Download 1-1 The Scope of Physics

Part One
MECHANICS
I
1 Fundamental Quantities
1-1 The Scope of Physics
Physics is a fundamental science dealing with matter and energy.
By convention, the subject matter of physics has been divided into such topics as mechanics, heat, sound, light, and electricity. In
addition to these general classifications, present-day physics includes atomic physics, nuclear physics, solid- state physics,
chemical physics, biophysics, and many other subdivisions. It is
impossible to include all aspects of physics in a single definition
or paragraph, and to distinguish physics clearly from its nearest
neighbors, the other physical sciences—astronomy, chemistry and
geology.
Like other scientists, the physicist studies nature, and, although the scientist is himself part of the world, he attempts to describe nature as it exists without his interference. Inspired by the
conviction that nature is orderly and “rational”, the physicist has
sought to find that order and to express it as elegantly and as concisely as possible. Physics has attempted to study the least
complex aspects of nature, or, if you will, the most fundamental,
and has found that quantitative descriptions have been most fruitful. Physics has thus become a quantitative science involving precise measurements of certain quantities and the formulation of
mathematical relationships among them..
Like other sciences, physics is an ahistorical subject. This
does not mean that there is no history of physics but rather that, at
a particular moment in time, the history of the development of
physical ideas is not pertinent to the question of the accuracy of
the description of nature. Because we can devote but little space
to the history of physical ideas, the students will find that the time
spent in consulting an encyclopedia on some of the topics studied
in this text, and in reading biographies of physicists, will enrich
his knowledge of the subject and may even stimulate his own inventiveness.
II
5 Force and Motion
5-1 Starting and Stopping Motion
All of us have many times had the experience of setting a body in
motion. If we analyze any of these experiences, we readily recall
that in each case some force was required to start the object moving. In throwing a ball, moving a piece of furniture, or pulling a
sled, the force needed to start the object moving is supplied by
one’s muscular effort as a push or a pull. In more complex cases,
such as setting a car or an airplane in motion, the analysis, although more complicated, will also show that a force is required to
start the body moving.
There are many cases in which the force that acts on the body
to produce the motion is not directly discernible. It was Newton
who first showed that the acceleration of a freely falling body is
produced by a force which acts between the earth and the body,
called the force of gravitation. We shall encounter other such action-at-a-distance forces in electricity and magnetism, and in molecular and atomic physics.
Once a body has been set in motion by the action of a force, it
will not necessarily stop moving when the force is removed. A
sled in motion along a level road will continue to move in a
straight line along the road, although with diminishing speed. The
reduction in speed is due to the force of friction between the runners of the sled and the ground. If there is clean snow on the
ground, the force of friction will be very small; if ashes or sand
have been dumped on the snow, the force of friction will be
greater, and the sled will come to rest much sooner.
The above examples illustrate the fact that a force is required
to change the state of motion of a body. It was Isaac Newton
3
(1642-1727) who first recognized the relationship between force
and the state of motion of the body on which it is acting. He epitomized the entire science of mechanics in the form of three
statements which have become known as Newton’s laws of motion. Although the first and third laws have been previously stated
and discussed, they are sufficiently important to bear repetition.
NEWTON’S FIRST LAW
5-2 Newton’s Laws of Motion
Newton’s three laws of motion can be stated as follows:
First law: A body at rest will remain at rest, and a body in
motion will continue in motion with constant speed in a straight
line, as long as no net force acts on it.
Second law: If a net force acts on a body, the body will be accelerated; the magnitude of the acceleration is proportional to the
magnitude of the force, and the direction of the acceleration is in
the direction of the force.
Third law: Whenever one body exerts a force on another, the
second body exerts a force equal in magnitude and opposite in direction on the first body.
Sir Isaac Newton is one of the greatest physicists of all time, he
developed the law of universal gravitation; epitomized the subject
of mechanics in the three laws of motion which bear his name;
made important contributions to optics. The publication of his
Principia, the Mathematical Principles of Natural Philosophy, in
1687, was an epoch-making event for science. (Courtesy of Scripta Mathematica.)
III
5-3 Newton’s First Law
Newton’s first law states that a body at rest will remain at rest,
and a body in motion will continue in motion with constant speed
in a straight line, as long as no net force acts on it.
