Practical lesson 1, 2. Solving problems according to equilibrium thermodynamics(4 hours)

Lesson plan: Second law of thermodynamics. Formulations of the second law (law) of thermodynamics. Entropy and its properties. Calculation of entropy changes in various equilibrium processes. Entropy in nonequilibrium processes. Examples of problem solving.

The concept of entropy

Entropy, denoted by the letter S, was introduced by R. Clausius when analyzing material on heat engines, initially in the form of the so-called “reduced heat”

where Q is the amount of heat exchanged between the system and the environment during a reversible process; T - temperature.

From (1) it follows that entropy is measured in joules per Kelvin (J/K). Entropy is a function of state, i.e. its change does not depend on the path along which this change occurs, but is determined only by the difference in entropy values ​​in the final and initial states. Also, entropy. extensive property of the system. This means that the entropy of the entire system (S) can be found by summing the entropy values ​​of all components systems () :

Clausius proved that, despite the fact that heat is a function of the transition and depends on the path of the process, the value of the ratio of heat to absolute temperature does not depend on the process path, i.e. it is a function of the state. This state function is entropy. From the position that entropy is a function of state it follows that an infinitesimal change in entropy is a total differential, and a finite change in entropy as a result of some process can be found as

(3)

Let us write expression (1) for infinitesimal quantities

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Formulas (3 −5) − basic formulas to calculate entropy. Having revealed the meaning, dQSΔ can be calculated in each specific case.

Following Clausius's reasoning, one can always introduce a new positive value, which is the difference between TdS and at irreversible change. This value is determined by the relation:

Equation (6) can be rearranged

where for reversible changes and 0=′Qd0>′Qd for irreversible.

Clausius called it uncompensated heat. IN classical thermodynamics given value was more of a qualitative nature. Usually it was simply stated that Qd′0=′Qd for reversible changes and for irreversible changes, and the value 0>′QdQd′ was not calculated.

By the way, the term “uncompensated heat” is not entirely appropriate. Heat is the energy that a system exchanges with the outside world and, therefore, that passes through the surface that limits the system from outside world. And that heat, which Clausius called uncompensated, arises as a result of processes occurring within the system itself.

It should be noted that on modern stage development physical chemistry possible quantification increase in entropy during irreversible processes.

One of the formulations of the second law of thermodynamics is as follows: for any isolated system that is in a nonequilibrium state, the entropy increases over time, and its growth continues until the system reaches an equilibrium state.

This law is also called the law of increasing entropy. Mathematically it can be written in the form

where the inequality sign refers to a nonequilibrium process, and the equal sign refers to equilibrium.

From expression (8) it follows that the entropy of an isolated system can only increase, but can never decrease. At equilibrium, entropy is maximum.

All calculations of entropy changes in various processes are based on the use of the Clausius inequality, which relates the change in entropy dS to the amount of heat that the system exchanges with the environment at temperature T

Sources of irreversible processes can be: diffusion, expansion of the system when there is a pressure difference between it and the environment, heat transfer when different temperatures, spontaneous chemical reactions in the volume of the system and other dissipative processes associated with the irreversible conversion of work into heat. Inequality (9) is satisfied regardless of the cause of the irreversible process; as a result, a release is observed within the system additional quantity warmth. As mentioned earlier, R. Clausius called this heat caused by nonequilibrium processes uncompensated heat.

It is known that if the process is carried out in equilibrium and reversibly, then the work done is maximum. If the process is irreversible, then the work is less than in reversible process, part of it seems to be lost. According to the first law of thermodynamics, “lost” work must appear in another form, for example, in the form of uncompensated heat, which is always non-negative: greater than zero in irreversible processes, is equal to zero in reversible processes.

At isothermal processes inequality (10) can be written as equalities

Where is the change in entropy caused by equilibrium heat exchange with the environment (index “e” from Latin external. external);

growth of entropy due to irreversible processes within the system (index “i” from the Latin internal - internal).

The entropy value of a given system cannot be measured directly experimentally, but it can be calculated using the formula

This formula allows you to find absolute value entropy, but the difference in entropy in two states of the system, i.e., the change in entropy during the transition of the system from state 1 to state 2.

In table Table 1 shows the basic relationships characterizing the change in entropy in various processes. Analysis of the table 1 shows that for any system (isolated, closed or open) the change in entropy due to internal reasons not negatively, that is, fair. 0≥Sdi

Nonequilibrium thermodynamics (thermodynamics of nonequilibrium processes) studies general patterns systems in which nonequilibrium processes occur: heat transfer, diffusion, chemical reactions, transfer electric current etc.

Classical thermodynamics as a science of mutual transformation work and energy studies equilibrium processes. Let us briefly dwell on the features of nonequilibrium thermodynamics.

When considering the second law of thermodynamics for open systems, too great attention is paid to the change in entropy. Entropy change open system can occur either due to the occurrence of internal irreversible processes within the system itself (), or due to processes of exchange between the system and the external environment (). SdiSde

For systems considered in chemistry, a change can be caused, for example, by the flow chemical reaction inside the system, and the magnitude. supply or removal from the system of both heat and reagents and products. SdiSde

In the thermodynamics of nonequilibrium processes, it is postulated that the components and are independent, and overall change the entropy of an open system is equal to their sum:

If only thermally reversible changes occur in the system, then = 0. If there are irreversible changes > 0.

In isolated systems there is no heat and mass exchange with the environment and value = 0, then equation (13) is transformed to the following form:

i.e., to the classical formulation of the second law of thermodynamics for isolated systems.

Any nonequilibrium process in the system: mixing of gases, spontaneous straightening of a compressed spring, chemical reaction - leads to increased disorder molecular state systems. This is quantitatively expressed in the growth thermodynamic probability state of the system and in increasing entropy.

An important characteristic of nonequilibrium processes is that they occur with terminal speed. Their study is essentially the field of kinetics. Time in nonequilibrium thermodynamics is a parameter.

So, if entropy arises in time τ, then we need to talk about the rate of its occurrence τdSdi. IN foreign literature this quantity is called entropy production. The quantity τdSde is called the rate of entropy exchange between the system and the environment. With the introduction of these concepts, from equation (13) we obtain an equation for the rate of total entropy change in the system τddS:

A positive value of τdSde corresponds to an increase in entropy as a result of the exchange of matter and/or energy with the external environment. A negative value of τdSde indicates that the outflow of entropy from the system external environment exceeds the influx of entropy from outside. This shows a fundamental difference in the thermodynamic properties of open and isolated systems: the total entropy of an open system can either increase or decrease, since the value τdSde can be either positive or negative.

The inequality τdSdi> 0 is always valid, but with respect to the general increase in entropy the following cases are possible:

IN the latter case installed in the system steady state, in which the production of entropy in the system due to irreversible internal processes is compensated by the outflow of entropy into the external environment.

