E4 Equilibrium of gases

Diffusion of gas. Partial pressure

Also the diffusion of gases is explained by the molecular motion of their particles. It is the process by which two gases, which are in contact, mix completely even while they are at the same pressure. Fill one vessel with carbon- dioxide, another with hydrogen, both at the same pressure, and then connect them as is shown in Fig. 201 (Dalton). After some time has elapsed, you discover that:
1. Every vessel contains the same amounts of hydrogen and Carbonic Acid, that is, the gases have mixed completely, irrespectively of the fact that Carbonic Acid, which is 22 times as heavy as hydrogen, was in the lower vessel. [The atmosphere owes its complete mixing of its gases to diffusion (apart from what happens at great heights), although its component gases have different weights.]
2. At the end of this process, you find the same pressure in both vessels, that is, both gases have distributed themselves throughout the two vessels, as if the other gas were not present. Naturally, corresponding to the increase in volume, the pressure of each single gas has dropped.

It is called partial pressure. But the combined pressure is, as experience shows, as large as the pressure of each single gas at the start. The same is true for mixing of arbitraily many gases which are chemically indifferent to each other.

In general, this observation means: If several ideal and mutually indifferent gases mix at a common temperature, are at the same pressure p and occupy the volumes v₁, v₂, v₃, · · ·, then the pressure p remains unchanged and the mixture occupies the volume V = v₁+ v₂+ v₃+ · · ·. Every single gas spreads throughout the volume V and thus attains a corresponding lower partial pressure, whence Dalton's Law follows from the Boyle-Mariotte Law: The partial pressures of gases in a mixture are p₁= p₁v₁/V, p₂= p₂v₂/V, p₃= p₃v₃/V, · · · and their sum equals the total pressure p.

For example, if you let diffuse into each other 21 parts of a space filled with oxygen and 79 parts of a space filled with nitrogen, keeping the temperature t and pressure p constant, then the oxygen and nitrogen have to a large approximation the partial pressures p₁= 21·p/100 and p₂= 79·p/100, respectively.

In an atmosphere without steam, the partial pressures in % are:

The partial pressure of water vapour in the air is at t = -20º up to 0.1 %, at t = 0º up to 0,6%, at t = 20º up to 2,3 % of the total pressure.

Naturally, the validity of Dalton's Law only extends as far as Boyle's Law .

With a molecular velocity of air of approximately 485 m/sec, it would appear to be incompatible that a cloud of smoke (smoke of a cigar in your room) and the smell of a gas in air at rest (no wind) diffuses very slowly.However, the gas molecules collide all the time and divert each other, whence it takes a long time until a single molecule traverses the room and can contribute noticeably to the spreading of the gas.

Weight of gases

The fact that Earth's atmosphere, in spite of the high velocities of its particles, does not distance itself from Earth and distributes through outer space is due to Earth's attraction, that is, due to the weight of the gas particles. Air has weight, and so has every gas and body. You can prove this with a balance. A hermetically lockable vessel without air weighs less than when it contains air or any other gas. Hence you can compare the masses of gases by their weight. The amount of mass, that is, how many grams are contained in a vessel filled with a gas, depends essentially on its pressure and temperature. For a comparison ofweights, you must relate them to a normal temperature and normal pressure. At 0ºC and 760 mm mercury you have:

Following Faraday, you can display these large differences in the weights of gases as follows (Fig. 202): Hang two glass beakers A and B, one erect, the other inverted, from a lever balance and establish equilibrium; fill into the first carbonic acid from above, into the second hydrogen from below. Both times, the balance will deflect along the arrow shown.

1 cm³ of air weighs 0.001293 g; this number states the density of air (as always) related to water. In order to avoid uncomfortably small numbers, one relates usually the density of a gas to that of hydrogen, the lightest gas, or to that of atmospheric air at the temperature and pressure of the gas concerned. The density of a gas, related to hydrogen (or air) does then not give the grams in 1 cm³, but states how often so many grams are contained in 1 cm³ of this gas as in 1 cm³ of hydrogen (or air) at equal temperature and pressure.

