Mixing of gases. Mixing gases and vapors having different temperatures
2. Mixing of gases and vapors having different temperatures.
This is how atmospheric fogs are formed. Most often, fog appears in clear weather at night, when the surface of the Earth, intensively giving off heat, is greatly cooled. Warm moist air comes into contact with the cooling Earth or with cold air near its surface and liquid droplets are formed in it. The same happens when the fronts of warm and cold air mix.
3. Cooling of a gas mixture containing steam.
This case can be illustrated by the example of a kettle in which water boils. Water vapor escapes from the spout, which is invisible because it does not scatter light. Further, the water vapor cools rapidly, the water in it condenses, and already at a short distance from the teapot spout we see a milky cloud - a fog that has become visible due to the ability to scatter light. A similar phenomenon is observed when we open the window on a frosty day. A stronger aerosol is formed when oil boiled in a frying pan creates a gas (oil aerosol) in the room, which can only be removed by well-ventilated room.
In addition, condensation aerosol can be formed as a result of gas reactions leading to the formation of non-volatile products:
During the combustion of fuel, flue gases are formed, the condensation of which leads to the appearance of furnace smoke;
When phosphorus is burned in air, white smoke is formed (P 2 O 5);
· when gaseous NH 3 and HC1 interact, smoke MH 4 C1 (tv) is formed;
· Oxidation of metals in air, which occurs in various metallurgical and chemical processes, is accompanied by the formation of fumes consisting of particles of metal oxides.
DISPERSION METHODS
Dispersion aerosols are formed during grinding (spraying) of solid and liquid bodies in a gaseous medium and during the transition of powdered substances in suspended states under the action of air flows.
The spraying of solids occurs in two stages:
grinding and then spraying. The transfer of a substance into an aerosol state must be carried out at the time of application of the aerosol, since, unlike other disperse systems - emulsions, suspensions, aerosols cannot be prepared in advance. In domestic settings, almost the only means of obtaining liquid and powdered aerosols is a device called "aerosol packaging" or "aerosol can". The substance in it is packed under pressure and sprayed using liquefied or compressed gases.
GENERAL CHARACTERISTICS OF AEROSOLS
The properties of aerosols are determined by:
The nature of the substances of the dispersed phase and the dispersion medium;
Partial and mass concentration of aerosol;
Particle size and particle size distribution;
The shape of primary (non-aggregated) particles;
Aerosol structure;
Particle charge.
To characterize the concentration of aerosols, as well as other dispersed systems, mass concentration and numerical (partial) concentration are used.
Mass concentration - the mass of all suspended particles in a unit volume of gas.
Numerical concentration - the number of particles per unit volume of the aerosol. No matter how great the numerical concentration at the time of aerosol formation, after a few seconds it cannot exceed 10 3 particles/cm 3 .
AEROSOL PARTICLE SIZES
The minimum particle size is determined by the possibility of the existence of a substance in the state of aggregation. Thus, one molecule of water cannot form either a gas, or a liquid, or a solid. Aggregates of at least 20-30 molecules are required for phase formation. The smallest particle of a solid or liquid cannot be smaller than 1 10 -3 µm. To consider a gas as a continuous medium, it is necessary that the particle sizes be much larger than the free path of the gas molecules. The upper limit of particle size is not strictly defined, but particles larger than 100 microns are not able to remain suspended in the air for a long time.
MOLECULAR-KINETIC PROPERTIES OF AEROSOLS
Features of the molecular-kinetic properties of aerosols are due to:
A low concentration of particles of the dispersed phase - so, if 1 cm 3 of gold hydrosol contains 10 16 particles, then in the same volume of gold aerosol there are less than 10 7 particles;
The low viscosity of the dispersion medium - air, therefore, the low coefficient of friction (B) arising from the movement of particles;
The low density of the dispersion medium, hence ρ part » ρ gas.
All this leads to the fact that the movement of particles in aerosols is much more intense than in lyosols.
Let us consider the simplest case, when the aerosol is in a closed vessel (i.e., external air flows are excluded) and the particles are spherical with radius r and density p. Such a particle is simultaneously affected by the force of gravity directed vertically downwards and the force of friction in the opposite direction. In addition, the particle is in Brownian motion, the consequence of which is diffusion.
To quantify the processes of diffusion and sedimentation in aerosols, one can use the values
specific diffusion flux i diff i
specific sedimentation flow i sed. .
To find out which flow will prevail, consider their ratio:
In this expression, (p - p 0) » 0. Therefore, the size of the fraction will be determined by the size of the particles.
If r > 1 μm, then i sed » i dif, i.e., diffusion can be neglected - rapid sedimentation occurs and particles settle to the bottom of the vessel.
If r< 0,01 мкм, то i сед « i диф. В этом случае можно пренебречь седиментацией - идет интенсивная диффузия, в результате которой частицы достигают стенок сосуда и прилипают к ним. Если же частицы сталкиваются между собой, то они слипаются, что приводит к их укрупнению и уменьшению концентрации.
Thus, both very small and very large particles quickly disappear from the aerosol: the first due to adhesion to the walls or sticking together, the second - as a result of settling to the bottom. Particles of intermediate sizes have the maximum stability. Therefore, no matter how great the numerical concentration of particles at the moment of aerosol formation, after a few seconds it does not exceed 10 3 parts/cm 3 .
ELECTRICAL PROPERTIES OF AEROSOLS
The electrical properties of aerosol particles differ significantly from the electrical properties of particles in lyosol.
1. DES does not appear on aerosol particles, since due to the low dielectric constant of the gaseous medium, electrolytic dissociation practically does not occur in it.
2. The charge on the particles arises mainly due to the indiscriminate adsorption of ions, which are formed in the gas phase as a result of gas ionization by cosmic, ultraviolet or radioactive rays.
3. The charge of particles is random, and for particles of the same nature and the same size it can be different both in magnitude and in sign.
4. The charge of a particle changes in time both in magnitude and in sign.
5. In the absence of specific adsorption, the particle charges are very small and usually exceed the elementary electric charge by no more than 10 times.
6. Specific adsorption is characteristic of aerosols, the particles of which are formed by a highly polar substance, since in this case a sufficiently large potential jump occurs on the interfacial surface, due to the surface orientation of the molecules. For example, on the interfacial surface of water or snow aerosols, there is a positive electric potential of about 250 mV.
It is known from practice that aerosol particles of metals and their oxides usually carry a negative charge (Zn, ZnO, MgO, Fe 2 0 3), and aerosol particles of non-metals and their oxides (SiO 2, P 2 O 5) are positively charged. NaCl and starch particles are positively charged, while flour particles carry negative charges.