Calculation of entropy change with temperature change

The change in entropy of a system whose temperature, for example, increases at constant volume from T1 to T2, is calculated by integrating the partial derivative of entropy with respect to temperature at constant volume

Where. isochoric heat capacity of the system, J/K. V.C.

If no phase transformations occur in the system in the specified temperature range, then integration (15) leads to the following expression:

Entropy change at constant pressure can be found by integrating the partial derivative of entropy with respect to temperature at constant pressure

Where. isobaric heat capacity of the system, J/K.

When solving equations (16) and (18), two cases are possible. Let us consider them using equation (18) as an example, i.e., with an isobaric temperature change.

Case 1. The heat capacity of a substance in the temperature range from to does not depend on temperature. Then after integration (18) we have

Constant heat capacity is most often characteristic of ideal gases, it is in this case (unless otherwise stated in the conditions of the problem) that the change in entropy during isobaric heating can be calculated using formula (19). According to classical theory heat capacities of ideal gases, it can be assumed that the molar isochoric heat capacities for monatomic and diatomic ideal gas are equal respectively

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However, it must be borne in mind that sometimes for ideal gases a dependence of the heat capacity on temperature is observed (this happens at high temperatures).

Case 2. Heat capacity is some function of temperature.

The temperature dependence of the molar isobaric heat capacity is usually expressed by power series of the form

where are empirically found coefficients. Their meanings are given in reference literature. Usually in something like this power series only three terms are taken into account: or c, c,b, a′c, b,ac, b,a′ - depending on which class the substance belongs to: class organic matter or class of inorganic. Exactly what coefficient or cс′ needs to be taken into account in equation (23) follows from the reference data table, which lists all the coefficients. It is obvious that if, for example, the coefficients are given, then the coefficient c, b, ac′ will be equal to zero.

Substituting expression (23) into (18) allows, after integration, to obtain the following expression

Using formula (24), the change in the entropy of a substance is calculated when its temperature changes from T1 to T2. This change applies to one mole; if found complete change entropy, you need to use the formula snSΔ=Δ, where n is the number of moles.

Calculation of entropy change during phase transition

During various phase transitions: crystallization, melting, evaporation, sublimation, etc., the degree of ordering of the system changes, i.e. phase transition accompanied by a change in entropy. For example, during evaporation, a compact condensed phase turns into a gas, which occupies a much larger volume. In this case, there should be a significant increase in the entropy of the substance. Entropy increases when going from crystalline state to liquid and from liquid to gas.

Let us consider a system that is clean water and the environment when normal temperature phase transition, i.e. at a temperature when two phases are in equilibrium at 1 atm. For the ice melting process (equilibrium: hard water− liquid water), this temperature is 273 K, and for the evaporation process (equilibrium: liquid water − steam) 373 K. Since both phases in each of the phase transitions under consideration are in equilibrium, then any heat exchange between the system and the environment occurs reversibly. At constant pressure, the amount of heat will correspond to the enthalpy, therefore the molar entropy of the phase transformation f. n.sΔ can be calculated using the formula

where is the molar enthalpy of phase transformation; f. p.hΔ

f. p.T is the temperature of phase transformation.

During crystallization or condensation, exothermic phase transformations are observed (< 0), характеризующиеся negative value entropy. In this case, there is a decrease in disorder during the transition from liquid to solid (crystallization process) or from vapor to liquid (condensation process). The change in entropy during the endothermic transition (> 0) is positive, and the system becomes more disordered. Melting and evaporation are endothermic processes accompanied by an increase in the entropy of the system. f. p.hΔ

In table Table 2 shows the values ​​of the molar entropy of phase transitions of some substances.

As can be seen from table. 3, which gives the standard molar entropy of evaporation of some liquids at boiling temperatures, many liquids have approximately the same standard entropy of evaporation of about 85 J/(mol. K). Such empirical pattern called Truton's rule. Trouton's rule can be explained if we assume that when various liquids evaporate and turn them into gas, the degree of disorder turns out to be practically the same, regardless of the substance. If this is true, then most liquid substances should have close values ​​of the standard molar entropies of evaporation.

Significant deviations from Trouton's rule are observed in liquids in which partial association of molecules occurs. For example, in water, in which there are hydrogen bonds and structuring of the liquid, during evaporation the system becomes more disordered than if the molecules were randomly distributed in the volume liquid phase.

Calculation of the change in entropy of a system with a change in temperature and the presence of a phase transformation

Since entropy is a function of state, the path of transition from the initial state to the final state can be any. Possible option- a reversible isobaric temperature change to the phase transition temperature, then the phase transition itself, and then a reversible isobaric temperature change to the final temperature.

The total change in entropy will consist of three terms. For example, let a certain conditional substance A transition isobarically from a solid state at a temperature of liquid state at temperature. Already from the conditions of the problem it is clear that at a certain temperature a phase transformation (phase transition) - melting - takes place. We find this temperature in the reference book - we designate it in general view. The first contribution to the total entropy change is the heating of the solid system from temperature to the melting point

At the melting temperature, the solid state transforms into a liquid state, the entropy changes abruptly, i.e., the total change in entropy will include a contribution in the form of a change in entropy during the phase transformation

And finally, the last contribution to the overall change in entropy will be made by a change in entropy with further heating of the already formed liquid

Obviously,

Calculation of entropy during chemical reactions

Examples of problem solving

Example 1. Calculate the change in entropy of 1 kg of ethylene glycol when it is heated from a temperature of 100 to 300 °C.

Using the reference book, we determine that the temperature of the ethylene glycol (liquid)–ethylene glycol (gas) phase transition is 479.4 K. We convert the temperature of the problem condition into thermodynamic scale− Kelvin scale and we conclude that the temperature interval from the problem conditions must be divided into two intervals. The first from 373 K to 479.4 K characterizes the state of ethylene glycol in the form of a liquid, and the second interval from 479.4 K to 573 K will correspond gaseous state ethylene glycol.

Taking into account the above and the values ​​​​given in the reference book, we obtain

The value of the entropy of phase transformation, i.e. the value of the entropy of evaporation

Example 2. What is the molar entropy of ethylene glycol at 350 K?