You can determine the density of a gas by weighing of a definite volume or following a method due to Robert Wilhelm Bunsen.

Buoyancy of gases

Since gases have weight, those lying above press on those below and, since they share with drop forming fluids the all round propagation of pressure, they also share with them buoyancy. We are here only interested in the buoyancy of the atmosphere around us. As during the discussion of the buoyancy of drop forming fluids, we reach also here the conclusion that every body loses as much weight in the atmosphere as the weight of air displaced by it.

This is displayed well by a companion of the hydrostatic balance (Fig. 203 Baroscope of Schoentjes). A telescopic tube A is closed apart from a small opening which can be closed by a stopper B. If it is fully collapsed (short) it has half the volume from what it has fully extended. If it is short, it is in equilibrium with the weight C; however, if it is long. it weighs less - by the air volume displaced in the lengthening process. Lengthening it has doubled its volume, that is, halved the amount of air per cm³. (A special mechanical device stops the air pressure from contracting the cylinder.) If you now open the hole, closed by the stopper, air intrudes into the tube - exactly as much as the lengthening of the tube has displaced outside and the balance returns to equilibrium.

The existence of this buoyancy yields:
1. The same body is lighter in air than in a vacuum;
2. Two bodies which have the same weight in air, but do not have the same size, have different weights in a vacuum.

The weights of a large sphere and a small sphere, which have the same weight in air, do not balance each other in a vacuum. In a vacuum, the large sphere weighs more by the weight of the air volume it displaces in air, the small sphere by the weight of a smaller air volume; the balance sinks on the side of the larger sphere.

Effect of buoyancy on body weight, reduction of weight to that in vacuum

For scientific purposes, weights taken in air demand corrections which take the loss of weight through buoyancy into account. It is indifferent for routine weighings! Let m denote the apparent weight of a body (the weight on a balance in air), l the density of air (in the mean 0,0012), s the density of the body, s the density of the weights (brass = 8.4); its weight M in vacuum is then M=m(1 + l/s - l/s ), that is, ml(1/s - 1/s ) must be added to the apparent weight. Hence the correction of the apparent weight w of a mass of water weighed with brass pieces (s = 8.4) is: w·0.0012(1/1-1.8.4) =w·0.00106, that is, you must add 1.06 mg for each gram.

Balloon, Airship

As in fluids, the state of motion of a freely moving body is determined by the ratio of its buoyancy to its weight relative to its environment. If the buoyancy of a body is larger than its weight, it rises in air, for example, a balloon.

A balloon's ability to rise becomes the larger the larger is its volume and the smaller its weight. In order to make its weight as small as possible, it is filled with hydrogen. At ordinary temperature and air pressure, 1 m³ of air weighs approximately 1.29 kg*, 1 m³ of hydrogen about 0.09 kg*. A 100 m³ balloon, filled with hydrogen, has a buoyancy of 129 - 9 = 120 kg*, a 5000 m³ balloon around 6000 kg*. One part of it is used up by its equipment (cover, gondola, ropes, etc). The weight of ropes and cover grows proportionally to the balloon's surface, the gondola does so approximately, whence a larger balloon is more practical than a small one, because its dead weight is only a smaller part of its buoyancy. The gas diffuses continuously through its cover ( two layers of cotton cloth with a rubber layer in between), although slowly (no material is totally gas tight!), whence its buoyancy decreases slowly. Again a larger balloon is superior to a small one: A spherical balloon with twice as large diameter has four times the surface area as the smaller balloon.

If a balloon reaches layers of air of lower pressure, its gas expands. In order to protect its cover against tearing, you let the balloon's filling tube hang below and open. (Also the Sun causes expansion by heating!) As a balloon rises, the displaced air weighs less than down below, it loses buoyancy until its weight equals the buoyancy - it then floats in equilibrium. In order to make it rise further, one throws ballast over board, as a rule, bags filled with sand.

