Change of aggregate state of substances through change of heat content
In order to dissolve a solid substance in a solvent. you require heat, latent solution heat. The substance takes it out of the solvent and its own heat store. Thereby drops the temperature of the solution below the initial temperature of the solvent and the substance while the solution is being formed.
If you dissolve nitrate of ammonia (NH4)NO3 in an equal weight of water and the initial temperature of each of them was 10ºC, the temperature of the solution sinks to about - 15º. Mixtures of solid substances, which become liquid as a result of their mixing, act similarly: Ice melts when it is mixed with sodium chloride (or spread on top of it). The salt dissolves in the melted water. A salt solution freezes at a much lower temperatures than pure water. (One application is to unthaw ice and snow on rails of tramways.) Snow and potassium chloride act in the same manner. The mixture has essentially three components: Ice, watery solution of the salt and undissolved salt. At 0ºC, the mixture is not in equilibrium, because at 0ºC ice evaporates much more strongly than the water from the solution, so that ice must convert itself into water. The heat required for this step is withdrawn from the entire solution. It cools it down until it has reached a temperature at which the ice and the solution evaporate equally strongly. Then equilibrium has been established. If the entire mixture were taken to a yet lower temperature, the solution would evaporate more strongly than the ice, and eventually the entire mixture would rigidify with development of heat and a simultaneous rise in temperature.
During the formation of crystals of a dissolved substance, the heat which became latent during solution is freed. This is most readily shown for super-saturated solutions. A water soluble salt - the best is sulphate of caustic soda or acetic soda - can dissolve in hot water in larger quantities than in cold water. If you dissolve it in hot water as much as is possible - you say: Until the solution is saturated - and then cool it, it becomes over saturated. Avoiding concussion or other sudden disturbance, you can maintain the oversaturated state, that is, prevent slow cystallization during slow cooling. However, if a oversaturated solution suffers a very slight disturbance, it crystallizes suddenly throughout the entire mass, accompanied by a strong rise in temperature.
Oversaturation of a solution is matched by undercooling of a fluid: With certain precautionary measures, you can cool water down to - 15ºC without it freezing. If it then suddenly freezes, for example due to concussion of the container, not the entire mass will freeze, but only a part of it, the remainder being heated by the freed, initially latent heat of the rigidified part to 0ºC and becomes fluid - this was for Black the clearest proof that not only mere cooling "below 0ºC, but the loss of a definite quantity of heat is the condition for solidification" (Mach).
As long as the vapour pressure of a fluid is less than the external surface pressure, it only evaporates at the surface and without appreciable motion of the surface and the vapour diffuses into the air. Also solid substances in contact with the free atmosphere evaporate (mostly only in small quantities) gradually at their surface; it is known to everyone from musk, camphor, naphthalene, ammonium chloride by their fragrance as well as from ice, which also evaporates at the severest cold into the air. Thus, also every solid has a definite vapour pressure (however, often it is so small that is can only be measured indirectly. The evaporation of a solid and the solidification of its vapour, without the solid of the vapour passing through the liquid state, is called sublimation. Every solid substance sublimes very slowly in the atmosphere; in contrast, it sublimes very violently when the sublimation pressure exceeds the atmospheric pressure. If the sublimation point (which can be compared to the boiling point of fluids) is below the melting point of the solid substance, it sublimes when it is heated; in order to be able to heat it to melting, it must be heated in a closed container. However, as a rule, the sublimation pressure of solids lies close to the melting point and far below the atmospheric pressure. The quantity of heat, which 1 g of a substance uses for sublimation, is called its sublimation heat; at the melting point, corresponding to the energy principle, it is equal to the melting heat plus the evaporation heat of the melted substance. For the preservation of thermodynamic equilibrium between the solid and the melted part of the substance, the substance, solid or melted, must have (according to the theory) at the melting point the same vapour pressure. In the cases of benzol and water, the theory and measurements agree satisfactorily.
The Clausius-Clapeyron equation also applies to evaporation. The symbols l T p v are then transferred to the evaporation process, whence the evaporation heat can also be computed for any temperature, if you know the specific volumes of the saturated vapour and fluid and the dependence of the pressure of the saturated vapour on the temperature (dp/dT). Thus, you obtain for the evaporation heat of water at the normal boiling point 539 cal (with v = 1674 cm³ as volume of 1 g saturated water steam at 100ºC and dp/dT = 27.12 mm Hg per degree); its measurement yielded 538.7 (Henning).