Let's use the basic calculation formula

Moreover, as the lower level of reference we will take the absolute value of entropy at 298 K and a pressure of 1 atm (298) stks (this is a reference value equal to 167.32 J/(mol. K)). It is known that at 298 K and 350 K ethylene glycol is in a liquid state, its molar isobaric heat capacity is constant and equal, and as the lower level of reference we take the absolute value of entropy at 298 K and a pressure of 1 atm (298) stks (this is a reference value equal to 167 .32 J/(mol⋅K)). It is known that at 298 K and 350 K ethylene glycol is in a liquid state, its molar isobaric heat capacity is constant and equal to https://pandia.ru/text/80/204/images/image047_2.jpg" width="474" height=" 71">

Example 3. There is an ideal gas - benzene, its mass is 0.4 kg. It is in state 1 at a temperature of 600 K and a pressure of 2.5 atm. It was transferred to state 2, at which the temperature is 298 K and the pressure is 1 atm. What are the changes in the molar and total entropy of the system during this transition?

From the reference book we determine the coefficients of the temperature dependence of the molar isobaric heat capacity

Find the change in molar entropy of an ideal gas at this temperature

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Since the expansion is reversible, the total change in entropy of the Universe is 0, so the change in entropy environment equal to the change in gas entropy with the opposite sign:

b) Entropy. state function, therefore the change in the entropy of the system does not depend on how the process occurred. reversible or irreversible. The change in gas entropy during irreversible expansion against external pressure will be the same as during reversible expansion. But the entropy of the environment will be different. It can be found by calculating the heat transferred to the system using the first law:

In this derivation we used the fact that ΔU = 0 (temperature is constant). Work done by the system against constant external pressure 38

is equal to A = − p(V2−V1), and the heat accepted by the environment is equal to the work done by the system, with the opposite sign.

The total change in entropy of the gas and the environment is greater than 0:

as expected for an irreversible process.

Example 5. Calculate the change in entropy of 1000 g of water as a result of its freezing at −5 °C. The enthalpy of melting of ice at 0°C is 6008 J/mol. The heat capacities of ice and water are considered constant in this temperature range and equal to 34.7 and 75.3 J/(mol⋅K), respectively. Explain why entropy decreases during freezing, although the process is spontaneous.

О The irreversible process of freezing water at a temperature of −5 C can be represented as a sequence of reversible processes:

1) heating water from −5°C to freezing temperature (0°C);

2) freezing of water at 0 °C;

3) cooling ice from 0 to −5 °C:

The change in the entropy of the system in the first and third processes (with a change in temperature) is calculated using formula (18):

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Since entropy is a function of state, the total change in entropy is equal to the sum over these three processes:

Entropy decreases during freezing, although the process is spontaneous. This is due to the fact that heat is released into the environment and the entropy of the environment increases, and the increase is greater than 1181 J/K; as a result, the entropy of the Universe increases when water freezes, as expected in an irreversible process.

Example 6. One vessel with a capacity of 0.1 m3 contains oxygen, and another vessel with a capacity of 0.4 m3. nitrogen In both vessels the temperature is 290 K and the pressure is 1.013.105 Pa. Find the change in entropy during mutual diffusion of gases from one vessel to another at constant pressure and temperature. Consider both gases to be ideal.

The change in entropy is determined by formula (74). The number of moles of each gas is found using the Mendeleev-Clapeyron equation:

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Example 7. The enthalpy of evaporation of chloroform is 29.4 kJ/mol in normal point boiling (334.88 K). Calculate the molar entropy of evaporation at this temperature.

We find the change in entropy using formula (25):

4. The system transitions from state 1 to state 2 in two ways: reversible (absorbs heat Qrev) and irreversible (absorbs heat Qrev). What is the relationship between Qrev and Qrev? What is the relationship between ΔSrev and ΔSrev?

5. Determine the number of microstates and the entropy of a sodium crystal weighing 2.3 g at 0 K if it contains three potassium atoms replacing sodium atoms in its structure.

6. For one mole of substance AB, calculate the number of microstates and entropy for an ideal crystal at 0 K, as well as for a crystal that has a single defect: one D− ion replaced one B− ion at a crystal lattice site.

7. In which of the following cases: a) in a process in which ΔCp = 0; b) during a reversible isothermal phase transition; c) in an adiabatic process; d) in the process at constant pressure?

8. Write down an expression to calculate the change in entropy during a phase transition. Formulate Truton's rule and indicate the scope of its application.

9. Why is the term “entropy of education” not used in practice? If we introduce such a term by analogy with the enthalpy of formation, then what sign will the entropy of formation have: a) sugar, b) molecular oxygen?

10. Estimate the change in entropy in the reaction CO2(g) + 4H2(g) → CH4(g) + 2H2O(l) at 298 K and partial pressures gaseous substances, equal to 2 atm, if ΔrS o 298 = − 98 cal mol−1 K−1. Please indicate the assumptions used.

11. Find the change in entropy during the isothermal expansion of one mole of van der Waals gas from volume V to volume 2V.

12. In the temperature range from 298 to 1000 K, the heat capacity of substance A is described by the following equation: Cp = a + bT + cT 2, where a, b and c are constants for substance A. Calculate the change in enthalpy and entropy of the substance when heated from 300 to 500 K.

13. Calculate the change in entropy when heating 1.5 mol Ni from 25°C to 1450°C. The molar heat capacities of Ni are given by the equations: Cp(α-Ni) = 16.99 + 29.46 ⋅10−3T , J mol−1 K−1, Cp(β-Ni) = 25.19 + 7.53 ⋅ 10−3T , J mol− 1·K−1. The temperature of the polymorphic transition of α-Ni to β-Ni is 360°C, the heat of the α→β transition ΔHo = 0.38 kJ mol−1.

14. In the temperature range from 0 to 12 K, the heat capacity of silver is well described by Debye’s “law of cubes” CV = αT 3, and it is known experimental value heat capacity of silver at 12 K Cp, 12. How to calculate the change in enthalpy and entropy of silver when heated from 0 to 12 K? Is it possible to calculate the absolute values ​​of H12 and S12 for silver?

16. Write down an equation to calculate the entropy of mixing of two ideal gases at constant temperature. Why is this value always positive?

17. Calculate the change in entropy in isolated system by adding 100 g of ice with a temperature of 0°C to 1000 g of water with a temperature of 20°C. The heat capacity of liquid water is 4.184 J g−1 K−1, the heat of fusion of ice is 6.0 kJ mol−1.

18. Calculate the temperature of the mixture and the change in entropy during the mixing of 3 kg of water at 353 K with 6 kg of water at 290 K. Assume that the heat capacity of water Cp is equal to 75.3 J mol−1 K−1 and does not depend on temperature.

19. Ice weighing 1 g at 0°C is added to 10 g of water, the temperature of which is 100°C. What is the final temperature of the mixture and what is the change in entropy during this process? The enthalpy of melting of ice is 80 cal g−1, Cp of water is 1 cal g−1 K−1.

20. The enthalpy of melting of ice at 273 K is 1436 cal mol−1, the Cp of ice and liquid water are 8.9 and 18.0 cal mol−1 K−1, respectively. Calculate the change in entropy in the nonequilibrium process of solidification of water at 263 K.

