Gas work

1. First law of thermodynamics

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

 U 1-2 = U 2 - U 1

in such a process is equal to the sum of the 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 )

WorkA’ 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: the heat imparted to the system is spent on changing the internal energy of the system and on the system performing work against external forces.

 Q = dU +  A (2. 3 )

dU - internal energy, is a total differential.

Qand Aare not complete 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 the machine would do work without receiving heat from outside and without spending any kind of energy. The first law of thermodynamics speaks of the impossibility of building 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 the gas pressure, V is the change in its volume.

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

thenV 2 - V 1 = (T 2 – T 1 ) and

A = R(T 2 – T 1 ); (2. 3 )

Thus, we get thatuniversal gas constant R is equal to the work done by a mole of an ideal gas when its temperature rises 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, Consequently,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 =  BUT

A = =
= RTln (2. 3 )

To ensure that the temperature of the gas does not decrease during expansion, to the gas during isothermal process it is necessary to supply the 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 the isothermal process is numerically equal to the area of ​​the shaded projection in Fig.

Comparing the areas of the figures under the sections of the isotherm and isobar, we can conclude that the expansion of gas from 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 capacityFROM 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 one, the heat capacity is called specific heat. It is denoted by a small letter s. It is measured in joules per kilogram. . kelvin (J/kg . K). There is a relation 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 )

Of particular importance are the heat capacities at constant volume FROM V and constant pressureFROM 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

U  = C V T (2. 3 )

Internal energy of an arbitrary mass of gast is determined by the formula

(2. 3 )

Considering that for 1 mole of an 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 relation is true:

 Q = dU = RdT; (2. 3 )

If the gas is heated 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 process is given the amount of heat Qthen introducing the concept of molar heat capacity at constant pressure С R = can be written

 Q = C p dT;

where C p is the molar heat capacity at constant pressure.

Because according to the first law of thermodynamics

 Q =  A+dU=RdT+RdT=

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

then

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

This ratio is calledMayer equation :

Expression for C R can also be written as:

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

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

With p =
(2. 3 )

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

Denoting 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 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 rather wide temperature range only for monatomic gases with only translational degrees of freedom. For diatomic gases, the number of degrees of freedom, which manifests itself 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 get: C V = .

It follows from the experimental dependence of the molar heat capacity of hydrogen that С V temperature dependent: at low temperature ( 50 K) FROM V = , at room temperature V = and very high - V = .

The discrepancy between theory and experiment is explained by the fact that when calculating the heat capacity, one must 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 energy distribution is not applicable to it). This explains the successive (at certain temperatures) excitation of the degrees of freedom that absorb thermal energy, and shown in Fig. 13 addiction C V = f ( T ).

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. 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 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

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

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.

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

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

Let's 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 a m = M(1 mol of ideal gas), then at Δ Τ = 1 K we get R = A. Hence follows 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.

On the chart p = f(V) in an isobaric process, the work is equal to the area of ​​the rectangle shaded 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 equal to 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 the work using the last formula only if it is known how the gas pressure changes with a change in its volume, i.e. the form of the function is known p(V).

Thus, when the gas expands, it does work. Devices and units, the actions of which are based on the property of gas in the process of expansion to do work, are called pneumatic. Pneumatic hammers, mechanisms for closing and opening doors in transport, etc. operate on this principle.

Literature

Aksenovich L. A. Physics in high school: Theory. Tasks. Tests: Proc. allowance for institutions providing general. environments, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsia i vykhavanne, 2004. - C. 155-156.

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

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

Ideal gas

Ideal gas is an idealization, like a material point. The molecules of such a gas are material points, and the collisions of molecules are absolutely elastic. We neglect the interaction of molecules at a distance. In problems of thermodynamics, real gases are often taken for ideal gases. 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 ideal gas molecules? 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 general state 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.

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

Here m is the mass of gas, M - its molecular weight (we find according to 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's 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 the gas pressure and the average kinetic energy E of its particles. This connection is called the basic equation of molecular kinetic theory and looks like:

Here n is the concentration of molecules (the ratio of their number to volume), E – medium kinetic energy. You can find them, as well as the root-mean-square velocity of molecules, respectively, using the formulas:

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

First law of thermodynamics. Formulas for isoprocesses

We remind you that the first law of thermodynamics says: the amount of heat transferred to a 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 this case, we are interested in such processes that occur at one constant parameter. Consider what the first law of thermodynamics looks like in each of them.

