Change in gas temperature during expansion. Change in gas temperature when its volume changes

Change in gas temperature when its volume changes. Adiabatic and isothermal processes

To do this, it is necessary to study what happens to the temperature of a gas if its volume changes so quickly that there is practically no heat exchange between the gas and surrounding bodies.

Figure 7 Let's do this experiment. In a thick-walled tube made of transparent material, closed at one end, we place cotton wool, slightly moistened with ether, and this will create a mixture of ether vapor and air inside the tube, which explodes when heated. Then quickly push a tightly fitting piston into the tube (Fig. 7). We will see a small explosion occur inside the tube. This means that when a mixture of ether vapor and air is compressed, the temperature of the mixture increases sharply. By compressing a gas with an external force, we produce work, as a result of which the internal energy of the gas should increase; this is what happened - the gas heated up.

Now let's give the gas the opportunity to expand and do work against external pressure forces. Let a large bottle contain compressed air at room temperature (Fig. 8). Let's give the air in the bottle the opportunity to expand, coming out of a small hole to the outside, and place a thermometer in the stream of expanding air. The thermometer will show a temperature lower than room temperature. Consequently, when a gas expands and does work, it cools and its internal energy decreases. It is clear that heating of a gas during compression and cooling during expansion are an expression of the law of conservation of energy.

If we turn to the microcosm, the phenomena of gas heating during compression and cooling, during expansion, will become quite clear. When a molecule hits a stationary wall and bounces off it, the speed A, therefore, the kinetic energy of the molecule is on average the same as before hitting the wall. But if a molecule hits and bounces off an advancing piston, its speed and kinetic energy are greater than before it hit the piston (just as the speed of a tennis sword increases when hit in the opposite direction by a racket). The advancing piston transfers additional energy to the molecule reflected from it. Therefore, the internal energy of a gas increases during compression. When rebounding from the retreating piston, the speed of the molecule decreases, because the molecule does work by pushing the retreating piston. Therefore, the expansion of the gas associated with the retraction of the piston or layers of surrounding gas is accompanied by work and leads to a decrease in the internal energy of the gas.

So, the compression of a gas by an external force causes it to heat up, and the expansion of the gas is accompanied by its cooling. This phenomenon always occurs to some extent, but is especially noticeable when the exchange of heat with surrounding bodies is minimized, since such exchange can compensate for temperature changes to a greater or lesser extent. Processes in which there is no heat exchange with the external environment are called adiabatic.

How to ensure a constant gas temperature despite changes in its volume? Obviously, to do this, it is necessary to continuously transfer heat to the gas from the outside if it is expanding, and to continuously remove heat from it, transferring it to surrounding bodies if the gas is compressed. In particular, the gas temperature remains almost constant if the expansion or compression of the gas is very slow, and heat exchange with the external environment occurs quite quickly. With slow expansion, heat from the surrounding bodies is transferred to the gas and its temperature decreases so little that this decrease can be neglected. With slow compression, heat, on the contrary, is transferred from the gas to the surrounding bodies, and as a result its temperature increases only negligibly. Processes in which the temperature is maintained constant are called isothermal.

Boyle-Marriott law. How are volume and pressure related to each other during an isothermal change in the state of a gas? Daily experience teaches us that when the volume of a certain mass of gas decreases, its pressure increases. But how exactly does the pressure of a gas increase as its volume decreases if the temperature of the gas remains constant?

The answer to this question was given by research carried out in the 17th century by the English physicist and chemist Robert Boyle (1627 - 1691) and the French physicist Edme Mariotte (1620 - 1684).

Experiments establishing the relationship between the volume and pressure of a gas can be reproduced using a device similar to the gas thermometer shown in Fig. 5. On a vertical stand equipped with divisions, there are glass tubes A and B, connected by rubber tube C. Mercury is poured into the tubes. Tube B is open at the top, tube A has a stopcock. Let us close this valve, thus locking a certain mass of air in tube A. As long as we do not move the tubes, the level of mercury in them is the same.

This means that the pressure of the air trapped in tube A is the same as the pressure of the outside air. Let us now slowly raise tube B. We will see that the mercury in both tubes will rise, but not equally: in tube B the mercury level will always be higher than in tube A. If we lower tube B, then the mercury level in both elbows decreases , but in tube B more than in tube A.

The volume of air trapped in tube A can be calculated by the divisions of tube A. The pressure of this air will differ from atmospheric pressure by the amount of pressure of the mercury column, the height of which is equal to the difference in the levels of mercury in tubes A and B. When raising tube B, the pressure of the mercury column is added to atmospheric pressure. The volume of air in tube A decreases. When tube B is lowered, the level of mercury in it is lower than in tube A, and the pressure of the mercury column is subtracted from the atmospheric pressure, the volume of air in tube A increases accordingly.

