The force acting on the piston from the side of the gas. Work in thermodynamics. Internal energy. First law of thermodynamics. adiabatic process

>>Physics: Work in thermodynamics

As a result of what processes can internal energy change? You already know that there are two kinds of such processes: doing work and transferring heat. Let's start with work. What is it equal to during compression and expansion of gas and other bodies?
Work in mechanics and thermodynamics. IN mechanics work is defined as the product of the modulus of force, the modulus of displacement of the point of its application, and the cosine of the angle between them. When a force acts on a moving body, work is equal to the change in its kinetic energy.
IN the movement of the body as a whole is not considered, we are talking about the movement of parts of a macroscopic body relative to each other. As a result, the volume of the body can change, and its velocity remains equal to zero. Work in thermodynamics is defined in the same way as in mechanics, but it is not equal to a change in the kinetic energy of a body, but to a change in its internal energy.
Change in internal energy when doing work. Why does the internal energy of the body change when a body contracts or expands? Why, in particular, does the air heat up when you inflate a bicycle tire?
The reason for the change in gas temperature during its compression is as follows: during elastic collisions of gas molecules with a moving piston, their kinetic energy changes. So, when moving towards the gas molecules, the piston transfers part of its mechanical energy to them during collisions, as a result of which the gas heats up. The piston acts like a football player kicking a flying ball. The foot gives the ball a speed that is much greater than that which it had before the impact.
Conversely, if the gas expands, then after colliding with the receding piston, the velocities of the molecules decrease, as a result of which the gas cools. The same applies to the footballer, in order to reduce the speed of the flying ball or stop it - the footballer's foot moves away from the ball, as if giving way to it.
During compression or expansion, the average potential energy of interaction of molecules also changes, since the average distance between molecules changes in this case.
Work calculation. Let us calculate the work depending on the change in volume using the example of gas in a cylinder under a piston ( fig.13.1).

The easiest way is to first calculate not the work of the force acting on the gas from the side of the external body (piston), but the work that the gas pressure force does, acting on the piston with the force. According to Newton's third law . The modulus of force acting from the side of the gas on the piston is equal to , Where p is the pressure of the gas, and S is the surface area of ​​the piston. Let the gas expand isobarically and the piston is displaced in the direction of the force by a small distance . Since the gas pressure is constant, the work done by the gas is:

This work can be expressed in terms of a change in the volume of the gas. Its initial volume V 1 \u003d Sh 1, and the final V 2 \u003d Sh 2. That's why

where is the change in the volume of gas.
When expanding, the gas does positive work, since the direction of the force and the direction of movement of the piston coincide.
If the gas is compressed, then formula (13.3) for the work of the gas remains valid. But now , and therefore (fig.13.2).

Job A, performed by external bodies on the gas, differs from the work of the gas itself A´ only sign: , since the force acting on the gas is directed against the force and the displacement of the piston remains the same. Therefore, the work of external forces acting on the gas is equal to:

When gas is compressed, when , work external force turns out to be positive. This is how it should be: when a gas is compressed, the directions of the force and the displacement of the point of its application coincide.
If the pressure is not maintained constant, then during expansion the gas loses energy and transfers it to the surrounding bodies: the rising piston, air, etc. The gas is then cooled. When a gas is compressed, on the contrary, external bodies transfer energy to it and the gas heats up.
Geometric interpretation of the work. work A´ gas for the case of constant pressure can be given a simple geometric interpretation.
We construct a graph of the dependence of gas pressure on the volume occupied by it ( fig.13.3). Here is the area of ​​the rectangle abdc, limited by schedule p1=const, axis V and segments ab And cd, equal to the gas pressure, is numerically equal to the work (13.3):

In general, the gas pressure does not remain constant. For example, in an isothermal process, it decreases inversely with volume ( fig.13.4). In this case, to calculate the work, you need to divide general change volume into small parts and calculate the elementary (small) work, and then add them all up. The work of the gas is still numerically equal to the area of ​​the figure bounded by the dependence graph p from V, axis V and segments ab And cd, equal to the pressures p1, p2 in the initial and final states of the gas.

???
1. Why do gases heat up when compressed?
2. Positive or negative work is done by external forces during the isothermal process shown in Figure 13.2?

