks up a certain mean irregularity in speed, and it is
our problem to ﬁnd it. When we do ﬁnd it, we can solve our problem better, be
cause the gases will adjust their velocities until the rate at which they are trying
to pour energy into each other through the piston will become equal. It is quite difﬁcult to ﬁgure out the details of the piston in this particular cir
cumstance; although it is ideally simple to understand, it turns out to be a little
harder to analyze. Before we analyze that, let us analyze another problem in
which we have a box of gas but now we have two different kinds of molecules in it,
having masses m1 and m2, velocities v1 and 02, and so forth; there is now a much
more intimate relationship. If all of the No. 2 molecules are standing still, that
condition is not going to last, because they get kicked by the No. 1 molecules and
so pick up speed. If they are all going much faster than the No. 1 molecules, then
maybe that will not last either—they will pass the energy back to the No. 1 mole
cules. So when both gases are in the same box, the problem is to ﬁnd the rule that
determines the relative speeds of the two. 39—7 Fig. 39—2. Atoms of two different monatomic gases are separated by a
movable piston. Fig. 393. A collision between un
equal molecules, viewed in the CM system. This is still a very difficult problem, but we will solve it as follows. First we
consider the following subproblem (again this is one of those cases where—never
mind the derivation—in the end the result is very simple to remember, but the
derivation is just ingenious). Let us suppose that we have two molecules, of differ—
ent mass, colliding, and that the collision is viewed on the centerof—mass (CM)
system. In order to remove a complication, we look at the collision in the CM.
As we know from the laws of collision, by the conservation of momentum and en
ergy, after the molecules collide the only way they can move is such that each
maintains its own original speed—and they just change their direction. So we have
an average collision that looks like that in Fig. 39—3. Suppose, for a moment, that
we watch all the collisions with the CM at rest. Suppose we imagine that they are
all initially moving horizontally. Of course, after the ﬁrst collision some of them
are moving at an angle. In other words, if they were all going horizontally, then
at least some would later be moving vertically. Now in some other collision, they
would be coming in from another direction, and then they would be twisted at
still another angle. So even if they were completely organized in the beginning,
they would get sprayed around at all angles, and then the sprayed ones would get
sprayed some more, and sprayed some more, and sprayed some more. Ultimately,
what will be the distribution? Answer: It will be equally likely to ﬁnd any pair
moving in any direction in space. After that further collisions could not change the
distribution. They are equally likely to go in all directions, but how do we say that? There
is of course no likelihood that they will go in any speciﬁc direction, because a speciﬁc
direction is too exact, so we have to talk about per unit “something.” The idea is
that any area on a sphere centered at a collision point will have just as many
molecules going through it as go through any other equal area on the sphere.
So the result of the collisions will be to distribute the directions so that equal areas
on a sphere will have equal probabilities. Incidentally, if we just want to discuss the original direction and some other
direction an angle 0 from it, it is an interesting property that the diﬁerential area
of a sphere of unit radius is sin 0 d0 times Zr, and that is the same as the differ
ential of cos 0. So what it means is that the cosine of the angle 0 between any two
directions is equally likely to be anything from —1 to +1. Next, we have to worry about the actual case, where we do not have the
collision in the CM system, but we have two atoms which are coming together with
vector velocities v1 and vz. What happens now? We can analyze this collision
with the vector velocities v1 and v2 in the following way: We ﬁrst say that there
is a certain CM; the velocity of the CM is given by the “average” velocity, with
weights proportional to the masses, so the velocity of the CM is vCM = (m lvl +
m2v2)/(m1 + m2). If we watch this collision in the CM system, then we see a
collision just like that in Fig. 39—3, with a certain relative velocity w coming in.
