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Aerothermodynamics of high speed flowsAERO 0033–1
Lecture 10: Quantum Mechanics and Statistical Physics (introduction)
Thierry Magin, Greg Dimitriadis, and Johan [email protected]
Aeronautics and Aerospace Departmentvon Karman Institute for Fluid Dynamics
Aerospace and Mechanical Engineering DepartmentFaculty of Applied Sciences, University of Liege
Room B52 +1/433Wednesday 9am – 12:00pm
February – May 2019
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Outline I
1 The Dawn of the quantum theoryThe photoelectric effectThe Hydrogen atomic spectrumThe de Broglie wavesThe Bohr theory of the hydrogen atomHeisenberg’s uncertainty principle
2 The Schrodinger equation and a particle in a boxThe classical wave equationThe Schrodinger equationA particle in a boxPostulates of quantum mechanics
3 Rovibrational energy levels of diatomic moleculesHarmonic oscillatorRigid rotator
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Outline II
4 Statistical physicsPure perfect gasMixture of inert perfect gasesMixture of reacting perfect gases
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First Solvay conference in 1911 at the Hotel Metropole (Brussels)
Seated (L-R): W. Nernst, M. Brillouin, E. Solvay, H. Lorentz, E. Warburg, J. Perrin, W. Wien,M. Curie, and H. Poincare, standing (L-R): R. Goldschmidt, M. Planck, H. Rubens, A.
Sommerfeld, F. Lindemann, M. de Broglie, M. Knudsen, F. Hasenohrl, G. Hostelet, E. Herzen,J.H. Jeans, E. Rutherford, H. Kamerlingh Onnes, A. Einstein and P. Langevin
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The Dawn of the quantum theory
Outline
1 The Dawn of the quantum theoryThe photoelectric effectThe Hydrogen atomic spectrumThe de Broglie wavesThe Bohr theory of the hydrogen atomHeisenberg’s uncertainty principle
2 The Schrodinger equation and a particle in a box
3 Rovibrational energy levels of diatomic molecules
4 Statistical physics
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The Dawn of the quantum theory The photoelectric effect
The photoelectric effect
Hertz’s experiment (1886)UV light causes electrons to be emitted from a metallic surface:
Electron KE independent of incident radiation intensityThreshold frequency ν0 below which no electrons are ejectedAbove ν0, electron KE varies linearly with frequency ν
Einstein’s interpretation (1905)Planck’s hypothesis for the energy of electrons in the constituents ofradiating blackbody: E = hν, with h = 6.626 10−34 J sEinstein: the photoelectric effect is caused by absorption of quanta oflight (photons)
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The Dawn of the quantum theory The photoelectric effect
The photoelectric effect
Kinetic Energy of ejected electron
KE = hν − φ
energy of incident photon - minimum energy required to remove anelectron from the surface of the metal
φ = hν0
minimum frequency ν0 that will eject an electron: frequency requiredto overcome the work function of the metal
Einstein obtained a value of h in close agreement with Planck’s valuededuced from the blackbody radiation formula
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The Dawn of the quantum theory The photoelectric effect
Exercise
When lithium is irradiated with light, the kinetic energy of the ejectedelectrons is 2.935 ×10−19 J for λ = 300 nm and 1.280 ×10−19 J forλ = 400 nm. Calculate
1 The Planck constant,
2 The threshold frequency,
3 The work function of lithium,
for these data. The speed of light is 2.998× 108 m/s.
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The Dawn of the quantum theory The Hydrogen atomic spectrum
The Hydrogen atomic spectrum
Every atom, when subjected to high temperatures or electricaldischarge, emits electromagnetic radiation of characteristic frequencies
Characteristic emission spectrum of an atom: certain discretefrequencies called linesHydrogen atomic spectrum: Lyman (1906), Balmer (1885), andPaschen (1908) series
The Balmer and Rydberg formulae for the frequency of the lines(1/λ = ν/c , RH = 109 677.581 cm−1)
νBalmer = 8.2202×1014
(1− 4
n2
)Hz ;
1
λ= RH
(1
n21
− 1
n22
)cm−1
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The Dawn of the quantum theory The de Broglie waves
The de Broglie waves
de Broglie (1924): wave-particle duality of light and matterEinstein’s eq. for photons: λ = h
p , with p = mvde Broglie: both light and matter obey this equation
Experimental observationWhen a beam of X rays / electrons is directed at a crystalline substance, the beamis scattered in a definite manner characteristic of the atomic structure of the crystalDiffraction occurs because the interatomic spacings in the crystal are about thesame as the wavelength of the X-rays / electrons
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The Dawn of the quantum theory The de Broglie waves
Exercise
1 Calculate the de Broglie wavelength of an electron traveling at 1% ofthe speed of light. Data:
The mass of an electron is 9.109× 10−31 kg,The speed of light is 2.998× 108 m/s.