Comparing the values of pressure obtained in this way and the volume of air locked in tube A, we will be convinced that when the volume of a certain mass of air increases by a certain number of times, its pressure decreases by the same number and vice versa. The air temperature in the tube during these experiments can be considered constant.

So, the pressure of a certain mass of gas at a constant temperature is inversely proportional to the volume of the gas(Boyle-Mariotte law).

For rarefied gases, the Boyle–Mariotte law is satisfied with a high degree of accuracy. For highly compressed or cooled gases, noticeable deviations from this law are found.

Formula expressing the Boyle–Mariotte law. Let us denote the initial and final volumes by letters V 1 And V 2 and initial and final pressure in letters p 1 And p2. Based on the results of the above experiments, we can write

p 1 / p2 = V 2 / V 1 (3) p 1 V 1=p 2 V 2 (4)

Formula (4) is another expression of the Boyle–Mariotte law. It means that for a given mass of gas, the product of the gas volume and its pressure during an isothermal process remains unchanged.

Formulas (3) and (4) can also be applied if the process of changing the volume of gas was not isothermal, but the temperature changes were such that at the beginning and at the end of the process the temperature of a given mass of gas was the same.

For rarefied gases, the Boyle–Mariotte law is satisfied with a high degree of accuracy, and provided that the temperature remains constant, the product pV for a given mass of gas can be considered strictly constant. But in the case of transition to very high pressures, a noticeable deviation from it is detected. With a gradual increase in the pressure of a certain mass of gas, the product pV at first it decreases significantly and then begins to increase, reaching values several times higher than those corresponding to a rarefied gas.

In the middle of the cylinder, closed at both ends, there is a piston (Fig. 9). The gas pressure in both halves is 750 mm Hg. Art. The piston moves so that the volume of gas on the right is halved. What is the pressure difference? (Answer: 1000 mmHg)

Two vessels with a capacity of 4.5 l and 12.5 l are connected by a tube with a tap. The first contains gas at a pressure of 20 kgf/cm2. In the second there is a small amount of gas that can be neglected. What pressure will be established in both vessels if the tap is opened? (Answer: 5.3 kgf/cm2)

In technology, graphs are often used showing the dependence of gas pressure on its volume. You can draw a graph like this for an isothermal process. We plot the gas volume along the abscissa axis, and its pressure along the ordinate axis. Let the pressure of a given mass of gas with a volume of 1 m 3 be equal to 3.6 kgf/cm 2. Based on the Boyle-Mariotte law, we calculate that with a volume of 2 m 3 the pressure is 3.6´0.5 kgf/cm 2 =
1.8 kgf/cm2. Continuing these calculations, we obtain the following table:

Table 5

V, m 3 1,2 1,5 1,8 2,3 2,7 3,5 4,5 5,5 R, kgf/cm 2 3,6 3,0 2,4 2,0 1,8 1,6 1,3 1,2 1,03 0,9 0,8 0,72 0,65 0,6

If we plot this data in the form of points, the abscissas of which are the values V, and the ordinates are the corresponding values R, we get a curved line (hyperbola) - a graph of an isothermal process in a gas.

The relationship between gas density and its pressure. The density of a substance is the mass contained in a unit volume. If, for example, the volume of gas is reduced by five times, then the density of the gas will also increase by five times. At the same time, the gas pressure will increase. If the temperature has not changed, then, as the Boyle-Mariotte law shows, the pressure will also increase fivefold. From this example it is clear that in an isothermal process, the gas pressure changes in direct proportion to its density.

If the gas density at pressures p 1 and p 2 are equal to ρ 1 and ρ 2, then we can write

ρ 1 / ρ 2 = p 1 / p2 (5)

This important result can be considered another and more significant expression of the Boyle–Mariotte law. The fact is that instead of the volume of gas, which depends on a random circumstance - on what mass of gas is chosen - formula (5) includes the density of the gas, which, like pressure, characterizes the state of the gas and does not depend at all on the random choosing its mass.

The density of hydrogen at a pressure of 1.00 kgf/cm2 and a temperature of 16 °C is 0.085 kg/m3. Determine the mass of hydrogen contained in a 20 liter cylinder if the pressure
80 kgf/cm2 and the temperature is 16 °C. ( Answer: 0.136 kg).

Molecular interpretation of the Boyle–Mariotte law. If the density of a gas changes, then the number of molecules per unit volume changes by the same factor. If the gas is not too compressed and the movement of the molecules can be considered completely independent of each other, then the number of blows N per unit time per unit surface area of the vessel wall is proportional to the number of molecules n per unit volume. Consequently, if the average speed of molecules does not change over time (in the macrocosm this means constant temperature), then the gas pressure must be proportional to the number of molecules n per unit volume, i.e. gas density. Thus, the Boyle-Mariotte law is an excellent confirmation of our ideas about the nature of gas.