The relative velocity is just v1 — v2. Now the idea is that, ﬁrst, the whole CM is
moving, and in the CM there is a relative velocity w, and the molecules collide
and come off in some new direction. All this happens while the CM keeps right on
moving, without any change. Now then, what is the distribution resulting from this? From our previous
argument we conclude this: that at equilibrium, all directions for w are equally
likely, relative to the direction of the motion of the CM.* There will be no particular
correlation, in the end, between the direction of the motion of the relative velocity
and that of the motion of the CM. Of course, if there were, the collisions would
spray it about, so it is all sprayed around. So the cosine of the angle between w
and VCM is zero on the average. That is, (W ' VCM> = * This argument, which was the one used by Maxwell, involves some subtleties. Al
though the conclusion is correct, the result does not follow purely from the considerations
of symmetry that we used before, since, by going to a reference frame moving through
the gas, we may ﬁnd a distorted velocity distribution. We have not found a simple proof
of this result. 39—8 But w  VCM can be expressed in terms of VI and v2 as well: (V1 — V2)'(m1V1 + m2V2)
m + m2 _ (mivi — "12723) + (m2 — m1)(V1'V2)I
m1 + m2 W'VCM— (39.20) First, let us look at the v1  v2; what is the average of VI  v2? That is, what is
the average of the component of velocity of one molecule in the direction of an
other? Surely there is just as much likelihood of ﬁnding any given molecule moving
one way as another. The average of the velocity v2 in any direction is zero. Certainly,
then, in the direction of v1, v2 has zero average. So, the average of VI  v2 is
zero! Therefore, we conclude that the average of m Iv? must be equal to the average
of mzvg. That is, the average kinetic energy of the two must be equal: §m1v§ = %mgv§. (3921) If we have two kinds of atoms in a gas, it can be shown, and we presume to have
shown it, that the average of the kinetic energy of one is the same as the average of
the kinetic energy of the other, when they are both in the same gas in the same box
in equilibrium. That means that the heavy ones will move slower than the light
ones; this is easily shown by experimentation with “atoms” of diﬁerent masses
in an air trough. Now we would like to go one step further, and say that if we have two diﬂerent
gases separated in a box, they will also have equal average kinetic energy when
they have ﬁnally come to equilibrium, even though they are not in the same box.
We can make the argument in a number of ways. One way is to argue that if we
have a ﬁxed partition with a tiny hole in it (Fig. 39—4) so that one gas could leak
out through the holes while the other could not, because the molecules are too
big, and these had attained equilibrium, then we know that in one part, where they
are mixed, they have the same average kinetic energy, but some come through the
hole without loss of kinetic energy, so the average kinetic energy in the pure gas
and in the mixture must be the same. That is not too satisfactory, because maybe
there are no holes, for this kind of molecule, that separate one kind from the other. Let us now go back to the piston problem. We can give an argument
which shows that the kinetic energy of this piston must also be %mgv§. Actually,
that would be the kinetic energy due to the purely horizontal motion of the piston,
so, forgetting its up and down motion, it will have to be the same as %m2v§z.
Likewise, from the equilibrium on the other side, we can prove that the kinetic
energy of the piston is %m1v¥z. Although this is not in the middle of the gas, but
is on one side of the gas, we can still make the argument, although it is a little
more difﬁcult, that the average kinetic energy of the piston and of the gas molecules
are equal as a result of all the collisions. If this still does not satisfy us, we may make an artiﬁcial example by which
the equilibrium is generated by an object which can be hit on all sides. Suppose
that we have a short rod with a ball on each end sticking through the piston, on a
frictionless sliding universal joint. Each ball is round, like one of the molecules,
and can be hit on all sides. This whole object has a certain total mass, m.
Now, we have the gas molecules with mass m1 and mass m2 as before. The
result of the collisions, by the analysis that was made before, is that the kinetic
energy of m because of collisions with the molecules on one side must be %m 10?, on
the average. Likewise, because of the collisions with molecules on the other side,
it has to be %m2v§ on the average. So, therefore, both sides have to have the
same kinetic energy when they are in thermal equilibrium. So, although we only
proved it for a mixture of gases, it is easily extended to the case where there are
two diﬁerent, separate gases at the same temperature. Thus when we have two gases at the same temperature, the mean kinetic energy
of the CM motions are equal. The mean molecular kinetic energy is a property only of the “temperature.”