2 Calculate the de Broglie wavelength of a ball of 45 g traveling at avelocity of 40 m/s.
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The Dawn of the quantum theory The Bohr theory of the hydrogen atom
The Bohr theory of the hydrogen atom
Bohr (1911)Hydrogen atom: rather massive nucleus with one associated electronNucleus assumed to be fixed with electron revolving about it
Classical physics:Because the electron is constantly accelerated, it should emit electromagneticradiation and lose energyIt will spiral into the nucleus and a stable orbit is classically forbidden
Two nonclassical assumptions:Existence of stationary electron orbitsThe de Broglie waves of the orbiting electron must be in phase as the electronmakes one complete revolution as shown in case (a), as opposed to cases (b) and (c)
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The Dawn of the quantum theory The Bohr theory of the hydrogen atom
The Bohr theory of the hydrogen atom
Bohr orbits are quantized
2πr = nλ, n = 1, 2, 3, . . .
The angular momentum of the electron about the proton is quantized
λ = hmev⇒
mevr = n~, n = 1, 2, 3, . . . , with ~ =h
2π
Bohr radius: a0 = 4πε0~2
mee2 = 0.529× 10−10m
Coulomb attraction force = centrifugal force
e2
4πε0r2=
mev2
r⇒ r = a0n
2
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The Dawn of the quantum theory The Bohr theory of the hydrogen atom
Application: derivation of Rydberg’s formula
Total energy of the electron in a hydrogen atom
En = KE + V (r)
=1
2mev
2 − e2
4πε0r
=1
2
(e2
4πε0r
)− e2
4πε0r
= − e2
8πε0r
= − mee4
8πε20h
2
1
n2= −13.598 eV
1
n2
Observed spectrum of the hydrogen atom (RH = mee4
8πε20h
2 )
∆E = En2 − En1 = RH
(1
n21
− 1
n22
)= hν
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The Dawn of the quantum theory The Bohr theory of the hydrogen atom
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The Dawn of the quantum theory Heisenberg’s uncertainty principle
Heisenberg’s uncertainty principle
The position and the momentum of a particle cannot be specifiedsimultaneously with unlimited precision: ∆x∆p & h
e.g ., to measure the position of an electron within a distance ∆x , we can use light with a
wavelength λ ∼ δx . For the electron to be “seen” a photon must interact or collide in
some way with the electron. During the collision, some of the photon momentum
p = h/λ will be transferred to the electron, changing its momentum. A smaller
wavelength leads to a smaller ∆x and a larger ∆p.
If we wish to locate any particle within a distance ∆x , then weintroduce atomically an uncertainty in its momentum ∆pThis uncertainty does not stem from poor measurement orexperimental technique, but it is a fundamental property of the act ofmeasurement itselfThe Heisenberg uncertainty principle is of no consequence formacroscopic bodies but it has very important consequences in dealingwith atomic and subatomic particles
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The Dawn of the quantum theory Heisenberg’s uncertainty principle
Exercise
1 Calculate the uncertainty in velocity if we wish to locate an electronwithin an atom so that ∆x is approximately 0.5× 10−10m.
The mass of an electron is 9.109× 10−31 kg.
2 Calculate the uncertainty in the position of a baseball thrown at 40m/s velocity if we measure its momentum with a relative error of10−8.
