Father of statistical thermodynamics

65

Transcript of Father of statistical thermodynamics

Page 1: Father of statistical thermodynamics
Page 2: Father of statistical thermodynamics

The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 2

Father of statistical thermodynamics

Ludwig Boltzmann (1844-1906)• Academical legacy- Maxwell-Boltzmann distribution for molecular speed in a gas.- Opinion and belief in the reality of atom and molecule.- To quote Plank, "The logarithmic connection between

entropy and probability was first stated by L. Boltzmannin his kinetic theory of gases".

𝑆 = 𝐾 ∙ 𝑙𝑜𝑔𝜔

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 3

Various statistical distribution

Maxwell-Boltzmann Statistics have to be distinguishable each other and one energy state can be occupied by two or more particles. Ex) gas molecules

Bose-Einstein Statistics have to be indistinguishable each other and one energy state can be occupied by two or more particles. Ex) phonon, photon

Fermi-Dirac Statistics have to be indistinguishable each other and one energy state can be occupied by only one particle. Ex) electron, hole

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 4

Interpretation of entropy as a mixed-up-ness of the system

Solid, 𝑆𝑠𝑜𝑙𝑖𝑑 liquid, 𝑆𝑙𝑖𝑞𝑢𝑖𝑑 Vapor, 𝑆𝑉𝑎𝑝𝑜𝑟

- the more “ mixed up” the constituent particles of a system, the larger is the value of its entropy.

𝑆𝑠𝑜𝑙𝑖𝑑 < 𝑆𝑙𝑖𝑞𝑢𝑖𝑑 < 𝑆𝑔𝑎𝑠

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 5

The concept of microstate

Terminology1. Isolated systems, where S ′ = S ′(U′,V′,N ) or U ′ = U ′(S′,V′,N) - microcanonical2. Closed systems, where S ′=S′(T,V′,N) - canonical3. Open systems, where S ′ = S ′(T,V′,μ) - grand canonical

Statistical thermodynamics postulates that the equilibrium state of a system is simplythe most probable of all of its possible (i.e., accessible) microstates. Therefore,statistical thermodynamics is concerned with

• The determination of the most probable microstate• The criteria governing the most probable microstate• The properties of this most probable microstate

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 6

The concept of microstate

- To derive relationship between entropy and mixed-up-ness, the quantization of the mixed-up-ness is necessary.

- Both Boltzmann (Ludwig Eduard Boltzmann, 1844– 1906) and Gibbs found it convenientto examine the distribution of energies among the particles of the system by placing the energy of the particles into discrete compartments .

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 7

The Microcanonical approach

- Considering a hypothetical system comprising a perfect crystal in which all of the distinguishable sitesare occupied by identical particles; the theoretical condition for Maxwell-Boltzmann distribution

- The crystal contains n particles and has the fixed energy U ʹ and fixed volume V ʹ, and whole system is

considered isolated one.

- Statistical thermodynamics asks the following questions:

• In how many ways can the n particles be distributed over the available energy levelssuch that the total energy of the crystal (i.e., Uʹ ) remains the same?

• Of the possible distributions, which is the most probable?

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 8

The Microcanonical approacha. All three particles on level 1

b. One particle on level 3, and the other two particles on level 0

c. One particle on level 2, one particle on level 1, and one particle on level 0

• Distribution a . There is only one microstate of this distribution, since the interchangeof the particles among the three sites does not produce a different microstate.

• Distribution b . Any of the three distinguishable sites can be occupied by any ofthe three particles of energy 3u , and the remaining two sites are each occupied bya particle of zero energy. Since the interchange of the particles of zero energy doesnot produce a different arrangement, there are three microstates in distribution b .

• Distribution c . Any of the three distinguishable sites can be occupied by the particleof energy 2u . Either of the two remaining sites can be occupied by the particleof energy 1u , and the single remaining site is occupied by the particle of zeroenergy. The number of distinguishable microstates in distribution c is thus 3 × 2× 1 = 3! = 6.

- Probability: 1/10

- Probability: 3/10

- Probability: 6/10

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Configurational entropy of differing atoms in a crystal

- Consider two crystals, one containing white atoms and the other containing gray atoms.- There is no difference in the energy of white/white, white/gray, and gray/gray bonds.(no mixing enthalpy)- When the two crystals are placed in physical contact with one another, a spontaneous process occurs in which the white atoms diffuse into the crystal of the gray atoms and the gray atoms diffuse into the crystal of the white atoms. (Configurational entropy)

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 10

Configurational entropy of differing atoms in a crystal

The mixing process can be expressed as

If 𝑛𝐴 atoms of A are mixed with 𝑛𝐵 atoms of B,

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 11

Configurational entropy of magnetic spins on an array of atoms

- The spin may take values of ±1

2, which called up and down spins for convenient.