Being a property of the “temperature,” and not of the gas, we can use it as a deﬁni—
tion of the temperature. The mean kinetic energy of a molecule is thus some 39—9 Fig. 39—4. Two gases in a box with a
semipermeable membrane. function of the temperature. But who is to tell us what scale to use for the tempera
ture? We may arbitrarily deﬁne the scale of temperature so that the mean energy
is linearly proportional to the temperature. The best way to do it would be to
call the mean energy itself “the temperature.” That would be the simplest possible
function. Unfortunately, the scale of temperature has been chosen differently,
so instead of calling it temperature directly we use a constant conversion factor
between the energy of a molecule and a degree of absolute temperature called a
degree Kelvin. The constant of proportionality is k = 1.38 X 10‘23 joule for
every degree Kelvin.* So if T is absolute temperature, our deﬁnition says that the
mean molecular kinetic energy is 325 kT. (The ; is put in as a matter of convenience,
so as to get rid of it somewhere else.) We point out that the kinetic energy associated with the component of motion
in any particular direction is only §kT. The three independent directions that are
involved make it ng. 39—5 The ideal gas law Now, of course, we can put our deﬁnition of temperature into Eq. (39.9)
and so ﬁnd the law for the pressure of gases as a function of the temperature: it is
that the pressure times the volume is equal to the total number of atoms times the
universal constant k, times the temperature: PV = NkT. (39.22) Furthermore, at the same temperature and pressure and volume, the number of
atoms is determined; it too is a universal constant! So equal volumes of different
gases, at the same pressure and temperature, have the same number of molecules,
because of Newton’s laws. That is an amazing conclusion! In practice, when dealing with molecules, because the numbers are so large,
the chemists have artiﬁcially chosen a speciﬁc number, a very large number, and
called it something else. They have a number which they call a mole. A mole is
merely a handy number. Why they did not choose 1024 objects, so it would come
out even, is a historical question. They happened to choose, for the convenient
number of objects on which they standardize, N0 = 6.02 X 1023 objects, and
this is called a mole of objects. So instead of measuring the number of molecules
in units, they measure in terms of numbers of moles.’r In terms of N 0 we can write
the number of moles, times the number of atoms in a mole, times kT, and if we
want to, we can take the number of atoms in a mole times k, which is a mole’s
worth of k, and call it something else, and we do—we call it R. A mole’s worth of
k is 8.3l7joules: R = Nok = 8.317jmole“1 °K_1. Thus we also ﬁnd the gas
law written as the number of moles (also called N) times RT, or the number of
atoms, times kT: PV = NRT. (39.23) It is the same thing, just a different scale for measuring numbers. We use 1 as a
unit, and chemists use 6 X 1023 as a unit! We now make one more remark about our gas law, and that has to do with the
law for objects other than monatomic molecules. We have dealt only with the
CM motion of the atoms of a monatomic gas. What happens if there are forces
present? First, consider the case that the piston is held by a horizontal spring,
and there are forces on it. The exchange of jiggling motion between atoms and
piston at any moment does not depend on where the piston is at that moment, of
course. The equilibrium conditions are the same. No matter where the piston is,
its speed of motion must be such that it passes energy to the molecules in just * The centigrade scale is just this Kelvin scale with a zero chosen at 273.16 °K, so
T = 273.16 + centigrade temperature. J[What the chemists call molecular weights are the masses in grams of a mole of a
molecule. The mole is deﬁned so that the mass of a mole of carbon atoms of isotope 12
(i.e., having 6 protons and 6 neutrons in the nucleus) is exactly 12 grams. 39—10 the right way. So it makes no difference about the spring. The speed at which
the piston has to move, on the average, is the same. So our theorem, that the mean
value of the kinetic energy in one direction is %kT, is true whether there are forces
present or not. Consider, for example, a diatomic molecule composed of atoms m A and m B.