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The Schrodinger equation and a particle in a box
Outline
1 The Dawn of the quantum theory
2 The Schrodinger equation and a particle in a boxThe classical wave equationThe Schrodinger equationA particle in a boxPostulates of quantum mechanics
3 Rovibrational energy levels of diatomic molecules
4 Statistical physics
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The Schrodinger equation and a particle in a box The classical wave equation
The classical wave equation
The one dimensional wave equation describes the motion of avibrating string
Newton’s law (T=tension of the string and µ its linear mass density)
µ dx∂2u
∂t2= T2 sinβ − T1 sinα
∼(T∂u
∂x
)x+dx
−(T∂u
∂x
)x
⇒ ∂2u(x , t)
∂t2= v2∂
2u(x , t)
∂x2with the propagation speed v =
√T
µMagin (AERO 0033–1) Aerothermodynamics 2018-2019 17 / 80
The Schrodinger equation and a particle in a box The Schrodinger equation
The Schrodinger equation (1925)
The Schrodinger eq. is the fundamental equation of quantummechanics
This eq. cannot be derived, just as Newton’s laws are fundamentalpostulates of classical mechanicsIt gives a description of a quantum system evolving with time, such asatoms, molecules, and subatomic particlesThe Schrodinger equation was developed principally from the deBroglie hypothesisThe solutions to this eq. are called wave functions
The time-independent Schrodinger eq. for finding the wave functionof a particle
Even though we cannot derive the Schrodinger eq., we can at leastshow that it is plausible
⇒ Finding the wave function of a particle
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The Schrodinger equation and a particle in a box A particle in a box
A particle in a box
Classical 1D wave equation
u(x , t): amplitude of displacement of the string from its equilibriumhorizontal position
∂2u(x , t)
∂x2=
1
v2
∂2u(x , t)
∂t2
Time-independent eq.: separation of variables
u(x , t) = ψ(x) cosωt
⇒ ∂2ψ
∂x2+ω2
v2ψ(x) = 0
Seeing that ω = 2πν and v = νλ
∂2ψ
∂x2+
(2π
λ
)2
ψ(x) = 0
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The Schrodinger equation and a particle in a box A particle in a box
A particle in a box
Time-independent Schrodinger eq.
Idea of de Broglie: matter waves
λ =h
p
Total energy of the particle: kinetic energy + potential energy
E =p2
2m+ V (x)
Momentum: p = mv ⇒ λ = hp
= h√2m[E−V (x)]
Substituting λ the wave eq., one obtains
− ~2
2m
∂2ψ
∂x2+ V (x)ψ(x) = Eψ(x)
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The Schrodinger equation and a particle in a box A particle in a box
Exercise
Calculate the wavefunction and energy of a particle in 1D box of size a.
The particle is free in the box (V (x) = 0),
It experiences no potential energy in the region 0 ≤ x ≤ a,
The wave function vanishes at the box boundary:ψ(x = 0) = 0 and ψ(x = a) = 0.
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The Schrodinger equation and a particle in a box A particle in a box
Geometry of theproblem
Schrodinger eq. for 0 ≤ x ≤ a:
d2ψ
dx2+
2mE
~2ψ(x) = 0
with ψ(x = 0) = 0 and ψ(x = a) = 0
General solution
ψ(x) = A cos kx + B sin kx
with k =√
2mE~
Boundary conditionsA = 0 and B sin ka = 0
⇒ ka = nπ, n = 1, 2, . . .
Energy and wave function
En =h2n2
8ma2, ψn(x) = B sin
nπx
a, n = 1, 2, . . .
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The Schrodinger equation and a particle in a box A particle in a box
Interpretation of the wave function
Born’s interpretation of the wave functionψ∗(x)ψ(x)dx is the probability that the particle is located between x and x + dxThe probability that the particle is found outside of 0 ≤ x ≤ a is zero: ψ(x) = 0ψ(x) is assumed to be a continuous function ⇒ ψ(x = 0) = 0 and ψ(x = a) = 0
The wave functions must be normalizedThe particle is certain to be found in the region 0 ≤ x ≤ a:∫ a
0ψ∗(x)ψ(x)dx = 1
|B|2∫ a
0sin2 kx dx = 1
|B|2[
12x −
sin 2kx
4k
]a0
= 1 (2 sin2 kx = 1− cos2kx)
⇒ B =
(2
a
)1/2
ψn(x) =(
2a
)1/2sin nπx
a, n = 1, 2, . . .
Probability of finding a particle between x1 and x2Prob(x1 ≤ x ≤ x2) =
∫ x2x1ψ∗(x)ψ(x)dx
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The Schrodinger equation and a particle in a box A particle in a box
Quantization of the energy
Quantized energy: En = h2n2
8ma2 n = 1, 2, . . .
The energy of the particle is quantized and the integer n is called aquantum numberQuantization arises naturally from the boundary conditions, beyond thestage of Bohr and Planck where the are introduced in an ad hocmanner
Schrodinger, Annalen der Physik 79, 361 (1926)“In this communication, I wish to show that the usual rules of quantization can bereplaced by another postulate (the Schrodinger equation) in which there occurs nomention of whole number. Instead, the introduction of integers arises in the samenatural way as, for example, in a vibrating string, for which the number of nodes isintegral. The new conception can be generalized, and I believe that it penetratesdeeply into the true nature of the quantum rules.”
