954

THE COLLIERY GUARDIAN.

May 9, 1913.

balance of wood resting on needle points can be made
for the purpose.

The velocity of the drops on striking the balance
scale can be calculated if we know h the height of the
level from which they fall, for

v = */%gh. = V2 x 981 X h.

The balance can be made of wood with an arm 0 D
and a transverse piece AB transfixed by needles resting
on two little squares of glass covered with paper
gummed upon them to prevent the needle points from
slipping. A curved iron wire rises from C and is turned
over downwards, terminating in a cardboard disc H.
At the other end a pointer Q points to a mark R. A
central arm P K dips down into a dashpot to check
oscillation. A beaker surrounds the cardboard disc H
to prevent splashing of the mercury. The stream of
mercury is now allowed to fall on the card disc, whereby
the disc is depressed and the end of the arm C D is
raised to a certain mark on the scale R. Then the
stream of mercury is stopped, and enough weight P is
put on the disc H to depress it to the same extent as
before. In this way the pressfir4 can be measured.

Fig. 1.

o Q

R

In one experiment P was found to equal the weight of
1*45 grammes; m n the mass of mercury falling from the
jet in 1 second was 4’6 grammes; h the height of the
level of mercury above H was 45 cm.; whence v the velocity
of fall = >/2 x 981 X 45 = 297 cm. per second; mn = 4'6,
whence by calculation P = 4’6 X 297 = 1,366 dynes—
1 366

that is, the weight of = 1’42 grammes, which
nearly accords with the pressure P found above by
experiment.

Another and simpler manner of experimentally
proving the law is to suspend a lead bullet about i in.
diameter and mass b grammes—that is. of weight w =
b X 981 dynes, by means of two threads of equal

Fig. 2.

the mass of each. But we know n m, the mass of the

whole aggregate. The number and force of the blows
given in a second will depend on the number, mass and
velocity of the molecules, and hence if G- be the velocity
in centimetres per second, which we will assume for the
time being to be uniform, then the velocities resolved in

the directions of the six faces of the box will give

G blows per second for each

molecule, for

is the

length, the upper parts of which are about 4 in.
apart, so as to form a sort of pendulum capable of
swinging in a fixed vertical plane (fig. 2). A jet of water
from a nozzle C strikes horizontally the centre of the bullet
and drives it back through a distance D B. If w be the
weight of the bullet, then P, the horizontal pressure of
the jet upon the bullet, = w P jA = 981 x b (1).

AD	A D v '

But if C E be the height of the nozzle from the ground
and F the spot where the jet when unimpeded strikes the
ground, Ven if a be the time of fall of a body from

C to E, and v be the horizontal velocity of the jet at the
nozzle,

EF < EF EFa/osI

a /2CE V2 C E

\/ ~98T

and if g be the quantity in grammes of water that issues
from the nozzle in one second—that is to say m n, when
n is the number of drops that issues in a second, and m
the mass of each drop, then

mnv = gv = X EF x ^981

V2 C E

If then the formula P t = m n v be true, the values of
the right-hand sides of equations (1) and (2) should be
found by experiment to be equal.

In one experiment b was 13 07 grammes, DB = 3’3cm.,
and A D was 20 8 cm. Whence, taking t as one second,
we have from (1)

P = 2,033 dynes.

In the same experiment g was found to be 3’8 grammes,
E F was 240 cm., and C E was 95 cm. Whence from (2)
we have m nv = 2,072 gramme centimetres, thus showing
P t — m n v very nearly. Another experiment with

another jet and at a different height gave the numbers
3,198 dynes and 3,276 dynes respectively. Of course,
this method is not susceptible of accuracy.

It will readily be understood that the modern theory
of dynamics does not rest upon rough experiments of a
nature such as those given above. It reposes on a
series of most delicate astronomical and other observa-
tions. The above rough experiments have therefore
only been cited to illustrate the general character of
the principles involved in the formula, and to show that
even the rudest experiments are in approximate
accordance with them.

The law P t = n m v for the force of impact of n
molecules per second of mass m and velocity v may
therefore be considered established by experiment, as
well as sanctioned by theory.

Section II.

From the foregoing considerations we may adopt the
law P t =. n m v as governing the law of the pressure of
n blows in a second each of mass m and velocity v. But
it has been established only for non-elastic bodies. For
the mercury drops do not bound off the disc H, but
simply stop and fall. If instead of non-elastic bodies
we were dealing with elastic balls, then a perfectly
elastic ball falling on a plane, in addition to producing
a pressure P, would on its rebound cause another
pressure P, just as a diver in springing off a spring-
board kicks it away from him. So that the pressure is
really = 2 P, and for perfectly elastic molecules our
formula becomes P t = 2mnv.

Let us now imagine a cubical box of 1 centimetre side
(0’39 in.) containing a number = n of air molecules each
of mass min a state of motion. The motions will all be in
straight lines, but the molecules will be constantly
bounding and rebounding from the sides of the box and
from one another.

