Let us determine the work of force F statically applied to some elastic system (Fig. 20, a), the material of which follows Hooke’s law.

For small deformations, the principle of independent action of forces is applicable to this system; therefore, the movements of individual points and sections of the structure are directly proportional to the load causing them:

where is the displacement in the direction of force F; - a certain coefficient depending on the material, design and size of the structure. An increase in force F by an infinitesimal amount dF will cause an increase in displacement by .

Let us formulate an expression for the elementary work of an external force on displacement , discarding infinitesimal quantities of the second order of smallness: .

Let's replace using (2.2):

Integrating this expression within the limits of the total change in force from zero to its final value, we obtain a formula for determining the work done by a statically applied external force F:

or, taking into account (2.2):

that is, the work of an external force during its static action on any elastic structure is equal to half the product of the value of this force and the value of the corresponding displacement.

To generalize the conclusion obtained, force is understood as any impact applied to an elastic system, that is, not only a concentrated force, but also a moment or a uniformly distributed load; Displacement is understood to be the type of motion on which a given force produces work: a concentrated force corresponds to linear displacement, a concentrated moment corresponds to angular displacement, and a uniformly distributed load corresponds to the area of ​​the displacement diagram in the area of ​​the load.

With static action on the group structure external forces the work of these forces is equal to half the sum of the products of each force by the amount of its corresponding displacement caused by the action of the entire group of forces. For example, when a beam (Fig. 20, b) is acted upon by concentrated forces F 1, F 2 and concentrated moments M 1 and M 2, the work of external forces is:

The work of external forces on the movements caused by them can be expressed in another way - through internal power factors(bending moments, longitudinal and transverse forces) arising in the cross sections of the system.

Let us select an infinitesimal element dz from a straight rod with two sections perpendicular to its axis (Fig. 21, a).

The rod consists of infinitely large number such elements. In the general case of a plane problem, a longitudinal force N z, a bending moment M x and a transverse force Q y are applied to each element dz.

For the selected element dz, the forces N, M, Q are external forces, therefore the work can be obtained as the sum of the works performed by statically increasing forces N, M, Q on the corresponding deformations of the elements dz.

Let us consider the element dz, which is only under the action of longitudinal forces N (Fig. 21, b). If its left section is considered motionless, then the right section is under the influence longitudinal force will move to the right by an amount. On this displacement the force N will do the work:

If the left section of the element dz, which is under the influence of only bending moments M, is fixedly fixed (Fig. 22,a), then the mutual angle of rotation of the end sections of the element will be equal to the angle of rotation of its right section:

At this displacement, the moment M will do the work:

Let us fix the left section of the element dz, which is under the action of only transverse forces Q (Fig. 22, b, c), and apply tangential forces to the right, the resultant of which is the transverse force Q. Let us assume that the tangential stresses are uniformly distributed over the entire area A cross section, that is, then the displacement is defined as: .

AND historical information.

1) M.V. Lomonosov, having carried out harmonious reasoning and simple experiments, came to the conclusion that “the cause of heat is 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 hits steel, iron heats up when its particles are forged with strong blows »

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

3) Davy in 1799 carried out interesting experience. Two pieces of ice, when rubbed against one another, 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.

Calculation of work.

In mechanics, work is defined as the product of the moduli 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 body volume, i.e. movement of parts of a macrobody relative to each other. This process leads to a change in the distance between particles, and also often to a change in the speed of their movement, therefore, to a change internal energy bodies.

Let there be a 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 to position 2 to a distance Δ l. The gas pressure force will do work on the external bodies. Because p= const, then the pressure force F = pS also constant. Therefore, the work of this force can be calculated using the formula A=F Δ l=pS Δ l=p Δ V, A= p Δ V

where Δ 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 when it contracts or expands? Why does a gas heat up when compressed and cool when expanding?

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 a gas is compressed, then during a collision a piston moving towards the molecules transfers part of its mechanical energy, causing the gas to heat up;

• If a gas expands, then after a collision with a retreating piston, the speed of the molecules decreases. As a result, the gas cools down.