AGGREGATIVE STABILITY. COAGULATION
Unlike other dispersed systems, aerosols do not have any interaction between the surface of particles and the gaseous medium, which means that there are no forces that prevent the adhesion of particles to each other and to macroscopic bodies upon impact. Thus, aerosols are aggregatively unstable systems. Coagulation in them occurs according to the type of rapid coagulation, i.e., each collision of particles leads to their sticking together.
The rate of coagulation increases rapidly with increasing numerical concentration of the aerosol.
Regardless of the initial concentration of the aerosol, after a few minutes there are 10 8 -10 6 particles in 1 cm 3 (for comparison - in lyosols ~ 10 15 particles). Thus, we are dealing with highly diluted systems.
| Dependence of the coagulation rate on the increase in the number of aerosol concentration | |
| Initial numerical concentration in 1 cm 3 | The time required to reduce the aerosol concentration by 2 times |
| Fractions of a second | |
| 15-30 s | |
| 30 minutes | |
| Several days | |
METHODS FOR DESTROYING AEROSOLS
Despite the fact that aerosols are aggregatively unstable, the problem of their destruction is very acute. The main problems, in the resolution of which it becomes necessary to destroy aerosols:
Purification of atmospheric air from industrial aerosols;
Capturing valuable products from industrial smoke;
Artificial sprinkling or dissipation of clouds and fog.
Aerosols are broken down by
scattering under the action of air currents or due to the same charges of particles;
· sedimentation;
Diffusion to vessel walls
· coagulation;
· Evaporation of particles of the dispersed phase (in the case of aerosols of volatile substances).
Of the treatment facilities, the most ancient is the chimney. They try to release harmful aerosols into the atmosphere as high as possible, since some chemical compounds, getting into the surface layer of the atmosphere under the influence of sunlight and as a result of various reactions, turn into less dangerous substances (at the Norilsk Mining and Metallurgical Combine, for example, a three-channel pipe has a height 420 m).
However, the current concentration of industrial production requires that flue emissions undergo pre-treatment. Many methods have been developed for the destruction of aerosols, but any of them consists of two stages:
the first is the capture of dispersed particles, their separation from the gas,
the second is to prevent the re-entry of particles into the gaseous medium, this is due to the problem of adhesion of the trapped particles, the formation of a strong deposit from them.
AEROSOL CANS
The principle of operation of an aerosol can is that the drug placed in the package is mixed with an evacuating liquid, the saturated vapor pressure of which is higher than atmospheric pressure in the temperature range at which the package is operated.
The mixture is ejected from the cylinder under the action of saturated vapor pressure above the liquid.
It is known that the saturation vapor pressure of any stable substance is determined only by temperature and does not depend on volume. Therefore, during the entire time of operation of the cylinder, the pressure in it will remain constant, therefore, the range of particles and the angle of the spray cone will remain practically constant.
Depending on the nature of the interaction of the sprayed substance with the evacuating liquid and its state of aggregation, systems in aerosol packaging will consist of a different number of phases. In the case of mutual solubility of the components, a homogeneous liquid solution is formed, in other cases, an emulsion or suspension, and, finally, a heterogeneous system, when the drug and the evacuating liquid form a macroscopically heterogeneous system. Obviously, in the first case, the aerosol package contains a two-phase system - liquid and saturated vapor. When an emulsion or suspension is released into the atmosphere, only the dispersion medium is crushed - the resulting particles, at best, will have the dimensions that they had in the liquid phase.
When the drug and the evacuating liquid do not mix or mix with each other to a limited extent, and one of the liquids is dispersed in the other in the form of small droplets, emulsions are formed.
The nature of the system formed when the product exits the package into the atmosphere depends on which of the liquids is the dispersed phase. If the dispersed phase is a preparation, then an aerosol is formed. If the dispersed phase is an evacuating liquid, then foam is obtained. The size of particles obtained using aerosol cans depends on the physicochemical properties of the substances that make up the preparation, the ratio of components, the design features of the can and the temperature conditions of its operation.
The degree of dispersion can be adjusted: “by varying the size of the outlet;
By changing the pressure of the saturated vapor of the evacuating liquid;
By changing the quantitative ratio of the drug and the evacuating agent.
EVACUATE SUBSTANCES
The most important auxiliary component is a substance that ensures the release of the drug into the atmosphere and its subsequent dispersion. These substances are called propellants (Latin "pro-peilere" - to drive). The propellant must perform two functions:
Create the necessary pressure to release the drug;
Disperse the product released into the atmosphere. Freons and compressed gases are used as propellants. Freons are low molecular weight organofluorine compounds of the aliphatic series.
The following freon designation system is adopted: the last digit (the number of units) means the number of fluorine atoms in the molecule, the previous digit (the number of tens) is the number of hydrogen atoms increased by one, and the third (the number of hundreds) is the number of carbon atoms reduced by one. For example: F-22 is CHC1F 2 , F-114 is C 2 C1 2 F 4 .
Substances consisting of molecules of a cyclic structure also have a numerical designation, but the letter “C” is placed before the numbers, for example: C318 - C 4 F 8 (octafluorocyclobutane).
As compressed gases, N 2, N 2 O, CO 2, etc. are used.
ADVANTAGES OF AEROSOL PACKS
1. The transfer of the drug to a finely dispersed state occurs due to the potential energy of the liquefied propellant and does not require the use of any extraneous devices.
2. No nozzles are needed to create aerosols.
3. In a unit of time, a significant amount of a substance can be dispersed to obtain particles of small size - if other methods were used, much more energy would be required.
4. The fogging mode is stable: the size of the obtained particles, their flight range, the angle at the top of the cone during the entire period of operation change little.
5. You can fix the dosage of the sprayed substance in advance.
6. You can set the particle size.
7. The degree of polydispersity of the aerosol is low.
8. All particles have the same chemical composition.
9. The sterility of sprayed preparations is ensured.
10. The drug in the package does not come into contact with atmospheric oxygen, which ensures its stability.
11. Self-closing valve eliminates the possibility of loss due to spillage or evaporation of the unused part of the product.
12. Packaging is always ready to go.
13. The packaging is compact. Allows individual or group use.
The first aerosol packages appeared in the 80s. 20th century in Europe. During the Second World War, the United States took the lead in developing them. In 1941, an aerosol package was created - an insecticide packaged in a glass vessel. Freon-12 served as the propellant.
On an industrial scale, production began after World War II in the United States, and then in other countries of the world.