The pressure on a side of the box will be the ordinary
atmospheric pressure, 1,013,200 dynes per square centi-
metre, and the mass of the air = m n = p = 0 00129
gramme, where p is the density of the gas as compared
with water. If we assume that the average pressure is
the same in all directions, the statical pressure effect
produced by the bombardment of the molecules when
resolved in three directions at right angles to one
another, amounts to six equal pressures P = 1,013,200
dynes, one on each of the six faces of the box, or a
total pressure of 6 P dynes.

Now we do not know the number of the molecules nor

average time they take to cross the box. Whence,
then, the total number of blows given by the molecules
will be n G per second, and since m is the mass
of each molecule and the velocity = G, the total
momentum will be n m G X G, or n m G2, where
n is the number of molecules in 1 cm.3, m the
mass of each, G the velocity in centimetres per second,
and t = one second, but since the molecules are perfectly
elastic the total pressure produced by this momentum
will equal twice the momentum, and therefore we shall
have P = | x 2nm G2.

= ip(P.

where p is the density of the gas. For air at standard
temperature 0 deg. Cent, and pressure, viz., 760 mm. of
mercury, the normal atmospheric pressure is 1,013,200

1

8

dynes per cm.2 The mass of a cubic centimetre of air
at 0 deg. Cent, and 760 mm. is 0’0012928 gramme.
Whence

G = a /8~xp - a /3x 1.013.200 = 484 m

V	V 0 0012928	P

It may be instructive to do the same problem in English
measures. P == the pressure of a mass of 14’7 lb. on the
square inch = 2,116 lb. on square foot, or, reckoning
in poundals—that is,* in English units of pressure=
14’7 X 32’2 = 68,161 poundals per square foot—a cubic
foot of air has a mass — 0’0807 lb. Whence

G = ./3 x P = a/3*^1 = 1,590 ft. per sec.

y/ n m y< 0’0807

For oxygen at 0 deg. Cent, and 760 mm., G- = 461 m.
per second. This is the velocity which the oxygen
molecules would have if they all moved at the same
velocity perpendicularly to the sides of the containing
box, but, inasmuch as they are moving in all direc-
tions, most of them move obliquely. This does not
affect the result, for an oblique motion may be con-
sidered as compounded of motions in three directions at
right angles to one another. Hence, since we are
dealing with squares of velocities, the square of the
resultant will always be equal to the squares of its
rectangular components, and we may thus treat a
velocity in any oblique direction as if it were simply
composed of rectangular components in some definite
directions. The above formula, however, depends on
the incorrect assumption that all the molecules have
the same velocity. But the molecules have all sorts of
velocities. We shall, however, see that it is possible
to represent their various velocities by a properly-
constituted average.

Inasmuch as the molecules, for chemical reasons, may
all be treated as of uniform size and composition, the
average velocity G might be supposed to be the
arithmetical mean of all the various velocities.

But this is not the case ; for according to the law of
conservation of energy it is not the total sum of all the
momenta that remains unchanged, but the total energy,
so that if M be the total mass of all the molecules of
gas in a cubic centimetre, and m3 be the
masses, and gx g2 g3 the velocities, then, however,
gr g2 . . . may change,

iMG2 =	+ %m2g22 +	.........

And since m1 m2 m3 are all equal; if n be the number
of the molecules in a cubic centimetre, and m the
average mass of a molecule, then M = m n and

SMG’=£(?1* + y.’+S'»! + .............)

whence G = a /A (g^ + g^ + g^ + ...),

V n

and therefore G is not equal to the arithmetical mean
of the velocities, that is A -f- g2 -f- g3 -f- <j4),but to the
n

square root of the mean of the’squares of the velocities.
It may also be observed, inasmuch as the kinetic energy
of a mass n m with a velocity G is | nm G2, that since
P = | n m G2 ; = K, the pressure of a gas is always a
measure of the kinetic energy K contained in it, and,
Q

numerically, K = P.

From this it follows that the kinetic energy in a cubic
centimetre of air = | X 1,013,200 = 1,519,800 ergs,
and the energy in a cubic foot of air at the earth’s
surface = 3,024 foot lb.

In the above calculation we have made an assumption,
which, however, is warranted by experiment. We have
assumed that on the average the kinetic condition of the
gas is uniform. This means that, at any time, the total
amount of kinetic energy resolved in any direction is
uniform, and equal to the total amount of kinetic energy
resolved in any other direction. The motion of the gas is
therefore taken as quite homogeneous ; there is no more
motion in any one direction than in any other. It is for
this reason that we have been able to treat the energy
as divisible into six equal portions, one along each of
the six radiating rectangular axes of co-ordinates, and
to assume this as true for any direction in which those
axes are taken.

The justification of this assumption is the experi-
mental fact that the pressures produced by a gas are
equal in all directions—but this points to general
uniformity of molecular motion, not necessarily equality
of motion everywhere, but general similarity everywhere
of any conditions of change of motion that may exist.
In other words, whatever differences of velocity or
direction of motion may exist between individual
molecules, yet on the whole their behaviour indicates
that they move on the average in a similar manner.

The temperature of a gas simply depends on the
energy in it, and this again on the square of the speed
of its molecules. In fact, heat is nothing but the energy
of molecules, and the effect of heating a gas is merely-