In order to become dirigible, that is, an airship, a balloon must have motion of its own (otherwise the wind will carry it away), indeed, a velocity which is larger than that of the moving air, if it wants to overcome it. The problem of dirigibility coincides largely with having an engine which is large enough for the required propulsion and has a small enough weight for use in a balloon. The invention of petrol engines (Gottlieb Daimler 1834-1900 1883) and the reduction in the price of aluminium (1890) pointed out the way to a solution of this task; today, we have usable engines the weights of which are 2 kg* for 1 PS¹. The shape of a balloon is important for its dirigibility. In its direction of motion, the resistance it experiences should be as small as possible, whence it is shaped like a cigar.

¹ On 4th and 5th August 1908, Count Ferdinand von Zeppelin made the first long air trip in Germany from Friedrichshafen to Mainz. The body of his airship had 15000 m³; it had 2 Daimler engines of 110 PS with a weight of 560 kg* each. The airship taken by Hugo Eckener 1868-1954 between the 12th and 15th October 1924 in 18 hours and 17 minutes from Friedrichshafen (Germany) to Lakehurst (USA) had a length of 200 m, a diameter of 27.64 m, a total volume of 70 000 m³, a total buoyancy at 0º C and 760 mm of 84.5 tons, an empty weight of 42 tons, whence a carrying capability of 42.5 tons for each 1000 kg*, five engines of 400 PS at 1400 revolutions /minute and a crew with effects of 4.5 tons. The largest airship built in 1935 was the Graf Zeppelin with a length of 236 m, largest diameter of 30 m, gas volume of 105000 m³, 5 Maybach engines of 530 PS each, own velocity at maximum engine poer of 128 km per hour and normal travel speed of 117 km per hour.

An air balloon rises like every gas bubble by lift. On the ground, you must hold on to it to stop it from rising and if its cover were completely gas proof, it would, having lifted to a certain height, stay there (aerostatic lift). The floating of the balloon corresponds to natural swimming. Like there exists artificial swimming - a lasting struggle with sinking and which is only possible by swimming motions - there also exists artifical flying: This too is a lasting struggle with sinkingm that is, falling down, and it is only possible through flying motion. A bird in flight, when it is hit by a bullet, drops like every other heavy body. It only can start from the ground by performing work to rise and stay aloft (aerodynamic lift).

Water level gauge

A gas bubble in a fluid rises like a balloon due to lift. If the fluid is bounded all around by a solid wall, the bubble cannot escape from the fluid. It remains in contact with the wall and the fluid, adjusts its shape to the bounding surfaces (the wall and the fluid) and in the process, independently of the position of the container, moves to the highest possible position. It displays its great sensitivity to changes in the position of the container by visible and measurable displacements. This property is employed in the water level gauge (Thévenot 1620-1692 1661) for establishment of horizontal planes or vertical axes. It is very slightly curved glass tube which is completely filled, apart froma small air bubble, with a as movable as pissible fluid (alcohol, ether). The tube is fixed to a metal plate so that, when the plate is exactly horizontal, the bubble oo is located at the centre aa' of a scale (Fig. 204).

Atmospheric pressure. Torricelli's experiment

Since air has weight, it presses on Earth's surface and everything on it. You demonstrate this phenomenon by the experiment of Torricelli 1643. Fill a straight, about 80 cm long, glass tube, which is closed at one end, up to its top with mercury (so that it does not contain air), close the other end and dip it upright into a vessel, filled with mercury, and then open the end . Then the level of the mercury in the tube will drop, but it will stop as soon as the upper end lies between 70 and 80 cm above the level of the mercury (Fig. 205).