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

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

runs at 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 pressure, the volume of a gas is directly proportional to its temperature. In an isobaric process, heat goes both to change the internal energy and to do work on the gas.

. An adiabatic process is a process that takes place without heat exchange with environment. This means that the formula for 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 is equal to the amount of heat required to raise one kilogram of a substance by one degree Celsius.

Apart from specific heat, there is molar heat capacity (the amount of heat required to raise the temperature of 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 a monatomic gas i=3, for a diatomic gas - 5.

Thermal machines. Efficiency formula in thermodynamics

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

Coefficient useful action heat engine is calculated by the formula

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

Work in thermodynamics

In thermodynamics, unlike mechanics, it is not the motion of a body as a whole that is considered, but only a relative change in the parts of a thermodynamic system, as a result of which its volume changes.

Consider the work of a gas during isobaric expansion.

Let us calculate the work done by the gas when it acts on the piston with a force $(F")↖(→)$ equal in magnitude and opposite in direction to the force $(F")↖(→)$ acting on the gas from the piston: $ (F")↖(→)=-(F")↖(→)$ (according to Newton's third law), $F"=pS$, where $p$ is the gas pressure and $S$ is the surface area of ​​the piston. If the displacement of the piston $∆h$ as a result of the expansion is small, then the gas pressure can be considered constant and the work of the gas is:

$A"=F"∆h=pS∆h=p∆V$

If the gas expands, it does positive work, since the movement of the piston coincides in direction with the force $(F")↖(→)$. If the gas is compressed, then the work of the gas is negative, since the movement of the piston is opposite to the force $(F")↖ (→)$. A minus sign will appear in the formula $A"=F"∆h=pS∆h=p∆V$: $∆V

The work of external forces $A$, on the contrary, is positive when the gas is compressed and negative when expanding:

Doing positive work on the gas, external bodies transfer part of their energy to it. When the gas expands, external bodies take away part of its energy from the gas - the work of external forces is negative.

On the plot of pressure versus volume $p(V)$, work is defined as the area bounded by curve $p(V)$, axis $V$, and segments $ab$ and $cd$ equal to pressures $p_1$ in the initial ($V_1 $) and $р_2$ in the final ($V_2$) states, both for isobaric and isothermal processes.

First law of thermodynamics

The first law (first law) of thermodynamics is the law of conservation and transformation of energy for a thermodynamic system.

According to the first law of thermodynamics, work can only be done by heat or some other form of energy. Therefore, work and the amount of heat are measured in the same units - joules (as well as energy).

The first law of thermodynamics was formulated by the German scientist J. L. Mayer in 1842 and experimentally confirmed by the English scientist J. Joule in 1843.

First law of thermodynamics is formulated like this:

The change in the internal energy of the system during its transition from one state to another is equal to the sum of the work of external forces and the amount of heat transferred to the system:

where $∆U$ is the change in internal energy, $А$ is the work of external forces, $Q$ is the amount of heat transferred to the system.

From $∆U=A+Q$ it follows law of conservation of internal energy. If the system is isolated from external influences, $A=0$ and $Q=0$, and hence $∆U=0$.

For any processes occurring in an isolated system, its internal energy remains constant.

If the work is done by the system, and not by external forces, then the equation ($∆U=A+Q$) is written as:

where $A"$ is the work done by the system ($A"=-A$).

The amount of heat transferred to the system is used to change its internal energy and to perform work on external bodies by the system.

The first law of thermodynamics can be formulated as the impossibility of the existence of a perpetual motion machine of the first kind, which would do work without drawing energy from any source, i.e. only due to internal energy.

Indeed, if the body does not receive heat ($Q=0$), then the work $A"$, according to the equation $Q=∆U+A"$, is performed only due to the loss of internal energy $A"=-∆U$ • After the energy supply is exhausted, the engine stops working.

It should be remembered that both work and the amount of heat are characteristics of the process of changing internal energy, so it cannot be said that the system contains a certain amount of heat or work. The system in any state has only a certain internal energy.

Application of the first law of thermodynamics to various processes

Consider the application of the first law of thermodynamics to various thermodynamic processes.

isochoric process. The dependence $p(T)$ on the thermodynamic diagram is shown isochore.