However, as was said, the Boyle-Mariotte law ceases to be justified if we move to high pressures. And this circumstance can be explained, as M.V. believed. Lomonosov, based on molecular concepts.

On the one hand, in highly compressed gases the sizes of the molecules themselves are comparable to the distances between them. Thus, the free space in which the molecules move is less than the total volume of the gas. This circumstance increases the number of impacts of molecules on the wall, since it reduces the distance that a molecule must fly to reach the wall.

On the other hand, in a highly compressed and therefore denser gas, molecules are noticeably attracted to other molecules much more of the time than molecules in a rarefied gas. This, on the contrary, reduces the number of impacts of molecules on the wall, since in the presence of attraction to other molecules, gas molecules move towards the wall at a lower speed than in the absence of attraction. At not very high pressures, the second circumstance is more significant, and the product pV decreases slightly. At very high pressures, the first circumstance and product play a greater role pV increases.

So, both the Boyle-Mariotte law and deviations from it confirm the molecular theory.

Change in gas volume with temperature change. Now let's establish how a gas behaves if its temperature and volume change, but the pressure remains constant. Let's consider this experience. Let us touch with our palm a vessel in which a horizontal column of mercury traps a certain mass of air. The gas in the vessel heats up, its pressure rises, and the mercury column begins to move. The movement of the column will stop when, due to an increase in the volume of air in the vessel, its pressure becomes equal to the external one. Thus, the volume of air increased when heated, but the pressure remained unchanged.

If we knew how the temperature of the air in the vessel changed in our experiment, and measured how the volume of the gas changes, we could study this phenomenon from a quantitative perspective.

Gay-Lussac's law. A quantitative study of the dependence of gas volume on temperature at constant pressure was carried out in 1802 by the French physicist and chemist Joseph Louis Gay-Lussac (1778–1850).

Experiments have shown that the increase in gas volume is proportional to the increase in temperature. Therefore, the thermal expansion of a gas can, as for other bodies, be characterized using temperature coefficient of volumetric expansion β. It turned out that for gases this law is observed much better than for solids and liquids, so that the temperature coefficient of volumetric expansion of gases is an almost constant value even with very significant changes in temperature (while for liquids and solids this constancy is observed only approximately):

b= (V " –V) /V 0 (t " – t) (6)

The experiments of Gay-Lussac and others revealed a remarkable result. It turned out that the temperature coefficient of volumetric expansion β for all gases is the same (more precisely, almost the same) and equals 1/273 °C -1. Volume of a certain mass of gas when heated to 1 °C at constant pressure increases by 1/273 of the volume that this mass of gas had at 0 °C (Gay-Lussac's Law).

As can be seen, the temperature coefficient of volumetric expansion of gases β coincides with their temperature pressure coefficient α .

It should be noted that the thermal expansion of gases is very significant, so the volume of gas V 0 at 0 °C is noticeably different from the volume at another, for example, room temperature. Therefore, in the case of gases, it is impossible to replace the volume in formula (6) without a noticeable error V 0 volume V. In accordance with this, it is convenient to give the expansion formula for gases the following form. For the initial volume we take the volume V 0 at 0 °C. In this case, the gas temperature increment τ equal to temperature t measured on the Celsius scale. Therefore, the temperature coefficient of volumetric expansion

β = (V – V 0) /V 0 t, Þ V = V 0 (1+βt). (7) Because β = 1/273 °C -1, then V = V 0 (1+t/273). (8)

Formula (7) can be used to calculate volume at temperatures as high as
0 °C and below 0 °C. In the latter case t will have negative values. It should be borne in mind, however, that Gay-Lussac's law does not hold true when the gas is highly compressed or so cooled that it approaches a state of liquefaction. In this case, formula (8) cannot be used.

Odds match α And β , included in Charles's law and Gay-Lussac's law, is not accidental. It is easy to see that since gases obey the Boyle–Mariotte law, then α And β must be equal to each other. Indeed, let a certain mass of gas have a volume at a temperature of 0 °C V 0 and pressure p 0 . Let's heat it up to temperature t at a constant volume. Then its pressure, according to Charles’s law, will be equal to p = p 0 (1+α t). On the other hand, let us heat the same mass of gas to a temperature t at constant pressure. Then, according to Gay-Lussac's law, its volume will become equal V = V 0 (1+βt). So, a given mass of gas can have at a temperature t volume V 0 and pressure p = p 0 (1+ αt) or volume V = V 0 (1+βt) and pressure p 0 .

According to the Boyle–Mariotte law V 0 p = Vp 0, i.e.

