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© 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) www.asminternational.org 2 CHAPTER Free Energy of Pure Substances Why does a ball tumble down a slope? This question can be answered easily using the idea of ‘‘energy’’ and conservation of energy. Why does water become steam when it is heated? The ability to answer this question requires knowledge of another basic thermodynamic concept ‘‘entropy,’’ besides energy. For example, the stability of a water and steam system composed of an immense number of H2O molecules depends on not only the energy level of each molecule, but also the stability of H2O molecules as a group. This chapter describes the basics of energy and entropy, and ‘‘free energy’’ including both. 2.1 Energy Relating to Microstructure 2.1.1 Units of Energy There are various kinds of energies such as thermal energy, chemical reaction energy, electric energy, nuclear energy, and action energy of a living cell. However, it was confirmed by the experiments of James Prescott Joule in 1843 that these different energies can be exchanged with each other, and the first law of thermodynamics, which is the energyconservation law, was founded. In the modern metric system, the SI unit of energy is appropriately the joule (J), and it has the following relation to other SI units: energy J joule power distance newton meter ¼ N m ¼ magnetic flux Wb weber mass acceleration distance ¼ Kg m=s2 m current A ¼ ampere electric potential V volt pressure volume ¼ Pa m3 current pascal time A s ampere second ¼ charge C coulomb potential V volt ðEq 2:1Þ Heat energy is expressed as the thermochemical calorie, which by definition is exactly 4.184 J. The calorie used is the field of nutrition (Calorie) is, in fact, a kilocalorie. A nutritional Calorie is 4184 J (Ref 1). © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) 14 / Thermodynamics of Microstructures www.asminternational.org The concept of an energy field is useful to describe how the position of an object dictates the energy ascribed to that object. In a gravitational field, the kilogram force (kgf ) is the force applied over one meter to move the kilogram object that meter. That energy 1 kgf m equals 9.80665 J. By definition, the Newton (N) equals 0.101972 kgf and 1 N m is 1 J. Likewise in an electric field, a charge moving across a voltage requires energy. A Coulomb (C) of charge moved one volt is a Joule. At the atomic level, the electron volt is the amount of energy needed to move one electron through one volt, that is, increase its potential one volt. The charge on the electron is 1.602 1019 C, so 1 eV equals 1.602 1019 J. The energy per electron can be converted to the energy per quantity of matter with the concept of the mole. A mole is the mass (in grams) numerically equal to the molecular mass (weight) of a substance. It is the amount of atoms in a system that contains as many elementary units (Avogadro’s number, 6.02 1023) as there are atoms of carbon in 0.012 kg of the pure nuclide 12C. The elementary unit may be an atom, molecule, ion, electron, photon, or specified group of such units (Ref 2). In Fig. 2.1 the energy content of matter is expressed both as eV per atom and kJ per mole. The conversion factor is 1 eV=atom ¼ 1:602 1019 C V 6:02 1023 =mol ¼ 96:48 kJ=mol (Eq 2:2) The comparison of various kinds of energies is shown in Fig. 2.1. (Fission) (Lattice defects) Energy per mole (Ionization) Dislocation Vaporization Vacancy (Chemical reaction) Grain boundary Energy per atom (or nucleon) 26protons+30neutrons (Phase transformations) Fusion Allotropic transformation Fig. 2.1 Comparison of energies concerning microstructures of a material (converted into values per atom for energies of dislocation and grain boundary) with chemical energies © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) www.asminternational.org Chapter 2: Free Energy of Pure Substances / 15 [Exercise 2.1] The energy required for severe plastic deformation of pure iron pieces (about 8 kJ/mol) is mostly consumed as thermal energy (about 95%), and only about 5% of the energy (about 400 J/mol) is stored in the pieces as lattice defects. The breakdown is presumed to be (i) about 75% to multiplied dislocation (DEd,), (ii) about 10% to excess vacancy (DEv,), and (iii) the remainder of about 15% to strain energy. Solve for the density of dislocation (rd), and density of vacancy (rv), formed by severe plastic deformation, assuming the energy of dislocation per line of unit length* to be Ded ¼ 5 109 J/m, and the energy of vacancy formation to be Dev ¼ 