Application of the equation of state of an ideal gas. Air flow during immersion How many times will the air flow change

Tasks

Decision.

Decision.

Examples

20 liter oxygen cylinder is under pressure
  10 MPa at 15 ºС. After the consumption of part of the oxygen, the pressure dropped to 7.6 MPa, and the temperature dropped to 10 ºС.

Determine the mass of oxygen consumed.

From the characteristic equation (2.5)

Therefore, before oxygen was consumed, its mass consisted of

kg

and after using up

kg

Thus, oxygen consumption

ΔM \u003d M 1 –M 2  \u003d 2.673 - 2.067 \u003d 0.606 kg.

Determine the density and specific volume of carbon monoxide With  at a pressure of 0.1 MPa at a temperature of 27 ºС.

The specific volume is determined from the characteristic equation (2.6)

m 3 / kg .

Carbon Monoxide Density (1.2)

kg / m 3.

In the cylinder with a movable piston is oxygen at
t\u003d 80 ºС and discharge (vacuum) equal to 427 hPa. At constant temperature, oxygen is compressed to overpressure
p huts\u003d 1.2 MPa. Barometric pressure IN\u003d 933 hPa.

How many times will the volume of oxygen decrease?

Answer:V 1 / V 2 \u003d22,96.

In a room with an area of \u200b\u200b35 m 2 and a height of 3.1 m, the air is at t\u003d 23 ºС and barometric pressure IN\u003d 973 hPa.

How much air will enter the room from the street if the barometric pressure rises to IN\u003d 1013 hPa. The air temperature remains constant.

Answer:  M \u003d5.1 kg .

In a vessel with a volume of 5 m 3 there is air at a barometric pressure IN\u003d 0.1 MPa and a temperature of 300 ºС. Then the air is pumped out until a vacuum pressure of 80 kPa is formed in the vessel. The air temperature after pumping remains the same.

How much air is pumped out? What will be the pressure in the vessel after pumping, if the remaining air is cooled to a temperature t\u003d 20 ºС?

Answer:2.43 kg of air were pumped out. After cooling the air, the pressure will be 10.3 kPa.

A fan of 130,000 m 3 / h of air is supplied to the air heater of the steam boiler at a temperature of 30 ºС.

Determine the volumetric flow rate of air at the outlet of the air heater, if it is heated to 400 ° C at constant pressure.

Answer:V\u003d 288700 m 3 / h.

How many times will the gas density in the vessel change if at a constant temperature the pressure gauge decreases from p 1  \u003d 1.8 MPa to p 2  \u003d 0.3 MPa?

Take the barometric pressure equal to 0.1 MPa.

Answer:

In a vessel with a volume of 0.5 m 3 there is air at a pressure of 0.2 MPa and a temperature of 20 ºС.

How much air must be pumped out of the vessel so that the vacuum in it is 56 kPa, provided that the temperature in the vessel does not change? Atmospheric pressure by a mercury barometer is 102.4 kPa at a mercury temperature in it equal to 18 ºС. The vacuum in the vessel was measured with a mercury vacuum gauge at a mercury temperature of 20 ° C.

Answer: M\u003d 1.527 kg.

Often it is necessary to solve problems in which not individual gases, but their mixtures are considered. When mixing chemically non-interacting gases having different pressures and temperatures, it is usually necessary to determine the final state of the mixture. In this case, two cases are distinguished (table 1).

Table 1

Gas mixing *

	Temperature, K	Pressure, Pa	Volume, m 3 (volumetric flow, m 3 / h)
Gas mixing at V \u003d const
Gas flow mixing **
* - all equations related to gas mixing are derived under the condition that there is no heat exchange with the environment; ** - if mass expenses ( M 1, M 2, ... M nkg / h) miscible flows are equal.

Here k i  Is the ratio of the heat capacities of gases (see formula (4.2)).

By gas mixtures is meant a mechanical mixture of several gases that do not chemically interact with each other. The composition of the gas mixture is determined by the amount of each of the gases entering the mixture, and can be set by mass m i  or voluminous r iin shares:

m i \u003d M i / M; r i \u003d V i / V, (3.1)

where M i  - weight icomponent

V i  - partial or reduced volume i-th component;

M, V  - mass and volume of the entire mixture, respectively.