PRACTICAL APPLICATIONS OF AEROSOLS
The widespread use of aerosols is due to their high efficiency. It is known that an increase in the surface of a substance is accompanied by an increase in its activity. A small amount of a substance sprayed in the form of an aerosol occupies a large volume and has a high reactivity. This is the advantage of aerosols over other dispersed systems.
Aerosols are used:
In various fields of technology, including military and space;
In agriculture; « in healthcare;
In meteorology; in everyday life, etc.
Recently, in pharmaceutical practice, the preparation of dosage forms in the form of aerosols has been widely used. The use of medicinal substances in the form of aerosols is convenient in cases where it is necessary to act on large surfaces with the drug (acute respiratory diseases, burns, etc.). A great effect is given by dosage forms containing liquid film-forming substances in their composition. When such a drug is sprayed onto the affected area, it is covered with a thin, transparent film that replaces the bandage.
Let us dwell in more detail on the use of aerosol packaging.
Currently, there are more than 300 types of products in aerosol packages.
The first group: household chemicals.
Insecticides are preparations for the destruction of insects.
Means against moths.
Insecticides for pets.
Means for protecting indoor plants and fruit and berry crops from fungal diseases and pests.
Lacquers and paints.
Air fresheners.
c Polishing and cleaning compounds.
Second group:
Perfumes and cosmetics. « Hair care products (varnishes, shampoos, etc.).
Shaving foams and gels.
Creams for hands and feet.
Oil for and against sunburn.
Deodorants.
Perfumes, colognes, toilet water.
Third group: medical aerosols.
Fourth group: technical aerosols.
Lubricating oils.
Anti-corrosion coatings.
Protective films. dry lubricants.
Emulsions for cooling cutters on drilling machines.
Fifth group: food aerosols.
FOOD AEROSOLS
The first food containers appeared in 1947 in the USA. They contained creams for decorating cakes and pastries and were only used by restaurants who returned them to be refilled. Mass production of this type of aerosol packaging began only in 1958.
Aerosol food packaging can be divided into three main groups:
packaging requiring low temperature storage;
packaging with subsequent heat treatment;
packaging without further heat treatment.
Three types of food products are produced in aerosol packages: creams, liquids, pastes. You can buy salad dressings, processed cheese, juices, cinnamon, mayonnaise, tomato juice, 30% whipped cream, etc. in aerosol packages.
The growth in the production of food aerosols is explained by the following:
advantages over conventional types of packaging;
development of new propellants;
improvement in filling technology.
Advantages of food aerosol packaging:
the convenience of use;
saving time;
food is packaged in a state prepared for consumption and is dispensed from the package in a homogeneous form;
no leakage of products;
moisture is not lost and does not penetrate into the packaging;
aroma is not lost;
the product is kept sterile.
The following requirements are imposed on food aerosol formulations:
1. Propellants must be of high purity, non-toxic, tasteless and odorless. Currently, carbon dioxide, nitrous oxide, nitrogen, argon and C318 freon are used.
2. Compressed gases, which have a very limited solubility in aqueous solutions, cannot participate in the formation of foam, which is necessary for whipped cream, decorative creams, mousses, etc. It is preferable to use C318 freon with these products, although it is much more expensive.
Table 18.4 Example formulations of various food aerosols
| Aerosol Ingredients | Quantity, % mass | |
| 1. Whipped Cream for Snack Sandwiches | ||
| Curd with cream | 50-60 | |
| 25-30 | ||
| Vegetable oil and aromatic additives | 6-10 | |
| Freon С318 | 7 | |
| 2. Icing sugar for confectionery decoration | ||
| Sugar | 55-60 | |
| Water | 15-25 | |
| Vegetable oil | ||
| solid | 9-14 | |
| liquid | 3-5 | |
| Salt | 0,1-0,3 | |
| Microcrystalline cellulose | 1,0 | |
| fragrances | 1-4 | |
| Emulsifiers | 0,5-1 | |
| Freon С318 | 7 | |
| 3. Mousse | ||
| Honey or fruit syrup | 78-83 | |
| Water | 7-9 | |
| Vegetable oil (solid) | 3-5 | |
| Microcrystalline cellulose | 1-2 | |
| Monoglycerides | 0,5-1 | |
| Sorbitol polyesters | 0,05-1 | |
| Freon SZ18 | 7 | |
| Continuation of table 18.4 | ||
| Aerosol Ingredients | Quantity, % mass | |
| 4. Decorative sauce in the form of foam | ||
| Mustard (finely ground powder) | 0,94 | |
| Lemon juice | 4,72 | |
| Vinegar | 9,44 | |
| Water | 34 | |
| Polysorbate 80 | 0,5 | |
| emulsifying mixture | 2,25 | |
| Microcrystalline cellulose | 2,5 | |
| Additives - foam stabilizers | 4,59 | |
| Freon С318 + nitrous oxide (Р=8 atm) | 7 | |
| 5. Oil-vinegar dressing in the form of foam | ||
| Water | 11,80 | |
| Salt | 1,96 | |
| Sugar | 1,47 | |
| Wine vinegar | 22,81 | |
| Olive oil | 61,75 | |
| Polysorbate 80 | 0,10 | |
| garlic oil | 0,12 | |
| black pepper oil | 0,10 | |
| Freon С318 | 10,0 | |
| 6. Dressing for roasted corn kernels | ||
| Salt (extra) | 10,00 | |
| Vegetable oil | 58,97 | |
| Other oil additives | 0,03 | |
| Dye | 1,00 | |
| Freon-S318 | 10,00 | |
3. The use of freons gives one more advantage: liquefied gases are introduced into the formulations of products that are released in the form of foam, in an amount of not more than 10% by weight, while they occupy a relatively small volume. This allows you to load significantly more products into the cylinder - 90% of the capacity of the cylinder (in packages with compressed gas, only 50%) and guarantees the complete release of the product from the package.
4. The choice of propellant is dictated by the type of food product and the intended form of its delivery (cream, liquid, paste). Mixtures of high purity CO2 and nitrous oxide have proven themselves well. To obtain foam, mixtures of C318 freon with nitrous oxide are used. Cake decorating cream packed with this mixture produces a stable foam that retains color well. For syrups, CO2 is considered the most suitable propellant.
The quality of the dispensing of the contents from the cylinder depends on the following factors:
Product preparation technologies;
Stabilizer (microcrystalline cellulose is widely used);
Proper selection of cylinder and valve.
For cinnamon and lemon juice, a controllable spray head has been developed, which can dispense products as desired either in the form of drops or in the form of a jet. For artificial sweeteners, dosing valves are used, one dose they dispense corresponds to one piece of sawn sugar, etc.