You will understand the significance of this process by comparing it with the behaviour of two fluids of different specific weight in communicating vessels. In this case, the two liquids are mercury and air, their common interface is the level of the mercury S (Fig. 206), to the communicating vessels correspond the tube and free atmosphere. On top of the common plane of separation S, you have, first of all, the column of mercury at a measurable height and also the column of air, of which you know that, while it cannot be measured with a tape, it extends to the bounds of the atmosphere, whence it does not experience a pressure from above, because there is nothing beyond.

What about the column of mercury? The space above it in the tube was also filled initially only with mercury. But the mercury dropped out of it and left behind a vacuum, a space with nothing in it. Hence also the pressure on the column of mercury is zero. The air column of the height of the atmosphere and that mercury column of measurable height are therefore in equilibrium, that is, the pressure of the air column on S equals the pressure of the mercury column on S (said more simply: The air pressure is so many centimetres of mercury.)

Hence, for example, the air presses on 1 cm², when the mercury column is h cm high, with the weight of h cm³ of mercury; thus, if r is the density of mercury, it presses with h ·r · g gram*. The pressure, which a 76 cm high column of mercury exerts at 0º C at sea level at 45ºN on 1 cm², is called one atmosphere (1 atm). Since the density of mercury at 0º C is 13.596 and Earth's acceleration g at 45ºN is 980.26, one atmosphere corresponds to a pressure of 1.033 kg*/cm². Earth's air mantle has in the mean this pressure at sea level. This is the origin of the use of the words atmosphere.

This pressure is called one physical atmosphere and has been abbreviated above and here by Atm. Technical instruments for the meaurement of pressures are calibrated in kg*/cm² (not 1,033 kg*/cm²!); this pressure is called one technical atmosphere and abbreviated 1 atm. In practice, both have the same meaning. The atmosphere around us presses on 1 cm² with about 1.033 kg and on the surface of an adult, which, according to Max Rubner 1854-1932, is 1¾ - 2 m², with about 1.75·10⁴ - 2·10⁴ kg.

As a rule, we do not feel the pressure, because in what ever direction it acts there acts a pressure of the same magnitude simultaneously in the opposite direction. (Inside a fluid, which is only exposed to gravity, for the same reason, the pressure of a fluid does not disturb the equlilibrium.) However, we will sense a sudden, very strong change of air pressure, especially when it comes from one side, for example, from an explosion.

If the atmospheric pressure changes, the mercury column indicates it by a change of its length. For example, let the air pressure be such that 76 cm of the mercury column establish eqilibrium. If the pressure rises, the weight of that column is no longer sufficient to balance the additional pressure, which drives the mercury in the column higher up. On the other hand, if the pressure sinks, it is not sufficient to maintain the column at that height and it sinks. Torricelli's device forms the basis of all instruments for the measurement of pressures in the atmosphere, the mercury barometer.

Barometer

Figs. 207 a,b,c show the most frequently occurring forms of mercury barometers. They differ by the accuracy with which they measure the height of the air column above the external mercury level. If the height of the mercury column changes, the lower level changes and, in fact, more strongly, the narrower is the reservoir compared with the tube diameter. No consideration is given to this aspect in the vial barometer a, but in the tube barometer b and the siphon barometer c it changes the design. In the vial barometer, this is achieved by shifting, before reading the result, the mercury level by means of an easily achieved change of shape to the zero of the scale; in the siphon barometer, this is done by shifting the scale along the mercury column until its zero lies at the level of the lower mercury meniscus.

In order to measure the pressure very exactly, you must correct the result which you read off as height of the column, that is, make allowance for the change with temperature which may be above or below 0º C. Thus, if you read 760 mm and the temperature is 20ºC, you must subtract about 2.5 mm. Moreover: The space above the mercury contains mercury vapour, which presses on the column and makes it somewhat lower (at 20ºC about 0.003

mm) than it would be if the space were totally empty. You see that thereadings from a mercury barometer are not readily interpreted. For less accurate measurements ( also in an aeroplane), you use the aneroid barometer (Vidie 1847, Greek:a = not, nhros = wet).