It's obvious that

M 1 + M 2 + ... + M n \u003d M; m 1 + m 2 + ... + m n \u003d1, (3.2)

V 1 + V 2 + ... + V n \u003d V ;r 1 + r 2 + ... + r n \u003d1, (3.3)

The relationship between the pressure of the gas mixture r  and partial pressure of individual components p iincluded in the mixture is set dalton's law

Compression of air in a vessel immersed in water

Consider the following situation. An empty open glass bottle is immersed in water to a depth of h.

1. Explain why when the bottle is immersed upside down, air comes out of it with bubbles and the bottle is filled with water (Fig. 46.1).

2. Why does the bottle sink immediately?

3. Explain why when the bottle is immersed upside down, air does not come out of it (Fig. 46.2).

4. Explain why, when the bottle is immersed upside down, the air volume in it decreases with increasing depth.

Denote the density of water ρ in, the internal volume of the bottle V 0, the volume of air contained in it V air, atmospheric pressure p a. We assume that the air temperature in the bottle remains constant.

5. Explain why, when immersing the bottle to a depth h, the equation

V air (p a + ρ in gh) \u003d V 0 p a. (1)

6. How many times will the air volume in the bottle decrease when immersed to a depth of 10 m?

7. How does the force of Archimedes acting on a bottle with air change with increasing depth?

8. Explain why in this case, when the Archimedes force is found, the volume of the body immersed in water should be considered equal to the total volume of glass and air in the bottle.

At a certain depth of immersion, the force of Archimedes will become equal to the force of gravity. When immersed to even greater depths, the force of Archimedes will be less than the force of gravity, so the air bottle will begin to sink.

We pose the question: can gravity acting on air be neglected compared to gravity acting on a bottle?

9. How many times is the mass of air contained in a half-liter bottle less than the mass of the bottle? Take the weight of the bottle equal to 0.5 kg; air density at 20 ºС is approximately 1.2 kg / m 3.

So, we see that the mass of air in the bottle can be neglected with good accuracy compared to the mass of the bottle.

Denote the density of the glass ρ s and the volume of the glass V s.

10. Explain why, when a bottle of air completely immersed in water is in equilibrium, the following equation holds:

ρ with V with g \u003d ρ in g (V air + V s). (2)

Equations (1) and (2) can be considered as a system of two equations with two unknowns. For example, if the values \u200b\u200bof all quantities included in these equations are known, except for Vzd and h, they can be found using these equations.

11. The open bottle containing air at atmospheric pressure is lowered upside down in water. Bottle capacity is 0.5 l, glass volume is 0.2 l, glass density is 2.5 times higher than water density, atmospheric pressure is 100 kPa.
  a) What is the volume of air in a bottle equal to when a bottle immersed in water is in equilibrium?
  b) At what depth will the bottle be?

In the situation considered, the mass of air can be neglected, because at a pressure close to atmospheric, the density of air is much less than the density of water and solids.

But in cases when it comes to lifting goods from great depths using compressed air, the mass of compressed air can be significant.

Consider an example.

12. Researchers of the deep ocean discovered a sunken treasure chest at a depth of 1 km. The mass of the chest is 2.5 tons, the volume is 1 m 3. The chest was tied with a rope to a sturdy, empty, waterproof bag and began to pump air into the bag until it began to emerge with the chest. To simplify the calculations, we assume that the density of sea water is equal to the density of fresh water. We assume that the water is incompressible, and the volume of the bag shell is negligible. The water temperature at great depths can be considered close to 0 ºС.
  a) Is atmospheric pressure necessary to determine the air pressure in the bag?
  b) Let ρ denote the density of water, m s and m in the mass of the chest and the mass of air in the bag, V s and V in the volume of the chest and the volume of air at the beginning of the ascent, M in - molar mass of air, T - absolute temperature of the water. Write down a system of two equations with two unknowns (m in and V in), assuming that atmospheric pressure can be neglected.
  c) What is the volume of air in the bag at the moment when the bag with the chest began to float?
  d) What is the mass of air in the bag when the bag with the chest began to float?
  e) Is it possible not to let air out of the bag until the bag with the chest floats to the surface?

Air in mercury column tube

There is air in a glass tube sealed at one end. This air is separated from atmospheric air by a column of mercury of length l Hg (Fig. 46.3).