V 0 p 0 (1+ α t) = V 0 p 0 (1+βt), Þ α = β

The volume of a balloon at 0 °C is 820 m 3 . What will be the volume of this ball if, under the influence of the Sun's rays, the gas inside it is heated to 15 °C? Neglect the change in gas mass due to its leakage from the shell and the change in its pressure. ( Answer: 865 m 3).

Clayperon–Mendeleev law: pV=RT , Where R– gas constant 8.31 J/mol´deg. This law is called the ideal gas equation of state. It was obtained in 1834 by the French physicist and engineer B. Clayperon and generalized in 1874 by D.I. Mendeleev for any mass of gas (initially Clayperon derived this equation only for 1 mole of an ideal gas substance).

pV=RT, Þ pV/T=R=const.

There are two cylinders. One contains compressed gas, the other liquefied. The pressure and temperature of both gases are the same. Determine which cylinder has accumulated more energy? And, therefore, which of the cylinders is more dangerous? Ignore the chemical properties of gases. (Answer: with liquefied gas).

Let us explain the solution to the problem with an example.

Uncontrolled depressurization of pressure vessels creates a risk of physical or chemical explosion. Let's explain this using the water-steam system.

At atmospheric pressure, water boils at 100 °C in an open container. In a closed vessel in a steam boiler, for example, water boils at 100 °C, but the steam generated presses on the surface of the water and boiling stops. In order for the water to continue to boil in the boiler, it must be heated to a temperature corresponding to the steam pressure. For example, a pressure of 6´10 5 Pa corresponds to a temperature of +169 °C,
8´10 5 Pa – +171 °C, 12´10 5 Pa – +180 °C, etc.

If, after heating the water, for example, to 189 °C, you stop supplying heat to the boiler furnace and consume steam normally, then the water will boil until the temperature drops below 100 °C. Moreover, the sooner the pressure in the boiler decreases, the more intense the boiling and steam formation will be due to the excess thermal energy contained in the water. This excess thermal energy, when the pressure drops from maximum to atmospheric, is entirely spent on vaporization. In the event of a mechanical rupture of the walls of a boiler or vessel, the internal equilibrium in the boiler is disrupted and a sudden drop in pressure to atmospheric pressure occurs.

In this case, a large amount of steam is formed (from 1 m 3 of water - 1700 m 3 of steam, at normal pressure), which leads to the destruction of the vessel and its movement due to the resulting reactive force, which causes destruction. Consequently, regardless of the operating pressure in the boiler, the danger lies not in the steam filling the steam space of the boiler, but in water heated above 100 °C, which has a huge reserve of energy and is ready to evaporate at any moment with a sharp decrease in pressure.

The volume of 1 kg of dry saturated steam (specific volume) depends on pressure: the higher the pressure, the less the volume of 1 kg of steam.

At 20 kgf/cm2, the volume occupied by 1 kg of steam is almost 900 times greater than the volume of 1 kg of water. If this steam, without changing the temperature, is compressed by 2 times, i.e. up to 40 kgf/cm 2, then its volume will also decrease by 2 times. Water cannot be compressed; it is almost incompressible.

Obviously, the same processes occur in a cylinder filled with liquefied gas. The greater the difference between the boiling point of a given gas under normal conditions and the boiling point at a given pressure in the cylinder, the higher the danger in case of mechanical damage to the integrity of the cylinder.

In this case, the danger lies not in the amount of gas pressure in the cylinder, but in the energy that was expended on liquefying the gas.

We have established how gas pressure depends on temperature if the volume remains unchanged. Now let's see how the pressure of a certain mass of gas changes depending on the volume it occupies if the temperature remains unchanged. However, before moving on to this issue, we need to figure out how to maintain the temperature of the gas constant. To do this, it is necessary to study what happens to the temperature of a gas if its volume changes so quickly that there is practically no heat exchange between the gas and surrounding bodies.

Let's do this experiment. In a thick-walled tube made of transparent material, closed at one end, we place cotton wool, slightly moistened with ether, and this will create a mixture of ether vapor and air inside the tube, which explodes when heated. Then quickly push the tightly fitting piston into the tube. We will see a small explosion occur inside the tube. This means that when the mixture of ether vapor and air was compressed, the temperature of the mixture increased sharply. This phenomenon is quite understandable. By compressing a gas with an external force, we produce work, as a result of which the internal energy of the gas should have increased; This is what happened - the gas heated up.

Now let's allow the gas to expand and do work against external pressure forces. This can be done. Let the large bottle contain compressed air at room temperature. By connecting the bottle with outside air, we will give the air in the bottle the opportunity to expand, leaving the small one. holes outward, and place a thermometer or flask with a tube in the stream of expanding air. The thermometer will show a temperature noticeably lower than room temperature, and a drop in the tube attached to the flask will run towards the flask, which will also indicate a decrease in the temperature of the air in the stream. This means that when a gas expands and at the same time does work, it cools and its internal energy decreases. It is clear that heating of a gas during compression and cooling during expansion are an expression of the law of conservation of energy.