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The Schrodinger equation and a particle in a box A particle in a box
The energy levels, wavefunctions (a), and
probability densities (b)for the particle in a box
Application: excitation of πelectrons in butadiene
H2C = CHCH = CH2 is assumedto be a linear moleculeLength a= 578 pm with 2 C=Cbonds (2×135 pm), 1 C-C bond(154 pm), 2 C radii (2×77 pm)Pauli exclusion principle: eachenergy state can hold only twoelectrons (with opposite spins)The four π electrons fill the firsttwo energy levelsFirst excited state (3rd level):
∆E = h2
8mea2 (32 − 22)Wavenumber:∆Ehc = 1
λ = 4.54× 104cm−1
1/λ ∼ 4.61× 104cm−1, butadieneabsorption band
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The Schrodinger equation and a particle in a box A particle in a box
Exercise
For a particle moving in a 1D box of size a, show that:
The mean value of x is < x >= a/2, and the mean square deviation isσ2x =< x2 > − < x >2= (a2/12)[1− 6/(π2n2)].
⇒ As n becomes large, this value agrees with the classical value. Theclassical probability distribution is uniform.
Correspondance principle: quantum mechanical results and classicalmechanical results tend to agree in the limit of large quantumnumbers.
Show that Heisenberg uncertainty principle is satisfied:
σxσp =~2
(π2n2
3− 2
) 12
≥ ~2.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 26 / 80
The Schrodinger equation and a particle in a box A particle in a box
A particle in a 3D box
Time-independent Schrodinger eq. for 0 ≤ x ≤ a, 0 ≤ y ≤ b,0 ≤ z ≤ c :
− ~2
2m∇2ψ = Eψ(x , y , z)
with the Laplacian operator ∇2ψ = ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
Boundary conditions
ψ(x = 0, y , z) = ψ(x = a, y , z) = 0, for all y and z
ψ(x , y = 0, z) = ψ(x , y = b, z) = 0, for all x and z
ψ(x , y , z = 0) = ψ(x , y , z = c) = 0, for all x and y
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The Schrodinger equation and a particle in a box A particle in a box
A particle in a 3D box
Wavefunction for 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c :
ψnxnynz (x , y , z) =
(8
abc
)1/2
sinnxπx
asin
nyπy
bsin
nzπz
c
with nx , ny , nz = 1, 2, . . .
Energy levels
En =h2
8m(n2x
a2+
n2y
b2+
n2z
c2), nx , ny , nz = 1, 2, . . .
Particular case: cube → a = b = c
En =h2
8ma2(n2
x + n2y + n2
z), nx , ny , nz = 1, 2, . . .
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The Schrodinger equation and a particle in a box A particle in a box
A particle in a 3D box
The energy levels for a particle in a cube (a=b=c)
The energy levels of a particle in a cube are degenerate (same energyvalue for different index triplets) because of the symmetry introducedwhen a general rectangular box becomes a cube
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 29 / 80
The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Postulates
The development of quantum mechanics is now sufficiently complete that we can reduce
the theory to a set of postulates. MacQuarrie and Simon have reviewed the following
postulates in their book Physical Chemistry, a molecular approach
Postulate 1. The state of a quantum-mechanical system is completelyspecified by a function ψ(x) that depends on the coordinate of theparticle. All possible information about the system can be derivedfrom ψ(x). This function, called the wave function or the statefunction, has the important property that ψ∗(x)ψdx is the probabilitythat the particle lies in the interval dx , located at the position x .
For simplicity of notation, we have assumed, that only one coordinate (x) is needed to
specify the position of one particle, as in the case of a particle in a one-dimensional box
e.g., particle in 1D box: ψn(x) =(
2a
)1/2sin nπx
a , n = 1, 2, . . .
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 30 / 80
The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Postulates
Postulate 2’. To every observable in classical mechanics corresponds alinear Hermitian operator A in quantum mechanics.
e.g., particle in 1D box: total energy operator (Hamiltonian)
H = − ~2
2m
d2
dx2+ V (x)
Definition: A is Hermitian if and only if∫all space
f ∗(x) A g(x) dx =
[∫all space
g∗(x) A f (x) dx
]∗for f (x) and g(x) any two state functions.
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The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Postulates
Postulate 3. In any measurement of the observable associated withthe operator A, the only values that will ever be observed are theeigenvalues an, which satisfy an eigenvalue equation
Aψn = anψn.
Property: The eigenvalues of Hermitian operators are real and theireigen functions are orthonormal.
e.g., particle in 1D box: Hψn(x) = Enψn(x), with energy
En =h2n2
8ma2, n = 1, 2, . . .