Let us consider how the length of the tube-filled part of the air depends on the position of the tube and the air temperature in it. We assume that the length of the tube is large enough so that mercury does not spill out of the tube at any position.

Let us denote atmospheric pressure p a, mercury density ρ rt, and the length of the tube-filled part of the tube when it is horizontal, let l 0.
  First, we assume that the air temperature in the tube is constant.

13. Write down the equation that relates the quantities l pt, l 0 and the length l of the air-filled part of the tube when it is located:
  a) vertically open end up;
  b) vertically open end down.

14. At the initial moment, the tube is open end down. When it was turned upside down, the length of the tube-filled portion of the tube decreased by 10%. What is the length of the column of mercury, if the atmospheric pressure is 760 mm Hg. Art.?

Let us now consider the case when the temperature in the cabin changes.

15. At the initial moment, the tube with air and a column of mercury is located horizontally. When it was lowered into boiling water with its open end up, the length of the tube-filled portion of the tube increased by 20%. What is the initial air temperature in the tube equal to if the length of the column of mercury is 5 cm? Atmospheric pressure is 760 mm Hg. Art.

2. Two gases in a cylinder with a piston or baffle

The cylinder is located horizontally

Let us first consider the case when a cylinder with different gases is located horizontally (in figure 46.4 different gases are schematically indicated by different colors). In this case, the weight of the piston may not be taken into account.

  A piston can have various properties that must be taken into account when solving problems.

16. What can be said about the pressure and temperature of two gases separated by a piston if it:
  a) heat-conducting and can move without friction?
  b) does not conduct heat, but can move without friction?
  c) heat-conducting, but it is necessary to take into account friction between the piston and the walls of the vessel?

17. In a horizontal cylinder with a piston, hydrogen and oxygen are located on opposite sides of the piston.
  a) What is the relationship between the volumes of gases and the amount of substance in them, if the piston is movable and heat-conducting?
  b) What is the ratio of the volumes and masses of gases in this case?
  c) How are the volumes, masses and temperatures of the gases connected if the piston is movable but does not conduct heat?

If it is said that the vessel is not separated by a piston, but by a partition, then it is understood that the volumes of the parts of the vessel remain constant. The partition can also have various properties.

18. What can be said about the temperature and partial pressure of two gases separated by a partition, if it:
  a) thermally conductive?
  b) porous (this usually means that the molecules of one gas can penetrate the septum, and the molecules of another gas cannot)?

19. A thermally insulated vessel is divided by a porous partition into two equal parts. At the initial moment, 2 moles of helium are on the left side of the vessel, and 1 mole of argon is on the right side. The initial temperature of helium is 300 K, and the initial temperature of argon is 600 K. Helium atoms can freely penetrate through the pores in the septum, and argon atoms cannot.
  a) Does it matter if the partition conducts heat or not?
  b) What gas atoms at the initial moment have a higher average kinetic energy? How many times bigger?
  c) The internal energy of which gas at the initial moment is greater? How many times more?
d) Explain why the average kinetic energies of atoms of different gases are equal after reaching thermal equilibrium.
  e) What temperature will be in the vessel under thermal equilibrium?
  f) How many times the average kinetic energy of helium atoms in thermal equilibrium will be greater than their average kinetic energy in the initial state?
  g) How will the pressure of helium in the left part of the vessel change compared to the initial one after equilibrium is established?
  h) How will the argon pressure change from the initial one after equilibrium is established?
  i) The pressure in which part of the vessel will be greater after equilibrium is established? How many times more?

The cylinder is located vertically

If the cylinder is located vertically (Fig. 46.5), then the weight of the piston, which presses on the gas located in the lower part of the cylinder, must be taken into account. Because of this, the pressure in the lower part of the cylinder is greater than in the upper. Consider an example.

20. A vertically arranged cylindrical vessel of height l is divided by a movable piston into two parts. In the upper part with a height of l in there are ν moles of helium, and in the lower part with a height of l n there are as many moles of hydrogen. The gas temperature remains the same all the time T. The mass of the piston m, area S, and the thickness of the piston can be neglected in comparison with the height of the vessel.
  a) Express the pressure in each part of the vessel through other quantities. Does the type of gas in the parts of the vessel matter for this?
  b) Write an equation relating the gas pressure in each part of the vessel to the mass of the piston and its area.
  c) What is the mass of the piston equal if l \u003d 50 cm, ν \u003d 0.22 mol, T \u003d 361 K, l in \u003d 30 cm?
  Hint. Use the ideal gas equation of state.