An approximate estimate of the evaporation heat in cal/mol is given by the rule of Mark August Pictet 1752-1825, which was renamed on rediscovery after Frederic Thomas Trouton 1863-1922: In first approximation, the evaporation heat of one mol at the normal boiling point is proportional to the absolute boiling temperature Ts: V ~ 21 Ts. This rule only applies more or less to substances which boil between about 0ºC and 100ºC. Evaporation heats at the normal boiling point in cal/g are:
In order to measure the evaporation heat, you can feed the vapour through a spiral tube through the water in a calorimeter, where it loses its latent heat to the water, the temperature of which rises and thereby condenses. The quantity of the calorimeter water and its rise in temperature yield the supplied heat, the quantity of condensate the quantity of the steam, passed through the calorimeter. You obtain more reliable results by generation of the heating energy, required to convert a given fluid into vapour, using an electrical current and measuring it. The evaporation heat (l) of all fluids drops with rising temperature (t). At the critical temperature (for water 374ºC), the evaporation heat is always zero. The most accurate values for water are those of the German Physical-Technical Government Institute (in 1935):
Measurement of steam pressure
There exist dynamic and a static methods for the determination of saturation pressures. In the dynamic one, you heat the fluid under a given pressure until the steam rises in bubbles out of it, that is, it boils regularly, and then measure the temperature. In the static method, you give the fluid at a given temperature into a space, which does not contain any other substance, so that the pressure in it is solely due to the steam of the fluid, and measure the pressure. Since there belongs to each temperature a definite saturation pressure and to each such pressure a definite temperature, both methods must yield the same result.
An example of a static method is that of Dalton You place some of the fluid to be tested into the vacuum of a Toricelli-tube A (Fig. 402), by introducing it into the mercury with a bent pipette and let it rise there. In the space above the mercury, as much of it will evaporate, as is required for its saturation, The vapour lowers the mercury (and the remaining fluid) in the tube. The number of millimetres, by which it depresses it, yields the saturation pressure of the vapour in mm mercury at the temperature, displayed by the thermometer T. B is a barometer. At 10ºC, the mercury column is caused to sink for : Water by 9.2 mm, alcohol by 24.4 mm, ether by 433.0 mm.
In order to measure vapour pressure at different temperatures, you surround the upper part of the tube by a bath (Regnault). Depending on the temperature, there forms above the mercury the amount of vapour, which corresponds to saturation, with the saturation pressure corresponding to this temperature. At the boiling temperature - by definition, the temperature at which the saturation pressure equals atmospheric pressure - the vapour depresses the column to the level of the mercury outside the tube. (This method fails therefore for temperatures above the boiling point.) The vapour then balances a column of mercury of the same height (760mm), which is held in equilibrium by the atmospheric pressure, whence one says: The steam of water at 100ºC has the pressure of one atmosphere (atm). Correspondingly, a pressure of 1520 mm is that of 2 atm, etc.
Regnault measured the pressure of steam up to 230ºC with an arrangement, which is comparable to his experiment, described earlier: He generated with a pressure pump a manometrically assessed pressure on the water surface (dynamic method) and measured the temperature of the steam which arose as the water boiled at this pressure. The manometer had the form of a U and was filled with mercury; the mercury in the one leg was under atmospheric pressure, that of the other leg under the same pressure as the water. At a 760 mm height difference lies the boiling point at 100ºC, at 1520 mm at 120.5ºC, whence the saturated steam at 120.5ºC balances a 1520 mm high column of mercury, that is, it exerts a pressure of 2 atm. As a rule, a pressure in excess of 760 mm is stated in atmospheres (atm), not in mm mercury.
|Temperature||Saturation pressure||Temperature||Saturation pressure||Temperature||Saturation pressure|
* Beyond 374 ºC, even at the highest pressure, water cannot exist as a fluid, whence there is no boiling point for it above 374ºC (its critical temperature).
The pressure, at which water boils, and its boiling temperature determine the usefulness ofboiling water for processes which demand a certain temperature, like for the extraction from plant and animal materials (tea, coffee, glue, etc.), fully cooked nourishment, etc. In order to raise its boiling pressure and thereby its boiling temperature, you use the pressure cooker of Papin, a pressure tight pot in which you place the substances under the pressure of its saturated steam, the intended temperature.