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

Work of external forces acting on gas

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

Let us 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 ideal gas), then at Δ Τ = 1 K we get R = A. It follows from this physical meaning 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:

On the graph p = f(V) for an isobaric process, the work is equal to the area of ​​the shaded rectangle in figure a).

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 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.

>>Physics: Work in thermodynamics

As a result of what processes can internal energy change? You already know that there are two types of such processes: work and heat transfer. 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 force modulus, the displacement modulus of the point of its application and the cosine of the angle between them. When a force acts on a moving body, the work is equal to the change in its kinetic energy.
IN the movement of the body as a whole is not considered, we're talking about about the movement of parts of a macroscopic body relative to each other. As a result, the volume of the body may change, but its speed remains equal to zero. 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 the body change when a body contracts or expands? Why, in particular, does the air heat up when inflating 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 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 meeting an incoming ball with a kick. The foot imparts a speed to the ball that is significantly greater than that which it possessed before the impact.
Conversely, if the gas expands, then after colliding with the retreating piston, the velocities of the molecules decrease, as a result of which the gas cools. A football player acts in the same way, in order to reduce the speed of a flying ball or stop it - the football player’s leg moves away from the ball, as if giving way to it.
During compression or expansion, the average potential energy of interaction between molecules also changes, since the average distance between the molecules changes.
Calculation of work. Let's calculate the work depending on the change in volume using the example of gas in a cylinder under the piston ( Fig.13.1).

The easiest way is to first calculate not the work done by the force acting on the gas from the external body (piston), but the work done by the gas pressure force acting on the piston with a force . According to Newton's third law . The modulus of force acting from the gas on the piston is equal to , Where p- gas pressure, and S- 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 the change in gas volume. Its initial volume V 1 = Sh 1, and the final V 2 = Sh 2. That's why

where is the change in gas volume.
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 gas work remains valid. But now , and therefore (Fig.13.2).

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

When gas is compressed, when , the work of the 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 surrounding bodies: a 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. Work A´ gas for the case of constant pressure can be given a simple geometric interpretation.
Let's construct a graph of the dependence of gas pressure on the volume it occupies ( Fig.13.3). Here is the area of ​​the rectangle abdc, limited by schedule p 1=const, axis V and segments ab And CD, equal to the gas pressure, is numerically equal to work (13.3):

In general, the gas pressure does not remain constant. For example, during an isothermal process it decreases in inverse proportion to the volume ( Fig.13.4). In this case, to calculate the work, you need to divide the total change in volume into small parts and calculate the elementary (small) works, and then add them all up. The work done by the gas is still numerically equal to the area of ​​the figure limited by the dependence graph p from V, axis V and segments ab And CD, equal to pressure p 1, p2 in the initial and final states of the gas.

???
1. Why do gases heat up when compressed?
2. Is external forces performing positive or negative work during the isothermal process shown in Figure 13.2?

G.Ya.Myakishev, B.B.Bukhovtsev, N.N.Sotsky, Physics 10th grade

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··· Oryol issue ···

G.A.BELUKHA,
School No. 4, Livny, Oryol region.

Gas work in thermodynamics

When studying the work of gas in thermodynamics, students inevitably encounter difficulties due to poor skills in calculating the work of a variable force. Therefore, it is necessary to prepare for the perception of this topic, starting with the study of work in mechanics and, for this purpose, solving problems on the work of a variable force by summation basic work all the way through integration.

For example, when calculating the work of the Archimedes force, the elastic force, the force of universal gravity, etc. one must learn to summarize elementary quantities using simple differential relations like dA = Fds. Experience shows that high school students easily cope with this task - the arc of the trajectory along which the force increases or decreases must be divided into the following intervals ds, on which force F can be considered a constant value, and then, knowing the dependence F = F(s), substitute it under the integral sign. For example,

The work of these forces is calculated using the simplest table integral

This technique makes it easier for future students to adapt to a physics course at a university and eliminates methodological difficulties associated with the ability to find the work of a variable force in thermodynamics, etc.

After students have learned what internal energy is and how to find its change, it is advisable to give a general diagram:

Having learned that work is one of the ways to change internal energy, tenth graders can easily calculate the work of a gas in an isobaric process. At this stage, it is necessary to emphasize that the gas pressure force does not change along the entire path, and according to Newton’s third law | F 2 | = |F 1 |, we find the work sign from the formula A = Fs cos. If = 0°, then A> 0, if = 180°, then A < 0. На графике зависимости r(V) work is numerically equal to the area under the graph.