If we turn to the microcosm, the phenomena of gas heating during compression and cooling during expansion will become quite clear. When a molecule hits a stationary wall and bounces off it, the speed, and therefore the kinetic energy of the molecule, is on average the same as before hitting the wall. But if a molecule hits and rebounds from an advancing piston, its speed and kinetic energy are greater than before it hit the piston (just as the speed of a tennis ball increases when it is hit in the opposite direction with a racket). The advancing piston transfers additional energy to the molecule reflected from it. Therefore, the internal energy of a gas increases during compression. When rebounding from the retreating piston, the speed of the molecule decreases, because the molecule does work by pushing the retreating piston. Therefore, the expansion of the gas associated with the retraction of the piston or layers of surrounding gas is accompanied by work and leads to a decrease in the internal energy of the gas.
So, compression of a gas by an external force causes it to heat up, and expansion of the gas is accompanied by its cooling. This phenomenon always occurs to some extent, but I notice it especially sharply when the exchange of heat with surrounding bodies is minimized, because such an exchange can, to a greater or lesser extent, compensate for the change in temperature.

Processes in which the transfer of heat is so negligible that it can be neglected are called adiabatic.

Let's return to the question posed at the beginning of the chapter. How to ensure constant gas temperature despite changes in its volume? Obviously, to do this, it is necessary to continuously transfer heat to the gas from the outside if it is expanding, and to continuously remove heat from it, transferring it to surrounding bodies if the gas is compressed. In particular, the temperature of the gas remains fairly constant if the expansion or compression of the gas is very slow, and the transfer of heat from outside or outside can occur with sufficient speed. With slow expansion, heat from the surrounding bodies is transferred to the gas and its temperature decreases so little that this decrease can be neglected. With slow compression, heat, on the contrary, is transferred from the gas to the surrounding bodies, and as a result its temperature increases only negligibly.

Processes in which the temperature is maintained constant are called isothermal.

When we are dealing not with a gas, but with a solid or liquid body, we do not have such direct methods at our disposal for determining the speed of the molecules of the body. However, even in these cases there is no doubt that with increasing temperature the speed of movement of molecules increases.

Change in gas temperature when its volume changes. Adiabatic and isothermal processes.

So, the compression of a gas by an external force causes it to heat up, and the expansion of the gas is accompanied by its cooling. This phenomenon always occurs to some extent, but I notice it especially sharply when the exchange of heat with surrounding bodies is minimized, because such an exchange can, to a greater or lesser extent, compensate for the change in temperature.

Processes in which the transfer of heat is so negligible that it can be neglected are called adiabatic.

Processes in which the temperature is maintained constant are called isothermal.

Boyle's Law - Mariotte

Let us now move on to a more detailed study of the question of how the pressure of a certain mass of gas changes if its temperature remains unchanged and only the volume of the gas changes. We have already found out that such an isothermal process is carried out under the condition that the temperature of the bodies surrounding the gas is constant and the volume of the gas changes so slowly that the temperature of the gas at any moment of the process does not differ from the temperature of the surrounding bodies.

We thus pose the question: how are volume and pressure related to each other during an isothermal change in the state of a gas? Daily experience teaches us that when the volume of a certain mass of gas decreases, its pressure increases. An example is the increase in elasticity when inflating a soccer ball, bicycle or car tire. The question arises: how exactly does the pressure of a gas increase with a decrease in volume if the temperature of the gas remains unchanged?

The answer to this question was given by research carried out in the 17th century by the English physicist and chemist Robert Boyle (1627-1691) and the French physicist Eden Marriott (1620-1684).

Experiments establishing the relationship between the volume and pressure of a gas can be reproduced: on a vertical stand equipped with divisions, there are glass tubes A and B, connected by a rubber tube C. Mercury is poured into the tubes. Tube B is open at the top, and tube A has a tap. Let us close this valve, thus locking a certain mass of air in tube A. As long as we do not move the tubes, the level of mercury in both tubes is the same. This means that the pressure of the air trapped in tube A is the same as the pressure of the surrounding air.

Let us now slowly raise tube B. We will see that the mercury in both tubes will rise, but not equally: in tube B the level of mercury will always be higher than in A. If we lower tube B, then the level of mercury in both elbows decreases, but in tube B the decrease is greater than in A.

The volume of air locked in tube A can be calculated by the divisions of tube A. The pressure of this air will differ from atmospheric pressure by the pressure of the mercury column, the height of which is equal to the difference in the levels of mercury in tubes A and B. When. As the tube is raised, the pressure of the mercury column is added to atmospheric pressure. The volume of air in A decreases. When tube B is lowered, the level of mercury in it is lower than in A, and the pressure of the mercury column is subtracted from the atmospheric pressure; the volume of air in A increases accordingly.