An isochoric (isochoric) process is a thermodynamic process that occurs in a system at a constant volume.

The isochoric process can be carried out in gases and liquids enclosed in a constant volume vessel.

In an isochoric process, the volume of the gas does not change ($∆V=0$), and, according to the first law of thermodynamics $Q=∆U+A"$,

i.e., the change in internal energy is equal to the amount of heat transferred, since work ($A=p∆V=0$) is not performed by the gas.

If the gas is heated, then $Q > 0$ and $∆U > 0$, its internal energy increases. When the gas is cooled $Q

Isothermal process graphically depicted isotherm.

An isothermal process is a thermodynamic process that occurs in a system at a constant temperature.

Since the internal energy of the gas does not change during an isothermal process ($T=const$), then all the amount of heat transferred to the gas goes to work:

When the gas receives heat ($Q > 0$), it does positive work ($A" > 0$). If the gas gives off heat to the environment, $Q

isobaric process the thermodynamic diagram shows isobar.

Isobaric (isobaric) process - a thermodynamic process occurring in a system with constant pressure$p$.

An example of an isobaric process is the expansion of a gas in a cylinder with a freely moving loaded piston.

In an isobaric process, according to the formula $Q=∆U+A"$, the amount of heat transferred to the gas goes to change its internal energy $∆U$ and to perform work $A"$ at constant pressure:

The work of an ideal gas is determined from the graph of $p(V)$ for an isobaric process ($A"=p∆V$).

For an ideal gas in an isobaric process, the volume is proportional to the temperature; in real gases, part of the heat is spent on changing the average interaction energy of particles.

adiabatic process

An adiabatic process (adiabatic process) is a thermodynamic process occurring in a system without heat exchange with the environment ($Q=0$).

Adiabatic isolation of the system is approximately achieved in Dewar vessels, in the so-called adiabatic shells. On adiabatically isolated system does not affect the change in temperature of the surrounding bodies. Its internal energy can change only due to the work done by external bodies on the system, or the system itself.

According to the first law of thermodynamics ($∆U=A+Q$), in an adiabatic system

where $A$ is the work of external forces.

With the adiabatic expansion of the gas $A

Consequently,

$∆U=(i)/(2)(m)/(M)R∆T

which means a decrease in temperature during adiabatic expansion. It leads to the fact that the gas pressure decreases more sharply than in an isothermal process.

In the figure, the adiabat $1-2$, passing between two isotherms, clearly illustrates what has been said. The area under the adiabat is numerically equal to the work done by the gas during its adiabatic expansion from volume $V_1$ to $V_2$.

Adiabatic compression leads to an increase in gas temperature, because as a result of elastic collisions of gas molecules with a piston, their average kinetic energy increases, in contrast to expansion, when it decreases (in the first case, the speeds of gas molecules increase, in the second they decrease).

Rapid heating of air during adiabatic compression is used in Diesel engines.

The principle of operation of heat engines

A heat engine is a device that converts the internal energy of a fuel into mechanical energy.

According to the second law of thermodynamics, a heat engine can continuously perform a periodically repeating mechanical work due to the cooling of the surrounding bodies, if it not only receives heat from a hotter body (heater), but at the same time gives off heat to a less heated body (refrigerator). Consequently, not all the amount of heat received from the heater is used to perform work, but only part of it.

Thus, the main elements of any heat engine are:

1. working fluid (gas or steam) that performs work;
2. a heater that imparts energy to the working fluid;
3. a refrigerator that absorbs part of the energy from the working fluid.

Heat engine efficiency

According to the law of conservation of energy, the work done by the engine is:

$A"=|Q_1|-|Q_2|$

where $Q_1$ is the amount of heat received from the heater, $Q_2$ is the amount of heat given to the refrigerator.

Efficiency(Efficiency) of a heat engine is the ratio of the work $A "$ performed by the engine to the amount of heat received from the heater:

$η=(A")/(|Q_1|)=(|Q_1|-|Q_2|)/(|Q_1|)=1-(|Q_2|)/(|Q_1|)$

Since in all engines some amount of heat is transferred to the cooler, then $η

thermal efficiency engine is proportional to the temperature difference between the heater and the cooler. With $T_1 - T_2=0$ the motor cannot run.