G.Ya.Myakishev, B.B.Bukhovtsev, N.N.Sotsky, Physics Grade 10

If you have corrections or suggestions for this lesson,

AND historical reference.

1) M.V. Lomonosov, having carried out coherent reasoning and simple experiments, came to the conclusion that “the cause of heat lies in the internal movement of particles of bound matter ... It is very well known that heat is excited by movement: hands warm up from mutual friction, wood lights up, sparks fly out when silicon strikes steel, iron glows when its particles are forged with strong blows »

2) B. Rumford, working at a cannon factory, noticed that when drilling a cannon barrel, it gets very hot. For example, he placed a metal cylinder weighing about 50 kg in a box of water and, drilling the cylinder with a drill, brought the water in the box to a boil in 2.5 hours.

3) Davy made an interesting experiment in 1799. Two pieces of ice, when rubbing one against the other, began to melt and turn into water.

4) The ship's doctor Robert Mayer in 1840, while sailing to the island of Java, noticed that after a storm the water in the sea is always warmer than before it.

Work calculation.

In mechanics, work is defined as the product of the modules of force and displacement: A=FS. When considering thermodynamic processes, the mechanical movement of macrobodies as a whole is not considered. The concept of work here is associated with a change in the volume of the body, i.e. moving parts of the macrobody relative to each other. This process leads to a change in the distance between the particles, and also often to a change in the speed of their movement, therefore, to a change in the internal energy of the body.

Let there be gas in a cylinder with a movable piston at a temperature T 1 (Fig.). We will slowly heat the gas to a temperature T 2. The gas will expand isobarically and the piston will move from position 1 into position 2 distance Δ l. In this case, the pressure force of the gas will do work on external bodies. Because p= const, then the pressure force F = PS also constant. Therefore, the work of this force can be calculated by the formula A=F Δ l=PS Δ l=p Δ V, A=p Δ V

where ∆ V- change in gas volume. If the volume of the gas does not change (isochoric process), then the work done by the gas is zero.

Why does the internal energy of a body change during contraction or expansion? Why does a gas heat up when compressed and cool when it expands?

The reason for the change in gas temperature during compression and expansion is the following: during elastic collisions of molecules with a moving piston, their kinetic energy changes.

• If the gas is compressed, then upon collision, the piston moving towards the molecules transfers part of its mechanical energy to the molecules, as a result of which the gas heats up;

• If the gas expands, then after colliding with the receding piston, the velocities of the molecules decrease. as a result of which the gas is cooled.

During compression and expansion, the average potential energy of interaction of molecules also changes, since the average distance between molecules changes in this case.

The work of external forces acting on the gas

• When a gas is compressed, whenΔ V = V 2 - V 1 < 0 , A>0, the directions of force and displacement are the same;
• When expanding, whenΔ V = V 2 - V 1 > 0 , A<0, направления силы и перемещения противоположны.

Let's write the Clapeyron-Mendeleev equation for two gas states:

pV 1 = m/M*RT 1 ; pV 2 =m/M* RT 2 ⇒

p(V 2 − V 1 )= m/M*R(T 2 − T 1 ).

Therefore, in an isobaric process

A= m/M*RΔ T.

If m = M(1 mol of ideal gas), then at Δ Τ = 1 K we get R = A. Hence follows physical meaning of the universal gas constant: it is numerically equal to the work done by 1 mole of an ideal gas when it is heated isobarically by 1 K.

Geometric interpretation of the work:

If the process is not isobaric (Fig. b), then the curve p = f(V) can be represented as a broken line consisting of a large number of isochores and isobars. The work on isochoric sections is equal to zero, and the total work on all isobaric sections will be equal to the area of ​​the shaded figure. In an isothermal process ( T= const) the work is equal to the area of ​​the shaded figure shown in figure c.

In thermodynamics, the movement of particles of a macroscopic body relative to each other is considered. friend. When work is done, the volume of the body changes. The speed of the body itself remains zero, but speed

Rice. 1. A' = p∆V

body molecules change! Therefore, the temperature also changes.body. The reason is that when colliding with a moving piston (gas compression), the kinetic energy of the molecules changes - the piston gives up part of its mechanical energy. When colliding with a receding piston (expansion), the velocities of the molecules decrease, the gas cools. When work is done in thermodynamics, the state of macroscopic bodies changes: their volume and temperature.