Let's do this experiment. In a thick-walled tube made of transparent material (plexiglass or glass), closed at one end, we place cotton wool, slightly moistened with ether, and this will create a mixture of ether vapor and air inside the tube, which explodes when heated. Then quickly push a tightly fitting piston into the tube (Fig. 378). We will see a small explosion occur inside the tube. This means that when the mixture of ether vapor and air was compressed, the temperature of the mixture increased sharply. This phenomenon is quite understandable. By compressing a gas with an external force, we produce work, as a result of which the internal energy of the gas should increase; This is what happened - the gas heated up.

Rice. 378. By quickly pushing the piston into a thick-walled glass tube, we cause the highly flammable cotton wool inside the tube to flare up.

Now let's give the gas the opportunity to expand and do work against external pressure forces. This can be done, for example, like this (Fig. 379). Let the large bottle contain compressed air at room temperature. Let's give the air in the bottle the opportunity to expand, coming out of a small hole to the outside, and place a thermometer or a flask with a tube, shown in Fig., in the stream of expanding air. 384. The thermometer will show a temperature lower than room temperature, and a drop in the tube connected to the flask will run towards the flask, which will also indicate a decrease in the temperature of the air in the stream. This means that when a gas expands and at the same time does work, it cools and its internal energy decreases). It is clear that heating of a gas during compression and cooling during expansion are an expression of the law of conservation of energy.

Rice. 379. Thermometer 2, placed in a stream of expanding air, shows a lower temperature than thermometer 1

Let us return to the question posed at the beginning of the paragraph. How to ensure constant gas temperature despite changes in its volume? Obviously, to do this, it is necessary to continuously transfer heat to the gas from the outside if it is expanding, and to continuously remove heat from it, transferring it to surrounding bodies if the gas is compressed. In particular, the gas temperature remains almost constant if the expansion or compression of the gas is very slow, and heat exchange with the external environment occurs quite quickly. With slow expansion, heat from the surrounding bodies is transferred to the gas and its temperature decreases so little that this decrease can be neglected. With slow compression, heat, on the contrary, is transferred from the gas to the surrounding bodies, and as a result its temperature increases only negligibly. Processes in which the temperature is maintained constant are called isothermal.

In production processes involving the use of gases (dispersion, mixing, pneumatic transport, drying, absorption, etc.), the movement and compression of the latter occurs due to the energy imparted to them by machines, which have the general name compression. At the same time, the productivity of compression plants can reach tens of thousands of cubic meters per hour, and the pressure varies within the range of 10–8–10 3 atm, which determines a wide variety of types and designs of machines used to move, compress and rarefy gases. Machines designed to create high pressures are called compressors, and machines that work to create a vacuum are called vacuum pumps.

Compression machines are classified mainly according to two criteria: the principle of operation and the degree of compression. Compression ratio is the ratio of the final gas pressure at the outlet of the machine R 2 to initial inlet pressure p 1 (i.e. p 2 /p 1).

According to the principle of operation, compression machines are divided into piston, vane (centrifugal and axial), rotary and jet.

According to the degree of compression, they are distinguished:

– compressors used to create high pressures, with a compression ratio R 2 /R 1 > 3;

– gas blowers used to move gases with high resistance of the gas pipeline network, while 3 > p 2 /p 1 >1,15;

– fans used to move large quantities of gas during p 2 /p 1 < 1,15;

– vacuum pumps that suck gas from a space with reduced pressure (below atmospheric) and pump it into a space with increased (above atmospheric) or atmospheric pressure.

Any compression machines can be used as vacuum pumps; deeper vacuums are created by piston and rotary machines.

Unlike droplet liquids, the physical properties of gases are functionally dependent on temperature and pressure; the processes of movement and compression of gases are associated with internal thermodynamic processes. At small differences in pressure and temperature, changes in the physical properties of gases during their movement at low speeds and pressures close to atmospheric are insignificant. This makes it possible to use all the basic provisions and laws of hydraulics to describe them. However, when deviating from normal conditions, especially at high gas compression ratios, many hydraulic positions undergo changes.

Thermodynamic fundamentals of the gas compression process

The influence of temperature on the change in gas volume at constant pressure, as is known, is determined by the Gay-Lussac law, i.e., when p= const the volume of a gas is directly proportional to its temperature:

Where V 1 and V 2 – volumes of gas, respectively, at temperatures T 1 and T 2 expressed on the Kelvin scale.

The relationship between gas volumes at different temperatures can be represented by the relationship

, (4.1)

Where V And V 0 – final and initial volumes of gas, m3; t And t 0 – final and initial gas temperature, °C; β t– relative coefficient of volumetric expansion, deg. -1 .