AEROSOL TRANSPORT
Pneumatic transport is widely used in the flour-grinding, cereal, feed industry, which creates conditions for the introduction of automation, increasing labor productivity and reducing costs. However, the use of pneumatic transport is associated with a large expenditure of electricity to move a large volume of air (1 kg of air moves 5-6 kg of bulk material).
More progressive is aerosol transport, in which a high concentration of material in the air flow is achieved due to flour aeration at the beginning of transportation and high air pressure. Aeration breaks the adhesion between flour particles, and it acquires the property of fluidity, like a liquid, as a result, 1 kg of air moves up to 200 kg of flour.
The aerosol transport plant consists of a feeder, a supercharger, a material pipeline and an unloader. The main element is the feeder, in which air is mixed with the material and the mixture is given an initial speed, which ensures its supply to the material pipeline.
The introduction of aerosol transport makes it possible to increase the productivity of mills and reduce the specific power consumption.
Aerosol transport is the future not only in flour milling, but also in other industries associated with the use of bulk materials and powders.
Aerosols are microheterogeneous systems in which solid particles or liquid droplets are suspended in a gas (S/G or L/G),
According to the state of aggregation of the dispersed phase, aerosols are divided into: fog (F/G); smoke, dust (T/G); smog [(W+T)/G)].
According to dispersion, aerosols are: fog, smoke, dust.
Like other microheterogeneous systems, aerosols can be obtained from true solutions (condensation methods) or from coarse systems (dispersion methods).
Water droplets in fogs are always spherical, and particulate smoke can have different shapes depending on their origin.
Due to the very small size of the particles of the dispersed phase, they have a developed surface on which adsorption, combustion, and other chemical reactions can actively proceed.
The molecular-kinetic properties of aerosols are due to:
low concentration of particles of the dispersed phase; low viscosity of the dispersion medium; low density of the dispersion medium.
Depending on the size of the particles of the dispersed phase, they can either rapidly sediment (at r » 1 µm) or stick to the walls of the vessel or stick together (at r » 0.01 µm). Particles of intermediate sizes have the greatest stability.
Aerosols are characterized by the phenomena of thermophoresis, thermoprecipitation, photophoresis.
The optical properties of aerosols are similar to those of lyosols; however, the scattering of light by them is much more pronounced due to large differences in the refractive indices of the dispersed phase and the dispersion medium.
The specificity of the electrical properties of aerosols is that DES does not appear on the particles, the charge of the particles is random and small in magnitude. When particles approach each other, electrostatic repulsion does not occur and rapid coagulation occurs.
The destruction of aerosols is an important problem and is carried out by sedimentation, coagulation, dust collection and other methods.
Powders are highly concentrated disperse systems in which the dispersed phase is solid particles, and the dispersion medium is air or another gas. Symbol: T/G.
In powders, the particles of the dispersed phase are in contact with each other. Traditionally, most bulk materials are referred to as powders, however, in a narrow sense, the term “powders” is used for highly dispersed systems with a particle size smaller than a certain critical value at which the forces of interparticle interaction become commensurate with the mass of particles. The most common are powders with particle sizes from 1 to 100 microns. The specific interfacial surface area of such powders varies from a few minutes on September 11, 2011 (soot) to fractions of m2/g (fine sands).
Powders differ from aerosols with a solid dispersed phase (also T/G) by a much higher concentration of solid particles. The powder is obtained from an aerosol with a solid dispersed phase during its sedimentation. The suspension (S/L) also turns into powder when it is dried. On the other hand, both an aerosol and a suspension can be made from a powder.
POWDER CLASSIFICATION
1. According to the shape of the particles:
Equiaxial (have approximately the same dimensions along three axes);
Fibrous (the length of the particles is much greater than the width and thickness);
Flat (length and width are much greater than the thickness).
2. By interparticle interaction:
Connectedly dispersed (particles are linked to each other, i.e., the system has a certain structure);
Freely dispersed (shear resistance is due only to friction between particles).
3. Classification by particle size of the dispersed phase:
Sand (2≤10 -5 ≤ d ≤ 2∙10 -3) m;
Dust (2∙10 -6 ≤ d ≤ 2∙10 -5) m;
Powder (d< 2∙10 -6) м.
POWDER PRODUCTION METHODS
Powders, like any other disperse system, can be obtained by two groups of methods:
On the part of coarse-dispersed systems - by dispersion methods;
From the side of true solutions - by condensation methods.
The choice of method depends on the nature of the material, the purpose of the powder, and economic factors.
DISPERSION METHODS
Raw materials are crushed on roller, ball, vibration or colloid mills, followed by separation into fractions, since as a result of grinding polydisperse powders are obtained (for example, flour of the same grade may contain particles from 5 to 60 microns).
Efficient dispersion can be achieved by grinding highly concentrated suspensions.
To facilitate dispersion, hardness reducers are used, which are surfactants. In accordance with the rule of polarity equalization, adsorbing on the surface of the ground solid, they reduce the surface tension, reducing energy consumption during dispersion and increasing the fineness of the ground phase.
In some cases, pre-treatment of the material is carried out before dispersion. So, titanium or tantalum is heated in a hydrogen atmosphere, converting into hydrides, which are crushed and heated in a vacuum - pure metal powders are obtained.
When obtaining flake powders, which are part of paints and pyrotechnic compositions, ball mills are used for grinding. The balls flatten and roll the particles of the crushed material.
Powders with particles of spherical shape from refractory metals (tungsten, molybdenum, niobium) are obtained in low-temperature plasma of an arc and high-frequency discharge. Passing through the plasma zone, the particles melt and take on a spherical shape, then cool and solidify.
During dispersion, the chemical composition of the material does not change.
CONDENSATION METHODS
These methods can be divided into two groups.
The first group of methods is associated with the deposition of particles due to the coagulation of lyophobic sols. As a result of evaporation of the solution or partial replacement of the solvent (reduction of solubility), a suspension is formed, and after it is filtered and dried, powders are obtained.
The second group of methods is associated with chemical reactions (chemical condensation). Chemical condensation methods can be classified based on the type of reaction used:
1. Exchange reactions between electrolytes. For example, precipitated chalk (tooth powder) is obtained as a result of the reaction:
Na 2 CO 3 + CaC1 2 \u003d CaCO 3 + 2 NaCl.
2. Oxidation of metals.
For example, highly dispersed zinc oxide, which is the main component of zinc oxide, is obtained by oxidation of zinc vapor with air at 300°C.
3. Oxidation of hydrocarbons.
Various types of soot, which is used in the production of rubber, plastics, printing ink, are obtained by burning gaseous or liquid hydrocarbons with a lack of oxygen.
