Gas work

1. First law of thermodynamics

The existence of two ways of transferring energy to a thermodynamic system allows us to analyze from an energy point of view the equilibrium process of transition of a system from any initial state 1 to another state 2 . Change in internal energy of the system

 U 1-2 = U 2 - U 1

in such a process is equal to the amount of workA’ 1-2 performed on the system by external forces and heatQ 1-2 reported system:

 U 1-2 = A ’ 1-2 + Q 1-2 (2. 3 )

JobA’ 1-2 numerically equal and opposite in sign to workA 1-2 committed by the system itself against external forces in the same transition process:

A’ 1-2 = - A 1-2 .

Therefore, expression (2.6) can be rewritten differently:

Q 1-2 =  U 1-2 + A 1-2 (2. 3 )

The first law of thermodynamics: heat imparted to a system is spent on changing the internal energy of the system and on doing work by the system against external forces.

 Q = dU +  A (2. 3 )

dU – internal energy, is a total differential.

QAnd Aare not total differentials.

Q 1-2 =
(2. 3 )

.

Historically, the establishment of the first law of thermodynamics was associated with the failure to create a perpetual motion machine of the first kind (perpetuum mobile), in which a machine would do work without receiving heat from outside and without expending any type of energy. The first law of thermodynamics says that it is impossible to build such an engine.

Q 1-2 =  U 1-2 + A 1-2

1. Application of the first law of thermodynamics to isoprocesses.

   1. Isobaric process.

R= const

A = = p ( V 2 - V 1 ) = p  V ,

where p is gas pressure, V – change in its volume.

BecausePV 1 = RT 1 ; PV 2 = RT 2,

ThatV 2 - V 1 = (T 2 – T 1 ) And

A = R(T 2 – T 1 ); (2. 3 )

Thus, we get thatuniversal gas constant R equal to the work done by the mole ideal gas when its temperature increases by one Kelvin at constant pressure.

Taking into account expression (2.10), the equation of the first law of thermodynamics (2.8) can be written as follows

 Q = dU + pdV. (2.3)

1. Isochoric process

V = const, hence,dV = 0

 A =p V = 0

 Q =  U.

 Q =  U = R T (2. 3 )

1. Isothermal process

T =const,

U = 0 the internal energy of an ideal gas does not change, and

Q =  A

A = =
= RTln (2. 3 )

To ensure that the gas temperature does not decrease during expansion, the gas should be isothermal process it is necessary to supply an amount of heat equivalent to the external work of expansion, i.e. A = Q.

In practice, the slower the process proceeds, the more accurately it can be considered isothermal.

G Graphically, the work during an isothermal process is numerically equal to the area of ​​the shaded projection in Fig.

Comparing the areas of the figures under the isotherm and isobar sections, we can conclude that the expansion of the gas from the volumeV 1 up to volumeV 2 at the same initial value of gas pressure, in the case of isobaric expansion, it is accompanied by the performance of more work.

1. Heat capacity of gases

Heat capacityWITH of any body is the ratio of an infinitesimal amount of heatd Q , received by the body, to the corresponding incrementdT its temperature:

C body = (2. 3 )

This value is measured in joules per kelvin (J/K).

When the mass of a body is equal to unity, the heat capacity is called specific heat. It is denoted by the small letter c. It is measured in joules per kilogram . kelvin (J/kg . K).There is a relationship between the heat capacity of a mole of a substance and the specific heat capacity of the same substance

(2. 3 )

Using formulas (2.12) and (2.15), we can write

(2. 3 )

Heat capacities at constant volume are of particular importanceWITH V and constant pressureWITH R . If the volume remains constant, thendV = 0 and according to the first law of thermodynamics (2.12) all the heat goes to increase the internal energy of the body

 Q = dU (2. 3 )

From this equality it follows that the heat capacity of a mole of an ideal gas at constant volume is equal to

(2. 3 )

From heredU  = C V dT, and the internal energy of one mole of an ideal gas is equal to

U  = C V T (2. 3 )

Internal energy of an arbitrary mass of gasT determined by the formula

(2. 3 )

Considering that for 1 mole of ideal gas

U = RT,

and counting the number of degrees of freedomi unchanged, for the molar heat capacity at constant volume we obtain

C v = = (2. 3 )

Specific heat capacity at constant volume

With v = = (2. 3 )

For an arbitrary mass of gas, the following relation is valid:

 Q = dU = RdT; (2. 3 )

If gas heating occurs at constant pressure, then the gas will expand, doing positive work on external forces. Therefore, the heat capacity at constant pressure must be greater than the heat capacity at constant volume.

