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 theLaw of Boyle-Mariotte1. Let v0 be the volume of the gas under the pressure p0; if the gas is exposed to pressure p1 and the corresponding volume is v1, then

v1/v0 = p0/p1 or v1p1= v0p0 or v0p0/v1p1 = 1.

For example, if p1 = 2p0, - you double the initial pressure - then v1=v0/2, that is, the initial volume is halved - however, only when the gas at pressure p1 has the temperature which it had at p0. 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 v1 and v0, which the gas has at the pressures p1 and p0, are inversely proportional to the pressures. Since v0p0 = v1p1= ··· = vnpn,, that is, the product of volume and pressure has a constant value C, you have v =/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, d1: d2 = p1 : p2. Also this law only applies when the Law of Boyle-Mariotte is valid and, as already been indicated, at constant temperature.

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 v0p0/v1p1 = 1. However, for p1 <p0, 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 v1p1> v0p0, that is, v1 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 p1 - p0=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 a³) 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 a³/8). The molecular velocity now yields 2n 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 ½mc² 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.

Ideal Gas

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 2mc, 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 2z between two consecutive impacts on the same wall in 2z/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 2mc. Altogether, during 1 second, the energy at the wall changes by (c/2z)·(N/3)·2mc = (N/3)(mc²/2). This is the force which acts on the entire wall xy. The pressure on the wall (force/unit area) is therefore

p = (N/3)·(mc²/z)·(1.xy) = Nmc²/3V, 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 3k/2, we find

mc²/2 = 3kT/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 M = Nm shows that the Mol masses of the chemical elements are interrelated like the masses of the individual molecules. Hence N, the number of molecules in one Mol, is for all elements the same. Hence Equation (3) says: At a given pressure p and a given temperature T, the volume V, which contains one Mol, is the same for all gases. 2g Hydrogen at 0ºC and 1 Atm fill 22.41 l, and the same volume , the Mol Volume, is filled under the same conditions by 32 g Oxygen, etc. N(=6.06·1023) is called Loschmidt Number. The constant k, introduced above, is

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 = T0 = 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 107 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 = 1000·0.10132·107 = 101.32·107erg = 101.32 Joule, whence R = 0.0820·101.32·107 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/)(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 = 3p/ r. If you measure p in dyn/cm³ (1 Atm = 1.0132·106) 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 · 109 Oxygen 102 · 10-7 cm 4.2 · 109 Hydrogen 180 · 10-7 cm 9.4 · 109 Helium 283 · 10-7 cm 4.3 · 109

The free path length is therefore of the order of 10-5 cm, the number of impacts of the order of 109 . 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/VG·s ²p and l = G/z = (V/N)/(2)1/2s 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)·ps3.

Hence you can write l = V·s/6·b21/2 and obtain for the diameter of the molecule

s = 6·b·21/2·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·21/2·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 = 60,6·1023, 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 = 8.313·107/60.6·1023= 1.37· 107erg/degree.