4. Recovery of metal oxides.
Reduction with natural gas, hydrogen, or solid reducing agents is used to produce highly dispersed metal powders.
And much more, without which life itself is unthinkable. The entire human body is a world of particles that are in constant motion strictly according to certain rules that obey human physiology. Colloidal systems of organisms have a number of biological properties that characterize a particular colloidal state: 2.2 Colloidal system of cells. From the point of view of colloid-chemical physiology...
A natural question arises: what equations describe mixtures of ideal gases? After all, we rarely encounter pure gases in nature. For example, our natural habitat - the air - is made up of nitrogen. N 2 (78,08 % ), oxygen O2 (20,95 % ), inert gases ( 0,94 % ), carbon dioxide CO2 (0,03 % ).
Let in some volume V at some temperature T contains a mixture of gases (which we will number
index i). The role of each component of the mixture will be characterized mass fraction:
where m i - weight i-th component. Our task - write an equation similar to Clapeyron's equation - Mendeleev, and deal with the effective number of degrees of freedom of a mixture, which can contain both monatomic and polyatomic molecules.
First of all, note that we are considering ideal gases. Molecules do not interact with each other, and therefore each component does not interfere with any other "live" in the same common vessel. The various gases in the vessel, by virtue of their supposed ideality, simply "do not notice" each other. Therefore, for each of the components, the same Clapeyron equation is valid - Mendeleev:
|
|
where n i - number of moles of substance in i-th component. Full number n moles in a mixture is equal to the sum of the number of moles n i in each of the components:
Similarly, the total mass of the mixture is equal to the sum of the masses of each of the components
and naturally determine the molar mass of the mixture m how mass of one mole of the mixture:
We introduce a quantity called partial pressure.
Takes place dalton's law for gas mixture:
|
The total pressure of the gas mixture is equal to the sum of all partial pressures |
Summing up the left and right sides of (1.21), we arrive at the standard form of the Clapeyron-Mendeleev equation
|
|
where m,μ, n are determined from the conditions of a specific task. For example, if the mass fractions of the components are given, then the molar mass of the mixture is found from the relation
Internal energy U i i th component of the mixture is determined in accordance with formulas (1.16) and (1.19):
|
|
On the one hand, the total internal energy of the mixture is equal to the sum of the energies of each component:
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1. Gas mixing at V=const. If the total volume occupied by the gases before and after mixing remains unchanged and the gases before mixing occupy volumes V 1, V 2, ... .. V n m 3 at pressures p 1, p 2, p n and temperatures T 1, T 2, Т n , and the ratio of heat capacities of these gases with р /с v are equal to k 1 , k 2 ,…. k n , then the parameters of the mixture are determined by the formulas:
temperature
pressure
(5.15)
For gases whose molar heat capacities are equal, and therefore the values of k are equal, formulas (62) and (63) take the form:
2. Mixing of gas streams. If the mass flow rates of the mixing flows are equal to M 1, M 2, ... M n, kg / h, volumetric flow rates are V 1, V 2, ... .. V n m 3 / h, gas pressures - p 1, p 2, p n and temperature - T 1 , T 2 ,…T n , and the ratios of heat capacities of individual gases are respectively k 1 , k 2 ,…. k n , then the temperature of the mixture is determined by the formula:
(5.18)
The volumetric flow rate of the mixture per unit time at temperature T and pressure p:
(5.19)
For gases whose k values are equal, the temperature of the mixture is determined by formula (64). If gas flows, in addition to the same values of k, also have pressures, then formulas (66) and (67) take the form:
(5.21)
Tasks
5.1. Find the change in the internal energy of 1 kg of air during its transition from the initial state t 1 \u003d 300 0 C to the final state at t 2 \u003d 50 0 C. Take the dependence of heat capacity on temperature as linear. Give your answer in kJ.
The change in internal energy is found by formula (5.9):
Du \u003d C vm (t 2 -t 1).
Using table. 4.3, we get for air
(С vm) 0 t =0.7084+0.00009349t kJ/(kg K);
(С vm) 50 300 =0.7084+0.00009349(50+300)=0.7411 kJ/(kg K).
Consequently,
Du=0.7411(50-300)= - 185.3 kJ/kg
Answer: DU = - 185.3 kJ / kg
5.2. Find the change in internal energy of 2 m 3 of air if its temperature drops from t 1 \u003d 250 0 C to t 2 \u003d 70 0 C. Accept the dependence of heat capacity on temperature as linear. Initial air pressure Р 1 =0.6 MPa.
Answer: DU=-1063 kJ.
5.3. To the gas enclosed in a cylinder with a movable piston, 100 kJ of heat is supplied from the outside. The amount of work done in this case is 115 kJ. Determine the change in the internal energy of the gas if its amount is 0.8 kg.
Answer: DU = - 18.2 kJ.
5.4. 2 m 3 of air at a pressure of 0.5 MPa and a temperature of 50 0 C are mixed with 10 m 3 of air at a pressure of 0.2 MPa and a temperature of 100 0 C. Determine the pressure and temperature of the mixture.
Answer: t cm \u003d 82 0 C; P cm \u003d 0.25 MPa.
5.5. The flue gases of three boilers, which have atmospheric pressure, are mixed in the boiler house flue. For simplicity, it is assumed that these gases have the same composition, namely: CO 2 =11.8%; O 2 \u003d 6.8%; N 2 \u003d 75.6%; H2O=5.8%. Hourly gas consumption is V 1 =7100 m 3 /h; V 2 \u003d 2600 m 3 / h; V 3 \u003d 11200 m 3 / h, and the temperatures of the gases, respectively, t 1 \u003d 170 0 C, t 2 \u003d 220 0 C, t 3 \u003d 120 0 C. Determine the temperature of the gases after mixing and their volume flow through the chimney at this temperature.
Answer: t=147 0 С; V=20900 m 3 /h.
5.6. Exhaust gases from three steam boilers at a pressure of 0.1 MPa are mixed in the gas collection duct and removed to the atmosphere through the chimney. The volumetric composition of flue gases from individual boilers is as follows: from the first
CO 2 =10.4%; About 2 \u003d 7.2%; N 2 =77.0%; H2O=5.4%;
from the second
CO 2 =11.8%; O 2 \u003d 6.9%; N 2 \u003d 75.6%; H2O=5.8%;
from the third
CO 2 =12.0%; O 2 \u003d 4.1%; N 2 \u003d 77.8%; H 2 O=6.1%.
