**Tendency to expand.
Compressibility. Law of ****Boyle-Mariotte**

The *mobility* of gas particles
is so much larger than that of fluid particles that their basic
properties, which follow directly from mobility - pressure transmission in all directions
and buoyancy - are quite different. Gas particles expand, because
they move at all times throughout their volume.
Due to this tendency to *expand*, you must close tightly containers in which you store
gas. (Vessels with fluid need not be closed above!) This tendency is known to
you from the spreading of escaping, but not lighted domestic gas
in a room or from the scent
of a perfume which you can sense throughout
a room. You see already that a gas can spread out in a container
which already holds another gas, say, air. (*diffusion of gases*).

Gases differ even more from fluids by their compressibility. The ability to change volume is a basic property of gases.

The instrument
in Fig. 198 lets you examine the *law between the pressure and volume
of gases* in the range of 1 and 2
atmospheres. You fill the pipe which can be closed by the tap *S*
- the tap *H *lets you close the pipes *s* and S
against the atmosphere - with gas, the leg *s* with
mercury; its weight generates the *pressure* to which you
want to expose the gas in *S*. If you connect *S*
and *s *by using the tap* H*, the mercury moves
partly into the leg *S *and *compresses* the gas by its
weight. The magnitude of the pressure exercised from *s*
and that of the gas volume in *S* are obtained from the
mercury columns in correspondingly calibrated pipes. The
apparatus is surrounded by water in order to keep the temperature
constant.

The relationship between the
pressure and volume of gases is formulated in the*Law of **Boyle-Mariotte*^{1}. Let *v*_{0} be the volume of the gas
under the pressure *p*_{0}; if the gas is exposed
to pressure *p*_{1} and the corresponding volume
is *v*_{1}, then

*v*_{1}/*v*_{0}
= *p*_{0}/*p*_{1} or *v*_{1}*p*_{1}*=
v*_{0}*p*_{0}* *or *v*_{0}*p*_{0}/*v*_{1}*p*_{1}*
=* 1.

For example, if *p*_{1}* *= 2*p*_{0}*,
- *you double the initial pressure - then *v*_{1}=*v*_{0}/2,
that is, the initial volume is halved - *however*,
only when the gas at pressure *p*_{1} has the *temperature** *which it had at *p*_{0}.
However, compression of a gas is always accompanied by a rise in temperature. This relation
ship applies *only* after the temperature of the compressed gas has sunk to
its initial value, .

^{1}.Boyle
discovered the Law in 1662, not Mariotte in 1679

You describe therefore the behaviour of a gas, which is far enough from its state** **of**
**liquefaction and the *temperature of which is held constant*: The volumes *v*_{1} and *v*_{0},
which the gas has at the pressures *p*_{1} and*
p*_{0}, are inversely proportional to the pressures.
Since *v*_{0}*p*_{0}* = v*_{1}*p*_{1}=
··· = *v*_{n}*p*_{n},, that is,
the product of volume and pressure has a constant value *C*,
you have* v *=* C·*/*p*, wence we can also say: The volume of a
gas is inversely proportional to the pressure to which it is
exposed.

The pressure
only changes the volume of a gas, not the number of the particles in it. An increase in pressure
is therefore accompanied by an increase in
density, a decrease in pressure by rarefaction of the gas. If you
double, triple, etc, the *pressure*, the *density** *of the gas is doubled, tripled, etc. In other
words, the *density
increases proportional to the pressure*,
that is, *d*_{1}:* d*_{2}* *=
*p*_{1}* *: *p*_{2}* .
*Also this law only applies when the Law of Boyle-Mariotte is
valid and, as already been indicated, at

The law *pv = const*
can be described geometrically. If you plot in a rectilinear
coordinate system
corresponding values of *p* and* v *along the
ordinate and abscissa, the rectangles formed by them have always
the same area. The equation *pv = *const is an equal-sided
hyperbola (Fig. 199).