Let the gas expand or contract isothermally. For example, gas is compressed under a piston, the pressure changes, and at any time

With an infinitesimal displacement of the piston by dl we get an infinitesimal change in volume dV, and the pressure r can be considered constant. By analogy with finding the mechanical work of a variable force, let’s create the simplest differential relation dA = pdV, then and, knowing the dependence r (V), let's write This is a table integral of the type The gas work in this case is negative, because = 180°:

because V 2 < V 1 .

The resulting formula can be rewritten using the relation

To consolidate, let's solve problems.

1. The gas changes from the state 1 (volume V 1, pressure r 1) in a state 2 (volume V 2, pressure r 2) in a process in which its pressure depends linearly on volume. Find the work done by the gas.

Solution. Let's build an approximate dependence graph p from V. The work is equal to the area under the graph, i.e. trapezoid area:

2. One mole of air, located under normal conditions, expands with volume V 0 to 2 V 0 in two ways - isothermal and isobaric. Compare the work done by air in these processes.

Solution

In an isobaric process A p = r 0 V, But r 0 = RT 0 /V 0 , V = V 0 therefore A p = RT 0 .

In an isothermal process:

Let's compare:

Having studied the first law of thermodynamics and its application to isoprocesses and having reinforced the topic of work in thermodynamics by solving problems, students were prepared to perceive the most complex part of thermodynamics, “Work of cycles and efficiency of heat engines.” I present this material in the following sequence: work of cycles – Carnot cycle – efficiency of heat engines – circular processes.

A circular process (or cycle) is a thermodynamic process, as a result of which a body, having gone through a series of states, returns to its original state. If all processes in a cycle are in equilibrium, then the cycle is considered to be in equilibrium. It can be depicted graphically as a closed curve.

The figure shows a graph of pressure dependence p from volume V(diagram p, V) for some cycle 1–2–3–4–1. At the sites 1–2 And 4–1 gas expands and does positive work A 1, numerically equal to the area of ​​the figure V 1 412V 2. On the site 2–3–4 gas compresses and does work A 2, the module of which is equal to the area of ​​the figure V 2 234V 1. Full gas work per cycle A = A 1 + A 2, i.e. positive and equal to the area of ​​the figure 12341 .

If the equilibrium cycle is represented by a closed curve on r, V- a diagram that moves clockwise, then the work of the body is positive, and the cycle is called direct. If a closed curve on r, V- diagram goes counterclockwise, then the gas does negative work per cycle, and the cycle is called reverse. In any case, the modulus of gas work per cycle equal to area figure limited by the cycle schedule on r, V-diagram.

In a circular process, the working fluid returns to its original state, i.e. into a state with initial internal energy. This means that the change in internal energy per cycle is zero: U= 0. Since, according to the first law of thermodynamics, for the entire cycle Q = U + A, That Q = A. So, the algebraic sum of all amounts of heat received per cycle is equal to the work of the body per cycle: A ts = Q n + Q x = Q n – | Q x |.

Let's consider one of the circular processes - the Carnot cycle. It consists of two isothermal and two adiabatic processes. Let the working fluid be an ideal gas. Then at the site 1–2 isothermal expansion, according to the first law of thermodynamics, all the heat received by the gas goes to perform positive work: Q 12 = A 12. That is, there is no heat loss to the surrounding space and no change in internal energy: U= 0, because T 12 = const (because the gas is ideal).

On the site 2–3 adiabatic expansion, the gas does positive work due to changes in internal energy, because Q hell = 0 = U 23 + A g23 A g23 = – U 23. There is also no heat loss here, by definition of an adiabatic process.

On the site 3–4 Positive work is done on the gas by an external force, but it does not heat up (isothermal process). Thanks to the fairly slow process and good contact with the refrigerator, the gas has time to transfer the energy obtained through work in the form of heat to the refrigerator. The gas itself does negative work: Q 34 = A g34< 0.