- Assume that there is eight sight in the system, So there will be nine distinct groups.

- The most probable microstate has a total magnetization of zero, and this is considered to be the equilibrium state. It can also be seen that the average value of M over all microstates is also zero. A material in this state is called a paramagnet .

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The Boltzmann distribution

- If n particles are distributed among the energy levels such that 𝑛0 are on levelε0 , 𝑛1 on level ε1 , 𝑛2 on level ε2 ,… , and 𝑛𝑟 on ε𝑟 , the highest level of occupancy, thenthe number of arrangements, Ω , is given by

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 13

The Boltzmann distribution

or

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 14

The Boltzmann distribution

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 15

The influence of temperature

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The Third Law .Heat capacity, Enthalpy, Entropy .Fundamental equations .Statistical interpretion of entropy . 16

The influence of temperature

Helmholtz free energy

- Consider now a system of particles in thermal equilibrium with a constant-temperatureheat bath .U ′ = 𝑈′𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 + 𝑈′ℎ𝑒𝑎𝑡 𝑏𝑎𝑡ℎV′ = 𝑉′𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 + 𝑉′ℎ𝑒𝑎𝑡 𝑏𝑎𝑡ℎn = the number of particles in the system + the heat bath of fixed size

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Heat flow and the production of entropy

- Consider two closed systems, A and B . Let the energy of A be 𝑈′𝐴and the numberof complexions of A be Ω𝐴. Similarly, let the energy of B be 𝑈′𝐵 and its number ofcomplexions be Ω𝐵.

- When thermal contact is made between A and B , the product Ω𝐴 Ω𝐵 will, generally, not have its maximumpossible value, and thermal energy will be transferred either from A to B or from B to A.

- The flow of thermal energy ceases when Ω𝐴 Ω𝐵 reaches its maximum value

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Heat flow and the production of entropy

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Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Application and practicality of thermodynamic method

- The main power of the thermodynamic method stems from its provision of criteria for equilibrium.

- The practical usefulness of this power is determined by the practicality of the equations of state for the system.

- It is important to establish variables that is easy to measure and easy to control.

𝑑𝑈 = 𝛿𝑞 − 𝛿𝑤 = 𝑇 ∙𝛿𝑞

𝑇− 𝑃 ∙ 𝑑𝑉 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉

- From a practical point of view, the choice of S and V as the independent variables is inconvenient.

- Entropy can be neither simply measured nor simply controlled.

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Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Practical thermodynamic variables

- From a practical point of view, a convenient pair of independent variables would be T and P, since these variables are easily measured and controlled.

- From the theoretician’s point of view, a convenient choice of independent variables would be V and T,since they are easily examined by the methods of statistical mechanics.

- These equations are the fundamental equations that can be obtained by using the first law of thermodynamic,then we will define more practical thermodynamic variables H, A, G, 𝜇𝑖.

H: the enthalpy A: the Helmholtz free energyG: the Gibbs free energy 𝜇𝑖: the chemical potential

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21The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The enthalpy, H

At the condition of constant pressure,

This equation shows that the change of state of a simple closed system at constant pressure, during which only P -V work is done, is the change in the enthalpy of the system and equals the thermal energy entering or leaving the system, 𝑞𝑃 . For this reason, it was called the heat function at constant pressure by Gibbs

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Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The Helmholtz free energy, A

- If process is done at constant volume,∆𝑈 = 𝑇∆𝑆𝑠𝑦𝑠

𝛿𝑞𝑠𝑦𝑠 = −𝛿𝑞𝑠𝑢𝑟

- For a system undergoing a change of state from state 1 to state 2

If system is closed,

And if the process is also isothermal,

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Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The Helmholtz free energy, A

- During a reversible isothermal process, for which ∆𝑆𝑖𝑟𝑟 is zero, the amountof work done by the system is a maximum.

- The amount of work done is equal to the decrease in the value of the Helmholtz free energy.

- 𝑑𝑆′𝑖𝑟𝑟 > 0 means the process is spontaneous(forward), then 𝑑𝐴′ would be negativein the spontaneous process.