**Deviation
from ****p·v-****law**** **

If this law were
strictly applicable, corresponding values of pressure and volume
would satisfy strictly the equation *v*_{0}*p*_{0}/*v*_{1}*p*_{1}
= 1. However, for *p*_{1} <*p*_{0},
this fraction is up to very high pressures at medium temperatures *larger* than 1 for all gases except hydrogen
and helium. It is smaller than 1 for Hydrogen and Helium, that
is, for these gases *v*_{1}*p*_{1}>
*v*_{0}*p*_{0}, that is, *v*_{1}
is larger that it would be by the law. Hence these two gases are *less compressible* than the law requires, all other gases
are more. Experience tells us that the fraction deviates more
from 1, the closer the pressure approaches the point at which
they become liquid. It is for *p*_{1} - *p*_{0}=1atm
at small pressure and at 0º C:

Helium | 0.99955 | Nitric Oxide | 1.00117 | |||

Hydrogen | 0.99922 | Hydrogen Chloride | 1.00737 | |||

Nitrogen | 1.00074 | Ammonia | 1.01499 | |||

Oxygen | 1.00097 | Sulphuric Acid | 1.02341 |

The deviation from unity is so small that it is normally neglected.

**Mechanical foundation of ****p·v-****Law.
Kinetic Theory of Gases**

This law can be explained by means of the *hypothesis* that gases do not have appreciable
cohesion and that their particles move along straight lines at
high speeds. The theory, based on this hypothesis, is the *Kinetic Theory of Gases*. (August
Karl Krönig 1822-1879, Clausius, Maxwell, O.E.Meyer, Boltzmann ). It explains the pressure in a gas as
follows: On their way, the *gas
particles collide* with
every obstacle, with each other, rebound, change their path until
they collide again, etc. Naturally, they also collide with the
wall of the confining vessel and the *totality of their impacts** *against it expresses itself as
the *pressure** *of the gas against the wall.
With the aid of several simplifying assumptions, this concept
leads to Boyle's Law:

Let a gas be confined in a cube
with edge length *a* (volume *a*³) and its
molecular velocity be so large that a gas particle, which moves
parallel to a wall in each second, flies to and fro between two
opposite walls *n-times*, that is, *it impacts **n*-times against the same wall. (We disregard
the collisions with gas particles and assume that they merely fly
to and fro between opposite walls and perpendicularly to them -
these are assumptions which, while not fulfilled, are not
contradictory.)

Now *compress* the cube,
so that it has the edge length *a*/2 (Volume *a*³/8).
The molecular velocity now yields 2*n *return trips per
second. Hence each wall experiences twice as many impacts, but
has only one quarter of its original area. The impacts are now
against a four times smaller area than before and repeat
themselves *twice as
often*, whence it is
eight times as intensive on each cm² as before, that is, the
wall experiences eight times as large a pressure. But the space
has only one eighth of its initial value. The result corresponds
to the Law of Boyle, assuming that the unchanged velocity of the
molecules corresponds to an unchanged temperature.

The *Gas Theory* gives you insight into the *nature of gases* after
introduction of the concept of temperature, in fact, of the *absolute temperature** T. *We use here the concept *temperature* only as far as it is known to everyone through use of a
mercury thermometer. You arrive at the *absolute temperature *of Guillaume
Amontons 1663-1705 as follows:

Most substances increase
(decrease) their volume while the temperature increases (decreases). All gases
distinguish themselves by *increasing
their volumes* equally: For each 1ºC
by 0.003665 = 1/273 of their volume which they have at 0ºC,
whence a volume of gas which at 0º is 1, has at 100º the volume
1 + 100·0.003665 = 1.3665. The uniformity of the expansion for
every degree and the accuracy to which it is known allows you to
employ gases
themselves as *thermo-metric* substances (similarly to the use of mercury in every
day thermometers). In order to explain the significance of the *absolute *temperature,
we will employ *air *as *thermo-metric
*substance.