On the site 4–1 the gas is adiabatically (without heat exchange) compressed to its original state. At the same time, it does negative work, and external forces do positive work: 0 = U 41 + A g41 A g41 = – U 41 .

Thus, during the cycle the gas receives heat only in the area 1–2 , expanding isothermally:

Heat is transferred to the refrigerator only during isothermal compression of the gas in the area 3–4 :

According to the first law of thermodynamics

A ts = Q n – | Q x |;

The efficiency of a machine operating according to the Carnot cycle can be found using the formula

According to the Boyle–Mariotte law for processes 1–2 And 3–4 , as well as the Poisson equation for processes 2–3 And 4–1 , it is easy to prove that

After the reductions, we obtain the formula for the efficiency of a heat engine operating according to the Carnot cycle:

It is methodically correct, as experience shows, to study the operation of heat engines operating in a reverse cycle using the example of the operation of a reverse Carnot cycle, because it is reversible and can be carried out in the opposite direction: expand the gas as the temperature decreases from T n to T x (process 1–4 ) and at low temperatures T x (process 4–3 ), and then compress (processes 3–2 And 2–1 ). The engine now does work to drive the refrigeration machine. The working fluid takes away the amount of heat Q x food inside at low temperature T x, and gives off the amount of heat Q n surrounding bodies, outside the refrigerator, at higher temperatures T n. Thus, a machine operating on a reverse Carnot cycle is no longer a heat machine, but an ideal refrigeration machine. The role of a heater (giving off heat) is performed by a body with a lower temperature. But, keeping the names of the elements, as in a heat engine operating in a direct cycle, we can present the block diagram of the refrigerator in the following form:

Please note that heat from a cold body passes in a refrigeration machine to a body with a more high temperature not spontaneously, but due to the work of an external force.

The most important characteristic of a refrigerator is the refrigeration coefficient, which determines the efficiency of the refrigerator and is equal to the ratio of the amount of heat removed from the refrigerating chamber Q x to the expended energy of the external source

During one reverse cycle, the working fluid receives an amount of heat from the refrigerator Q x and releases the amount of heat into the surrounding space Q n, what's more Q x to work A movement performed by an electric motor over gas per cycle: | Q n | = | Q x | + A dv.

The energy expended by the engine (electricity in the case of compressor electric refrigerators) is used for useful work on gas, as well as for losses when heating the engine windings with electric current QR and for friction in the circuit A tr.

If we neglect losses due to friction and Joule heat in the motor windings, then the coefficient of performance

Considering that in the forward cycle

after simple transformations we get:

The last relationship between the coefficient of performance and the efficiency of a heat engine, which can also operate in a reverse cycle, shows that the coefficient of performance can be greater than one. In this case, more heat is removed from the refrigeration chamber and returned to the room than the energy used by the engine for this purpose.

In the case of an ideal heat engine operating on a reverse Carnot cycle (ideal refrigerator), the refrigeration coefficient has a maximum value:

In real refrigerators because not all the energy received by the engine goes to work on the working fluid, as described above.

Let's solve the problem:

Estimate the cost of making 1 kg of ice in a home refrigerator if the temperature of freon evaporation is – t x °C, radiator temperature t n °C. The cost of one kilowatt-hour of electricity is equal to C. Temperature in the room t.

Given:

m, c, t, t n, t x, , C.
____________
D – ?

Solution

The cost D of making ice is equal to the product of the work of the electric motor and the tariff C: D = CA.

To turn water into ice at a temperature of 0 °C, it is necessary to remove an amount of heat from it Q = m(ct+ ). We assume approximately that a reverse Carnot cycle occurs over freon with isotherms at temperatures T n and T X. We use formulas for the coefficient of performance: by definition, = Q/A and for an ideal refrigerator id = T X /( T n – T X). It follows from the condition that id.

We solve the last three equations together:

When discussing this problem with students, it is necessary to pay attention to the fact that the main work of the refrigeration device is not to cool food, but to maintain the temperature inside the refrigerator by periodically pumping out the heat penetrating through the walls of the refrigerator.

To consolidate the topic, you can solve the problem:

Efficiency of a heat engine operating in a cycle consisting of an isothermal process 1–2 , isochoric 2–3 and adiabatic 3–1 , is equal to , and the difference between the maximum and minimum gas temperatures in the cycle is equal to T. Find the work done on a mole of a monatomic ideal gas in an isothermal process.