Its most important part is an air tight, closed, thin-walled metal capsule, a vacuum, as a rule a flat box or a bent tube (Fig. 208). As the air pressure changes, the pressure in the container changes. The accompanying deformation of its wall is transferred (magnified by a lever) to a pointer which turns on top of a scale. You calibrate the scale by comparing the readings with a mercury barometer.

Barometric measurement of height

The air pressure is not constant, so that if a barometer at the same location is higher or lower, you say that it has risen or fallen. At sea level, the central value at mid-latitudes is about 760 mm. Moreover, the reading of a barometer depends on the height of the location above sea level. The air column between the mercury level S of Fig. 206 and the bound of the atmosphere changes depending on whether the measurement is taken at sea level or above it. At sea level, it is highest, it is zero at the edge of the atmosphere and its height changes throughout the atmosphere. It must be calibrated with a mercury barometer at different heights. The mercury column of Torricelli's experiment will be higher on the roof of a house than in its cellar. In low lying regions, you must raise the barometer by about 11 m to cause it to sink by 1 mm, at 3000 m by about 15 m.

The difference in the heights of a point of observation and in the lengths of a mercury column are interrelated as follows (Halley 1686): If the heights change in an arithmetic sequence, the pressures drop in a geometric sequence. This relationship is explained by the following results of observations (with corrected barometer readings):

Mathematical formulation: Let h₁ and h₂ be the heights above sea level, B₁ and B₂ the barometer levels at the lower and higher points of observation, and k a constant to be determined, then

h₂ - h₁ = k (ln B₁ - ln B₂) (ln = logarithm to base e).

It has been shown that k = 7991 m. Using Brigg's logarithms and this constant (taking into consideration the mean temperature between above and below), the basic formula for barometric height determination is: h₂ - h₁ = 18400(log B - log B)(1 + a t). This basic formula demands also corrections for air moisture and for the dependence of gravity on the geographical latitude and the height over sea level. The constant k = 7991 m is the reduced height of the atmosphere, that is: The height of a simulated homogeneous atmosphere whicb balances a 760 mm high mercury column. If there is no wind, - strong winds tend to make measurements useless - a barometric height measurement is as exact as a geodetic one. In very many cases, it provides the only useable method.

Submarine barometric depth measurement

You can employ the barometer (in a diving bell) to measure the depth of a point below sea level. You can simulate a diving bell in water by means of a drinking glass which you dip upside down into water. However deep you lower it into the water - the water will never fill it completely, that is, it will always contain an air volume. The water can only intrude as far as it can compress the air by its pressure (the weight of the water column up to the water level). The same process occurs in a diving bell. The pressure of the water column decreases the volume of air and thus increases the pressure of the air and causes a barometer in the bell to rise, the more, the deeper the bell is submerged. If the barometer is at the depth T below the sea level and has risen by h cm, and r is the density of the mercury, s that of the sea water (it is larger than 1 due to its salt content), then T = h·r /s

Mercury air pump (Heinrich Johann Wilhelm Geissler 1815-1879 1857)

Torricelli's experiment proves that it is possible to make a space void of air. Until the discovery of this fact, it was considered to be impossible to do so, because of Nature's horror vacui (Latin: fear of vacuum - Aristotle). For many scientific and technical purposes, the possibility of creating in a space a vacuum or at least to reduce its pressure arbitrarily below the atmospheric pressure is very important; for example, for electrical light bulbs, vacuum brakes, thermos flasks, steam condensers ,electronic valves.

Torricelli's experiment provides the basis of one of the most effective methods for the creation of a vacuum: You generate in a space A (Fig. 209) above a mercury column a vacuum (just as in Torricelli's experiment), then you link the vessel B which is to be evacuated by opening the tap A so that, at first confined to B, it spreads through A + B. Then you close the tap and evacuate again A and repeat the procedure. In this way, you can evacute B until you can call say that it is void of air.