We will follow a presentation of the air
thermometer (Fig. 200) in its most primitive form, due to Maxwell: A straight cylindrical, long
tube which is open at the top and closed at the bottom and
contains in its lower section *as a piston a drop of mercury* on top of air. As a result of a temperature
increase or decrease, the air in the lower part of the tube
expands or contracts and the mercury piston indicates through its
position the *temperature*, - provided
that the air pressure
outside remains constant.

Let the mercury at the
temperature of melting ice be at what we will denote by 0ºC. We then construct *a scale in both directions*
of 1/273 of the volume at 0ºC. If we continue this process
downwards far enough, we reach eventually the closed end of the
tube. What number will we
have there? We know that the distance of
the point of freezing of water from the end of the tube relates
to the distance of the point of boiling of water (100ºC) from
the end of the tube as 1:1.3665, because the volumes of air,
bounded by the drop of mercury, at 0ºC and 100ºC are so
interrelated. If we denote
the number of degrees from the boiling point to the bottom end of
the tube by *x*, then the number of
degrees from there to the boiling point of water is *x* +
100, whence

(*x* + 100) : *x*
= 1.3665 : 1, that is, *x* = 272.85 ~ 273 (exactly:
273,20).

Thus, you have at the bottom of the tube - 273ºC.

W will not discuss here to what extent this
result corresponds to reality, but , for the sake of simplicity,
will use in many cases the temperature point 273ºC as *zero of the air thermometer*. It is called the *absolute
zero of the temperature *and temperature measurements, related to it, are called *absolute temperatures*. The zero of the Celsius-scale, related to the air thermometer scale, is then
273, and *t*°*C *means (273 +* t*)°. You
write *T *for (273 +* t*), that is, you denote the absolute temperature of a body by *T*°*C*.

In order to use this
thermometer for our purposes, we must make - as will be confirmed
later on by conclusions drawn from it - the assumption: The *kinetic energy ½mc*² of a gas
particle, where *m *is its mass and *c* its
velocity, is proportional
to the absolute temperature *T* of the gas. Since every particle
changes all the time its velocity, and all possible velocities
are present *simultaneously* in a gas*, *we set, more
exactly, the *mean value
of the kinetic energies of all gas particles** *proportional to the absolute
temperature.

Let a prism with edges *x, y, z*, that
is, with the volume *V* = *x·y·z*, contain *N equal particles*, each with the mass* m*, whence the total mass
in *V* is *N*·*m *and the density *r **= Nm*/*V*.
In reality, all gas particles change all the time the direction
and magnitude of their velocity. In order to simplify the
presentation, we will assume (a strict computation
can show that this assumption does not
influence the result): 1/3 of the *N *particles move
parallel to each of the three edges of the prism and all *particles at the same velocity** c. *Moreover, we will assume that the gas is
*ideal*, that is, its particles are like points, elastic and do
not influence each other by any forces. If a gas particle impacts
on the wall *x*·*y* of the prism, its momentum *mc*
changes by 2*mc*, because on impact the velocity changes
its direction by 180º; if its momentum was +*mc *before
the impact, it is* -mc *afterwards. Everyone of the *N*/3
gas particles, which impact parallel to the edge *z* on
the wall *xy*, covers the distance 2*z* between two
consecutive impacts on the same wall in 2*z*/*c *seconds,
whence: Everyone of the *N*/3
gas particles hits the wall in 1 second *c*/*2z
*times and transfers to it the impulse 2*mc*.
Altogether, during 1 second, the energy at the wall changes by (*c*/2*z*)·(*N*/3)·2*mc
= *(*N*/3)(*mc*²/2). This is the *force*
which acts on the *entire* wall *xy*. The *pressure* on the wall
(force/unit area) is therefore