Balloon lift

A balloon (Fig. 46.6) can be in equilibrium in the air only if the Archimedes force acting on it from the air side is equal in absolute value to the total gravity acting on the balloon and the load suspended from it:

F A \u003d \u200b\u200bF t.sh + F t.gr. (3)

  In the case of a balloon, Archimedes silt is equal to the weight of the surrounding air in the volume occupied by the balloon and the load. We highlighted the word “surrounding” in italics, because the density of atmospheric air during rise varies for two reasons: firstly, its pressure decreases, and secondly, its temperature decreases.

We denote the volume of the ball V. The volume of the load and the shell of the ball are usually neglected in comparison with the volume of the ball itself, but the mass of the load and the shell of the ball are of great importance! The mass of the cargo is m gr, and the mass of the shell is m r. Then

F t.sh \u003d (m int + m about) g,

where m int is the mass of gas with which the ball is filled.

We denote the density of the air surrounding the ball ρ externally, and the density of the gas inside the ball ρ int.

21. Explain why the following equations are true:

F A \u003d \u200b\u200bρ ext. GV,
  m int \u003d ρ int V,
  V (ρ ext - ρ int) \u003d m gr + m vol. (4)

Hint. Use equation (3) and the relationship between mass, volume, and density.

The underground power of a balloon is the weight of the load that this balloon can lift.

22. Explain why the module of the lifting force of the balloon is expressed by the formula

F under \u003d Vg (ρ out - ρ int) - m about g. (5)

It follows from formulas (4) and (5) that a balloon can lift a load only if the density of the gas with which the balloon is filled is less than the density of the surrounding air.

If the ball were rigid, this could be achieved by partially pumping air out of it: a rigid shell would be able to withstand the difference in air pressure inside and outside the ball. However, the shell of a hard ball would be too heavy. The soft shell, which is always used for balloons, can not withstand any significant pressure difference. Therefore, the gas pressure inside the ball is equal to the pressure of the surrounding air.

23. Explain why if the pressure inside the ball is equal to the pressure of the surrounding air, then the equality

ρ int / ρ ext \u003d (M ext * T ext) / (M ext * T ext). (6)

Hint. Use the ideal gas equation of state.

From formula (6) it can be seen that the density of the gas with which the ball is filled can be made less than the density of the surrounding air in two ways:
  - use heated air as an “internal” gas;
  - use gas with a lower molar mass.

The first method is used for walking balloons (Fig. 46.6), and the second - for meteorological probes (Fig. 46.7), which rise to a high height (in this case, the balloon is usually filled with helium).

24. Explain why from formulas (5) and (6) it follows that the module of the lifting force of a balloon is expressed by the formula

  ? 25. The balloon with a volume of 3000 m 3 has a hole in the lower part through which the air inside the balloon is heated by the burner to a temperature of 77 ºС. The ball is in equilibrium at a height where the ambient temperature is 7 ºС and its density is 1.2 kg / m 3. The mass of the shell of the ball is 300 kg. What is the mass of the cargo?

Additional questions and tasks

26. In the pontoon lying on the bottom of the lake at a depth of 90 m, air is pumped from above (Fig. 46.8). At an atom, water is displaced from the pontoon through an opening located in its lower part. How much atmospheric air must be supplied to the pontoon so that it can lift the load if the total mass of the pontoon with the load is 20 tons, and the total volume of the cargo and the walls of the pontoon is 5 m 3? Accept that the water temperature is close to 0 ºС, and atmospheric pressure is 10 5 Pa.

27. An air column 30 cm high is located in the sealed elbow of the U-shaped tube. The mercury in both elbows is at the same level. What will be the height of the air column if you slowly add mercury to the top? The pressure is equal to normal atmospheric pressure.

28. A helium-filled ball is in equilibrium in the air. The mass of one square meter of the shell of the ball is 50 g, the temperature of air and helium is 27 ºС, the pressure is equal to normal atmospheric pressure. What is the radius of the ball?