4
An examination of this first law shows that a body at rest and
a body moving with constant velocity have one characteristic in
common: there is no net external force acting upon either one.
This is the case when the resultant of all the external forces acting
on the body is zero. As we have already seen, this is the condition
for the equilibrium of a particle; this is also the condition for the
translational equilibrium of a rigid body.
According to Newton’s first law, a train moving at a constant
velocity along a level track is in equilibrium. It is acted upon by
several external forces whose resultant is zero. Consider the forces acting on a train of cars being pulled by a locomotive (see Figure 5-2). The weights W1, W2, W3, of the cars act vertically
downward through the respective centers of gravity. They are opposed by the forces N1, N2, N3, and so on, which the tracks exert
upward on the wheels of the train to support the weight. The sum
of these upward forces must equal the total weight of the train.
There are also frictional forces which oppose the motion of the
train. Some of these frictional forces occur between the wheels
and the tracks and in the wheel bearings; there is also another
type of frictional force owing to the
V=50 mi / hr
Fig.5-2
resistance of the air to motion through it. All of these frictional
forces are represented in the figure by the single force F. The effect of these frictional forces would be to reduce the speed of the
train; to prevent this reduction in speed, the locomotive supplies a
force P equal to F in magnitude but in the forward direction.
There is no net force acting on the train when it is moving with
constant velocity.
In this illustration we have taken a very liberal view of the
meaning of body and of net force. We have considered the collection of all the cars of the train as a body, or a system which could
be surrounded by an imaginary box. Everything within the walls
of the box was considered to be the body, and only the forces acting from outside the box upon objects inside the box were consid5
ered as forces acting on the body. In addition to the forces illustrated in the figure, each car exerts a force upon the two cars immediately adjacent to it. Nevertheless, these internal forces can be
disregarded in our analysis of the over-all motion of the system,
and our attention can be focused upon the external forces acting
upon the system. From Newton’s third law the sum of these internal forces must be zero. This procedure is analogous to what we
have already done in the study of the equilibrium of a rigid body
when we considered only the external forces. Acting on the rigid
body and paid no attention to the internal forces which connected
one particle to another particle of the body.
Implicit in the statement of Newton’s first law is a property
common to all objects – the property known as inertia. The inertia of a body is that property of a body associated with the first
law, that a body at rest will remain at rest unless acted on by a net
force, and that a body in motion will continue to move with uniform velocity unless acted on by a net force.
The magician who whisks a cloth from under the dinner dishes on a table, the prankster who places a brick under a hat on the
sidewalk, have a qualitative understanding of the concept of inertia. We shall attempt to systematize and formalize this concept in
the following sections of this chapter.
IV
7 Work and Energy
7-1 Work Done by Forces
An extremely important concept that has been developed in physics is that of the work done on a body by the action of some external agent which exerts a force on this body and produces motion. For example, whenever someone lifts a body, he does work
by exerting a force upward on it and moving it upward. Whenever
a steam locomotive pulls a train, a series of processes takes place
in the steam engine of the locomotive which enables it to exert a
6
force on the train and move it in the direction of the force. The
term work, as used in physics, is a technical term. Whenever work
is done by an external agent on a body, the work done is the
product of the force which acts on the body and the distance
through which the body moves while the force is acting on it, provided that the force and the distance through which the body
moves are parallel to each other.
In the initial development of the concept of work, we shall restrict our discussion to work done by a constant force. We shall
later remove this restriction and treat the more general case of
work done by a variable force.
If a constant force F acts on a body for a distance s in the direction of the force, then the work done W is, from the definition,
W = Fs.
If the force F and the distance s are not parallel, then only that
component of the force which is in the direction of the motion
does the work. For example, if a heavy block is to be moved, it
may be more convenient to pull with a force F at some angle θ
with respect to the ground. The component of F in the direction of
the motion is F cos θ; if the block is moved through a distance s
while this force is acting on it, the work done W is
W = Fs cos θ.
Work is a scalar quantity. The concept of physical work is often a confusing one, in part because of the way the word “work”
is used in everyday language, and in part because there is no direct physiological analogue to the physical concept of work. In
physics no force exerted, no matter how great nor for how long a
time, generates any work unless there is a displacement. No force
generates work unless the force has a component in the direction
of the displacement. The centripetal force which a string exerts on
a stone in uniform circular motion does no work upon the stone.