Calculate the change in entropy during the melting of three moles of supercooled benzene at 270 K, if at 278.65 K its enthalpy of melting is equal to 2379.5 cal mol−1, the heat capacities of liquid and solid benzene are equal to 30.4 and 29.5 cal mol−1 K−1, respectively, and the pressure constant and equal to 1 atm.

Literature

1. Shapovalov synergetics: Macroscopic approach M.: Firm "Ispo-Service" 2000. Ch. 2-3

2. Bazarov M.: Higher. school 1991. Chapter 3

3. Zelentsov “OOOFizikon” 2002 Media disc

(simple, complex).

Entropy (S) - state function, quantitatively characterizing the degree of disorder of the system. It is usually referred to as a mole of a substance. This statistical value, therefore it is associated with thermodynamic probability.

S= R *lnW [J/mol*K] (Louis Boltzmann formula)

R-gas constant =8.314 J/mol∙K,

W- thermodynamic probability (this is the number of microstates by which a given state of a macrosystem can be realized) or: the number of ways in which a given system can be constructed.

6 Particles (6 ions):

order state: 1,2,3,4,5,6 W=1 S=0

disorder state: W=6! -1 =719 S>>0

S of a real system is always greater than 0; a state of disorder is much more likely.

For the perfect crystal and S = 0. Δ S 0 reactions = ∑νn Δ S 0 products - ∑νn Δ S 0 original substances

For processes involving gaseous x in-TV sign S is determined by the ratio of gaseous moles in the reaction.

Standard entropy of matter– the absolute value of the entropy of a substance under standard conditions in any given state of aggregation.

Approximate sign score Δ S reactions: can be assessed by the change in the number of moles of gaseous substances in the reaction, since they make the main contribution to the entropy of the system.

Calculation of entropy change in a chemical reaction

The connection between entropy and composition of the substance

1) The more complex the composition and structure of a substance (more electrons, atoms, mass), the greater the entropy. S(UU 2) S(Li)

2) the stronger chemical bonds in a substance, the lower the entropy, the lower the mobility of particles. S(Sgraphite)>S(Salmaz)

3) With increasing T of the transition of particles from solid to liquid and then to gaseous, entropy increases.

4) Nernst's postulate. At T=0 the entropy of any pure substance= 0, since there is no movement => all substances take on the state of an ideal crystal.

Δ S reaction characterizes the tendency of systems to the most probable state, i.e. to a state with max entropy

Calculation of Δ S

Δ S 0 reactions = ∑νn Δ S 0 products - ∑νn Δ S 0 original substances

Conclusions:

1) The S dimension characterizes the system’s tendency to the most probable state with the greatest disorder (with the highest S)

2) A change in S is not an unambiguous criterion for the possibility of a spontaneous process.

Non-spontaneous Processes are those that require external work to complete.

The non-spontaneous process leads to a decrease in order in the system and is characterized by a decrease in S.

S<0 – несамопроизвольный процесс

Spontaneous Processes are processes that can occur without the expenditure of external work (exchange, neutralization).

A spontaneous process leads to an increase in order in the system and is characterized by an increase in S.

S>0 – spontaneous process

15. Determination of the direction of chemistry according to the thermodynamic functions of the state. Gibbs energy, calculation.

Determining the direction of a chemical reaction using thermodynamic functions of state .

The direction of the real reaction is the result of competition between two opposing entropic factors Δ S and energy Δ N. The predominance of a favorable factor (Δ S>0 and Δ N<0 ) and determines the possibility of spontaneous occurrence of the process

Gibbs energy - thermodynamic function of the state of the system. A generalized criterion that takes into account the enthalpy and entropy factors at p=const and V=const, characterizes the direction and limit of spontaneous reaction.

Gibbs equation:  G= H–T S. G 0 reaction, = Δ N 0 reaction, 298 -T Δ S reactions, 298 . where ‘-‘ is the counteraction of factors

H is the total energy; TS is a bond of energy that cannot be converted into other types of energy.

G reaction = G cont – G out.

G-Gibbs Energy is the part of the energy effect of a chemical reaction that can be converted into work.

G characterizes direction And spontaneous limit reaction under conditions p and V = const:

1) G< 0: самопроизвольно 1  2.

a) both factors are favorable

H<0,S>0(G<0) 1-2,при любой t

b) favorable energy factor (H<0)

if(H)> TS, then G<0 1-2

if the r-tion is determined by the energy factor, then it is most likely at lower t

c) favorable-probable factor (entropy)

S>0 (H>0 unfavorable)

 TS>H, then G<0 1-2

If the r-tion is determined by the entropy factor, then it most likely occurs at high t.

2) G > 0: spontaneously 2  1.

3) G = 0: equilibrium: 1  2. H = TS.  Estimation of the decomposition temperature of a substance produced based on equilibrium(ΔG=ΔH-TΔS)

In practice boundary condition: 40 KJ/mol. When G > 40, then  is impossible under any conditions.

Gibbs Energy Calculation

1 way (preferred at standard temperature) ν A A+ ν B B= ν C C+ ν D D

G 0 r-tion, 298 = ∑ν i G 0 arr -∑ν j G 0 arrj ( starting products substances)

G 0 arr. simple in-in =0 G 0 arr (H + solution) = 0 stable under standard conditions.

G is the change in the Gibbs energy in the reaction of the formation of 1 mole of a substance from simple substances taken under standard conditions in the most stable form.

2 way (approximate calculation method, at any T)

G 0 r-tion, = Δ H 0 r-tion, 298 -TΔS r-tion, 298

It can be used if the aggregative states of all components of the reaction do not change in a given interval T. In this case, Δ H r-tions = const Δ S r-tions = const

The mathematical expression of the second law of thermodynamics is written:

Here the > sign refers to irreversible processes, and the = sign to reversible ones. Since entropy is a function of state, its change during both reversible and irreversible processes the same. Therefore, when calculating the change in entropy, it is necessary to use formulas for reversible processes.

Entropy has additivity properties, so a change in entropy in complex process equal to the sum of entropy changes in its individual stages. Absolute value the entropy of any substance at any temperature can be calculated if the absolute entropy at a certain temperature is known, for example, at 298K and temperature coefficients heat capacity:

The change in entropy in various processes is calculated using the following equations:

When heated, n is a mole of substance from T 1 to T 2 at P = const:

Integration gives:

During phase transformation:

Where λ is the molar heat of phase transition (melting, evaporation, sublimation, modification transformation); T – phase transition temperature.

During the transition n – mole of an ideal gas from state 1 to state 2 at T=const:

When mixing ideal gases (T, P = const):

Where n 1 and n 2 are the numbers of moles of the first and second gas: V 1 and V 2 are their initial volumes:

V= V 1 + V 2 - final volume.