Storage of work in hot water
If you have heated water in a tightly closed container under high pressure until it boils and then let the steam above it escape, that is, reduce the pressure, steam will escape until equilibrium has been reinstalled. This method to obtain steam from highly heated water under high pressure (20 atm) by reduction in pressure allows to accumulate work in thermal storage, then available in cases of sudden, large demand for heat without renewed generation (Ruths thermal storage for industries with varying demands for steam). It consists essentially of a large, pressure resistant container, filled to 90% with water into which excess steam condenses and heats the water**.
** In this context, an experiment of Watt is of interest: For example, in an open Papin pot during half an hour, boil down 25 mm water. If it is refilled, the water again is made to boil, it is closed at the start of boiling and reopened after half an hour, then steam escapes for two minutes, during which time 25 mm of water disappear. The quantity of heat, which the already heated to 100ºC water takes in during a another half hour is therefore sufficient to evaporate gradually during the intake or quickly afterwards 25 mm water (Ernst Mach: "Principles of thermodynamics").
Steam pressure formula
The equation of Clapeyron-Clausius shows that there is a connection between the pressure p of saturated steam and its temperature and, if one were to know the dependence of the quantities l (evaporation heat) and v (saturation volume of the steam and the volume of the fluid) on T, it would yield a general steam pressure curve. In actual fact, this cannot be done and you must be satisfied with empirical formulae with more or less theoretical foundations. Because Nernst introduced his chemical constant, which enters into it, we quote here his steam pressure formula:
log p = - l0/(4.571·T) + 1.75 ·log T·e /(4.571·T) + i,
where i is called the chemical constant, because it is important for the computation of chemical equilibria. The theory yields for monatomic substances i = -1.587 + log A with A the atomic weight. Nernst's Heat Theorem states that the numerical value of i of a substance is the same for every aggregate state. You can also determine i empirically, that is, check the truth of the formula, and show that observations and computations agree well.
The vapour pressure curve (Fig. 403) demonstrates the link between the pressure and temperature of saturated vapour, but it has yet another meaning: The point M means that at this temperature (t-coordinate) and this pressure (p-coordinate) water and steam are in equilibrium (that is, neither evaporation nor condensation occur), that is, it also shows the lasting simultaneity of the presence of water and steam. If you raise at constant pressure the temperature from M to M', the equilibrium is disturbed - water evaporates - and equilibrium reoccurs only when all water has evaporated; in contrast, if you lower the temperature from M to M", all the steam condenses (for the same reason). Moreover: If you lower at constant temperature the pressure to M'1, then all the water evaporates gradually, if you raise it to M", gradually all the steam condenses. The points M and M' thus characterize the states in which only water is present.
The vapour pressure curve tells about the equilibrium state of the phases water and steam: Points on the curve correspond to states in which both these phases are in equilibrium with each other and the curve separates the merely fluid phase from the merely liquid phase. A partner to the vapour pressure curve is the melt pressure curve (Fig. 404): At the points of this curve (temperature, pressure), the fluid and solid phases - here water and ice - touch each other in equilibrium, that is, neither the ice melts nor the water freezes (this happens below a pressure of 760 mm at 0º, below the pressure of 4.62 mm at 0.0074º). It is simultaneously the border between the merely solid and merely fluid phases. Hence the point at which the steam pressure and melt-pressure curves intersect marks the state, in which all the three phases touch one another in equilibrium: It is called the triple point. Hence the three aggregate states can be in simultaneous equilibrium only at a definite temperature and pressure (fundamental temperature and pressure). For water, this point lies at t' = +0.0074ºC and p' = 4.62 mm, for carbonic acid at t'=-79ºC and p' = 5.1 atm. If you raise t at constant p, the solid substance transits only through the fluid state to the gaseous one, when p > p'. If p < p', it sublimes. For this reason, CO2 can only be fluid above 5.1 atm. There also exists a sublimation pressure curve, which also passes through the triple point.
So far, we have only talked about saturated steam - the steam was in equilibrium with its fluid or, in other words: There existed always excess fluid, so that at further heat input more steam could develop. However, what happens when, for example, during the steam pressure experiment in Toricelli's vacuum (Fig. 402) so little water reaches the vacuum that it evaporates totally? -When there is just sufficient water, in order to saturate the space with steam and now heat is added, the temperature of the steam rises, it expands and depresses the mercury column. The space above the column is now larger than at the time, when the steam just saturated it; in order to saturate the larger space, there would be a need for additional (unavailable) fluid. Hence the space contains less steam than it can hold, that is, it is not saturated. The pressure, which the steam exercises in order to be able to occupy this space, is only available due to its higher temperature - as saturated steam it would create the same pressure at a lower temperature - in the state, in which the steam occupies a larger space due to its higher temperature, it is said to be overheated. However, if the space becomes so small that the steam eventually saturates it - assuming all along that the temperature is constant - the steam condenses and its pressure remains constant in spite of a further reduction in space, because it is then again in contact with the fluid and again saturated.