*p* = (*N*/3)·(*mc*²/*z*)·(1.*xy*)
= *Nmc*²/3*V*, whence *pV* = *Nmc²/*3.
(1)

The kinetic energy of a single
particle is *mc²*/2. This, according
to our assumptions, is also the mean
value for the kinetic energy of *all* particles
which we have set proportional to the absolute temperature *T*
of the gas. Denoting the proportionality factor by 3*k*/2, we find

*mc*²/2
= 3*kT*/2 (2) and *pV = N·k·T. *(3)

However, the zero
of this temperature *T* does not lie at the melting point
of ice, but 273º lower, where the body does not contain any
energy of motion.

Equation (3) above
contains the most important *gas
laws*. All of them have
been found *empirically*. They only apply* approximately*, but the more so, the more rarefied is
a gas. A gas is very rarefied (from the mechanical point of
view), if the space filled by it has so few particles that in the
mean the relative *distances
*between the particles
are very large and the particles can be considered to be points.
But also the second condition of the ideal gas - the gas
particles do not interact - can then be considered to be correct,
since at large mutual distances of the mass points their interaction will be very small.

A *correct *theory of the *ideal *gas
*must* lead to the *gas laws*. The equation *pV=N·k·T *is
not related to a *definite** *gas; the mass *m* of
the gas particle, which we have introduced above and which
differs for each gas, has disappeared at the introduction of
temperature. Therefore the equation holds for every ideal gas whatever is the weight *m *of its
particles. It says: *All
ideal gases obey at constant temperature T the Boyle-Mariotte Law
pV = const*. If the
pressure *p* is constant, the *volume**
*increases proportionally to the temperature *T* (Law of Gay-Lussac
1778-1850 **335**). In contrast, if the *volume**
V* is constant, the pressure grows (Law of Jacques César Charles 1746-1823)
proportionally to the temperature *T*. Equation (3) also
contains the Law of Avogadro: A given volume *V* of any gas
contains always at a given pressure *p *and given
temperature *T* the *same
number* of molecules,
whence: The volume of a litre, for example, at 0ºC and 1 Atm,
contains the same number of molecules irrespectively of whether
it contains hydrogen, oxygen, nitrogen, etc. 1 atm (Atmosphere)
is the pressure, which a 76 cm high column of mercury exerts on 1
cm²**.**

**Gas Constant. ****Loschmidt****'s Number**

You can measure *p,
V *and *T *and* *compute *N·k*. The
last quantity is called the *gas
constant* (*R*),
when *N* is so large that the total mass *N*·*m
*= *M* of molecules, measured in grams, equals the molecular weight number of the gas. You
call the mass (*M*)
one *Mol** . *You
understand by 1 Mol hydrogen 2g of hydrogen, by 1 Mol oxygen 32 g
of oxygen, etc. The relation

*k = R*/*N
*(4)

(Boltzmann's Number), that
is, it is equal to the Gas Constant divided by the Loschmidtt
Number. Like *N*, also *R* has the same value for
all gases. Introduction of *R* into Equation (3) leads to

*pV = RT, *(5)

the** ***general
gas equation *for the
case that *V *contains one Mol of a gas. For the freezing point of ice: *T = T*_{0}*
= *273.20. Hence you find after substitution of the value of *V*,
valid for this temperature and the pressure *p*
= 1* *Atm,

*R = pV/T = *1·22.41/273.2
= 0.0820 Litre Atmosphere per degree

or, in general,

*pV = *0.0820·T
Litre Atmospheres.

One *litre atmosphere* is *work*,
like 1 meter kilogram. It is the work which, for example, you
must perform in order to shift by 1 dm a piston of 1 dm²
cross-section, which is movable in a cylinder and exposed to 1
Atm, *against *this pressure, in order to overcome the
pressure of 1 Atm over the space of 1 litre.