1.5 eV. (See Exercise 2.9 for the formation energy and the equilibrium concentration of vacancies.) [Answer] The total energy of multiplied dislocation is DEd ¼ 400 0.75 J/mol ¼ 300 J/mol. The density rd can be obtained by dividing this energy by the molar volume of pure iron V (7 106 m3/mol) to convert it into the energy per unit volume, and dividing it again by the energy of dislocation line per unit length Ded (5 109 J/m). rd ¼ V DEd 300 J=mol ¼ 9 1015=m2 ¼ Ded 7 106 m3 =mol 5 109 J=m This value is close to the limiting value of the dislocation density in a metal (1016/m2). Next, the total energy of vacancy is DEv ¼ 40 J/mol. Because the energy per vacancy is Dev ¼ 1.5 eV 9.648 104 J atom/mol according to Eq 2.2, the density is rv ¼ DEv 40 J=mol ¼ Dev 1:5 9:648 104 J atom=mol ¼ 3 10 4 =atom This value is about three times as much as the equilibrium density (104) at the temperature just below the melting point. *The elastic strain energy of a dislocation per unit length can be approximated by Eb2/2(1 þ n), where E denotes Young’s modulus, n denotes Poisson ratio, and b is Burgers vector. It can be estimated that Ded 5 109 J/m because E ¼ 200 109 Pa, n 0.28, b ¼ 2.5 1010 m for body-centered cubic (bcc) Fe. The value can be converted into the dislocation energy per atom length (7.7 eV) if it is multiplied by one interatomic distance 2.5 1010 m. The dislocation energy shown in Fig. 2.1 is the converted value. © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) 16 / Thermodynamics of Microstructures www.asminternational.org 2.1.2 Energy of Atoms and Molecules and Macroscopic Energy (Ref 3,4) Internal Energy of Gas Molecules. The total of kinetic energy of atoms or molecules that compose a material and potential energy such as the binding energy between atoms is called internal energy (U). A physicist, D. Bernoulli, in 1738 in Switzerland related pressure of the gas and internal energy of the gas molecule and opened the way to ‘‘the kinetic theory of gases’’ by James Clark Maxwell after about 100 years. [Exercise 2.2] Show that the internal energy of monatomic ideal gas can be expressed as Bernoulli’s equation:* Ugas ¼ (3=2)PV ¼ (3=2)RT (Eq 2:3) where P, V, and T are the system parameters (pressure, volume, and temperature) and R is the universal gas constant. [Answer] If a monatomic gas is considered a group of point particles, each with mass (m) flying about in a disorderly manner, the internal energy per mole is the total kinetic energy of these particles, and it can be approximated by Ugas ¼ (1=2) m( v )2 N0 (Eq 2:4) Here, ( v)2 is the mean (average) velocity of particles squared. N0 is Avogadro’s constant. As shown in Fig. 2.2(a), notice one of N0 particles (the black point) enclosed in a piston of height h, cross section A, and volume, V ¼ Ah. Let the velocity components in the directions X-Y-Z be vX vY vZ (m/s). The frequency at which particles collide with the ceiling side of the piston at the unit time is vZ /2h. The change in the momentum per collision is 2mvz. Therefore, the change of the momentum per unit time is 2mvz vz =2h ¼ mv2z =h ðkg m=s2 Þ. *Equation 2.3 is applicable to the case of monatomic molecules (inert gases such as He, Ar, or metallic gases of Fe, Cu, and so on), and its general formula can be expressed as follows because energies of atomic vibration and rotation in molecules are to be added in cases of gases of polyatomic molecules (H2, H2O, and so on) that have more than two atoms bound together. Ugas ¼ ( f =2)PV ¼ ( f =2)RT (Eq 2:3a) Here, f is the degrees of freedom of motion in a molecule, f 5 in case of diatomic molecules such as H2, and f ¼ 6 to 13 in cases of triatomic molecules such as H2O. © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) (a) Constant temperature constant pressure www.asminternational.org Chapter 2: Free Energy of Pure Substances / 17 (Heat energy) (Heat energy) (b) Heating at constant pressure (c) Heating at constant volume Fig. 2.2 The internal energy (DU) of monatomic ideal gases. (a) Gas at constant temperature (T) and pressure (P). The change in energy (Q) and temperature according to heating at (b) constant pressure and (c) constant volume. Single crystal Nucleus of solidification Dendrite Polycrystal Supercooled liquid phase (b) Enthalpy according to interface tension (a) Enthalpy according to grain-boundary tension Fig. 2.3 Enthalpy according to grain-boundary or interface tension, peculiar to microstructures. Because the force (N) ¼ the momentum change per unit time (kg m/s2) according to particle dynamics, the pressure (P) by collision of N0 molecules (the force