The hourly consumption of gases is
M 1 =12000 kg/h; M 2 =6500 kg/h; M 3 =8400 kg/h; and gas temperatures, respectively, t 1 \u003d 130 0 С; t 2 \u003d 180 0 С; t 3 \u003d 200 0 C.
Determine the temperature of the flue gases after mixing in the collection duct. Assume that the molar heat capacities of these gases are the same.
Answer: t 2 \u003d 164 0 C.
5.7. Three gas streams are mixed in the gas duct, having the same pressure equal to 0.2 MPa. The first stream is nitrogen with a volume flow V 1 = 8200 m 3 / h at a temperature of 200 0 C, the second stream is carbon dioxide with a flow rate of 7600 m 3 / h at a temperature of 500 0 C and the third stream is air with a flow rate of 6400 m 3 / h at a temperature of 800 0 C. Find the temperature of the gases after mixing and their volume flow in the common gas pipeline.
Answer: t 1 \u003d 423 0 C; V=23000 m3/h.
5.8. Combustion products from the steam boiler flue in the amount of 400 kg/h at a temperature of 900 0 C must be cooled to 500 0 C and sent to the drying plant. The gases are cooled by mixing the gas stream with the air stream at a temperature of 20 0 C. The pressure in both gas streams is the same. Determine the hourly air flow if it is known that R gas \u003d R air. The heat capacity of the combustion products is assumed to be equal to the heat capacity of air.
Answer: M air \u003d 366 kg / h.
13.7. Thermal transformers
Often, for the technological process it is necessary to maintain a certain temperature.
The simplest way of such support is to burn fuel and transfer heat from hot combustion products either directly to the consumer or to an intermediate coolant. In this case, heat exchange occurs naturally from a hot source with a temperature T 1 colder with temperature T 2. With this method, it is impossible to transfer more heat than that obtained during the combustion of fuel (and due to losses, it is much less).
However, it is fundamentally possible, having a certain amount of heat q" at high temperature T 1, get more heat at a lower temperature without the cost of work T 2. To do this, it suffices to carry out a reversible direct Carnot cycle between a source with a high temperature and an environment with a temperature T With, as a result of which the work will be obtained (see (7.7)):
Having spent this work in the reverse reversible Carnot cycle between the medium with temperature T With and consumer with temperature T 2 , we will transfer to the latter the amount of heat equal to
Substituting in this expression the value of work l With from the previous expression, we get:
where the coefficient of proportionality ψ 1.2 is called heat conversion factor temperature T 1 to temperature T 2 .
Therefore, having received q" the amount of heat from a source with a temperature T 1 , can be transferred to the body with temperature T 2 amount of heat ψ 1.2 q" .
Because T 2 T 1 , then q" >q" .
For example, let t 1 \u003d 1000 about C, t 2 \u003d 50 o C, t With \u003d 0 ° C. Coefficient. Thus, in order to obtain, suppose, 5 J of heat at a temperature of 50 ° C, only 1 J of heat at 1000 ° C should be expended, while in a conventional heating installation 1 J of heat at high temperature is converted into the same amount of heat at low temperature.
Therefore, from the point of view of thermodynamics, a heating installation is 5 times less economical than a reversible heat-converting installation.
A device that allows direct and reverse cycles of heat transfer from a source with one temperature to a consumer with a different temperature is called thermotransformer.
If the required temperature is lower than the initial temperature, then the thermotransformer is called lowering.
Assistance is required to maintain a higher temperature than the original increasing thermotransformer, for which, since T 2 > T 1 .
Rice. 13.7 Fig. 13.8
A thermal transformer is a combination of a heat engine and a heat pump.
On fig. 13.7 shows a diagram of a step-down thermal transformer, and in fig. 13.8 is his theoretical cycle.
On fig. 13.9 shows a diagram of a step-up thermal transformer, and in fig. 13.10 - his theoretical cycle.
In the figures: I - heat engine, II - heat pump.
If a thermal transformer is designed to maintain temperatures both lower and higher than the original, then it is called mixed type thermotransformer.
Rice. 13.9 Fig. 13.10
test questions
How does a reverse Carnot cycle work?
What parameter evaluates the thermodynamic efficiency of a heat pump installation?
What is the difference between heat pump and refrigeration circuit diagrams?
14. Mixing of gases and vapors
In various devices, it is often necessary to deal with the mixing of various gases, vapors or liquids. In this case, it is required to determine the state parameters of the mixture from the known state parameters of the components that make up this mixture.
The solution of this problem depends on the conditions under which this mixing process is carried out. All methods for the formation of mixtures can be divided into three groups:
mixing gases at constant volume,
mixing of gas streams,
mixing gases when filling the tank.
14.1. Mixing process at constant volume
This method of mixture formation consists in the fact that several gases with pressures R 1 , R 2 , …, R n, temperatures T 1 , T 2 , …, T n and the masses G 1 , G 2 , …, G n occupy different volumes V 1 , V 2 , …, V n(Fig. 14.1).
If you remove the separating partitions between the gases, then the gases will mix, and the volume of the mixture
V = V 1 + V 2 + …+ V n ,
and the mass of the mixture
G = G 1 + G 2 + …+ G n .
When an equilibrium state is established, the parameters of the mixture will be R, v, T, u.
Since the process is adiabatic and the volume has not changed, then, in accordance with the first law of thermodynamics, the internal energy of the system is conserved:
U = U 1 + U 2 + …+ U n or Gu=G 1 u 1 + G 2 u 2 + … + G n u n .
Hence, the specific internal energy of the mixture is determined as follows:
, (14.1)
where g i- mass fraction i th gas.
And the specific volume, by its definition, is equal to
. (14.2)
Other parameters ( R, T) for real gases, vapors and liquids are found from the diagrams for these substances.
In the particular case when ideal gases with constant heat capacities are mixed, for which du= c v dT, we get
In the case when portions of the same gas are mixed, the temperature of the mixture is calculated using a simpler formula:
.
The gas pressure after mixing is determined by the Claiperon–Mendeleev equation
where R is the gas mixture constant (defined in Section 1.4).
14.2. Mixing process
In this case, the mixing of gases occurs as a result of the connection of several flows in one channel.
Let the pipeline 1 (Fig. 14.2) gas with parameters enters the mixing chamber p 1 , v 1 , T 1 , h 1 , and through the pipeline 2 – gas with parameters p 2 , v 2 , T 2 , h 2 .
Gas flow through the pipeline 1 equals G 1, through the pipeline 2 – G 2. At the inlet to the mixing chamber, these gas flows are throttled so that the pressure in the chamber R was less than R 1 and R 2 (if, for example, R > R 1 , then the gas from the mixing chamber would rush into the pipeline 1 ).