You can replace
the Litre Atmosphere by any other unit of work, for example, by
erg or 10^{7 }erg (1 Joule). The pressure 1 Atm equals the weight of a 76 cm high column of
mercury over 1 cm², that is, 76·13.596·980.6 = 1013200 =
0,10132·dyn/cm². Hence 1 Litre Atmosphere = 1000·0.10132·10^{7}
= 101.32·10^{7}erg = 101.32 Joule, whence *R*
= 0.0820·101.32·10^{7} erg/degree = 8.313 Joule/degree.

**Supplement of ****p·v
- ****Law. (****van der Waals****)**

The Gas Theory also explains the *deviations *from Boyle's Law. Gas particles do not
have for their to and fro motion the entire width of the
container, because they themselves also occupy space. At
pressures, which do not raise the density of the gas above a
certain level, this is immaterial - but it is important when the
pressure reduces considerably the total space available to them.
This is taken into account by the Law of van der Waals 1873, which also allows for the mutual
attraction of the gas particles. Van de Waals obtained for the
dependence of the volume *v* on the pressure of a gas a
formula, which applies to the *gaseous*
as well as the *liquid* state. The corresponding numbers agree
well with measured results. This formula is: (*p* + *a*/*r²*)(*v*
- *b*) = *RT, *where *b* is the *correction of the volume* and *a* makes allowance for the
attraction between the gas particles. You can compute from the
constants *a* and *b *of this equation at what
temperature a certain gas *follows
exactly *Boyle's Law (Boyle Point).

The equations *pV = N·mc²*/3 and *N·m*/*V
*= *r* yield *c² = *3*p*/ *r**. *If you measure *p* in
dyn/cm³ (1 Atm = 1.0132·10^{6}) and *r *in g/cm³, you find *c* in m/sec.
Substituting the numbers for *r*, you obtain for 0º and 1 Atm:

c |
G |
c |
G |
|||||||

Nitrogen | 492 | (425) | Hydrogen | 1844 | (1692) | |||||

Oxygen | 461 | (454) | Helium | 1303 | (1204) |

The numbers for *c*
are the square roots of the mean value of the squares of the
velocity; *they do not
agree* exactly with the
mean *G* of the velocity itself. The velocities of gas
molecules are seen to approximate those of grenades of modern
guns. In spite of their high velocity, the molecules cover in the
same direction only *very
short distances*.
Unceasingly, they collide with their neighbouring molecules and
are deflected from their paths. The number of impacts, which a
molecule undergoes in the mean during 1 second, that is, also the
distance *l* = *G*/*z, *the *mean free path length*, which it covers in the mean in a
definite direction without collision, can be computed from the
diffusion, the heat conduction or the friction of gases. (*Diffusion** *is the spreading of a gas
within another gas**). **The velocity of diffusion
is proportional to the path length which a gas particle can cover
without being deflected from its path, that is, it is
proportional to its *mean
free path length*.

*Heat
conduction* can be
conceived as the diffusion of two volumes of gas of the same
kind, but at different temperatures. The colder gas spreads into
the warmer gas and conversely. Therefore the heat must transfer
at a rate which is proportional to the *free path length l*.

We will treat *friction** *in detail*. *If you
place a disk at rest parallel and close to a movable disk, a *larger **force *is required to move the
second disk parallel to the first disk when the space between
them contains a gas. The gas impedes the motion just as in the
case of a rough mechanical frictional process and the strength of
the *force through gas
friction* depends on
the kind of gas involved. The substance of the disks is
immaterial, because a thin layer of gas, which takes part in the
disks motion, covers each of the disks, whence the friction is
between the *layers of
gas* which move with
respect to each other. The *frictional
coefficient* *h *of a gas is linked to its molecular
properties; in fact, neighbouring horizontal layers, which shift
with respect to each other, exchange molecules with each other;
the layers *effectively
intrude* into each
other and attempt to hold each other back. The Kinetic Theory
yields for the coefficient of friction *h *the relation *h* = 0.350·*r*··G·l. You can measure the frictional coefficient *h *and the density *r*·, while *G *has already been
found; the free path length can so be computed. Division of the
mean molecular velocity *G* by the mean free path length *l
*yields the number *z* (= *G*/*l*) of
collisions, which in the mean the molecules experience per
second. At 0º and 1 Atm, you find :