Solution

When solving problems in which the efficiency of the cycle appears, it is useful to first analyze all sections of the cycle, using the first law of thermodynamics, and identify the sections where the body receives and releases heat. Let us mentally draw a series of isotherms on r, V-diagram. Then it will become clear that the maximum temperature in the cycle is on the isotherm, and the minimum temperature is also on the isotherm. 3 . Let us denote them by T 1 and T 3 respectively.

On the site 1–2 change in internal energy of an ideal gas U 2 – U 1 = 0. According to the first law of thermodynamics, Q 12 = (U 2 – U 1) + A 12. Since on the site 1–2 the gas expanded, then the work of the gas A 12 > 0. This means that the amount of heat supplied to the gas in this section Q 12 > 0, and Q 12 = A 12 .

On the site 2–3 the work done by the gas is zero. That's why Q 23 = U 3 – U 2 .

Using the expressions U 2 = c V T 1 and the fact that T 1 – T 3 = T, we get Q 23 = –c V T < 0. Это означает, что на участке 2–3 the gas receives a negative amount of heat, i.e. gives off heat.

On the site 3–1 there is no heat exchange, i.e. Q 31 = 0 and, according to the first law of thermodynamics, 0 = ( U 1 – U 3) + A 31. Then the gas work
A 31 = U 3 – U 1 = c V(T 3 –T 1) = –c V T.

So, during the cycle the gas did work A 12 + A 31 = A 12 – c V T and received heat only on the site 1–2 . Cycle efficiency

Since the work of the gas on the isotherm is equal to

Gennady Antonovich Belukha– Honored Teacher of the Russian Federation, 20 years of teaching experience, every year his students take prizes at various stages All-Russian Olympiad in Physics. Hobby: computer technology.

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 body volume, i.e. movement of parts of a macrobody relative to each other. This process leads to a change in the distance between 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 a gas in a cylinder with a movable piston at a temperature T 1 (Fig. 1). We will slowly heat the gas to a temperature T 2. The gas will expand isobarically and the piston will move from position 1 to position 2 to a distance Δ l. The gas pressure force will do work on the external bodies. Because p= const, then the pressure force F = pS also constant. Therefore, the work of this force can be calculated using the formula

\(~A = F \Delta l = pS \Delta l = p \Delta V, \qquad (1)\)

where Δ 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.

The force of gas pressure performs work only in the process of changing the volume of gas.

When expanding (Δ V> 0) of the gas, positive work is done ( A> 0); during compression (Δ V < 0) газа совершается отрицательная работа (A < 0), положительную работу совершают внешние силы A' = -A > 0.

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

\(~pV_1 = \frac mM RT_1 ; pV_2 = \frac mM RT_2 \Rightarrow\) \(~p(V_2 - V_1) = \frac mM R(T_2 - T_1) .\)

Therefore, in an isobaric process

\(~A = \frac mM R \Delta T .\)

If m = M(1 mol of ideal gas), then at Δ Τ = 1 K we get R = A. This implies the 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.

On the chart p = f(V) in an isobaric process, the work is equal to the area of ​​the shaded rectangle in Figure 2, a.

If the process is not isobaric (Fig. 2, b), then the curve p = f(V) can be represented as a broken line consisting of a large number of isochores and isobars. Work on isochoric sections is zero, and the total work on all isobaric sections will be

\(~A = \lim_(\Delta V \to 0) \sum^n_(i=1) p_i \Delta V_i\), or \(~A = \int p(V) dV,\)

those. 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 2, c.

It is possible to determine work using the last formula only if it is known how the gas pressure changes when its volume changes, i.e. the form of the function is known p(V).

Thus, the gas does work when expanding. Devices and units whose actions are based on the property of a gas to do work during the expansion process are called pneumatic. Pneumatic hammers, mechanisms for closing and opening doors on vehicles, etc. operate on this principle.

Literature

Aksenovich L. A. Physics in high school: Theory. Assignments. Tests: Textbook. benefits for institutions providing general education. environment, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsiya i vyakhavanne, 2004. - P. 155-156.