The gas in the vessel under the piston acts on the piston with a force F' = pS , Where p - gas pressure, S - area of ​​the piston. If the piston moves, the gas does work. Let us assume that the gas expands at a constant pressure p. Then strength F' , with which the gas acts on the piston, is also constant. Let the piston move a distance ∆x(Fig. 1). The work of the gas is: A’ = F’ ∆x = pS∆x = p∆V . is the gas work under isobaric expansion. If V 1 And V 2 - the initial and final volume of gas, then for the operation of the gas we have: A' = p(V2 − V1) . When expanding, the work done by the gas is positive. When compressed, it is negative. Thus: A' = pΔV is the work of the gas. A= - pΔV is the work of external forces.

In the isobaric process, the area under the graph in coordinates p, V is numerically equal to the work (Fig. 2). External work on the system is equal to the work of the system, but with the opposite sign A = - A'.

In an isochoric process, the volume does not change, therefore , no work is done in an isochoric process! A=0

Any body (gas, liquid or solid) has energy, even if the body has no speed and is on the Earth. This energy is called domestic, it is due to the chaotic (thermal) motion and interaction of the particles that make up the body. Internal energy consists of kinetic and potential energy of particles of translational and oscillatory motions of microparticles of the system. The internal energy of a monatomic ideal gas is determined by the formula: The internal energy of a body can only change as a result of its interaction with other bodies. Exists two ways to change internal energy: heat transfer and mechanical work(e.g. heating by friction or compression, cooling by expansion).
Heat transfer - this is a change in internal energy without doing work: energy is transferred from more heated bodies to less heated bodies. There are three types of heat transfer: conduction(direct exchange of energy between randomly moving particles of interacting bodies or parts of the same body); convection(energy transfer by liquid or gas flows) and radiation(transfer of energy by electromagnetic waves). The measure of the transferred energy during heat transfer is quantity of heat (Q).
These methods are quantitatively combined in law of energy conservation , which for thermal processes reads as : the change in the internal energy of a closed system is equal to the sum of the amount of heat transferred to the system and the work of external forces performed on the system., Where ΔU - change in internal energy, Q - the amount of heat transferred to the system, A - the work of external forces. If the system itself does work, then it is conventionally denoted A' . Then the law of conservation of energy for thermal processes, which is called first law of thermodynamics , can be written like this: the amount of heat transferred to the system is used to perform work by the system and change its internal energy).
Consider Application first law of thermodynamics to isoprocesses occurring with an ideal gas.

In an isothermal process, the temperature is constant, therefore, the internal energy does not change. Then the equation of the first law of thermodynamics will take the form: Q = A' , i.e., the amount of heat transferred to the system goes to doing work during isothermal expansion, which is why the temperature does not change.

In the isobaric process, the gas expands and the amount of heat transferred to the gas goes to increase its internal energy and to do work: Q \u003d ΔU + A '

In an isochoric process, the gas does not change its volume, therefore, no work is done by it, i.e. A = 0 . Equation I of the law has the form Q=ΔU (the transferred amount of heat goes to increase the internal energy of the gas).

The process is called adiabatic. flowing without heat exchange with surrounding bodies. An example of a heat-insulated vessel is a thermos. For an adiabatic process Q=0 , therefore, during expansion, the gas does work by reducing its internal energy, therefore, the gas cools, A' = - ΔU . If you force the gas to do a sufficiently large amount of work, then you can cool it very much. It is on this that the methods of liquefying gases are based. Conversely, in the process of adiabatic compression, A'< 0 , That's why ∆U > 0 : The gas is heated. Adiabatic air heating is used in diesel engines to ignite fuel

Almost all real processes involve heat transfer: adiabatic processes are a rare exception.

Illustrative examples of adiabatic processes:

1. There are drops of water in a vessel closed with a cork with a pump hose threaded through it. After forcing a certain amount of air into the vessel, the cork quickly flies out and fog is observed in the vessel (Fig.).
2. There is a small amount of fuel in the cylinder closed by the movable piston. When the piston is pressed quickly, the fuel ignites.