Change in gas pressure depending on temperature:

, (4.2)

Where R And R 0 – final and initial gas pressure, Pa;β R– relative temperature coefficient of pressure, degrees. -1 .

Gas mass M remains constant when its volume changes. If ρ 1 and ρ 2 are the densities of two temperature states of the gas, then
And
or
, i.e. The density of a gas at constant pressure is inversely proportional to its absolute temperature.

According to the Boyle-Mariotte law, at the same temperature the product of the specific volume of gas v on the value of its pressure R there is a constant quantity pv= const. Therefore, at constant temperature
, A
, i.e. gas density is directly proportional to pressure, since
.

Taking into account the Gay-Lussac equation, we can obtain a relationship connecting three parameters of a gas: pressure, specific volume and its absolute temperature:

. (4.3)

The last equation is called Clayperon equations. In general:

or
, (4.4)

Where R– gas constant, which represents the work done per unit mass of an ideal gas in an isobaric ( p= const) process; when the temperature changes by 1°, the gas constant R has the dimension J/(kgdeg):

, (4.5)

Where l R– specific work of volume change performed by 1 kg of ideal gas at constant pressure, J/kg.

Thus, equation (4.4) characterizes the state of an ideal gas. At gas pressure above 10 atm, the use of this expression introduces an error into the calculations ( pv≠RT), therefore it is recommended to use formulas that more accurately describe the relationship between pressure, volume and temperature of a real gas. For example, with the van der Waals equation:

, (4.6)

Where R= 8314/M– gas constant, J/(kg K); M– molecular mass of gas, kg/kmol; A And V - values that are constant for a given gas.

Quantities A And V can be calculated using critical gas parameters ( T cr and R cr):

;
. (4.7)

At high pressures the value a/v 2 (additional pressure in the van der Waals equation) is small compared to the pressure p and it can be neglected, then equation (4.6) turns into the equation of state of a real Dupre gas:

, (4.8)

where is the value V depends only on the type of gas and does not depend on temperature and pressure.

In practice, thermodynamic diagrams are more often used to determine the parameters of a gas in its various states: T–S(temperature–entropy), p–i(dependence of pressure on enthalpy), p–V(dependence of pressure on volume).

Figure 4.1 – T–S diagram

On the diagram T–S(Fig. 4.1) line AKV represents a boundary curve that divides the diagram into separate regions corresponding to certain phase states of the substance. The region located to the left of the boundary curve is the liquid phase, and to the right is the region of dry vapor (gas). In the area bounded by the curve AVK and the abscissa axis, two phases coexist simultaneously - liquid and vapor. Line AK corresponds to complete condensation of steam, here the degree of dryness x= 0. Line KV corresponds to complete evaporation, x = 1. The maximum of the curve corresponds to the critical point K, in which all three states of matter are possible. In addition to the boundary curve, the diagram shows lines of constant temperatures (isotherms, T= const) and entropy ( S= const), directed parallel to the coordinate axes, isobars ( p= const), lines of constant enthalpies ( i= const). Isobars in the wet vapor region are directed in the same way as isotherms; in the region of superheated steam they change direction steeply upward. In the region of the liquid phase, the isobars almost merge with the boundary curve, since liquids are practically incompressible.

All gas parameters on the diagram T–S referred to 1 kg of gas.

Since, according to the thermodynamic definition
, then the heat of change of state of the gas
. Consequently, the area under the curve describing the change in state of the gas is numerically equal to the energy (heat) of the change in state.

The process of changing gas parameters is called the process of changing its state. Each gas state is characterized by parameters p,v And T. During the process of changing the state of the gas, all parameters can change or one of them can remain constant. Thus, a process occurring at a constant volume is called isochoric, at constant pressure – isobaric, and at constant temperature – isothermal. When, in the absence of heat exchange between the gas and the external environment (heat is not removed or supplied), all three parameters of the gas change ( p,v,T) V the process of its expansion or contraction , the process is called adiabatic, and when changes in gas parameters occur with continuous supply or removal of heat – polytropic.

With changing pressure and volume, depending on the nature of heat exchange with the environment, the change in the state of the gas in compression machines can occur isothermally, adiabatically and polytropically.

At isothermal In the process, the change in the state of the gas follows the Boyle–Mariotte law:

pv = const.

On the diagram p–v this process is depicted by a hyperbola (Fig. 4.2). Work 1 kg gas l graphically represented by the shaded area, which is equal to
, i.e.

or
. (4.9)

The amount of heat that is released during isothermal compression of 1 kg of gas and which must be removed by cooling so that the gas temperature remains constant:

, (4.10)

Where c v And c R are the specific heat capacities of gas at constant volume and pressure, respectively.