- 𝑑𝑆′𝑖𝑟𝑟 = 0 means the process is reversible, then 𝑑𝐴′ would be zero in the reversible process.

- 𝑑𝑆′𝑖𝑟𝑟 < 0 means the process is backward, then 𝑑𝐴′ would be positive in the backward reaction.

※Remember that we are dealing with isochoric process.

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Statistical interpretion of

entropy .Fundamental equations .

The Helmholtz free energy, A

𝑇1 > 0K𝑇0 = 0K 𝑇2 > 𝑇1

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25The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The Gibbs free energy, G

- The Gibbs free energy is defined as

- For a system undergoing a change of state from state 1 to state 2

- If the process carried out isothermal and isobar,

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26The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The Gibbs free energy, G

- For an isothermal, isobaric process, and no form of work other than P-V work is done,

- 𝑑𝑆′𝑖𝑟𝑟 > 0 means the process is spontaneous(forward), then 𝑑𝐺 would be negativein the spontaneous process.

- 𝑑𝑆′𝑖𝑟𝑟 = 0 means the process is reversible, then 𝑑𝐺 would be zero in the reversible process.

- 𝑑𝑆′𝑖𝑟𝑟 < 0 means the process is backward, then 𝑑𝐺 would be positive in the backward reaction.

- Since ∆𝑆𝑖𝑟𝑟 is a criterion for thermodynamic equilibrium, then an increment of an isothermalisobaric process occurring at equilibrium requires that

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27The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Several forms of equilibrium

𝑇𝑚

𝑇𝑏

Gibbs free energy versus temperature

𝑑𝐺

𝑑𝑇=0, two phase can coexist

Gibbs free energy versusInteratomic distance

𝑑𝐺

𝑑𝑟=0,

𝑑2𝐺

𝑑𝑟2> 0, equilibrium

Gibbs free energy versusParticle radius during solidification

𝑑𝐺

𝑑𝑟=0,

𝑑2𝐺

𝑑𝑟2< 0, not equilibrium

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28The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The fundamental equation

- Since 𝑑𝑈 = 𝛿𝑞 − 𝛿𝑤,

𝒅𝑼 = 𝑻𝒅𝑺 − 𝑷𝒅𝑽 …①

- Since 𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃,

𝒅𝑯 = 𝑻𝒅𝑺 + 𝑽𝒅𝑷 …②

- Since 𝑑𝐴 = 𝑑𝑈 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇,

𝒅𝑨 = −𝑺𝒅𝑻 − 𝑷𝒅𝑽 …③

- Since 𝑑𝐺 = 𝑑𝐻 − 𝑆𝑑𝑇 − 𝑇𝑑𝑆,

𝒅𝑮 = −𝑺𝒅𝑻 + 𝑽𝒅𝑷 …④

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29The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The fundamental equation

- If the system is also under the influence of an applied magnetic field,

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30The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The chemical potential

- The chemical potential of the i th species in a homogeneous phase is theincremental change of the Gibbs free energy that accompanies an incrementalincrease of the species to the system at constant temperature, pressure, and numbersof moles of all of the other species.

= 𝜇𝑖

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31The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The chemical potential

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32The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The chemical potential

- The term 𝜇𝑖𝑑𝑛𝑖 is the chemical work done by the system, which was denoted as wʹ

∆ 𝐺𝐴𝑀 = 𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑜𝑓 𝐴 𝑎𝑡 𝑋𝐴 = 𝑋𝐵

∆ 𝐺𝐵𝑀 = 𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑜𝑓 𝐵 𝑎𝑡 𝑋𝐴 = 𝑋𝐵

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33The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Thermodynamic relations

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Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Maxwell's relations

- If Z is a state function and x and y are chosen as the independent thermodynamic variables in a closed system of fixed composition

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35The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The first 𝑻𝒅𝑺 equation- The term

𝜕𝑃

𝜕𝑇 𝑉can be shown to equal 𝛼/𝛽𝑇

- For an isothermal expansion,

- For an isentropic expansion

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36The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The first 𝑻𝒅𝑺 equation

- If the substance is 1 mole of an ideal gas

- the integration of which (assuming constant c v ) between the states 1 and 2,

- For an isothermal expansion of an ideal gas

- If the temperature of an ideal gas is raised at constant volume

- For an isentropic expansion of an ideal gas

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37The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The second 𝑻𝒅𝑺 equation

- For an isothermal reversible change of pressure

- For an isentropic process

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38The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The second 𝑻𝒅𝑺 equation

- For an isothermal reversible change of pressure of an ideal gas

- For an isentropic process of an ideal gas

- For an isobaric reversible change in temperature of an ideal gas

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39The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

S and V as dependent variables and T and P as independent variables

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40The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃

𝑈(𝑆, 𝑉) H(𝑆, 𝑃)

An energy equation

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41The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

Another important formula

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42The Third Law .Heat capacity, Enthalpy,

Entropy .