The sensation of tiredness has no direct relationship to physical
work. Suppose, for example, that you are called upon to support
one end of a car while the driver changes a tire. When the opera7
tion is concluded, you will have done no work, for, although you
were called upon to exert a large force, there was no displacement
of the car. Consider an even more unlikely situation. Suppose a
barge is being towed through a canal by an engine located alongside the barge canal. If the force exerted by the engine is F, the
component of the force parallel to the canal is F cos θ, while the
component perpendicular to the canal is F sin θ. The perpendicular component tends to urge the barge against the side of the canal. You are called upon to exert a force against the side of the
barge of magnitude F sin θ to keep the barge from scraping along
the walls of the canal, and you accompany the barge on its trip
through the canal, continually exerting the required force. When
the trip is completed, you will have done no work on the barge,
for the force exerted had no component in the direction of the
displacement.
7-2 Units for Expressing Work
There are several different units that are used for expressing the
work done. In every case the unit used must be equivalent to the
product of a force by a distance. In the British gravitational system the unit used is the foot pound (ft lb), the product of the unit
of distance by the appropriate unit of force. In the cgs system the
analogous unit would be the dyne centimeter, but this has been
given a special name, the erg; one erg is the work done by a force
of one dyne acting though a distance of one centimeter, or
1 erg = 1 dyne × 1 cm = 1 dyne cm.
In the mks system the unit of work is called the joule. A joule
is defined as the work done by a force of one Newton acting
through a distance of one meter; that is,
1 joule = 1 nt × 1 m = 1 nt m.
The relationship between the joule and the erg can be found
readily from the facts that 1 nt = 100,000 dynes and 1 m = 100
cm. By the usual conversion procedure, we write
8
1 joule =1 nt m x
100,000 dynes 100cm
x
1m
1nt
1 joule = 10,000,000 dyne cm = 10,000,000 ergs,
1 joule = 107 ergs
or 10-7 joule = 1 erg.
While both work and torque are compounded of the product
of a force by a displacement and therefore have the same units,
these are quite different things. Torque is a vector quantity, while
work is a scalar quantity. Torque is the product of a force by a
distance which is always measured in a direction perpendicular to
the force, while work is the product of a force by a distance parallel to the direction of the force. While there need be no displacement to generate a torque, on the other hand, no work can be done
without a displacement of the force.
Suppose that the body is pulled along a level floor by a rope
making an angle of 30˚ with the floor. If the body is moved a distance of 15 m, and if the force F is 40 nt, the work done is
W = Fs cos θ = 40 nt × 15 m × cos 30˚
= 40 × 15 × 0.866 = 520 nt m,
or W = 520 joules.
V
7-5 Work and Energy
An important question which arises from this discussion concerns
the result of the work done by the various forces which act on different bodies. In some cases the results are immediately obvious.
For example, the work done by a force which accelerates a body
produces a change in its speed; the work done in lifting a body
produces an increase in the height of the body with respect to its
former position; the work against frictional forces produces an increase in the temperature of one or more of the bodies involved.
9
In other cases the results may not be so obvious. Some bodies
may become charged electrically; others may become magnetized. These changes will be discussed at the appropriate places in
the text. One general conclusion can be drawn here; that is, that
whenever work is done, some change is produced in the body or
system of bodies on which the forces acted. To describe these
changes, another technical term is used. We say that the work
done produces a change in the energy of the body or system of
bodies. In the first case above, the energy of motion or the kinetic
energy of the body is changed; in the second case the positional
energy or potential energy of the body is increased. In each case
the change in energy is defined as equal to the work done on the
body or system of bodies. From this, it follows that the units used
in expressing the energy of a system are the same as the units of
work. We shall see that it is possible for a body to gain energy as
a result of work done upon it, and, conversely, a body may lose
energy by doing work upon a second object.
7-6 Kinetic Energy
Suppose that a constant force F acts on a body of mass m for a
distance s in the direction of F. The work done on the body by the
force is
W = Fs
Under the action of a constant force, the body will receive an acceleration a given by
a=
F
m
and the velocity of the body will be increased from its initial value u to some final v, given by Equation (2-27c) as
v2 = u2 + 2as,
and, substituting F/m for a, we find
10
v2 = u2 + 2
Fs
m
We now multiply the equation above by
m
, substitute the value
2
W for the product Fs, and transpose to find
W=
We call the quantity
1
1
m v 2 - m u2
2
2
1
m v2 the final kinetic energy of the body
2
1
m u2 its initial kinetic energy, and we say that the work
2
done on the body by the force F acting through a distance s has
produced a change in the kinetic energy of the body.
and
In general, the kinetic energy Ek of a body of mass m moving
with speed v is given by
Ek=
1
m v2
2
Just as work is a scalar quantity, so is kinetic energy a scalar
quantity.