What we have proved is that the motion of the CM of part A and that of part B
are such that @myﬁ) = <%mBU2B> = ng. How can this be, if they are held
together? Although they are held together, when they are spinning and turning
in there, when something hits them, exchanging energy with them, the only thing
that counts is how fast they are moving. That alone determines how fast they
exchange energy in collisions. At the particular instant, the force is not an es
sential point. Therefore the same principle is right, even when there are forces. Let us prove, ﬁnally, that the gas law is consistent also with a disregard of the
internal motion. We did not really include the internal motions before; we just
treated a monatomic gas. But we shall now show that an entire object, considered
as a single body of total mass M, has a velocity of the CM such that %Mv(23M = akT (39.24) In other words, we can consider either the separate pieces or the whole thing!
Let us see the reason for that: The mass of the diatomic molecule is M = mA +
m B, and the velocity of the center of mass is equal to vCM = (m AVA + m BvB)/ M.
Now we need (0201“). If we square VCM, we get 2 2 2 2
2 mm + ZmAmBVA 'VB + vaB. UCM — M2 Now we multiply %M and take the average, and thus we get mAng + ZmAmB(VA ‘VB> + "1ng71
M ZmAmB<VA 'VB>_
M %Mv(2:M —
= ng + (We have used the fact that (mA + mB)/M = 1.) Now what is (VA 'VB>? (It
had better be zero!) To ﬁnd out, let us use our assumption that the relative velocity,
w = vA — VB is not any more likely to point in one direction than in another—
that is, that its average component in any direction is zero. Thus we assume that (W ‘ VCM) = 0.
But what is w ~ VCM? It is (VA — VB) ' (mAVA + vaB)
M W'VCM = 771.4031 + (m3 — mA)(VA ‘VB) — vais.
M Therefore, since (mAvi) = (m3023>, the ﬁrst and last terms cancel out on the
average, and we are left with (m3  mA)<VA M3) = 0. Thus if mA ¢ m B, we ﬁnd that (VA  VB) 2 0, and therefore that the bodily motion
of the entire molecule, regarded as a single particle of mass M, has a kinetic energy,
on the average, equal to ng. Incidentally, we have also proved at the same time that the average kinetic
energy of the internal motions of the diatomic molecule, disregarding the bodily
motion of the CM, is ng! For, the total kinetic energy of the parts of the molecule
is §mAvi + §—va%, whose average is ng + ng, or 3kT. The kinetic energy
of the centerof—mass motion is ng, so the average kinetic energy of the rotational
and vibratory motions of the two atoms inside the molecule is the difference, ng. 3911 The theorem concerning the average energy of the CM motion is general:
for any object considered as a whole, with forces present or no, for every inde
pendent direction of motion that there is, the average kinetic energy in that motion
is é—kT. These “independent directions of motion” are sometimes called the
degrees of freedom of the system. The number of degrees of freedom of a molecule
composed of r atoms is 3r, since each atom needs three coordinates to deﬁne its
position. The entire kinetic energy of the molecule can be expressed either as the
sum of the kinetic energies of the separate atoms, or as the sum of the kinetic
energy of the CM motion plus the kinetic energy of the internal motions. The
latter can sometimes be expressed as a sum of rotational kinetic energy of the
molecule and vibrational energy, but this is an approximation. Our theorem,
applied to the ratom molecule, says that the molecule will have, on the average,
3rkT/2 joules of kinetic energy, of which ng is kinetic energy of the centerof—mass
motion of the entire molecule, and the rest, %(r — 1)kT, is internal vibrational and
rotational kinetic energy. 3912 ...
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 Spring '09
 LeeKinohara
 Physics, Thermodynamics, Kinetic Energy

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