mean free path
lengths l |
Number of
impacts per second z |
|||

Nitrogen | 95.5 · 10^{-7 }cm |
4.8 · 10^{9 } |
||

Oxygen | 102 · 10^{-7 }cm |
4.2 · 10^{9 } |
||

Hydrogen | 180 · 10^{-7 }cm |
9.4 · 10^{9 } |
||

Helium | 283 · 10^{-7 }cm |
4.3 · 10^{9 } |

The free path
length is therefore of the order of 10^{-5 }cm, the
number of impacts of the order of 10^{9
}. Using the *number of impacts* and the *mean free path lengths,* we can interrelate the diameter of the
particles as well as their number in a Mol. Two molecules with
the diameter *s* collide when their centres approaches
the distance *s* . If* *the centre of a molecule
covers in 1 sec the distance *G *(more correctly: If its
centre retains its direction for 1 sec, that is, it covers the *entire* distance *G*, etc.), the
molecule covers, since its active cross-section is *s* ²p, the space
G·*s* ²p. Within
this space, the molecule collides with every other present
molecule. If you know the number *A* of molecules in this
space, you obtain the required number of impacts at *z* = *A*.
The Mol volume *V* contains *N* molecules, the
space is *G*·*s* ²p, whence *A = G*·*s* ²p·*N*/*V.
*This is the number of impacts. An exact computation yields:

*z *= (2)^{1/2}(*N/V*)·*G*·*s* ²p and *l*
= *G*/*z* = (*V*/*N*)/(2)^{1/2}*s* ²p. (6)

The total volume *b*
of the *N *spheroidal molecules in one Mol is

*b = N*·(4/3)p(*s*/2)^{3}
or *b* = *N*·(1/6)·p*s*^{3}.

Hence you can
write *l* = *V*·*s*/6·*b*2^{1/2} and obtain for the
diameter of the molecule

*s* = 6·b·2^{1/2}·*l*/*V.*

Since we know
already the pressure *l* for 0º and 1 Atm, we only need
to compute under the same conditions the space *b*/*V *filled
by* *the molecule. We know that oxygen at 0º and 1 Atm
has 0.001429 g mass per 1 cm³. Moreover, certain measurements
have yielded: Oxygen has at the stage of solidification, that is,
in the state in which its molecules are already very close to
each other, the density 1.27. The mass of 0.0011429 then fills
the volume 0.001429/1.27 = 0.001124 cm³. In reality, the volume
of the molecule will yet be smaller, whence this value *b*/*V
*= 0.001124 cm³ must be considered to be an upper limit.
With *l* = 102·10^{-7 }cm for oxygen, the
diameter of its molecule has the upper bound *s* = 0.97·10^{-7 }cm*.* A
more accurate theory yields *s* =
0.29·10^{-7 }cm. - For other gases, molecular diameters
of the same order of magnitude have been found, that is, all are
several hundred times smaller than the free path length at 0º
and 1 Atm.

Once you know
the *mean free path
length* *l* and
the *diameter* *s* of
a molecule, Equation (6) yields the number *N *of
molecules in a Mol:

*N = V*/*l*·2^{1/2}·*s* ²p.

At 0º and 1 Atm, *V* = 22410 cm³. For example,
for Oxygen, you have (using the above values for *l* and *s* ) *N* = 60,6·10^{23},
which is exact within 1%.

Equation (4)
allows to compute *k*. If you substitute for the *Gas Constant* *R* its value in erg/degree, you
find

*k *=
8.313·10^{7}/60.6·10^{23}= 1.37· 10^{7}erg/degree.