7-7 Potential Energy in a Uniform Gravitational Field
When a mass m is placed in a uniform gravitational field in which
the intensity of the gravitational field is q, the gravitational force
is given by mg, the weight of the body. If such a body is lifted
through a height h, the work done by the agency which lifts it is
mgh. We say that the work done has increased the potential energy Ep of the body, and write as a defining equation
Ep = mgh.
11
The position at which the potential energy is zero is quite arbitrary and, in fact, makes no difference in the consideration of a
particular problem. We shall always be interested in the change in
potential energy associated with a change in position, and shall
not attempt to ascribe a meaning to the value of the potential energy itself. For this reason it is often convenient to take the initial
position of the body as the position of zero potential energy.
The potential energy in the earth’s gravitational field is sometimes called the gravitational potential energy, to distinguish it
from other forms of potential energy such as the energy of a
stretched spring or the energy of an electric charge in an electric
field, which will be discussed in the later chapters of this book.
VI
10 Momentum and Impulse
10-1 Momentum
An extremely important concept in the development of mechanics
is that of momentum. The momentum of a body is defined as the
product of its mass by its velocity. We shall use the symbol p to
denote the momentum of a body. The momentum of a body is a
vector quantity, for it is the product of mass, a scalar, by velocity,
a vector. While momentum and kinetic energy are compounded
of the same two ingredients, mass and velocity, they are quite different concepts, and one aspect of their difference may be seen in
the fact that momentum is a vector while energy is a scalar quantity.
Newton himself recognized the importance of momentum as a
mechanical concept, for a free translation of his second law of
motion expressed in modern terms would read: the rate of change
of momentum is proportional to the net force and is in the direction of that force. Expressed in the form of an equation, Newton’s
second law would read
12
F=
d (mv ) dp
=
dt
dt
(10-1)
In our discussion of Newton’s second law in Chapter 5, we treated mass as a constant and obtained the result
F=m
dv
= ma .
dt
This form of Newton’s second law is true for most problems in
mechanics, when the speed of the body is small in comparison
with the speed of light. Newton’s original formulation, as represented in Equation (10-1), remains correct even for bodies which
travel at speeds approaching the speed of light, when, according
to Einstein’s theory of relativity, the mass of a body may be expressed as
m=
m0
1  v2 / c2
(10-2)
where m0 is the mass of the body at rest, v is the speed of light
(3×1010 cm sec), and m is the mass of the moving body. For our
present purposes we shall not digress further into the relativistic
aspects of mechanics, but shall focus our attention upon Equation
(10-1).
From Equation (10-1) we see that if an unbalanced force is
applied to a body, its momentum will change at a rate determined
by the force. If the unbalanced force acting on a body is zero, the
change of momentum is zero; that is, the momentum of the body
remains constant. Thus, in the design of space ships for interplanetary flight, consideration must be given to the provision of fuel
to accelerate the ship and to decelerate it for landing at the destination, but no fuel need be provided for propelling the ship over
the major portion of its path, for, acted on by no appreciable external forces, the momentum of the ship will remain substantially
constant.
13
Our study of mechanics has thus far concentrated its attention
on the motion of a particle when acted upon by a force. Suppose
we consider a stream of particles of mass m, each moving with
velocity v, that strike a target and come to rest in it, and inquire
about the average force exerted on the target to hold it in place.
From another point of view we may ask what force the particles
exert on the target. Each time a particle is stopped by the target,
the momentum of the particle is changed from mv to zero. The
change in momentum of the particle is mv. If n particles strike the
target in each second, the average rate of change of momentum,
that is, the change in momentum per second, is
p
= nmv.
t
Rewriting Equation (10-1) in incremental form, we see that the
average rate of change of momentum is the average force. Thus
we have
F=
p
,
t
(10-3)
so that
F = nmv.