per unit area) can be expressed by P¼m N0 X v2z =Ah 1 P¼ m N0 v2 N=m2 ¼ Pa 3V Here, it is assumed that N0 X 1 by isotropy at the speed. v2z ¼ N0 X 1 v2x ¼ N0 X 1 v2y ¼ N0 2 v 3 (Eq 2:5) © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) 18 / Thermodynamics of Microstructures www.asminternational.org From Eq 2.4 and 2.5, the relation of P ¼ 2/3 Ugas/V can be obtained, and if the ideal gas equation PV ¼ RT is used, Bernoulli’s equation in question can be obtained. Because m N0 ¼ M corresponds to molecular weight, the following relation is obtained by Eq 2.3 and 2.4: v2 ¼ 2Ugas=M ¼ 3RT=M (Eq 2:6) The mean square velocity of ideal gas molecules is understood to be proportional to the absolute temperature T. Enthalpy of Gas Molecules. The energy H that is the combination of PV and internal energy U is called an enthalpy.* H ¼ U þ PV (Eq 2:7) The following exercise using the ideal gas helps explain the meaning of this energy. [Exercise 2.3] Investigate the energy balance when the ideal gas is heated on the condition of (i) constant pressure (DP ¼ 0) and (ii) constant volume (DV ¼ 0), and compare the increases in temperature. [Answer] (i) Because not only the internal energy of gas molecules increases from U to U þ DU but also the volume expands from V to V þ DV, the energy balance can be expressed at constant pressure: Heat quantity from outside DQ ¼ Change of internal energy DU Change of enthalpy Work to outside þ P DV ¼ DH (Eq 2:8) Therefore, the energy-conservation law at constant pressure holds with regard to H (Fig. 2.2b). (ii) On the other hand, because the work to outside is P DV ¼ 0 when it is heated without changing the volume, all of the energy by heating is converted into internal energy as in the following equation at constant volume: *The term PV in Eq (2.7) often plays the leading part in thermodynamics of microstructures. For example, a polycrystalline material of mean grain radius R has an excessive enthalpy DH ¼ 2sV/R by internal pressure DP ¼ 2s/R according to grain-boundary tension s to a single crystal (Fig. 2.3a). Crystal nuclei formed in a liquid phase (radius r) and apexes of crystals (radius of curvature r) also have an excessive enthalpy DH ¼ 2sV/ r according to interface tension s (Fig. 2.3 b). As a result, grain growth occurs in (a), and supercooling can be observed in (b). See Sections 5.2 ‘‘Gibbs-Thomson Effect’’ and 8.1 ‘‘Basic Subjects of Nucleation’’ for details. © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) Heat quantity from outside DQ ¼ www.asminternational.org Chapter 2: Free Energy of Pure Substances / 19 Change of internal energy DU (Eq 2:9) Therefore, the energy-conservation law at constant volume holds with regard to U (Fig. 2.2c). When Eq 2.8 and 2.9 are applied to Bernoulli’s equation (Eq 2.3) of monatomic ideal gas, the rise in temperature DT by heating can be obtained: (i) at constant pressure: DQ ¼ DH ¼ (5=2)R DT (Eq 2:8a) (ii) at constant volume: DQ ¼ DU ¼ (3=2)R DT0 (Eq 2:9a) Therefore, DT ’ ¼ (5/3)DT and the rise in temperature at constant volume is larger. 2.1.3 Heat Capacity and Enthalpy of Transformation (Ref 5) Heat Capacity at Constant Volume and Constant Pressure. The heat quantity (per mole) that is required to raise the temperature of a substance by 1 K is called the heat capacity. However, it is also called specific heat (per unit mass) or molar heat in physics. There are two kinds of heat capacity: Heat capacity at constant volume: CV ¼ DQ DT Heat capacity at constant pressure: CP ¼ ¼ V @U @T (Eq 2:10) V DQ @H ¼ DT P @T P (Eq 2:11) When these equations are integrated, one can obtain the equations to calculate U and H from the values of CV and CP respectively: Internal energy: U ¼ U0 þ ðT CV dT (Eq 2:12) 0 Enthalpy: H ¼ H0 þ ðT CP dT (Eq 2:13) 0 Here U0 and H0 are values at T ¼ 0 K. Because the value of CP under ordinary pressure has been measured precisely for years and an enormous store of data has accumulated, the value of H at each temperature can be obtained by Eq 2.13. On the other © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) 20 / Thermodynamics of Microstructures www.asminternational.org hand, because the analysis on the thermal vibration energy of a crystal is carried out on condition that interatomic distance is constant as described in Section 2.3, the obtained theoretical value is CV. CP and CV are converted mutually by* CP CV ¼ @U þP @V T @V @T ¼ P 2 a V T b (Eq 2:14) Here, a ¼ (@V=@T)P =V is the volumetric thermal expansion coefficient and b ¼ (@V=@T)T =V is the isothermal compressibility. [Exercise 2.4] Show that the following relation of