« Physics - Grade 10 "

What processes can change the internal energy?
How is work defined in mechanics?

Work in mechanics and thermodynamics.

IN mechanics work is defined as the product of the modulus of force, the modulus of displacement of the point of its application and the cosine of the angle between the vectors of force and displacement. When a force acts on a moving body, the work of this force is equal to the change in its kinetic energy.

Work in thermodynamics is defined in the same way as in mechanics, but it is equal not to the change in the kinetic energy of the body, but to the change in its internal energy.

Change in internal energy when doing work.

Why does the internal energy of a body change when it contracts or expands? Why, in particular, does the air heat up when you inflate a bicycle tire?

The reason for the change in gas temperature during its compression is as follows: during elastic collisions of gas molecules with a moving piston, their kinetic energy changes.

During compression or expansion, the average potential energy of interaction of molecules also changes, since the average distance between molecules changes in this case.

So, when moving towards gas molecules, the piston transfers part of its mechanical energy to them during collisions, as a result of which the internal energy of the gas increases and it heats up. The piston acts like a football player kicking a ball flying at him. The football player's foot gives the ball a speed that is much greater than that which he had before the impact.

Conversely, if the gas expands, then after colliding with the receding piston, the velocities of the molecules decrease, as a result of which the gas cools. The same applies to the footballer, in order to reduce the speed of the flying ball or stop it - the footballer's foot moves away from the ball, as if giving way to it.

We calculate the work of the force acting on the gas from the side of the external body (piston), depending on the change in volume, using the example of gas in a cylinder under the piston (Fig. 13.1), while the gas pressure is maintained constant. First, we calculate the work that the gas pressure force does, acting on the piston with the force ". If the piston rises slowly and evenly, then, according to Newton's third law, = ". In this case, the gas expands isobarically.

The module of the force acting from the side of the gas on the piston is F "= pS, where p is the gas pressure, and S is the surface area of ​​the piston. When the piston is lifted a small distance Δh = h 2 - h 1, the work of the gas is:

A" \u003d F "Δh \u003d pS (h 2 - h 1) \u003d p (Sh 2 - Sh 1). (13.2)

The initial volume occupied by the gas, V 1 = Sh 1, and the final V 2 = Sh 2. Therefore, it is possible to express the work of the gas through a change in volume ΔV \u003d (V 2 - V 1):

A" \u003d p (V 2 - V 1) \u003d pΔV\u003e 0. (13.3)

When expanding, the gas does positive work, since the direction of the force and the direction of movement of the piston coincide.

If the gas is compressed, then formula (13.3) for the work of the gas remains valid. But now v2< V 1 , и поэтому А < 0.

The work A performed by external bodies on the gas differs from the work A "of the gas itself only in sign:

A \u003d -A "\u003d -pΔV. (13.4)

When gas is compressed, when ΔV \u003d V 2 - V 1< 0, работа внешней силы оказывается положительной. Так и должно быть: при сжатии газа направления силы и перемещения точки её приложения совпадают.

If the pressure is not maintained constant, then during expansion the gas loses energy and transfers it to the surrounding bodies: the rising piston, air, etc. The gas cools down. When a gas is compressed, on the contrary, external bodies transfer energy to it and the gas heats up.

Geometric interpretation of the work. The work A" of a gas for the case of constant pressure can be given a simple geometric interpretation.

At constant pressure, the graph of the dependence of gas pressure on the volume it occupies is a straight line parallel to the abscissa axis (Fig. 13.2). It is obvious that the area of ​​the rectangle abdc, limited by the graph px = const, the axis V and the segments ab and cd equal to the gas pressure, is numerically equal to the work defined by formula (13.3):

A" = p1(V2 - V2) = |ab| |ac|.

In general, the gas pressure does not remain constant. For example, in an isothermal process, it decreases inversely with volume (Fig. 13.3). In this case, to calculate the work, you need to divide the total volume change into small parts and calculate the elementary (small) work, and then add them all up. The work of the gas is still numerically equal to the area of ​​the figure bounded by the plot of p versus V, the V axis, and the segments ab and cd, the length of which is numerically equal to the pressures p 1 p 2 in the initial and final states of the gas.