Atmospheric air and its properties. The layer of air surrounding the globe is called the atmosphere. The higher the earth’s surface, the lower the air density.

Atmospheric air is a mixture of gases. One liter of it weighs 1.29 g at atmospheric pressure and a temperature of 15 ° C.

The composition of the air (by volume) includes nitrogen-78.13%, oxygen - 20.90%, carbon dioxide - 0.03%, argon - 0.94%. In addition, there is a small amount of helium, hydrogen and other inert gases in the air.

In addition to these gases, air contains water vapor, the amount of which is variable.

Nitrogen  - Under normal conditions, a gas neutral to the body. It is colorless, has no smell and taste, does not burn and does not support burning. One liter of nitrogen weighs 1.25 g, its density is 0.967. About one liter of nitrogen is dissolved in the human body under normal atmospheric pressure.

Oxygen  - The most important gas for humans. Without it, life on Earth is impossible. Oxygen does not burn, but supports combustion. In its pure form, it is flammable. One liter of oxygen weighs 1.43 g. Pure medical oxygen (98.99%) is used for breathing.

Carbon dioxide  - the heaviest of all gases. One liter of it weighs 1.96 g. The density is 1.529 g. At a partial pressure of 0.03 atm, which corresponds to 3% in air, carbon dioxide has a toxic effect on the body.

Atmospheric pressure measurement. Air weighs on the earth and objects on it. The first to determine atmospheric pressure was the Italian scientist Toricelli (in the 17th century). To do this, he used a long glass tube with a cross-sectional area of \u200b\u200b1 cm 2, sealed at one end and filled with mercury.

Having lowered the unsealed end of the tube into an open vessel with mercury, he noticed that the latter in the tube dropped only to a certain level. She did not go lower, since this was prevented by the air pressure on the mercury in the vessel. When measured, it turned out that the height of the mercury column in the tube was 760 mm, and its weight was 1,033 kg (Fig. 2). Thus, it was determined that at the surface of the earth at sea level, atmospheric pressure is 760 mm Hg. Art., which corresponds to a pressure with a force of 1.033 kg per 1 cm 2 or 10.33 m of water. Art. This pressure is called atmospheric, normal or barometric and is indicated by atm. This is a physical atmosphere.

Fig. 2. Atmospheric air pressure

In practice, for the convenience of calculations, the technical atmosphere is accepted as the unit of pressure, which equals the pressure of 1 kg per 1 cm 2 area. It is designated at.

Water pressure on a diver. We have already said above that, plunging under water, a person experiences not only the pressure of atmospheric air, but also water. When diving for every 10 m, the pressure increases by 1 at. This pressure is called excess and is denoted by ati.

Total (absolute) pressure of water and air on a diver. Underwater, both the atmospheric and the overpressure of the water column act on the diver.

Their total pressure is called absolute pressure and is indicated by ata. For example, at a depth of 10 m, a diver is under pressure of 2 atm (1 ati + 1 atm), at a depth of 50 m - 6 atm, etc.

Gas compressibility and elasticity. Gases are composed of particles in continuous motion. Gas molecules of negligible size, but occupy a large volume. The force of attraction between individual gas molecules is much less than in liquids or solids. Gases do not have a constant volume and take the form and volume of the vessel in which they are located.

In contrast to liquids, gases are able to expand, and to compress under pressure, while reducing their volume and increasing elasticity.

The relationship between the volume and pressure of gases is established by the Boyle-Mariotte law, which states that the volume occupied by a gas varies inversely with the pressure acting on it at a constant temperature. The product of the gas volume (V) and the corresponding pressure (P) at a constant temperature does not change PхV \u003d const.

For example, if you take 2 liters of gas under a pressure of 2 atm and change this pressure, then the volume will change as follows:

In other words, how many times the pressure increases, the gas volume decreases by the same amount, and vice versa.

The significance of this law is (of practical importance. It explains why the air flow for breathing increases with depth (immersion. If the diver spends 30 liters of atmospheric air per minute on the surface, then at a depth of 20 m this air is compressed to 3 atm, which already corresponds to 90 liters of air consumption actually tripled.

Using this law, it is possible to make the necessary calculations related to diving descents.