Determine the change in entropy when 2 g of ice taken at a temperature of 253 K and a pressure of 1.013 * 10 5 n/m 2 is converted into steam at a temperature of 423 K, if the heat of fusion of ice at 273 K is 0.335 kJ/g, specific heat ice is equal to 2.02 J/g * K water – 4.2 J/g. K, the latent heat of vaporization of water is 2.255 kJ/g, the molar heat capacity of steam at constant pressure:

C p = 30.13+11.3*10 -3 T, J/mol. TO

This process consists of five stages:

1) heating of ice from 253 to 273 K – ∆ S 1 ;

2) ice melting at 273 K – ∆ S 2 ;

3) heating liquid water from 273 to 373 K – ∆ S 3 ;

4) transition of liquid water into steam at 373K – ∆ S 4 ;

5) heating water vapor from 373 to 473 K – ∆ S 5 .

One of the vessels with a capacity of 0.1 m3 contains oxygen, the other, with a capacity of 0.4 m3, contains nitrogen. In both vessels the temperature is 290 K and the pressure is 1.013 · 10 5 N/m 2. Find the change in entropy when gases are mixed, considering them ideal.

We find the numbers of moles of gases using the Mendeleev–Clapeyron equation:

Calculate the standard entropy change for the reaction: Cd+2AgCl = 2Ag+CdCl 2 if

2.2. Calculation of changes in isobaric and isochoric
potentials in various processes

In an isobaric-isothermal process ( R, T= const) the criterion for the direction of the process and equilibrium is the isobaric-isothermal potential or Gibbs free energy: ∆ G≤ 0. At equilibrium G minimal. In an isochoric-isothermal process ( V, T= const) the criterion for the direction of the process and equilibrium is the isochoric-isothermal potential or Helmholtz free energy: ∆ F≤ 0. At equilibrium F minimal.

Changes ∆ G and ∆ F at constant temperature are calculated using the formulas: ∆ G = ∆H – T∆S and ∆ F = ∆U – T∆S.

From these equations it is clear that the free energy G or F are part full stock system energy N or U less bound energy T S. Free energy can be taken out of the system and turned into work: -∆ G = A r max and -∆ F = = A V max, where A r max – maximum full time job; A V max – maximum useful work.

When an ideal gas expands or contracts at constant temperature

Dependence ∆ G and ∆ F on temperature is expressed by the Gibbs–Helmholtz equation. For ∆ G V integral form it is written like this:

or in the range from 298 to T:

here ∆ N = f(T).

For a chemical reaction

∆G = ∆F + ∆nRT,

LECTURE 11

Phase diagrams

Phase is a state of matter characterized by the fact that it occupies a certain region of space, and within this region the parameters and properties of the substance either remain constant or change continuously. This spatial region is separated from other parts of space by a boundary. The mass of a substance contained in one phase may change over time. In this case they talk about phase transition. The phase transition occurs through the phase interface. The following most common phase transitions are distinguished:

boiling (transition of a substance from liquid to vapor);

condensation (transition of a substance from vapor to liquid);

crystallization, hardening (transition of a substance from a liquid to a solid state);

melting (transition of a substance from solid to liquid).

It is convenient to depict phases on phase diagrams. A phase diagram is a plane with Cartesian system coordinates, along the axes of which the values ​​of a pair of basic thermodynamic parameters are plotted. This plane is divided into a number of regions, each of which represents a certain phase. The phase diagram also shows the main isolines (lines of constancy of the main thermodynamic parameters: isochores, isobars, isotherms, isentropes, isoetalpes and lines of constant dryness.

The most common phase diagrams are T-S, P-V, H-S, H-lgP. Let's consider phase T-S diagram. In Fig. Figure 31 shows the main phases and phase boundaries:

F - liquid

F + T - liquid + body

NC – supercritical region

G – gas regionVP – wet steam

bkc – saturation curve. Characterizes the saturated state of a substance.

bk – line saturated liquid. Saturated liquid- this is a liquid state of a substance, characterized by the fact that the supply of an arbitrarily small amount of heat leads to intense formation of steam.

ks – dry line saturated steam . This is the gaseous state of a substance, characterized by the fact that any slight cooling leads to the beginning of the condensation process.

abc - line of triple dots. Triple point- is a state of matter characterized by equilibrium coexistence at once three phases: solid, liquid and gas. Phase equilibrium characterized by the fact that there is no phase transition between phases. At constant external conditions phase equilibrium can coexist as long as desired. In order for the two phases to be in a state of equilibrium, it is necessary to perform three conditions: 1) the phases must have the same pressure; 2) the phases must have the same temperature; 3) phases must have chemical potential.

be – line of the beginning of the solidification process or the end of the melting process.

ad - line of the end of the solidification process or the beginning of the melting process.

dek is the critical temperature isotherm.

P=P cr – critical isobar.

k – critical point . It is characterized by the fact that at a temperature above the critical temperature, it is impossible to obtain liquid using isothermal compression. Critical pressure and temperature are pressures and temperatures below the critical point.

Region G – gas region. This region is located at a pressure below critical and a temperature above critical. The gas region is characterized by the fact that the state of the gas in this region is described by the equation of state of an ideal gas.

PP area - superheated steam region. Located at a temperature below critical and to the right of the kc line. This region is characterized by the fact that the behavior of matter in it is described by the van der Waals equation or the modified ideal gas equation

where z is the compressibility coefficient (a correction factor that takes into account the deviation of the behavior of real substances from an ideal gas).

Area F+P - wet steam area. Limited by saturation curve and triple point line. This is a two-phase region characterized by the equilibrium state of saturated vapor and saturated liquid. This is the area where condensation and boiling processes occur.

Region Zh. - supercooled liquid region. It is limited from above by the critical isotherm, on the right - by the line of saturated liquid, on the left - by the line of the beginning of crystallization.

The T+L region is a two-phase region of equilibrium coexistence of the liquid phase and the solid. This is the area where solidification (crystallization) and melting processes occur.

The T+P region is a two-phase region of equilibrium coexistence of saturated vapor and solid. From above, this area is limited by a line of triple points. Triple point is called a state of equilibrium of three states of aggregation at once. This is the area where the processes of sublimation and desublimation occur. Sublimation called the process of transition of the solid phase into the gaseous phase. Desublimation called the process of transition of saturated steam into the solid phase.

The NC region is the region of the supercritical state of the substance. Located at pressure and temperature above critical. It is characterized by the fact that a substance in this state has the properties of both a liquid and a gas.

In Fig. Figure 32 shows the main process lines.