Thermal phenomena can be described using quantities (macroscopic parameters) recorded by instruments such as a manometer and a thermometer. These devices do not respond to the impact of individual molecules. The theory of thermal processes, which does not take into account the molecular structure of bodies, is called thermodynamics. This has already been mentioned in chapter 1. In this chapter, we will study thermodynamics.

§ 5.1. Work in thermodynamics

In Chapter 3, we looked at various processes that change the state of a thermodynamic system. We have dealt mainly with the change in the state of an ideal gas during isothermal, isobaric, and isochoric processes.

For further consideration of thermodynamic processes, it is necessary to study in detail, as a result of which external influences the state of any thermodynamic system can change. There are two essentially different types of influences that lead to a change in the state of the system, i.e., to a change in thermodynamic parameters- pressure p, volumeV, temperature T characterizing the state. The first one- This doing work.

Work in mechanics and thermodynamics

In mechanics, the motion of macroscopic bodies is considered. Work is defined as the product of the modules of force and displacement and the cosine of the angle between the directions of force and displacement. Work is done under the action of a force or several forces on a moving macroscopic body and is equal to the change in its kinetic energy.

In thermodynamics, the motion of a body as a whole is not considered, and we are talking about the movement of parts of a macroscopic body relative to each other. When work is done, the volume of the body changes, and its velocity remains equal to zero. But the velocities of the molecules of a body, such as a gas, change. Therefore, the temperature of the body also changes.

The reason is as follows: during elastic collisions of molecules with a moving piston (for the case of gas compression), their kinetic energy changes. So, when moving towards molecules, the piston transfers part of its mechanical energy to them during collisions, as a result of which the gas heats up. The piston acts like a football player who meets a flying ball with a kick and imparts a speed to the ball that is much greater than that which he had before the kick *.

* The problem of changing the speed of a ball during its elastic collision with a moving wall is considered in detail in § 6.12 "Mechanics" (task 5).

Conversely, if the gas expands, then after colliding with the receding piston, the velocities of the molecules decrease, as a result of which the gas cools. A football player works in the same way: in order to reduce the speed of a flying ball or stop it, the football player's foot moves away from the ball, as if giving way to it.

So, when doing work in thermodynamics, the state of macroscopic bodies changes: their volume and temperature change.

Work calculation

We calculate the work depending on the change in volume using the example of gas in a cylinder under a piston (Fig. 5.1). The easiest way is to first calculate not the work of the force , acting on the gas from the side of the external body (piston), and the work done by the gas itself, acting on the piston with a force . According to Newton's third law
.

The modulus of force acting from the side of the gas on the piston is equal to F" = PS, Where R is the gas pressure, and S is the surface area of ​​the piston. Let the gas expand and the piston move in the direction of the force a small distance Δ h = h 2 – h 1 If the displacement is small, then the gas pressure can be considered constant.

The work of the gas is:

This work can be expressed in terms of a change in the volume of the gas. Initial Volume V 1 = Sh 1 , and final V 2 = Sh 2 . That's why

where ∆ V = V 2 - V 1 - change in gas volume.

When expanding, the gas does positive work, since the directions of force and displacement of the piston coincide.

If the gas is compressed, then the formula (5.1.2) for the work of the gas remains valid. But now V 2 < V 1 and therefore A"< 0 (Fig. 5.2).

The work A performed by external bodies on a gas differs from the work of a gas A" sign only: A= -A", because the strength , acting on the gas, is directed against the force
, and the movement remains the same. Therefore, the work of external forces acting on the gas is equal to:

(5.1.3)

The minus sign indicates that during gas compression, when Δ V = V 2 - V 1 < 0, работа внешней силы положительна. Понятно, почему в этом случае А >0: when gas is compressed, the directions of force and displacement are the same. When the gas expands, on the contrary, the work external bodies negative (A< 0), так как ΔV = V 2 – V 1 > 0. Now the directions of force and displacement are opposite.

Expressions (5.1.2) and (5.1.3) are valid not only for compression or expansion of gas in a cylinder, but also for a small change in the volume of any system. If the process is isobaric (p = const), then these formulas can be applied to large volume changes.