Calculation example:

Determine how many liters of compressed air a diver gets under a pressure of 4 atm according to a manometer if he is supplied with 150 liters of free air per minute?

According to the Boyle-Mariotte law, P1 V1 \u003d P2 V2.

In the example

These calculations are valid only for a constant temperature. In practice, it is necessary to take into account changes in volume and pressure at various temperatures. The dependence of the volume and pressure of air on its temperature is determined by the Gay-Lussac law, which states that a change in gas volume at constant pressure is directly proportional to the heating temperature. The change in gas pressure at a constant volume is also directly proportional to the heating temperature.

Accurate calculation of diving air is the second most important factor after the impeccable technical condition of the equipment. Since this problem has been standing since the moment of scuba inventions, special methods for calculating the required volume of air have long been developed. The volume of air required by one diver per minute is taken as the basis and then the resulting value is divided by the volume of gas in the cylinder.

These calculations are complicated by the fact that air consumption depends on physical activity. With quiet swimming, it is much less than with intensive work with flippers. Another factor that is also always taken into account is the depth of the dive. The greater the depth, the more air must be supplied under high pressure. All factors considered can be represented by a list:

1. Cylinder volume.
2. Cylinder pressure.
3. Air consumption per minute (denoted as RMV)
4. Immersion depth.

The first two parameters can be very accurate. Their accuracy depends only on how much they correspond to the indicated volume, as well as how precisely the valve on the pump that was used for refueling is adjusted. The compressor is switched off at the end of the charge by the pressure sensor. It is from responsible for the fact that the volume of air in the cylinder exactly matches the declared.

The hardest thing to calculate is RMV. Accurate data can only be obtained empirically. This is exactly what they do when training divers. The student remembers the readings of the manometer at various diving modes, drift with the flow, ascent or standing still. Further, based on the data obtained, an individual indicator of RMV is derived. Data is recorded in a table with three columns: immersion time and depth, and cylinder pressure according to the pressure gauge. Having recalculated the pressure in the cylinder by the volume (you just need to multiply the indicators) we get the exact value of air consumption per minute and derive corrections for the load and depth.

If there is no time for such measurements requiring trial dives with an instructor, then general indicators are taken. They are calculated with some margin, which is necessary to cover all individual characteristics. So the air consumption on the surface by a diver weighing 80 kg is 20 - 25 l / min. (Actually, a little less - 16 - 22 liters). Women have even less air consumption. Further adjustment for depth is made. With an increase in immersion depth, the volume of required air grows very quickly. At 50 meters (the maximum depth for amateur diving) it is already needed almost twice as much (about 40 l / min.).

For different mixtures, the maximum inhalation pressure is different. For oxygen, it is only 1.3 - 1.4 atm. For this reason, special mixtures are required for deep sea diving. When compiling, they try to make the oxygen content in them slightly different from the natural in ordinary air. The nitrogen content in the deep-sea mixture is also reduced, since if you use ordinary air, nitrogen anesthesia begins already at 30 meters. The helium-oxygen mixture is optimal for the deepest dives. In amateur diving, it is almost never used. Helium gas stationing is hindered by the fact that it is characterized by ultra-high permeability, however, in the mixture with oxygen this disadvantage is almost leveled.

When using clean air, it also matters where the tank was refilled. The main requirement here is one. Clean air is needed. Therefore, it is better with an electric drive. Then the risk of carbon monoxide and excess carbon dioxide is minimal. It is optimal that the filling of cylinders is carried out in an environmentally friendly place, for example, on the seashore or in the countryside.

Theme of the lesson: Gas laws. The laws of hydrostatics and hydrodynamics.

Gas is one of the aggregate states of matter in which its particles move freely, uniformly filling the space available to them. They exert pressure on the envelope enclosing this space. The gas density at normal pressure is several orders of magnitude lower than the density of the liquid.