(10-4)
Illustrative Example. A pitcher throws baseballs at a target
mounted on a helical spring at the rate of one ball every 2 sec.
The baseballs strike the target at a speed of 80 ft/sec and come to
rest in it. What is the average force exerted by the baseballs
against the target? A baseball weighs 5 oz.
When the baseball collides with the target, all of the momentum of the ball is absorbed by the target. Each ball has a momentum of
p=mv=
14
5 1
ft
slug ft
=0,78
 slug  80
16 32
sec
sec
Since one ball strikes the target every 2 sec, the number of balls
1
striking the target per second is n= . Substituting in Equation
2
(10-4), we have
F = nmv
slug ft
1
=  0,78
,
2
sec2
or
F = 0.39 lb.
Illustrative Example. A stream of water 102 cm in area, moving
horizontally with a speed of 25 m/sec, strikes the wall of a house
and splatters to the ground, losing all of its forward motion. What
is the force exerted on the wall of the house by the stream of water?
If Q is the volume of water that strikes the wall of the house
per second, then
Q = Av,
Where A is the cross-sectional area of the stream, and v is its velocity. If the water is of density p, the mass of water striking the
house in each second is
m = Qp.
The water has momentum mv = Qpv in the horizontal direction
before striking the wall, and zero momentum in this direction afterward, so that the change of momentum in each second is
p
= Qpv = Av2p.
t
From Equation (10-3) we have, for the force in the horizontal
direction,
F=
2
p
= Av p.
t
(10-5)
15
Substituting numbers into Equation (10-5), we find
F = 10 cm2  (2,500 ) 2
gm cm
cm2
gm
=
 1 3 = 625  10 5
2
sec2
sec
cm
= 6.25 × 107 dyne.
VII
10-2 Impulse
In many mechanical problems the applied force is not steady, nor
can the force be described in terms of simple mathematical functions. When a baseball bat strikes a ball, the force the bat exerts
against the ball is zero at the initial instant of collision, then rises
to some maximum value when the ball is violently deformed, and
finally returns to zero when the ball leaves the bat. The behavior
of materials and structures under such impulsive forces is quite
different from their behavior when subjected to steady forces, and
we speak of materials as being brittle when they are not capable
of withstanding impulsive loading.
In our earlier discussion of work and energy, we have seen the
usefulness of considering the effect of a force acting through a
distance. From such an analysis we derived our understanding of
the concept of work, and we saw that the result of doing work on
a particle was to change its energy. Another way to consider the
effect of a force on a body is to study the effects produced when a
force acts for a time interval. The product of a force by the time
interval during which the force acts is called the impulse. When a
force F acts for a time interval ∆t, the impulse ∆J is given by the
formula
∆J = F ∆t.
(10-6)
We see that impulse is a vector quantity, for it is given by the
product of force, a vector, by time, a scalar. We shall show that
16
the effect of an impulse acting upon a particle is to produced
change in its momentum.
Let us consider the incremental form of Newton’s second law,
as given by Equation (10-3). We have
F=
p
t
If we multiply both sides of this equation by the time interval ∆t
and call the force acting in this time interval F, we find
F ∆t = ∆p,
and, substituting from Equation (10-6), we have
∆J = ∆p,
(10-7a)
which relates the magnitude of the impulse to the magnitude of
the change in momentum. In vector form this equation becomes
∆J = ∆p
(10-7b)
Illustrative Example. A body of mass 10 gm moves along the
x axis with a speed of 3 cm/sec. A force of 400 dynes is applied in
the positive y direction for a time interval of 0.1 sec. Find the velocity of the particle produced by the impulse.
Since the impulse is applied in the y direction, there is no
change in the x momentum of the particle. The impulse in the y
direction is
∆Jy = 400 dynes × 0.1 sec = 40 dyne sec.
From Equation (10-7) we have
∆py = 40 dyne sec.
Since the momentum in the y direction was initially zero, the final
momentum in the y direction is equal to 40 dyne sec, and we have
∆py = mvy = 40 dyne sec,
17
vy =
cm
40 dyne sec
=4
sec
10 gm
Thus the body has a velocity of 4 cm/sec in the y direction and a
velocity of 3 cm/sec in the x direction. Its resultant velocity is
therefore 5 cm/sec, directed at an angle of 53˚ with the x axis into
the first quadrant.