J.R. Mayer (1842) is valid between CP and CV for the ideal gas. CP ¼ CV þ R (Eq 2:19) [Answer] From the ideal gas equation PV ¼ RT, the volumetric thermal expansion coefficient a and the isotropic compressibility b of the ideal gas are *Equation 2.14 can be derived; at first, from Eq 2.11: CP ¼ (@H=@T)P ¼ (@U=@T)P þ P(@V=@T)P (Eq 2:15) If U is regarded as a function of V and T, (@U=@T)P ¼ (@U=@V)T ¼ (@U=@V)T (@V=@T)P þ (@U=@T)V (@V=@T)P þ CV (Eq 2:16) The first part of Eq 2.14 can be obtained by rearranging Eq 2.15 and 2.16. Next, if the relation between entropy S and internal energy U (to be described later) DU ¼ T DS P DV and one of Maxwell relations (@S/@V)T ¼ (@P/@T)V are used: (@U=@V)T þ P ¼ T(@S=@V)T ¼ T(@P=@T)V (Eq 2:17) Because (@V/@T)P dT þ (@V/@P)T dP ¼ 0 at constant volume (DV ¼ 0), (@P=@T)V ¼ (@V=@T)P =(@V=@P)T ¼ a=b (Eq 2:18) Rearranging Eq 2.17 and 2.18, the second part of Eq 2.14 can be obtained. © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) a¼ www.asminternational.org Chapter 2: Free Energy of Pure Substances / 21 1 @V 1 ¼ V @T P T 1 @V 1 ¼ b¼ V @P T P When these are substituted into Eq 2.14, Mayer’s equation can be obtained: CP CV ¼ a2 V b ¼R T ¼ PV T Because the heat capacity at constant volume of the ideal gas is CV ¼ ( f/2) R according to Eq 2.3a, the difference (R) between CP and CV is quite large. However, in case of a solid, because CV 3R as mentioned in Section 2.3 but a2V/b 103 J/K2 mol in Eq 2.14, the difference between CP and CV is only about 10% even at temperatures near the melting point. Enthalpy of Transformation. When a substance is heated or cooled, transformations such as fusion, solidification, evaporation, or condensation will occur, and absorption or emission of a specific heat quantity can be observed. This heat quantity was called ‘‘latent heat’’ in the past. However, because latent heat is the change of enthalpy in accordance with phase transformation as described, it has recently been called the enthalpy of melting, the enthalpy of evaporation, or the enthalpy of transformation as a generic name. [Exercise 2.5] The enthalpy of boiling for H2O is DHb ¼ 40.7 kJ/mol, and the enthalpy of melting is DHm ¼ 6.0 kJ/mol. Investigate the components of these heat quantities. [Answer] First, the enthalpy change in accordance with the phase transformation from water to water vapor can be classified as (i) the energy required to cut the mutual bindings of H2O molecules and to change them into vapor molecules DUb and (ii) the energy required for the expansion of volume P DV. Because DV(water ! vapor) Vvapor, the value for (ii) can be estimated: P DV P Vvapor R Tb 3:1 kJ=mol Therefore, most (about 90%) of DHb of boiling enthalpy is consumed by the internal energy change DUb to cut the intermolecular bindings of H2O. DHb (water ! vapor, 40:7 kJ=mol) ¼ DUb (37:6 kJ=mol) þ P DV(3:1 kJ=mol) © 2008 ASM International. All Rights Reserved. Thermodynamics of Microstructures (#05232G) 22 / Thermodynamics of Microstructures www.asminternational.org water ice water Heat capacity Enthalpy water vapor water vapor ice Temperature Temperature (b) (a) Fig. 2.4 (a) Enthalpy (DH) at ordinary pressure and (b) heat capacity at constant pressure of H2O. The heat capacity of water is abnormally large. In case of mercury, for instance, CP is 27.7 J/K mol. Next, the volume change is DV (ice ! water) < 0 in the case of phase transformation from ice to water. However, because P DV ¼ 0.17 J/mol and it is very small, the enthalpy of melting DHm is approximately equal to the change of internal energy DUm. DHm (ice ! water, 6:0 kJ=mol) DUm As is generally known, the heat capacity of water is CP (water) ¼1 cal/g K ¼ 75.3 J/K mol. This value is about twice as large as the heat capacity of ice as shown in Fig. 2.4(b), and it is remarkably large compared to the values of heat capacity for ordinary liquids (0.9 to 1.2 times that of solids). The cause of this anomaly is thought to be the hydrogen bonds between H2O molecules. The enthalpy of melting and the enthalpy of evaporation of a metal is in general approximately proportional to the melting point Tm and the boiling point Tb and the following general rules for them are: R.E. Richards’ rule (1897) for enthalpy of melting: DHm RTm (Eq 2:20) F.T. Trouton’s rule (1884) for enthalpy of boiling: DHb 92Tb (Eq 2:21) Here, the coefficient R in the Richards rule is the gas constant (8.314 J/K mol). According to Eq 2.22 in the next section, the ratio of the enthalpy change DH and the temperature T corresponds to the change of ‘‘entropy’’ DS. Therefore, Eq 2.20 and Eq 2.21 can be expressed as