Download Free Energy of Pure Substances

© 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