Carnot cycle

The Carnot cycle is a circular reversible process consisting of two isothermal and two adiabatic processes.

This process was first considered by the French engineer and scientist N. L. S. Carnot in 1824 in the book Reflections on driving force fire and about machines capable of developing this force.

The purpose of Carnot's research was to find out the reasons for the imperfection of heat engines of that time (they had an efficiency of $< 5%$)и поиски путей их усовершенствования.

The choice of two isothermal and two adiabatic processes was due to the fact that the work of the gas during isothermal expansion is performed due to the internal energy of the heater, and when adiabatic process due to the internal energy of the expanding gas. In this cycle, contact of bodies with different temperatures, therefore, heat transfer without work is excluded.

The Carnot cycle is the most efficient of all. Its efficiency is maximum.

The figure shows the thermodynamic processes of the cycle. In the process of isothermal expansion ($1-2$) at a temperature $Т_1$, work is done due to a change in the internal energy of the heater, i.e., due to the supply of heat $Q_1$ to the gas:

$A_(12)=Q_1.$ Cooling of gas before compression ($3-4$) occurs during adiabatic expansion ($2-3$). The change in internal energy $∆U_(23)$ in an adiabatic process ($Q=0$) is completely converted into mechanical work:

$A_(23)=-∆U_(23)$

The temperature of the gas as a result of adiabatic expansion ($2-3$) decreases to the refrigerator temperature $T_2

The cycle is completed by the process of adiabatic compression ($4-1$), during which the gas is heated to a temperature of $T_1$.

The maximum value of the efficiency of heat engines operating on ideal gas, according to the Carnot cycle:

$η=(T_1-T_2)/(T_1)=1-(T_2)/(T_1)$

The essence of the formula $η=(T_1-T_2)/(T_1)=1-(T_2)/(T_1)$ is expressed in the theorem proved by S. Carnot that the efficiency of any heat engine cannot exceed the efficiency of the Carnot cycle carried out at the same temperature of the heater and refrigerator.

··· Oryol Issue ···

G.A.BELUKHA,
school number 4, Livny, Oryol region

The work of a gas in thermodynamics

When studying the work of a 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 already with the study of work in mechanics and, for this purpose, solving problems for the work of a variable force by summing elementary works along the whole path using integration.

For example, when calculating the work of the Archimedes force, the elastic force, the universal gravitational force, etc. one must learn to sum elementary quantities with the help of simple differential relations of the type dA = fds. Experience shows that high school students can easily cope with this task - the arc of the trajectory on which the force increases or decreases must be divided into such intervals ds, on which the 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 facilitates the adaptation of future students to the perception of a physics course at a university and eliminates the methodological difficulties associated with the ability to find the work of a variable force in thermodynamics, etc.

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

Having learned that work is one of the ways to change internal energy, tenth graders easily calculate the work of a gas in an isobaric process. At this stage, it must be emphasized that the gas pressure force does not change all the way, and according to Newton's third law | F 2 | = |F 1 |, we find the sign of work 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, a gas is compressed under a piston, the pressure changes, and at any given time

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

because V 2 < V 1 .

The resulting formula can be rewritten using the relation

Let's solve the problem to fix it.

1. The gas passes from the state 1 (volume V 1 , pressure R 1) in state 2 (volume V 2, pressure R 2) in a process in which its pressure depends linearly on volume. Find the work of the gas.

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

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

Solution

With an isobaric process Ap = R 0 V, but R 0 = RT 0 /V 0 , V = V 0 , so Ap = RT 0 .

In an isothermal process:

Compare:

Having studied the first law of thermodynamics and its application to isoprocesses and having fixed the topic of work in thermodynamics by solving problems, the students prepared themselves for the perception of the most difficult part of thermodynamics, “Operation of cycles and efficiency of heat engines”. I present this material in the following sequence: the work of cycles - the Carnot cycle - the efficiency of thermal engines - circular processes.