Isobars corresponding to pressures Р 1 , Р 2 , Р 3 = Р кр and Р 4 are depicted by solid lines. In this case, the relations P 1 are satisfied between these pressures<Р 2 <Р 3 <Р 4 . Следует отметить, что процессы, что процессы, протекающие в двухфазных областях, изображаются горизонтальными линиями, т.е. эти изобарные процессы одновременно являются изотермическими. Изобара с давлением Р 4 лежит выше критической точки не проходит через область влажного пара, а сразу из области надкритического состояния попадает в область переохлаждённой жидкости. Изобара с давлением Р 1 лежит ниже линии тройных точек, также не проходит через область влажного пара, а из области перегретого пара попадает в область твёрдого состояния вещества посредством процесса десублимации. Изобара с давлением Р 3 касается критической точки. Изобара с давлением Р 2 , проходя через область влажного пара, реализует процесс кипения или конденсации.

Isochores in v 1 and v 2 (v 1 >v 2) depicted by dashed lines are located in T-S chart steeper than isobars. It should be noted that in two-phase regions isochores do not coincide with isotherms, i.e. not horizontal.

Isoenthalpes h 1 , h 2 and h 3 (h 1 >h 2 >h 3) are depicted by dotted lines. You can pay attention to the fact that as the temperature decreases, the angle of inclination of the isenthalpe to the S axis increases.

Wet steam

Wet vapor is a state of matter in which saturated vapor and saturated liquid are in equilibrium. Equilibrium is due to the equality of their temperatures and pressures. The wet steam region finds greatest application in thermal power and low-temperature devices, because in this area it is most easy to implement processes that are important in technical applications (isothermal).

The wet steam region depicted in the T-S diagram is shown in Fig. 33.

Point “a” characterizes the state of wet steam, in which saturated liquid and saturated steam are in equilibrium in certain mass fractions.

Saturated vapor is in a state, and the state of a saturated liquid is characterized by a point. Let wet steam a in the state of point a occupy a certain volume, where m is the mass of wet steam; v a - specific volume of wet steam. The same volume can be considered as the sum of the volumes of saturated liquid and saturated vapor

where is the volume of saturated liquid;

Volume of saturated steam;

Mass of saturated liquid;

Saturated steam mass;

Specific volume of saturated liquid at point state;

Specific volume of saturated steam at point state.

In this case, the relationship is obvious

Dividing both sides of the equation by m in the last expression, we obtain an equation expressing the specific volume of wet steam through the specific volumes of saturated liquid and saturated steam

In this expression, the degree of dryness of wet steam, which shows mass fraction saturated steam in wet steam. If x=1, then the wet steam consists entirely of saturated steam. If x=0, then the wet steam consists entirely of saturated liquid. The degree of dryness can take any value from 0 to 1. The set of all points of the wet vapor area on the T-S diagram that have the same value for the degree of dryness are called lines of constant dryness (see Fig. 33).

Using exactly the same reasoning, using the property of additivity of enthalpy and entropy, one can obtain the expressions

where is the specific enthalpy of the saturated liquid in the point state;

Specific enthalpy of saturated steam in the point state;

Specific entropy of a saturated liquid in a point state;

Specific entropy of saturated steam in the point state.

Let us express x from the last equation

From this formula it follows that to increase the degree of dryness, you need to increase entropy, i.e. add heat to wet steam. In this case, the proportion of saturated liquid will decrease, and the proportion of saturated vapor will increase. The parameters of the saturated liquid and saturated vapor will not change. This process is called boiling. If heat is removed from wet steam, then the entropy will decrease, which means the degree of dryness will decrease, i.e. the substance will pass from the saturated vapor state to the saturated liquid state. This process is called condensation.

In order to completely convert 1 kg of saturated liquid into the state of dry saturated vapor, it is necessary to supply a certain amount of heat, which is called specific heat vaporization r, .

In an isobaric process, such as boiling or condensation, the heat added or removed is equal to the change in enthalpy. Therefore the relation is valid

LECTURE 12

Thermodynamic cycle

Thermodynamic cycle is called a closed thermodynamic process, i.e. a process as a result of which the system returns to its original state. Another definition can be given to the thermodynamic cycle as a sequence of thermodynamic processes, the implementation of which leads the system to its initial state. Let us write the first law of thermodynamics for closed system in the form

Since the system returns to its original state, then . The result is a generalized thermodynamic cycle equation

where Q 1 is the total heat supplied to the system in the cycle;

Q 2 is the total heat removed from the system in the cycle.

Substituting (140) into (139), we get

In this expression, the removed heat is taken to be positive, because the sign of the heat removed is taken into account in the formula with a minus in front of Q 2 .

Equation (141) allows us to classify thermodynamic cycles into two types:

1. if , then the cycle is called direct;

2. if , then the cycle is called inverse.

Direct cycle

Direct cycle also called thermal power. This is a cycle, as a result of which the system produces, i.e. performs work due to the heat supplied to the system.

Schematic diagram a device that implements a direct, or thermal power, cycle is shown in Fig. 34.

In this picture:

TDS(M) – thermodynamic system(machine) which makes a cycle;

GI – hot spring with temperature T GI. It is understood as a set of environmental bodies that transfer heat Q 1 to the thermodynamic system.

CI is a cold source, or refrigerator, with a temperature of T CI. This is a set of environmental bodies to which the system, completing a cycle, gives off heat Q 2. In order for the diagram presented in Fig. 34 could be implemented, the cold source should have a temperature T CI lower than the temperature of the hot source T CI (T CI<Т ГИ). Кроме того, температура холодного источника должна быть меньше минимальной температуры системы в цикле, а температура горячего источника должна быть больше максимальной температуры системы.

Rice. 35. Fig. 36.

In Fig. Figure 35 shows the thermal power cycle in the T-S diagram. Process 1a2 is accompanied by the supply of heat Q 1, because entropy increases. In this case, the heat supplied is equal to the area under line 1a2. In process 2b1, heat Q 2 is removed, because entropy decreases, and this heat is equal to the area under the 2b1 line. From the figure it can be seen that the area of ​​the figure is m1a2n more area m1b2n, so Q 1 >Q 2, and this cycle is direct. As a result, the difference between the supplied and removed heat is equal to the work of the cycle, and is equal to the area of ​​the cycle.

In Fig. 36 shows the thermal power cycle in P-V diagram. Process 1a2 is accompanied by the performance of work L 1 a 2, because the volume in this process increases. In this case, the work done is equal to the area under line 1a2. In process 2b1, work L 2 b 1 is wasted, because the volume decreases, and this work is equal to the area under the line 2b1. From the figure it is clear that the area of ​​the figure m1a2n is greater than the area of ​​m1b2n, therefore L 1 a 2 >L 2 b 1, and this cycle is straight. As a result, the difference between the work done and the work expended is equal to the work of the cycle and equal to the area limited by the cycle.