If 1 mole of gas atisobaric the amount of heat is reported in the process Qthen introducing the concept of molar heat capacity at constant pressure C R = can be written down

 Q = C p dT;

where C p – molar heat capacity at constant pressure.

Because in accordance with the first law of thermodynamics

 Q =  A + dU = RdT +RdT =

=(R +R)dT = (R +WITH V )dT,

That

WITH R ==R+WITH V . (2. 3 )

This ratio is calledMayer's equation :

Expression for C R can also be written as:

WITH R = R + R =
. (2. 3 )

Specific heat capacity at constant pressureWith p define by dividing expressions (2.26) into :

With p =
(2. 3 )

In isobaric communication with a gas of massmamount of heat Qits internal energy increases by the amount U = C V  T, and the amount of heat transferred to the gas during an isobaric process is Q= C p  T.

Having designated the ratio of heat capacities letter , we get

(2. 3 )

Obviously,  1 and depends only on the type of gas (number of degrees of freedom).

From formulas (2.22) and (2.26) it follows that the molar heat capacities are determined only by the number of degrees of freedom and do not depend on temperature. This statement is valid in a fairly wide temperature range only for monatomic gases with only translational degrees of freedom. For diatomic gases, the number of degrees of freedom, manifested in heat capacity, depends on temperature. A diatomic gas molecule has three translational degrees of freedom: translational (3), rotational (2) and vibrational (2).

Thus, the total number of degrees of freedom reaches 7 and for the molar heat capacity at constant volume we should obtain: C V = .

From the experimental dependence of the molar heat capacity of hydrogen it follows that C V depends on temperature: at low temperature ( 50 K) WITH V = , at room C V = and very high - C V = .

The discrepancy between theory and experiment is explained by the fact that when calculating heat capacity, it is necessary to take into account the quantization of the energy of rotation and vibration of molecules (not any rotational and vibrational energies are possible, but only a certain discrete series of energy values). If the energy of thermal motion is insufficient, for example, to excite oscillations, then these oscillations do not contribute to the heat capacity (the corresponding degree of freedom is “frozen” - the law of uniform distribution of energy does not apply to it). This explains the sequential (at certain temperatures) excitation of degrees of freedom that absorb thermal energy, and shown in Fig. 13 addiction C V = f ( T ).

Basic formulas of thermodynamics and molecular physics, which will be useful to you. Another great day for practical classes in physics. Today we will put together the formulas that are most often used to solve problems in thermodynamics and molecular physics.

So, let's go. Let us try to present the laws and formulas of thermodynamics briefly.

Ideal gas

Ideal gas is an idealization, just like a material point. The molecules of such a gas are material points, and collisions of molecules are absolutely elastic. We neglect the interaction of molecules at a distance. In problems in thermodynamics, real gases are often taken to be ideal. It's much easier to live this way, and you don't have to deal with a lot of new terms in the equations.

So, what happens to the molecules of an ideal gas? Yes, they are moving! And it is reasonable to ask, at what speed? Of course, in addition to the speed of molecules, we are also interested in the general state of our gas. What pressure P does it exert on the walls of the vessel, what volume V does it occupy, what is its temperature T.

To find out all this, there is the ideal gas equation of state, or Clapeyron-Mendeleev equation

Here m – mass of gas, M – its molecular weight (we find it from the periodic table), R – universal gas constant equal to 8.3144598(48) J/(mol*kg).