When the impulsive force is given by a simple rectangular pulse,
the evaluation of the impulse is simple and straight-forward.
When the impulsive force is given as an arbitrary function of
time, we may follow the procedures of the integral calculus and
imagine the graph of F(t) to be divided into a number of rectangular pulses of different heights, in which each impulse serves to
change the momentum by a small amount. The total impulse is
the area under the curve. To find the over-all effect we add the
changes in momentum due to each impulse and write
J
t


o
o
J= dJ = F dt =
pf
 dp ,
pi
(10-8a)
J = pf - pi
Thus, when an arbitrary impulsive force strikes a body, the impulse is equal to the difference between the final momentum pf
and the initial momentum pi. Although Equation (10-8a) as an
equation in one unspecified component of the more general vector
equation
J = pf - pi .
(10-8b)
In general, it is quite difficult to measure an impulse, but it is
easy to observe a change in momentum. Thus the difference in
momentum between a pitched baseball and a batted ball may be
used to measure the impulse of the bat against the ball.
18
VIII
16 Kinetic Theory of Gases
16-1 General Gas Law
The behavior of a gas under various conditions of temperature
and pressure has already been studied in some detail. When the
pressure of a constant mass of gas is not too great, say less than
about 2 atm, we find that a gas obeys the following relationships:
at constant temperature PV = constant;
(16-1)
at constant volume
(16-2)
at constant pressure
P = KT;
V=K
T.
(16-3)
These three equations are special cases of a single experimental
equation which gives the relationship between the pressure P, the
volume V, and the absolute temperature T of a constant mass of
gas. We may derive the general form of the gas law from the
above equations.
Let us consider a gas contained in a cylinder with a closely
fitting piston. The initial condition of the gas may be described in
terms of its initial pressure Pi, its initial volume Vi, and its initial
temperature Ti. The gas is allowed to expand at constant temperature, say by keeping the cylinder immersed in a bath of melting
ice, until its new pressure is Pf and its new volume is V2. Since
the expansion was at constant temperature, we find from Equation
(16-1) that
Pi Vi = Pf V2.
Now suppose that the gas is heated to a higher temperature T f, the
volume being allowed to expand to a new value Vf, but the pressure on the piston being maintained at the same value P f throughout this process. Then, from Equation (16-3) we may write
19
V2 V f
=
= K',
Tf
Ti
or
V2=Vf
Ti
.
Tf
Substituting for V2 into the first of the above equations, we find
Pi Vi Pf V f
=
.
(16-4)
Tf
Ti
Equation (16-4) is one form of the general gas law. Since the initial state, described by the subscript I, and the final state, described by the subscript f, are entirely arbitrary, the only way in
which the quantities on the right and left-hand sides of the equation can be equal is for each quantity to be separately equal to the
same constant. Thus we may rewrite the gas law as
PV
=c
T
(16-5)
where c is a constant whose value depends upon the mass of the
enclosed gas. Any convenient units may be used for the pressure
and volume, but the temperature T must always be the absolute
temperature.
Illustrative Example. A given mass of air occupies a volume
of 2,000 cm3 at 27˚C when its pressure corresponds to the pressure at the base of a column of mercury 75cm high. The air is
compressed until its volume is 1,200 cm3, and its pressure corresponds to 225 cm of mercury. Determine the temperature of the
gas after it has been compressed.
From Equation (16-4) we have
Pi Vi Pf V f
=
Tf
Ti
20
The pressure at the base of a column of mercury h cm high is given by P=hpg, where p is the density of the mercury. Substituting
numerical values, we have
75 cm  pg  2,000 cm 3 225 cm  pg  1,200 cm3
=
Tf
300 .2 abs
from which
Tf = 540.4˚ abs.
16-2 The Universal Gas Constant R
The constant c appearing in Equation (16-5) can be evaluated for
any given mass of a gas. Let us designate the value of this constant for a gram molecular weight, or mole, of a gas by the symbol R. A gram molecular weight of any substance is an amount of
that substance whose mass, expressed in grams, is numerically
equal to the molecular weight of the substance. In the limit of low
pressures, the value of R is independent of the chemical nature of
the gas, so that R is known as the universal gas constant. In the
event that n moles of gas are present in a container, Equation
(16-5) may be rewritten as
PV = nRT.