This principle has led to the development of the earliest mercury pumps which are now only of historical interest. However, it was very important for the development of vaccum techniques. Fig. 209 shows its most important parts: The tube C with the expansion A is about 80- 90 cm long and corresponds to the barometer tube of Torricelli's device. It is connected below by a hose to the mercury container E, at the top, it has a tap o which, depending on its position, links the vessel A (through the tube d) to the container to be emptied or to the external air or closes simultaneously both.

In order to pump out the vessel B, you act as follows: Turn the tap o so that A is connected to the atmosphere, lift the container E so high, that the mercury in C and in A rises, displaces all air in it through the tap o into the atmosphere and eventually arrives at the tap itself, as Fig. 209 shows. You then close the tap and lower the container E. The mercury in the tube C falls, as in Torricelli's experiment, and, in fact, as deep, as corresponds to the air pressure; there is now a vacuum of considerable extent in A. Now you turn the tap o so that B, which is to be evacuated, and A, the already empty space, are connected. The air, hitherto confined to B, spreads into the space A and the tube and presses down the mercury, corresponding to the existing pressure downwards. Now you turn the tap, so that A is again separated from B and linked to the atmosphere, lift the vessel E, drive the air out of A into the atmosphere and recreate by lowering of E a vacuum, etc. The rarefaction in B grows, of course, the faster, the smaller is B and the larger is A.

How fast does the rarefaction in B grow? At first, the pressure of the air in B is at the same pressure p₀, the atmospheric pressure. When B is linked to the empty space A, the air, limited to B, spreads out into A + B and becomes rarefied. Denoting the pressure then by p₁, Boyle's law yields

p₁/p₀ = B/(A + B), whence p₁ = B·p₀/(A + B).

where p₁ is the pressure in B as you linked it for the first time to A. If the pressure has then sunk to p₂, then

p₂/p₁ = B/(A + B), whence p₂ = B·p₁/(A + B)·p₁ = p₀·[B/(A + B)]².

At the n-th time, the pressure has decreased to

p_n = [B/(A + B)]ⁿ.p₀.

The pump of Heinrich Geissler 1814-1879 was the lbeginning of enumerable similar pumps, also of the pump of A.J.J .Toepler, which employed instead of the tap o an automatic valve. All the mercury pumps, which rest on Toricelli's principle, have been displaced by the rotating mercury pump of Gaede.

Mercury air pump (Gaede 1905)

The rotating mercury air pump of Gaede has neither taps nor valves. Like Geissler, Gaede also fills a vessel completely with mercury, lets the entering mercury displace only a little of the air, as most of it has been removed by means of an auxiliary pump - in order to link it - without air in it - to the vessel which is to be pumped out; this cycle is repeated until the vessel is empty. However, in the case of Gaede's device, it is not always the same vessel which intermittently fills and empties. In fact, there are three of them which in turn perform the action, more or less like in Fig. 209, if there were not only the vessel A, but two more, which at the angle of 120º lie together like the wings of a wind mill and, in turn, act between the tube C and the vessel to be pumped out. And the mercury is not, like in Geissler's pump, filled in and drawn out: The vessel is more like a bucket which - almost like a dredge - fills and again pours out, but pours out in such a manner that at first the space at the bottom of the bucket empties and space is left behind by the mercury, that is, the empty space gets into contact with the vessel to be emptied.

The essential part of the pump of Wilhelm Gaede 1878-1945 (Fig. 210) is the drum T (made of porcelain or steel), which is subdivided into three chambers W and rotates about its horizontal axis in the mercury (about 30 revolutions per minute). During the rotation, the chambers empty and fill in through the narrow channels Z and connect through the slot openings L with the tube R leading to the space to be pumped out. The mercury presses through the slots z, coming from the chambers, the air drawn from the recipient into the previously by means of an auxiliary pump evacuated space G from which it is removed by an auxiliary pump. These pumps are among the best in existence, but Gaede himself has surpassed them by other air pumps of original design (molecular air pump, diffusion pump), which depend on other physical processes.

continue step back