The graphical presentation of Fig. 405 will make this clearer. The overheated steam fills a cylindrical tube, closed at one end by a solid wall, at the other end by a gas tight, moveable piston K. The cylinder is at a constant temperature; push the piston gradually inwards, in order to reduce the volume of steam, employ the distance of the bottom of the piston from O as abscissa, the pressure corresponding to the position of the piston as ordinate. You will then observe: The corresponding values of p and v for the pressure and volume form (almost) a hyperbola PQR as in Fig. 199, until the space of the steam and the corresponding pressure reach a certain magnitude (the piston is at M). From then on, the pressure remains constant in spite of a further reduction of the volume: The steam condenses until the piston arrives at T and the tube contains only fluid. Now the pressure will rise steeply at the smallest further movement of the piston, corresponding to the small compressibility of the fluid. The curve of state PQRS corresponds to a here specially selected temperature; however, if you repeat the experiment at a higher temperature, you will obtain another, similar curve. Only the straight segment is reduced, because then liquefaction only starts at a higher pressure and higher density (smaller volume) of the steam.
Density of steam
A comparison of the numbers for the latent melting heat with those for the latent evaporation heat indicates that, in general, a conversion of fluid into steam demands much more heat than the conversion of the corresponding solid body into a fluid. This is readily conceived. Latent heat is work. During evaporation, it comprises the total separation of fluid particles. While also its cohesion is small compared with that of the particles of the solid body (which must be overcome during melting), it requires for its conversion into steam, that the average distances of the molecules of the fluid become so large, the mutual attractions so small, that their reunion is impeded. While the heat only loosens the structure during melting, it bursts it apart during evaporation. This is demonstrated by the low density of vapours compared with that of fluids. For example, the densities, related to water at 4ºC, are:
|mercury||melt point or boiling point||density|
|solid||melt point (-39ºC)||14.38|
|fluid||boiling point (+358ºC)||12.83|
The density of vapours have here been related to water. It is frequently related to air. However, since the density of air differs with the temperature and pressure acting on it and since the same is true for vapours, you understand by vapour density the number, which states how much more mass is contained in a volume of vapour at a given pressure and temperature than contains an equal volume of air at the same pressure and the same temperature.
The ratio of the masses contained in unit volume to each other would also remain unchanged, if the volume were altered by pressure or temperature, provided the vapours - whether saturated or overheated - followed the laws of Boyle and Gay-Lussac. It would then be sufficient to measure the mass (in gram) of a given volume of steam at a given pressure and a given temperature and to compare it with the known mass (in gram) of an equal volume of air at the same pressure and the same temperature. However, only strongly overheated vapours follow Boyle's law; closer to saturation, they deviate more. Hence the vapour density, related to air, has only a constant magnitude for overheated vapours.
Measurement of vapour density
Chemistry employs density for the determination of molecular weight and the constitutional formulae of compounds. How are they measured?
We recall the number of grams a volume of air of v cm³ has at the pressure p mm mercury and the temperature tº. At the pressure of 760 mm and 0ºC, 1 cm³ air contains 0.001 293 2 g. At tº, the volume, which occupied 1 cm³ at 0º, occupies (1+a t) cm³, but if the pressure is not 760 mm, but p mm, the space is 760(1+a t)/p cm³. Hence the 0.001 293 2 g air are contained in this formula. Hence 1 cm³ air at tº and pressure p mm: 0.001 293 2 ·p/[760·(1+a t)] = m0g. Hence 1 cm³ air at tº and pressure p mm contains: v·m0 = m1 g. If now the same volume at the same pressure and the same temperature contains mg of vapour, then the vapour density, related to air, D = m/m1, where m1 is the mass of air computed above, whence
D = m/(0.0012932·p·v/[760(1+a t)] = m·760·(1+a t)/0.0012932·p·v.
Hence, in order to determine vapour density, you must measure: The pressure p, the volume v, the temperature t and the mass m of the vapour.
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