The laws of gas dynamics

• Boyle-Marriott Law (Isothermal Process)
• Charles's Law (Isochoric Process) and Gay-Lussac (Isobaric Process)
• Dalton's law
• Henry's Law

• Pascal's Law
• Archimedes Law
• Euler-Bernoulli Law

Boyle-Marriott Law (Isothermal Process)

• For a given mass of gas M at a constant temperature T, its volume V is inversely proportional to pressure P: PV \u003d const, P 1 V 1 \u003d P 2 V 2, P 1 and P 2 are the initial and final pressure values, V 1 and V 2 are the initial and final pressure value.
• Conclusion - How many times the pressure increases, so many times the volume decreases.
• Using this law, you can understand how many times with increasing depth the air consumption for breathing of an underwater swimmer increases, as well as calculate the time spent under water.
• Example: V cylinder \u003d 15l, P cylinder \u003d 200, Bar V lung \u003d 5l, D depth \u003d 40m How long does the cylinder last at this depth? If a person takes 6 breaths per minute? 15x200 \u003d 3000l of air in the cylinder, 5x6 \u003d 30l / min - air flow per minute on the surface. At a depth of 40m, P abs \u003d 5 bar, 30x5 \u003d 150 l / min at a depth. 3000/150 \u003d 20min. Answer: enough air for 30 minutes.

Charles's Law (Isochoric Process) and Gay-Lussac (Isobaric Process)

• For a given mass of gas M at constant volumeV  the pressure is directly proportional to the change in its absolute temperature T: P 1 xT 1 \u003d P 2 xT 2
• For a given mass of gas M at constant pressure P  the gas volume varies in direct proportion to the change in the absolute temperature T: V 1 xT 1 \u003d V 2 xT 2
• Absolute temperature is expressed in degrees Kelvin. 0 ° С \u003d 273 ° К, 10 ° С \u003d 283 ° К, -10 ° С \u003d 263 ° К
• Example: Assume that the cylinder was filled with compressed air at a pressure of 200 bar, after which the temperature rose to 70 ° C. What became the air pressure inside the cylinder? P 1 \u003d 200, T 1 \u003d 273, P 2 \u003d ?, T 2 \u003d 273 + 70 \u003d 343, P 1 xT 1 \u003d P 2 xT 2, P 2 \u003d P 2 xT 2 / T 1 \u003d 200 × 343/273 \u003d 251 bar

Dalton's law

• The absolute pressure of the gas mixture is equal to the sum of the partial (partial) pressures of the individual gases that make up the mixture.
• The partial pressure of the gas P g is proportional to the percentage n of the gas and the absolute pressure P abs of the gas mixture and is determined by the formula: P g \u003d P ABC n / 100. This law can be illustrated by comparing a mixture of gases in a confined space with a set of weights of various weights placed on a scale. It is obvious that each of the weights will exert pressure on the scale, regardless of the presence of other weights on it.

Henry's Law

• The amount of gas dissolved in a liquid is directly proportional to its partial pressure.  If the partial pressure of the gas increases by two, then the amount of dissolved gas increases by two. When the swimmer dives, P abs increases, therefore the amount of gas inhaled by the swimmer becomes larger and, accordingly, it dissolves in the blood in a larger amount. When it comes up, the pressure decreases and the gas dissolved in the blood comes out in the form of bubbles, as when opening a bottle of sparkling water. This mechanism underlies the CST.

The laws of hydrostatics and hydrodynamics

For water, as well as for gases, due to their fluidity, Pascal's law is fulfilled, which determines the ability of these media to transmit pressure. For a body immersed in a liquid, the Archimedes law is fulfilled, due to the action on the surface of the body of the pressure created by the liquid due to its weight (i.e., the action of gravity). For moving liquids and gases, the Euler-Bernoulli law is valid.

Pascal's Law

Pressure on the surface of a liquid (or gas) produced by external forces is transmitted by the liquid (or gas) equally in all directions.

The effect of this law underlies the work of all kinds of hydraulic apparatus and devices, including scuba gear (cylinders - gearbox - breathing machine)

Archimedes Law

Every body immersed in a liquid (or gas) is acted upon by the liquid (or gas), directed upward, applied to the center of gravity of the displaced volume and equal in magnitude to the weight of the liquid (or gas) displaced by the body.

Q= yV

at – specific gravity of liquid;

V- the volume of water displaced by the body (submerged volume).

Archimedes' law defines such qualities of bodies immersed in a fluid as buoyancy and stability.

Euler-Bernoulli Law

The pressure of the flowing liquid (or gas) is greater in those sections of the flow in which the speed is lower, and vice versa, in those sections in which the speed is higher, the pressure is less .