Any cycle, both direct and reverse, is characterized by an efficiency coefficient that evaluates the efficiency of the energy conversion process

Since, due to the definition of a direct cycle, the efficiency is always less than one. The process of converting thermal energy into useful work in a cycle is more efficient, the more closer value Cycle efficiency to one.

Reverse cycles

A reverse cycle is a cycle in which the heat supplied is less than the heat removed. As a result, the work of the reverse cycle is negative, i.e. To implement it, work must be expended.

A schematic diagram of a device that implements the reverse cycle is shown in Fig. 37.

Rice. 38. Fig. 39.

In Fig. 38 shown reverse cycle in the T-S diagram. Process 1a2 is accompanied by the supply of heat Q 1, because entropy increases. In this case, the heat supplied is equal to the area under line 1a2. In process 2b1, heat Q 2 is removed, because entropy decreases, and this heat is equal to the area under the 2b1 line. From the figure it can be seen that the area of ​​the figure is m1a2n less area m1b2n, so Q 1

In Fig. Figure 39 shows the reverse cycle in a P-V diagram. Process 1a2 is accompanied by the performance of work L 1 a 2, because the volume in this process increases. In this case, the work done is equal to the area under line 1a2. In process 2b1, work L 2 b 1 is wasted, because the volume decreases, and this work is equal to the area under the line 2b1. It can be seen from the figure that the area of ​​the figure m1a2n is less than the area m1b2n, therefore L 1 a 2

Reverse thermodynamic cycles are divided into three types:

1. refrigeration cycles;

2. heat pump cycles;

3. combined cycles.

The refrigeration cycle is shown in Fig. 40 under the Roman numeral I. This is a reverse cycle in which work is expended to remove heat Q 1 from a cooled object located at a temperature T OO below the ambient temperature T OC.

Refrigeration cycles are implemented in low-temperature units, in particular in household refrigerators. In this case, the heat Q 1 supplied to the working substance (freon) is the heat removed from the products in the freezer.

Heat pump cycle II is a reverse cycle in which work is expended to supply heat Q 2 to the heated object at a temperature THO higher than the ambient temperature TOS. This cycle is implemented by household air conditioners operating in room heating mode. The heated object in this case is room air. The temperature of the heated object is room temperature. The environment is low-temperature outside air. The heat Q 2 used to heat the room in this case and determined by expression (144) is greater than the heat that would be supplied when heating the room with an electric heater, in which electrical energy L is converted into thermal energy.

Combined cycle III is a reverse cycle in which work is expended to remove heat Q 1 from the cooled object, located at the temperature TOO below the ambient temperature, and simultaneously supply heat Q 2 to the heated object, located at the temperature TOO above the ambient temperature. The device that implements the combined cycle is a household refrigerator located in a residential area. In turn, from the outside of this room there is air with a low temperature. In this case, the heating object to which heat Q 2 is supplied (removed from the cycle) is the air in the room at room temperature. The object of cooling is the products in the freezer, from which heat Q 1 is removed and which is supplied to the freon circulating in the refrigerator.

The efficiency factor of a refrigeration cycle is called the coefficient of performance ε. The useful energy in this case is the heat Q 1 removed from the cooled object and supplied to the working substance completing the cycle. The energy expended is the work supplied L. Therefore

From this expression it is clear that the heating coefficient is always greater than unity, and the heat pump cycle is more efficient, the greater the value μ takes over unity.

The combined cycle efficiency factor does not have a special name and is denoted by k. The useful energy in this case is the heat Q 1 removed from the cooled object, and at the same time the heat Q 2 supplied to the heated object. The energy expended is the work supplied L. Therefore

From this expression it is clear that the efficiency coefficient of the combined cycle is obviously greater than one,

LECTURE 13

Reversible Carnot cycle

All cycles, both direct and reverse, are divided into 2 types: reversible and irreversible. Reversible cycle called a cycle consisting only of reversible processes. Irreversible cycle is a cycle in which at least one irreversible process is present. In order for the process to be reversible, it must be in equilibrium, i.e. must flow at an infinitely slow speed. This is only possible if the potential difference interacting with the system and the environment is infinitely small. For a thermodynamic system, this means that with reversible heat exchange with the environment, the temperature difference between the system and the environment should be an infinitesimal value, i.e. There must be no thermal resistance between the system and the environment. Reversible expansion and contraction is possible in the case of an infinitesimal pressure difference between the system and the environment. This is only possible when there is no friction in the system. It follows from this that in a thermomechanical system there are two sources of irreversibility:

1. the presence of thermal resistance between different parts of the system, which leads to a finite temperature difference during heat exchange;

2. the presence of friction in the system (or between the system and the environment), which leads to a finite pressure difference.

Of all thermodynamic cycles, the reversible Carnot cycle (direct) is distinguished on the basis that for a given temperature difference between hot and cold sources, the reversible Carnot cycle has the highest possible efficiency.

The reversible Carnot cycle shown in Fig. 41 and fig. 42, consists of two adiabats and two isotherms.

Rice. 41. Fig. 42

1-2 – process of adiabatic expansion. In this process, work L 12 is performed.

2-3 – process of isothermal compression. In this process, work L 23 is expended and heat Q 23 is removed.

3-4 – process of adiabatic compression. This process consumes work L 34.

4-1 – process of isothermal expansion. In this process, work L 41 is performed and heat Q 41 is supplied.

The main processes of the cycle are processes 4-1 and 1-2. They carry out the work of the cycle. The remaining processes are auxiliary and are aimed at returning the system to its original state with the least amount of energy.

Let us determine the efficiency of the reversible Carnot cycle η bcc:

By definition of efficiency (143)

Substituting these expressions into (148) and reducing by the entropy difference, we obtain

Based on the same considerations, we obtain

This formula shows that The efficiency of a reversible Carnot cycle does not depend on the properties of the working fluid performing the Carnot cycle and is determined only by the temperatures of the hot and cold sources . This conclusion is a formulation Carnot's first theorem.

9.9. Entropy. Physical meaning of entropy. Entropy and probability.

Considering the efficiency of a heat engine operating according to the Carnot cycle, it can be noted that the ratio of the temperature of the refrigerator to the temperature of the heater is equal to the ratio of the amount of heat given by the working fluid to the refrigerator and the amount of heat received from the heater. This means that for an ideal heat engine operating according to the Carnot cycle, the following relationship holds:
. Attitude Lorenz called reduced heat . For an elementary process, the reduced heat will be equal to . This means that when the Carnot cycle is implemented (and it is a reversible cyclic process), the reduced heat remains unchanged and behaves as a function of the state, then, as is known, the amount of heat is a function of the process.