Statistical interpretion of

entropy .Fundamental equations .

The Gibbs-Helmholtz equation

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43The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Heat capacity

∆𝑈 = 𝑈 𝑇1, 𝑉 − 𝑈 𝑇2, 𝑉 = 𝑇1

𝑇2

𝐶𝑣 𝑑𝑇

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44The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

History of heat capacity

Dulong-Petit DebyeEinstein

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45The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Dulong-petit law

- An empirical rule which states that the molar heat capacities (𝐶𝑣) of all solid elements have the value 3R.- Although the molar heat capacities of most elements at room temperature have values which are closeto 3R , subsequent experimental measurement showed that the heat capacity usually increases slightly with increasing temperature and can have values significantly lower values than 3R at low temperatures.

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46The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Einstein’s calculation

- Einstein considered the properties of a solid containing n atoms, each of which behaves as a quantumharmonic oscillator vibrating independently in three orthogonal directions about its position.

- the behavior of each of the 3n oscillators is not influenced by the behavior of its neighbors, and have a singlefrequency 𝑣 to each of the oscillators.(Einstein solid)

- Equipartition theorem: the average energy of each quadratic contribution to the energy is 𝐾𝑇/2

※Review

- Each atom has six degree of freedom.- It reveals 3R of specific heat for 1 mole.

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47The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Einstein’s calculation

- For a fixed frequency of vibration, the energy levels of a quantum harmonicoscillator take values of the 𝑖 th energy level as

- We defined the partition function 𝓩 as

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48The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Einstein’s calculation

Where ℎ𝑢

𝑘𝐵= 𝜃𝐸. 𝜃𝐸 is called Einstein characteristic temperature.

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49The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Einstein’s calculation

- Einstein’s model fits well when T is sufficiently high and 𝐶𝑣 is close to 3R, and when 𝑇 → 0, 𝐶𝑣 → 0.

- But the theoretical values of the Einstein model approach zero more rapidly thando the actual values.- This discrepancy is caused by the fact that the quantum oscillators do not vibrate with a single frequency.

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entropy .

Heat capacity, Enthalpy,

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Debye model

- Debye suggested the oscillation model that have the range of frequency of vibration.- The lower limit of the wavelength of these vibrations is determined by the interatomic distances in the solid.- Taking this minimum wavelength,𝜆𝑚𝑖𝑛 , to be in the order of 5 × 10– 10 m, and the wave velocity, 𝜈 , in the solidto be about 5 × 103 m/sec,

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51The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Debye model

- For very low temperature,

- This is called the 𝐷𝑒𝑏𝑦𝑒 𝑇3 𝑙𝑎𝑤 for low temperature heat capacities.

𝑦 ≅ 25.98

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52The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Debye model

- Debye’ s theory does not consider the contribution made to the heat capacityby the uptake of energy by free electrons at the Fermi level in a metal at lowtemperatures.- For a metal at low temperatures, the heat capacity varies as

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53The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

The empirical representation of heat capacities

Monoclinic, Below to 1478K

Monoclinic, 1478~2670K

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54The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Enthalpy as a function of temperature and composition

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55The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Enthalpy as a function of temperature and composition

① ②

Since 𝐻 is state function, 𝐻 = 0,

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56The Third Law .Fundamental equations .Statistical interpretion of

entropy .

Heat capacity, Enthalpy,

Entropy .

Enthalpy as a function of temperature and composition

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57Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

The dependence of entropy on temperature

and

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58Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

The Third law of thermodynamics

- If the value of 𝑆0 for a reaction could be determined, ∆𝐺 would be known as a function of temperature as well,and hence, the reaction thermodynamics would be known.