The universal gas constant can be expressed in terms of other constants ( Boltzmann constant and Avogadro's number )

Massat , in turn, can be calculated as the product density And volume .

Basic equation of molecular kinetic theory (MKT)

As we have already said, gas molecules move, and the higher the temperature, the faster. There is a relationship between gas pressure and the average kinetic energy E of its particles. This connection is called basic equation of molecular kinetic theory and has the form:

Here n – concentration of molecules (the ratio of their number to volume), E – average kinetic energy. They can be found, as well as the root mean square speed of molecules, accordingly, using the formulas:

Substitute energy into the first equation, and we get another form of the basic equation MKT

The first law of thermodynamics. Formulas for isoprocesses

Let us remind you that the first law of thermodynamics states: the amount of heat transferred to the gas goes to change the internal energy of the gas U and to perform work A by the gas. The formula of the first law of thermodynamics is written as follows:

As you know, something happens to gas, we can compress it, we can heat it. IN in this case we are interested in processes that occur under one constant parameter. Let's look at what the first law of thermodynamics looks like in each of them.

By the way! There is now a discount for all our readers 10% on any type of work.

Isothermal process occurs at a constant temperature. The Boyle-Mariotte law applies here: in an isothermal process, the pressure of a gas is inversely proportional to its volume. In an isothermal process:

proceeds at a constant volume. This process is characterized by Charles' law: At constant volume, pressure is directly proportional to temperature. In an isochoric process, all the heat supplied to the gas goes to change its internal energy.

runs at constant pressure. Gay-Lussac's law states that at constant gas pressure, its volume is directly proportional to the temperature. In an isobaric process, heat goes both to change the internal energy and to do work by the gas.

. An adiabatic process is a process that occurs without heat exchange with environment. This means that the formula of the first law of thermodynamics for an adiabatic process looks like this:

Internal energy of a monatomic and diatomic ideal gas

Heat capacity

Specific heat equal to the amount of heat required to heat one kilogram of a substance by one degree Celsius.

In addition to specific heat capacity, there is molar heat capacity (the amount of heat required to heat one mole of a substance by one degree) at constant volume, and molar heat capacity at constant pressure. In the formulas below, i is the number of degrees of freedom of gas molecules. For monatomic gas i=3, for diatomic – 5.

Thermal machines. Efficiency formula in thermodynamics

Heat engine , in the simplest case, consists of a heater, a refrigerator and a working fluid. The heater imparts heat to the working fluid, it does work, then it is cooled by the refrigerator, and everything repeats. O v. A typical example of a heat engine is an internal combustion engine.

Efficiency heat engine is calculated by the formula

So we have collected the basic formulas of thermodynamics that will be useful in solving problems. Of course, these are not all the formulas from the topic of thermodynamics, but knowledge of them can really serve you well. And if you have any questions, remember student service, whose specialists are ready to come to the rescue at any time.

AND historical information.

1) M.V. Lomonosov, having carried out harmonious 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 well known that heat is excited by movement: hands warm up from mutual friction, wood catches fire, 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 carried out an interesting experiment in 1799. 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 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.). 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 it transfers part of its mechanical energy to the molecules, as a result of which the gas heats 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 of ideal gas), then at Δ Τ = 1 K we get R = A. It follows from this 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:

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.

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 performed 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.

When a group of external forces acts statically on a structure, the work of these forces is equal to half the sum of the products of each force and 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 displacements they cause can be expressed in another way - through internal force 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 an infinitely large number of 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 stationary, then the right section, under the influence of the 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 of the cross section , that is, then the displacement is defined as: .

··· 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 summing up the elementary work along the entire path using 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. Location on 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).

Location on 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.

Location on 3–4 Positive work is done on the gas by an external force, but it does not heat up (isothermal process). Due to the rather slow process and good contact with a 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.

Location on 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 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:

Let us note that heat from a cold body is transferred in a refrigeration machine to a body with a higher 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 Q R 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.

Location on 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 .

Location on 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.

Location on 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 of the All-Russian Olympiad in Physics. Hobby: computer technology.