It should be emphasized that the pressure R in the mixing chamber can be selected differently (by adjusting the valves); In this respect, the process of mixing in a flow differs significantly from mixing in a constant volume, where the pressure is uniquely determined by the parameters of the mixed gases.
From the mixing chamber gas with parameters R,v, T discharged through the pipeline 3 . Gas consumption in the pipeline 3 is obviously equal to G = G 1 + G 2 .
Since the gas moves in pipelines, then, in addition to internal energy, it also has (as a whole) kinetic and potential energy. For simplicity (for most technical problems it is justified), we will assume that
pipelines are located horizontally, thus the change in potential energy can be neglected;
gas movement speeds are relatively small, i.e. the change in kinetic energy is also neglected.
Then, according to the first law for the adiabatic flow (9.3), under the above conditions, we have
From here we obtain an expression for the specific enthalpy of the mixture obtained as a result of mixing in the flow:
. (14.3)
Knowing the specific enthalpy h and pressure R gas after mixing, using state diagrams, you can find the remaining parameters of the mixture ( T, v, s and etc.).
For ideal gases, replacing the specific enthalpy by the expression With R T, we get
. (14.4)
In the case of mixing two flows of the same gas, the formula for the temperature of the mixture is simplified:
. (14.5)
Knowing the temperature determined in this way T, from the equation of state for an ideal gas, you can find the specific volume:
Formulas (14.3)–(14.5) are similarly written for an arbitrary number of mixing gas flows.
14.3. Mixing when filling the volume
Let in the tank 1 (Fig. 14.3) volume V there is a gas (steam, liquid) with a mass G 1 with options R 1 , T one . This tank is fed through a pipeline. 2 gas with parameters R 2 , v 2 , T 2 (obviously, R 2 > R 1) and weight G 2, after which the valve closes. The tank contains a mixture of gases with a volume V and weight G = G 1 + G 2. It is necessary to determine the parameters of the resulting mixture.
During the filling process, the work of pushing is done on the gas in the pipeline 2 equal to – p 2 v 2 G 2; no work takes place in the tank because the volume of the tank is constant.
In an adiabatic process, work is done due to a change in internal energy (as before, we neglect the kinetic energy of the incoming gas due to the smallness of the flow velocity):
Hence, the specific internal energy of the mixture in the vessel is equal to
The specific volume of a mixture is, by definition, equal to v = V/ G.
Knowing u and v, with the help of diagrams find the remaining parameters of the mixture ( R, T, s, h).
In the case of mixing the same ideal gas with constant heat capacities
where k is the adiabatic index.
The pressure in the tank after mixing is
Two portions of air are mixed, and the mass of the first component is 10 kg, and its temperature is 400 ° C, and the mass of the second component is 90 kg, and the temperature is 100 ° C. Determine the temperature of the mixture for various mixing methods.
Solution: The temperature of the mixture resulting from the mixing process at constant volume or the mixing process in the gas stream will be determined by the formula t = g 1 t 1 +g 2 t 2. And in our example it is t\u003d 0.1 ∙ 400 + 0.9 ∙ 100 \u003d 130 o C.
If the mixture is obtained as a result of filling the volume in which the first gas is already located, then its absolute temperature is calculated by the formula T = g 1 T 1 +kg 2 T 2. In the example under consideration, the air adiabatic index k= 1.4, and the temperature of the mixture is t\u003d 0.1 (400 +273) +1.4 ∙ 0.9 ∙ (100 +273) - 273 \u003d 264 o C.
14.4. Entropy change during mixing
The entropy of a mixture is the sum of the entropies of the constituents of this mixture, i.e.
or in specific units
Since the mixing process is an irreversible process, the entropy of the thermodynamic system (all substances participating in the adiabatic mixing) will increase in this process according to the second law of thermodynamics, i.e.
The irreversibility of the mixing process is explained by the diffusion of the mixing components accompanying this process. The increase in entropy during mixing is a measure of this irreversibility.
test questions
What are the main methods of mixing?
How is the mixture defined?
How to determine the temperature of the mixture with different mixing methods?
How can one explain the fact that with adiabatic mixing of gases or vapors, the entropy of the mixture increases?
15. Fundamentals of chemical thermodynamics
An inhomogeneous system is determined by the composition of its components. Under certain conditions, this composition can change due to the chemical and physicochemical transformations taking place in the system, under which the destruction of old and the emergence of new bonds between atoms occurs. These processes are accompanied by the release or absorption of energy as a result of the forces of these bonds.
Chemical thermodynamics considers the application of the first and second laws of thermodynamics to chemical and physicochemical phenomena.
15.1. chemical reactions
Chemical substance is a macroscopic body of a certain chemical composition, i.e. a body, in relation to which it is known not only what chemical elements and in what proportion it consists ( individual chemical), but it is also known from which compounds of chemical elements it is formed ( mixture or solution).
A chemical substance (compound) is usually characterized by a chemical formula showing what elements it consists of and in what ratio the atoms of these elements combine during its formation.
The processes of interaction between individual chemicals, leading to the formation of new substances, are called chemical reactions.
Any chemical reaction can occur in both forward and reverse directions.
In closed systems, chemical reactions occur in such a way that the total amount of each of the chemical elements present in the system does not change. For this reason, not arbitrary quantities of substances participate in chemical reactions, but their stoichiometric quantities, i.e. quantities corresponding to the chemical formulas of substances. Therefore, chemical reactions are written as equalities between the chemical formulas of the substances involved in the reaction and the chemical formulas of the products of this reaction. Let BUT 1 , BUT 2 , …, BUT n are the starting materials, and AT 1 , AT 2 , …, AT m are the end products of the reaction. Then the chemical reaction between substances BUT 1 , BUT 2 , …, BUT n, leading to the formation of substances AT 1 , AT 2 , …, AT m, will be written as an equality:
in which α 1 , α 2 , … α n, β 1 , β 2 … β m are stoichiometric coefficients. For example, combustion of methane produces carbon dioxide and water:
CH 4 + 2O 2 \u003d CO 2 + 2H 2 O.
1 is taken as a unit of the amount of a substance in chemistry. mol. This quantity contains a strictly defined number of molecules (atoms) of a given substance, equal to the Avogadro constant N A= 6.02204∙10 23 . In other words: 1 mole of a substance is defined as the amount of a substance whose mass in grams is equal to its molecular (atomic) mass M.
The composition of complex systems formed from many substances, the amount of each of which is n i moles, in chemistry is given mole fractions system component.