(16-6)
The numerical value of the gas constant R can be determined
by noting that 1 mole of any gas occupies a volume of 22.4 liters
at a pressure of 76 cm of mercury at 0˚C; putting these values into
Equation (16-6), we get
R = 8.31  10 7
2 liter atm
erg
cal
=1,99
=8.21  10 


mole K
mole K
mole K
21
IX
16-3 Kinetic Theory of Gases
From the preceding discussion we have seen that all gases exhibit
similar thermal and mechanical properties, regardless of their
chemical composition, as long as their pressure is sufficiently
small. This behavior is quite unlike that of the same substances in
liquid or solid form, where these substances exhibit widely different thermal and elastic properties. We are led to infer that the
molecules of a gas are sufficiently far apart so that they rarely interact with each other. The pressure of a gas then results from the
collisions of the molecules of the gas with the walls of the container. The moving molecules of the gas completely fill every
container in which the gas is placed.
We may construct a theory of an ideal gas which is in good
agreement with the experimental results described in the preceding sections on the basis of a few simple assumptions. We shall
assume that a gas is composed of molecules that are so small that,
to a first approximation, they may be considered as point masses.
We assume further that the molecules do not exert forces on each
other except during collisions. We shall further assume that the
molecules of the gas are perfectly elastic, and that the container is
made of perfectly elastic, rigid walls. This implies that mechanical energy is conserved in collisions between molecules. If this
were not the case, we would expect to observe that the pressure of
a tank of gas would diminish with time, as the molecules lost mechanical energy in inelastic collisions. For the sake of simplicity
we shall assume that the gas is in a cubical container of edge d
and of volume V = d3.
The pressure exerted by the gas on the walls of the container
is due to the impact of the molecules on the walls, and, when in
equilibrium, is equal to the pressure throughout the gas. To calculate this pressure let us assume that the impact of a molecule with
a wall is an elastic impact; that is, if a molecule is approaching
the wall with a velocity v and momentum mv, then it will leave
the wall with a velocity – v and a momentum – mv. The change in
22
momentum of the molecule produced by this impact will thus be
– 2mv. To determine the pressure on the walls of the container, let
us first calculate the force exerted by the molecules on one of the
six faces of the cube and then divide by its area.
Let us consider those molecules which at some instant are
very close to this face. Only those molecules whose velocities
have components perpendicular to this face, and directed toward
it, will strike it and rebound. Suppose we consider a small number
of molecules which have the same value v1 for this velocity component. The number of these molecules which will strike this face
during a small time interval ∆t will be one half of the number
contained in a small volume A ∆l, where A is equal to the area of
the face of the cube and ∆l = v1 ∆t; the other half having a velocity component of magnitude v1 are moving away from the wall. If
n1 represents the number of molecules per unit volume which
have a velocity component of magnitude v1, then the number
striking this face of the cube in time ∆t will be
n1
Av1 t .
2
Since each such molecule will have its momentum changed
by – 2mv1 as a result of this impact, the impulse imparted to the
wall will be equal and opposite to it, or + 2mv1. The impulse F1 ∆t
on the wall produced by these collisions in time ∆t will then be
F1 t =
n1
Av 1 t  2mv1 ,
2
2
F1 = An 1mv1 .
from which
The pressure on the wall produced by the impact of these molecules is
p1=
F1 n1mv12
=
.
A
23
We can now consider another group of molecules, n2 per unit volume, which have a slightly different velocity component v2 in this
direction; they will produce an additional pressure p2 given by
p 2 = n 2 mv 22 .
In this way, we can break up the gas into different groups of
molecules, each group contributing a similar term to the pressure
on this face of the cube. The total pressure P due to all different
groups of molecules will therefore be of the form
P=n1mv12 + n2mv22 + n3mv33 + …
This equation can be simplified by introducing a new term called
the average of the squares of the components of the velocities of
all the molecules moving perpendicular to face A and defined by
the equation
v 2A =
n1v12  n 2 v 22  n3 v 32
…
n
in which n represents the total number of molecules per unit vol2
ume. Substituting this value v A in the equation for the pressure,
we get
P=nmvA2.
(16-7)
There will be a similar expression for the pressure on each of
2
the six faces of the cube, except that the factor v A will be replaced by the appropriate average of the squares of the components of the velocities of the molecules for that particular face.