On the diagram T–S process of isothermal compression of gas from pressure R 1 to pressure R 2 is represented by a straight line ab, drawn between isobars R 1 and R 2 (Fig. 4.3).


Figure 4.2 – Process of isothermal gas compression on the diagram	Figure 4.3 – Process of isothermal gas compression on the diagram T–S

Heat equivalent to the work of compression is represented by the area limited by the extreme ordinates and the straight line ab, i.e.

. (4.11)

Figure 4.4 – Gas compression processes on the diagram
:

A – adiabatic process;

B – isothermal process

Since the expression for determining the work expended in the isothermal compression process includes only volume and pressure, then within the limits of applicability of equation (4.4) it does not matter which gas will be compressed. In other words, the isothermal compression of 1 m 3 of any gas at the same initial and final pressures requires the same amount of mechanical energy.

At adiabatic In the process of gas compression, a change in its state occurs due to a change in its internal energy, and consequently, temperature.

In general form, the equation of the adiabatic process is described by the expression:

, (4.12)

Where
– adiabatic index.

Graphically (Fig. 4.4) this process is shown in the diagram p–v will be depicted as a steeper hyperbola than in Fig. 4.2., since k> 1.

If we accept

, That
. (4.13)

Because the
And R= const, the resulting equation can be expressed differently:

or
. (4.14)

By means of appropriate transformations, it is possible to obtain dependencies for other gas parameters:

;
. (4.15)

Thus, the temperature of the gas at the end of its adiabatic compression

. (4.16)

Work done by 1 kg of gas under conditions of an adiabatic process:

. (4.17)

The heat released during adiabatic compression of a gas is equivalent to the work expended:

Taking into account relations (4.15), the work on gas compression during an adiabatic process

. (4.19)

The process of adiabatic compression is characterized by a complete absence of heat exchange between the gas and the environment, i.e. dQ = 0, a dS = dQ/T, That's why dS = 0.

Thus, the process of adiabatic gas compression occurs at constant entropy ( S= const). On the diagram T–S this process will be represented by a straight line AB(Fig. 4.5).

Figure 4.5 – Representation of gas compression processes on the diagram T–S

If during the compression process the heat released is removed in less quantity than is necessary for an isothermal process (which happens in all real compression processes), then the actual work expended will be greater than during isothermal compression and less than during adiabatic:

, (4.20)

Where m– polytropic index, k>m>1 (for air m
).

Polytropic index value m depends on the nature of the gas and the conditions of heat exchange with the environment. In compression machines without cooling, the polytropic index may be greater than the adiabatic index ( m>k), i.e. the process in this case proceeds along a superadiabatic path.

The work spent on rarefaction of gases is calculated using the same equations as the work on compressing gases. The only difference is that R 1 will be less than atmospheric pressure.

Polytropic compression process gas pressure R 1 up to pressure R 2 in Fig. 4.5 will be depicted as a straight line AC. The amount of heat released during polytropic compression of 1 kg of gas is numerically equal to the specific work of compression:

Final gas compression temperature

. (4.22)

Power, spent by compression machines on compression and rarefaction of gases depends on their performance, design features, and heat exchange with the environment.

Theoretical power expended on gas compression
, is determined by the productivity and specific work of compression:

, (4.23)

Where G And V– mass and volumetric productivity of the machine, respectively;
– gas density.

Therefore, for various compression processes the theoretical power consumption is:

; (4.24)

; (4.25)

, (4.26)

Where – volumetric productivity of the compression machine, reduced to suction conditions.

The actual power consumed is greater for a number of reasons, i.e. The energy consumed by the machine is higher than that which it transfers to the gas.

To evaluate the effectiveness of compression machines, a comparison of this machine with the most economical machine of the same class is used.

Refrigerated machines are compared with machines that would compress the gas isothermally under given conditions. In this case, the efficiency is called isothermal,  from:

, (4.27)

Where N– actual power consumed by this machine.

If the machines operate without cooling, then the gas compression in them occurs along a polytrope, the index of which is higher than the adiabatic index ( m k). Therefore, the power expended in such machines is compared with the power that the machine would expend during adiabatic gas compression. The ratio of these powers is the adiabatic efficiency:

. (4.28)

Taking into account the power lost to mechanical friction in the machine and taking into account the mechanical efficiency. –  fur, power on the shaft of the compression machine:

or
. (4.29)

Engine power is calculated taking into account efficiency. the engine itself and efficiency transmission:

. (4.30)

The installed engine power is taken with a margin (
):

. (4.31)

The value  hell ranges from 0.930.97;  from, depending on the degree of compression, has a value of 0.640.78; mechanical efficiency varies within 0.850.95.

Change in gas temperature during expansion. Change in gas temperature when its volume changes

Change in gas temperature when its volume changes. Adiabatic and isothermal processes

Thermodynamic fundamentals of the gas compression process