Let in separate thermostated vessels under the same pressure p there are gases BUT and AT taken in quantities of imole. When these vessels are connected, spontaneous mixing of gases will occur until a homogeneous composition of the gas mixture is established throughout the entire volume of the system. We will assume that the initial gases and their mixtures obey the equations of state of ideal gases. Then, while maintaining a constant total gas pressure p partial pressures of gases in the resulting mixture will be equal to
When ideal gases are mixed, there are no thermal effects, so there is no heat exchange between the gases and the thermostat, and the change in the entropy of the system will be completely determined by the irreversibility of the processes inside the system.
To find the desired change in entropy, it is necessary to counter the described spontaneous process with a mental equilibrium transition between the same initial and final states of the system.
For equilibrium mixing of gases, we use a special hypothetical device, by analogy with a thermostat called a chemostat . This device consists of a thermostatically controlled cylinder equipped with a piston that moves without friction; at the base of the cylinder there is a membrane selectively permeable only for a given individual chemical substance; the latter separates the individual substance loaded into the chemostat from the studied mixture of substances located in another vessel. Unlike a thermostat, designed to maintain a given temperature of a body immersed in it, or to heat or cool the latter in an equilibrium mode, with the help of a chemostat, a certain value of the chemical potential of a given individual substance in the mixture of substances under study is maintained, as well as an equilibrium supply and removal of a substance from mixtures. Chemical potential i -gochemical component in a chemostat is uniquely determined by the temperature T and pressure on the piston. By changing the pressure on the piston, it is possible to change the direction of the transition of a given component through the selective membrane: if is the chemical potential of the component in the mixture under study, then at , the substance will be supplied to the mixture, at , it will be removed from the mixture, and at , chemical equilibrium is maintained between the chemostat and the mixture. The quasi-equilibrium change in the composition of the mixture corresponds to the diffusion transfer of a substance through the membrane under the action of a very small difference in the values of the chemical potential on both sides of the membrane.
The chemical potential of an ideal gas, regardless of whether this gas is in an individual state or in a mixture with other ideal gases, is expressed by a simple relation, where p i is either the pressure of a pure gas or its partial pressure in the mixture. Therefore, when an ideal gas is transferred through a semipermeable membrane, the equilibrium between the mixture and the chemostat is characterized by the equality of the pressure in the chemostat and the partial pressure of the gas in the mixture.
Rice. 2.3. Equilibrium mixing of two gases using chemostats: a is the initial state of the system; b– state of the system after isothermal expansion of gases; in– final state after mixing of gases through membranes; 1 – chemostats for individual gases A and B ; 2 – semi-permeable membranes; 3 - vessel for equilibrium mixing of gases.
Equilibrium mixing of ideal gases A and B will be carried out in a thermostatically controlled system consisting of two chemostats of individual components A and B, connected to the third vessel - a collector of the resulting mixture, equipped, like chemostats, with a movable piston (Fig. 2.3).
Let at the initial moment the chemostats contain, respectively, moles of the component A and moles of the component B under the same pressure p
; the piston in the mixture collector is in the zero position (the volume of gas under the piston is zero). The mixing process is carried out in two stages. At the first stage, we perform a reversible isothermal expansion of gases A and B; while the pressure A we reduce from p
up to the set pressure and pressure B respectively from p
before . The volumes occupied by gases in the first and second chemostats will change from to and from to respectively. The work done by the expanding gas in the first chemostat is 
; in the second 
. Thus, at the first stage, the total work is performed in our hypothetical device. Since during the isothermal expansion of an ideal gas its internal energy does not change, the specified work is carried out due to the equivalent heat supply from the thermostat. Hence, the reversible change in entropy in the system will be equal to
At the second stage of the process (actual mixing), we pass gases from chemostats through selective membranes into the mixture collector by synchronized movement of three pistons. At the same time, a constant pressure is maintained on each of the pistons, respectively, both in the chemostats and in the collector, which ensures the equilibrium passage of gases through the membranes (more precisely, a pressure is created in the collector that is slightly less than p , while maintaining a non-zero driving force for diffusion through membranes). The reversibility of the mixing process in this case is ensured by the possibility of a synchronous change in the direction of movement of all three pistons, which would lead to a reverse separation of the mixture into individual components. After the operation is completed, the mixture will obviously take up volume in the collector.
Since, in the case of ideal gases, mixing is not accompanied by any thermal effect, there is no heat exchange between our device and the thermostat at the second stage of the operation, . Consequently, there is no change in the entropy of the system at this stage, .
It is useful to verify by direct calculation that the work of the gases in the second stage is zero. Indeed, work is expended on moving pistons in chemostats, while at the same time, the same amount of work is produced in the collector by gases. From here.
So, the total increase in entropy when mixing gases is determined by expression (2.9), . If, under the equilibrium variant of mixing, this increase is associated with the reverse supply of heat and the production of an equivalent amount of work 
, then with direct (irreversible) mixing of gases, the same increase in entropy occurs due to its generation inside the system; the system does not do any work.
After substitution (2.8), expression (2.9) can be rewritten as
. (2.10)
This relation is given an obligatory place in thermodynamics courses due to its seeming paradox. It is noteworthy that for the change in entropy (when mixing ideal gases!) It does not matter what is mixed with what, and also at what pressure and temperature. In essence, here is an informal derivation of (2.10).
Let us supplement the derivation (2.10) with its useful corollaries. Introducing the mole fractions of the components 
and , we obtain an expression for the change in entropy per 1 mol of the resulting mixture:
. (2.11)
The maximum of this function falls on the equimolar mixture of gases, 0.5.
From the point of view of the theory of separation of mixtures of substances, it is of interest to trace the change in the production of entropy when a sufficiently large number of moles of the component is added B to one mole of the component A. Assuming in (2.10) and , we obtain
When deriving (2.12), we used the mathematical representation of the logarithmic function
.
Formula (2.12) shows that the successive dilution of the mixture is accompanied by an infinite increase in entropy per mole of the impurity component.
Formula (2.10) gives the integral value of the entropy increment when mixing finite amounts of gas. In order to arrive at a compact differential expression similar to formula (2.7) for heat transfer, we modify the component mixing model (see Fig. 2.4). We will assume that mixing occurs through a membrane that is permeable to both components, or through a sufficiently narrow valve separating vessels filled with mixtures A and B different composition. The system is thermostatically controlled, and constant pressure is maintained in both vessels by means of pistons. p . With a limited mixing rate, the composition of the mixture in each of the vessels can be considered uniform throughout the volume of the vessel. Thus, this system is similar to a heat exchange system with a weakly conductive baffle.