The velocity v of any one molecule may be in any direction; it
can be resolved into three mutually perpendicular components vx,
vy, vz. The magnitude of v in terms of the magnitudes of these
components is given by
v 2 = v x2  v 2y  v z2
24
There will be a similar equation for the square of the velocity of
each molecule of the gas in terms of the squares of its three mutually perpendicular components. If we add the squares of the component velocities in the x direction and divide this sum by the total number of molecules we will get the average value of the
square of this velocity component; it will be represented by v x2 .
Similarly, v 2y and v z2
will represent the average squares of the
velocities in the y and z directions, respectively. By adding these
average squares of the three velocity components, we get
v 2 = v x2  v 2y  v z2 ,
2
where v is the average of the squares of the velocities of all the
molecules. Since the velocities of the molecules have all possible
directions, the average value of the squares of the velocity in any
one direction should be the same as in any other direction, or
v x2  v 2y  v z2
so that
,
v 2  3v x2 .
If we take the x direction as perpendicular to the face A, we can
write
v 2  3v 2A
so that Equation (16-7) becomes
1
P= nmv 2
3
(16-8)
Recalling that the kinetic energy of a moving molecule is equal to
1 2
mv ,
2
25
Equation (16-8) may be written as
P=
2 1
n( mv 2 ) .
3 2
Since n is the number of molecules per unit volume, we see that
the pressure is numerically equal to two thirds the kinetic energy
of the molecules in a unit volume of gas.
Let us suppose that No is the total number of molecules in a
mole of gas, called Avogadro’ s number, which is contained in a
volume V. Then the number of molecules per unit volume n is
given by the expression
n
No
.
V
Substituting for n into Equation (16-8), we find
2
1
PV= N o  mv 2 .
3
2
(16-9)
Equation (16-9) is a theoretical result obtained from our hypotheses about an ideal gas, relating the pressure and volume of 1 mole
of an ideal gas.
If we compare this result to the experimental equation given in
Equation (16-6), which for 1 mole of gas becomes
PV = RT,
we find the theoretical and experimental results to be in agreement if
RT =
2
1
N o mv 2 ,
3
2
1 2 3 R
mv 
T
2
2 No .
It’s customary to define a new constant k, called Boltzmann’s
constant, such that
or if
26
k
R
No
(16-10)
Since R is the gas constant per mole, and No is the number of
molecules in a mole of gas, the constant k may be described as the
gas constant per molecule. In terms of k the preceding equation
becomes
1 2 3
mv  kT :
2
2
(16-11)
that is, the mean kinetic energy of translation of a molecule of gas
3
is given by 2 the product of Boltzmann’s constant by the absolute temperature.
This equation gives us some physical meaning of temperature for
an ideal gas. For such a gas the temperature is associated with the
kinetic energy of the random translational motion of the molecules of the gas. According to Equation (16-11) the average energy of each molecule, and therefore the total internal energy of an
ideal gas, is associated with its temperature. Thus the internal energy of an ideal gas is a function of its temperature only, and not
of its pressure or volume.
In our derivation of the gas law in the form of Equation
(16-9), we used the word “molecules” to describe the particles
with which we were dealing. These molecules were described by
the condition that they were small, relatively far apart, and perfectly elastic. Thus this equation might be used to describe the
behavior of any aggregate of particles whose physical dimensions
were small compared to their average separation, provided that
these particles were elastic and rarely interacted with each other.
The neutrons in a nuclear reactor satisfy these conditions. If the
neutrons are in equilibrium with the material of the reactor at a
temperature T, we speak of them as thermal neutrons. The mean
27
velocity v of these thermal neutrons may be obtained from Equation (16-1). The particles of a colloidal suspension may also be
thought of as though they were molecules of an ideal gas, and it is
found that these also obey the gas laws.
Equation (16-9) incorporates another result called Avogadro’s
hypothesis, first stated by Avogadro in 1811, that all gases occupying equal volumes at the same temperature and pressure contain equal numbers of molecules. The accepted value for the
number of molecules in a mole of gas No is
No = 6.023 × 1023 molecules/gm molecular wt.
As we have already seen, 1 mole of gas occupies a volume of
22.4 liters at 0˚C and at a pressure of 76 cm of mercury. If we
perform the calculation indicated in Equation (16-10) to find the
numerical value of Boltzmann’s constant, we obtain
k = 1.38 × 10-16 erg / ˚K.
28