Using the first law of thermodynamics for reversible processes,
and dividing both sides of this equality by temperature, we get:

(9-41)

Let us express from the Mendeleev-Clapeyron equation
, substitute into equation (9-41) and get:

(9-42)

Let's take into account that
, A
, substitute them into equation (9-42) and get:

(9-43)

The right side of this equality is full differential Therefore, in reversible processes, the reduced heat is also a total differential, which is a sign of the state function.

State function whose differential is , called entropy and is designated S . Thus, entropy is a function of state. After introducing entropy, formula (9-43) will look like:

, (9-44)

Where dS– entropy increment. Equality (9-44) is valid only for reversible processes and is convenient for calculating the change in entropy during finite processes:

(9-45)

If a system undergoes a circular process (cycle) in a reversible way, then
, and, therefore, S=0, then S = const.

Expressing the amount of heat through the increment of entropy for an elementary process, and substituting it into the equation for the first law of thermodynamics, we obtain a new form of writing this equation, which is usually called basic thermodynamic identity:

(9-46)

Thus, to calculate the change in entropy during reversible processes, it is convenient to use reduced heat.

In the case of irreversible nonequilibrium processes
, and for irreversible circular processes it holds Clausius inequality :

(9-47)

Let's consider what happens to entropy in an isolated thermodynamic system.

In an isolated thermodynamic system, with any reversible change in state, its entropy will not change. Mathematically, this can be written as follows: S = const.

Let us consider what happens to the entropy of a thermodynamic system during an irreversible process. Let us assume that the transition from state 1 to state 2 along path L 1 is reversible, and from state 2 to state 1 along path L 2 is irreversible (Fig. 9.13).

Then the Clausius inequality (9-47) is valid. Let us write the expression for the right side of this inequality corresponding to our example:

.

The first term in this formula can be replaced by a change in entropy, since this process is reversible. Then the Clausius inequality can be written as:

.

From here
. Because
, then we can finally write:

(9-48)

If the system is isolated, then
, and inequality (9-48) will look like:

, (9-49)

T That is, the entropy of an isolated system increases during an irreversible process. The growth of entropy does not continue indefinitely, but up to a certain maximum value characteristic of a given state of the system. This maximum entropy value corresponds to a state of thermodynamic equilibrium. An increase in entropy during irreversible processes in an isolated system means that the energy possessed by the system becomes less available for conversion into mechanical work. In a state of equilibrium, when entropy reaches its maximum value, the energy of the system cannot be converted into mechanical work.

If the system is not isolated, then entropy can either decrease or increase depending on the direction of heat transfer.

Entropy as a function of the state of the system can serve as the same state parameter as temperature, pressure, volume. By depicting a particular process on a diagram (T,S), one can give a mathematical interpretation of the amount of heat as the area of ​​a figure under the curve depicting the process. Figure 9.14 shows a diagram for an isothermal process in entropy – temperature coordinates.

Entropy can be expressed through the parameters of the gas state - temperature, pressure, volume. To do this, from the main thermodynamic identity (9-46) we express the entropy increment:

.

Let's integrate this expression and get:

(9-50)

The change in entropy can also be expressed through another pair of state parameters – pressure and volume. To do this, you need to express the temperatures of the initial and final states from the equation of state of an ideal gas through pressure and volume and substitute them into (9-50):

(9-51)

With isothermal expansion of the gas into the void, T 1 = T 2, which means the first term in formula (9-47) will be zeroed and the change in entropy will be determined only by the second term:

(9-52)

Despite the fact that in many cases it is convenient to use reduced heat to calculate the change in entropy, it is clear that reduced heat and entropy are different, not identical concepts.

Let's find out physical meaning of entropy . To do this, we use formula (9-52) for an isothermal process in which the internal energy does not change, and all possible changes in characteristics are due only to changes in volume. Let us consider the relationship between the volume occupied by a gas in an equilibrium state and the number of spatial microstates of gas particles. The number of microstates of gas particles, with the help of which a given macrostate of a gas as a thermodynamic system is realized, can be calculated as follows. Let us divide the entire volume into elementary cubic cells with a side of d~10 –10 m (of the order of magnitude of the effective diameter of the molecule). The volume of such a cell will be equal to d 3. In the first state, the gas occupies volume V 1, therefore, the number of elementary cells, that is, the number of places N 1 that molecules can occupy in this state will be equal to
. Similarly, for the second state with volume V 2 we obtain
. It should be noted that the change in the positions of the molecules corresponds to a new microstate. Not every change in the microstate will lead to a change in the macrostate. Suppose molecules can occupy N 1 places, then swapping the places of any molecules in these N 1 cells will not lead to a new macrostate. However, the transition of molecules to other cells will lead to a change in the macrostate of the system. The number of microstates of a gas corresponding to a given macrostate can be calculated by determining the number of ways of placing particles of this gas in elementary cells. To simplify the calculations, consider 1 mole of an ideal gas. For 1 mole of an ideal gas, formula (9-52) will look like:

(9-53)

The number of microstates of the system occupying volume V 1 will be denoted by Г 1 and determined by counting the number of placements N A (Avogadro's number) of molecules contained in 1 mole of gas in N 1 cells (locations):
. Similarly, we calculate the number of microstates G 2 of the system occupying volume V 2:
.

The number of microstates Г i with the help of which the ith macrostate can be realized is called thermodynamic probability of this macrostate. Thermodynamic probability Г ≥ 1.

Let's find the ratio Г 2 /Г 1:

.

For ideal gases, the number of free places is much greater than the number of molecules, that is, N 1 >>N A and N 2 >>N A. . Then, taking into account the expression of the numbers N 1 and N 2 through the corresponding volumes, we obtain:

From here we can express the volume ratio through the ratio of the thermodynamic probabilities of the corresponding states:

(9-54)

Substitute (9-54) into (9-53) and get:
. Considering that the ratio of the molar gas constant and Avogadro's number, there is Boltzmann's constant k, and also that the logarithm of the ratio of two quantities equal to the difference logarithms of these quantities, we obtain:. From this we can conclude that the entropy of that state S i is determined by the logarithm of the number of microstates through which a given macrostate is realized:

(9-55)

Formula (9-55) is called Boltzmann's formula who first received it and understood statistical meaning of entropy , How disorder functions . Boltzmann's formula has a more general meaning than formula (9-53), that is, it can be used not only for ideal gases, and allows one to reveal the physical meaning of entropy. The more ordered the system, the less number microstates through which a given macrostate is realized, the lower the entropy of the system. An increase in entropy in an isolated system, where irreversible processes occur, means the movement of the system in the direction of the most probable state, which is the state of equilibrium. It can be said that entropy is measure of disorder systems; the more disorder there is, the higher the entropy. This is physical meaning of entropy .