The comparison of various kinds of energies is shown in Fig. 2.1.
(Phase transformations)
(Fission)
Dislocation
VaporizationVacancy
Fusion
Allotropic transformation
Grain boundary
(Lattice defects)26protons+30neutrons
(Ionization)
(Chemical reaction)
Ene
rgy
per
mol
e
Ene
rgy
per
atom
(or
nuc
leon
)
Fig. 2.1 Comparison of energies concerning microstructures of a material (convertedinto values per atom for energies of dislocation and grain boundary) with
2.1.2 Energy of Atoms and Molecules and MacroscopicEnergy (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 physi-
cist, 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 tem-
perature) 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 kg � m=s2ð 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 for-
mula can be expressed as follows because energies of atomic vibration and
rotation in molecules are to be added in cases of gases of polyatomic mole-
cules (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
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
P ¼ mXN0
1
v2z=Ah
P ¼ m � N0�v2
3VN=m2 ¼ Pa� �
(Eq 2:5)
Here, it is assumed that
XN0
1
v2z ¼XN0
1
v2x ¼XN0
1
v2y ¼N0
3�v2
by isotropy at the speed.
(a) Constant temperature constant pressure (b) Heating at constant pressure (c) Heating at constant volume
(Heat energy)(Heat energy)
Fig. 2.2 The internal energy (DU) of monatomic ideal gases. (a) Gas at constant tempera-ture (T) and pressure (P). The change in energy (Q) and temperature according to
heating at (b) constant pressure and (c) constant volume.
Single crystal PolycrystalNucleus of solidification Dendrite
Supercooled liquid phase
(a) Enthalpy according to grain-boundary tension (b) Enthalpy according to interface tension
Fig. 2.3 Enthalpy according to grain-boundary or interface tension, peculiar tomicrostructures.
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 pro-
portional 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 þ
Work
to
outside
P � DV ¼
Change
of
enthalpy
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 � DV¼ 0 when it
is heated without changing the volume, all of the energy by heating is con-
verted 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 ofcrystals (radius of curvature �r) also have an excessive enthalpy DH ¼2sV/�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