- Consideration of the value of 𝑆0 lead to the statement of the Third Law of Thermodynamics.- The values of ∆𝐺 and ∆𝐻 asymptotically approached each other at low temperatures with slopesthat approached zero.

and

- 𝑵𝒆𝒓𝒔𝒕 𝒉𝒆𝒂𝒕 𝒕𝒉𝒆𝒐𝒓𝒆𝒎

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59Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

The Third law of thermodynamics

- 𝑃𝑙𝑎𝑛𝑐𝑘 (Max Karl Ernst Ludwig Planck, 1858– 1947) extended the Nernst’ s heattheorem by positing to the effect that the entropy of any homogeneous substancewhich is in complete internal equilibrium is zero at 0 K.

Point defect line defect planar defect

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60Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

Apparent contradictions to the third law of thermodynamics

Internal equilibrium Non-internal equilibrium

𝑃(𝐴 − 𝐵) = 1

𝑃 𝐴 − 𝐴 = 𝑃 𝐵 − 𝐵 = 0

0 < 𝑃(𝐴 − 𝐵) < 1

𝑃 𝐴 − 𝐴 + 𝑃(𝐴 − 𝐵)

2> 𝑃(𝐴 − 𝐵)

0 < 𝑃(𝐴 − 𝐵) < 1

𝑃 𝐴 − 𝐴 + 𝑃(𝐴 − 𝐵)

2= 𝑃(𝐴 − 𝐵)

① ②

Page 61: Father of statistical thermodynamics

61Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

Apparent contradictions to the third law of thermodynamics

③ Even chemically pure elements are mixtures of isotopes, and because of the chemicalsimilarity between isotopes, it is to be expected that completely random mixingof the isotopes occurs. For example, solid chlorine at 0 K is a solid solution of Cl35 – Cl35 , Cl35 – Cl37 ,and Cl37 – Cl37 molecules.

④ At any finite temperature, a pure crystalline solid contains an equilibrium numberof vacant lattice sites, which, because of their random positioning in the crystal

∆𝐺 = ∆𝐻𝑣 − 𝑇∆𝑆= ∆𝐻𝑣𝑋𝑣 − 𝑇(∆𝑆𝑣𝑋𝑣 − 𝑅𝑋𝑣𝑙𝑛𝑋𝑣 − 𝑅 1 − 𝑋𝑣 𝑙𝑛(1 − 𝑋𝑣))

𝑤ℎ𝑒𝑟𝑒 𝑇∆𝑆𝑣𝑋𝑣 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑛𝑡𝑟𝑜𝑝𝑦, 𝑅𝑋𝑣𝑙𝑛𝑋𝑣 − 𝑅 1 − 𝑋𝑣 𝑙𝑛(1 − 𝑋𝑣) = 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑒𝑛𝑡𝑟𝑜𝑝𝑦

𝑑𝐺

𝑑𝑋𝑣= ∆𝐻𝑣 − 𝑇∆𝑆𝑣 + 𝑅𝑇𝑙𝑛𝑋𝑣 + 𝑅𝑇 − 𝑅𝑇𝑙𝑛 1 − 𝑋𝑣 − 𝑅𝑇

= ∆𝐻𝑣 − 𝑇∆𝑆𝑣 + 𝑅𝑇𝑙𝑛𝑋𝑣 ∵ 1 − 𝑋𝑣 ≅ 1

𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑠 𝑖𝑔𝑛𝑜𝑟𝑎𝑏𝑙𝑒,𝑑𝐺

𝑑𝑋𝑣= ∆𝐻𝑣 + 𝑅𝑇𝑙𝑛𝑋𝑣 = 0 at equilibrium.

𝑿𝒗 = 𝒆𝒙𝒑(−∆𝑯𝒗𝑹𝑻)

Page 62: Father of statistical thermodynamics

62Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

Apparent contradictions to the third law of thermodynamics

⑤ Random crystallographic orientation of molecules in the crystalline state can alsogive rise to a nonzero entropy at 0 K

For the Third Law to be obeyed, ∆𝑆𝐼𝑉 = 0

Page 63: Father of statistical thermodynamics

63Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

Experimental verification of the third law

- With the constant-pressure molar heat capacity of the solid expressed in the form

Page 64: Father of statistical thermodynamics

64Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

The influence of pressure on enthalpy and entropy

- For one mole of a closed system of fixed composition undergoing a change of pressure at constant temperature

Page 65: Father of statistical thermodynamics

65Heat capacity, Enthalpy,

Entropy .Fundamental equations .

Statistical interpretion of

entropy .The Third Law .

The influence of pressure on enthalpy and entropy

- For a closed system of fixed composition undergoing changes in both pressure and temperature