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

The figure shows a pressure graph p from the volume V(diagram p, V) for some cycle 1–2–3–4–1. On the plots 1–2 and 4–1 gas expands and does positive work BUT 1, numerically equal to the area of ​​the figure V 1 412V 2. Location on 2–3–4 gas compresses and does work BUT 2 , whose modulus is equal to the area of ​​​​the figure V 2 234V one . Total gas work per cycle BUT = BUT 1 + BUT 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-diagram that goes around clockwise, then the work of the body is positive, and the cycle is called straight. If a closed curve on R, V-diagram is bypassed counterclockwise, then the gas does negative work per cycle, and the cycle is called reverse. In any case, the module of gas work per cycle is equal to the area of ​​the figure bounded by the cycle graph on R, V-diagram.

In a circular process, the working body returns to its original state, i.e. into a state of original 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, then Q = A. So, the algebraic sum of all the amounts of heat received per cycle is equal to the work of the body per cycle: A c = Q n + Q x = Q n - | Q x |.

Consider one of circular processes- Carnot cycle. It consists of two isothermal and two adiabatic processes. Let the working fluid be an ideal gas. Then on 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 a change in internal energy, because Q hell=0= U 23 + A g23 A r23 = - 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). Thanks to a rather slow process and good contact with the refrigerator, the gas has time to give the energy received 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 compressed adiabatically (without heat transfer) to its original state. At the same time, he does negative work, and external forces do positive work: 0 = U 41 + A g41 A r41 = - U 41 .

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

Heat is given off to the refrigerator only when the gas is isothermally compressed in the area 3–4 :

According to the first law of thermodynamics

A c = Q n - | Q x|;

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

According to the Boyle–Mariotte law for processes 1–2 and 3–4 , as well as the Poisson equation for the 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:

The work of heat engines operating on the reverse cycle is methodically correct, as experience shows, to study using the example of the work of the reverse Carnot cycle, because it is reversible and can be carried out in reverse direction: to expand the gas as the temperature drops from T n to T x (process 1–4 ) and at low temperature T x (process 4–3 ) and then compress (processes 3–2 and 2–1 ). The engine is now doing work to power the chiller. The working fluid takes away the amount of heat Q x for food inside at low temperature T x, and gives off the amount of heat Q on surrounding bodies, outside the refrigerator, at a higher temperature T n. Thus, a machine operating according to the reverse Carnot cycle is no longer a thermal machine, but an ideal refrigeration machine. The role of the 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 on a direct cycle, we can represent the block diagram of the refrigerator in the following form:

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

The most important characteristic of the refrigerator is the coefficient of performance, which determines the efficiency of the refrigerator and is equal to the ratio of the amount of heat taken from the refrigerator Q x to the expended energy of an external source

In one reverse cycle, the working fluid receives from the refrigerator the amount of heat Q x and gives off to the surrounding space the amount of heat Q n what more Q x to work A dv performed by the electric motor on the gas per cycle: | Q n | = | Q x | + BUT 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 the engine windings are heated by electric current Q R and friction in the scheme BUT tr.

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

Considering that in the direct cycle

after simple transformations we get:

The last relation between the coefficient of performance and the efficiency of a heat engine, which can also operate in reverse cycle, shows that the coefficient of performance can be greater than one. In this case, the heat is taken away from the refrigerator and returned to the room more than the energy used by the engine.

In the case of an ideal heat engine operating on the reverse Carnot cycle (an ideal refrigerator), the coefficient of performance 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 freon evaporation temperature is t x °С, radiator temperature t n °C. The cost of one kilowatt-hour of electricity is equal to C. The 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 the amount of heat from it Q = m(ct+ ). We consider approximately that the reverse Carnot cycle takes place over freon with isotherms at temperatures T n and T X. We use the 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 together the last three equations:

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

To fix 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 by a mole of a monatomic ideal gas in an isothermal process.

Solution

When solving problems in which the cycle efficiency is involved, it is useful to preliminarily analyze all sections of the cycle using the first law of thermodynamics and identify areas where the body receives and gives off 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 - incl. 3 . Let's denote them by T 1 and T 3 respectively.

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

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

Using 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 transfer, i.e. Q 31 = 0 and, according to the first law of thermodynamics, 0 = ( U 1 – U 3) + A 31 . Then the work done by the gas
A 31 = U 3 – U 1 = c V(T 3 –T 1) = –c V T.

So, for the cycle, the gas did the work A 12 + BUT 31 = BUT 12 – c V T and got heat only on the plot 1–2 . cycle efficiency

Since the work done by the gas on the isotherm is

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