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The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Postulates
Postulate 4. If a system in a state described by a normalized wavefunction ψ, then the average value of the observable corresponding toA is given by
< A >=
∫all space
ψ∗Aψ dx .
e.g., particle in 1D box: < x >=∫ a
0 ψ∗n x ψn dx = a/2
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 33 / 80
The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Application: commutators
Definition The commutator of two operators is [A, B] = AB − BA.These operators commute if [A, B] = 0.
Schrodinger’s uncertainty relation Consider two operators A andB. Their standard deviations are related by
σaσb ≥ 12
∣∣∣∣∫ ψ∗(x)[A, B]ψ(x) dx
∣∣∣∣Proof Using the relation AB = 1
2(AB + BA) + 1
2[A, B], one obtains the inequality
| < AB > |2 ≥ 14| < [A, B] > |2. Schwarz’s inequality < A2 >< B2 > ≥ | < AB > |2
yields to < A2 >< B2 > ≥ 14| < [A, B] > |2. The proof follows from the definition
σ2c =< (C− < C >)2 > and the following identity: [A− < A >, B− < B >] = [A, B].
There is an intimate connection between commuting operators and the UncertaintyPrinciple. If two operators A and B commute, then their eigen values a and b can bemeasured simultaneously to any precision. If two operators A and B do not commute,then their eigen values a and b cannot be measured simultaneously to any precision.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 34 / 80
The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Exercise
Considering the simultaneous measurement of the momentum andposition of a particle, so that Px = −i~ d
dx and X = x , show
[Px , X ] = −i~I .Based on the previous relation, show that the Heisenberg uncertaintyprinciple is satisfied:
σxσp ≥~2.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 35 / 80
The Schrodinger equation and a particle in a box Postulates of quantum mechanics
Postulates
Postulate 5. The wavefunction, or state function, of a system evolvesin time according to the time-dependent Schrodinger equation
i~∂
∂tΨ(x , t) = HΨ(x , t).
For most of the cases, the solution to this equation is
Ψ(x , t) = ψ(x) exp
(−iE~
t
),
where quantity ψ(x) is a solution to the following time-independent Schrodinger eq.
Hψ(x) = Eψ(x).
Notice that the probability density is then entirely given by the stationary state wave eigenfunctions:
|Ψn(x , t)|2 = |ψn(x)|2.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 36 / 80
Rovibrational energy levels of diatomic molecules
Outline
1 The Dawn of the quantum theory
2 The Schrodinger equation and a particle in a box
3 Rovibrational energy levels of diatomic moleculesHarmonic oscillatorRigid rotator
4 Statistical physics
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 36 / 80
Rovibrational energy levels of diatomic molecules Harmonic oscillator
Harmonic oscillator
Two masses connected by aspring: model used to describe
the vibrational motion of adiatomic molecule
Observable: total energy E
Operator: Hamiltonian
H = − ~2
2µ∂2
∂x2 + V (x)
Total energy conservation (classicalNewton law)
1
2µx2 +
1
2kx2 = E
Relative coordinate:x = x2 − x1 − l0Undistorted length: l0Force constant (Hooke’s law): kReduced mass: µ
Schrodinger eq.
− ~2
2µ
∂2ψ
∂x2+
1
2kx2ψ(x) = Eψ(x)
(solution based on Hermitepolynomials)
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 37 / 80
Rovibrational energy levels of diatomic molecules Harmonic oscillator
Harmonic oscillator
Left: harmonic-oscillator wavefunctions for the first four
energy levels, Right:probability densities
Vibrational frequency
ω =
(k
µ
)1/2
Vibrational energy levels (nodegeneracy)
Ev = ~ω(v +1
2)
Vibrational quantum number:v = 0, 1, 2, . . .
Zero-point energy (uncertaintyprinciple)
E0 =1
2~ω
Constant energy spacing:Ev+1 − Ev = ~ω
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Rovibrational energy levels of diatomic molecules Rigid rotator
Rigid rotator
Two masses rotating about theircenter of mass: model used to
describe the rotational motion ofa diatomic molecule
Observable: rotational kineticenergy E
Operator: Hamiltonian
H = L2
2I= − ~2
2Ir2∇2
Rotational kinetic energyconservation (classical Newton law)
L2
2I= E
Angular momentum: L = IωMoment of inertia: I = µr2
Reduced mass: µFixed separation of two masses:r = r1 + r2
Angular speed: ω
Schrodinger eq.
−~2
2I(r2∇2Y ) = EY (θ, φ)
(solution based on Legendrepolynomials)
Rigid-rotator wave function: YSpherical coordinates angles: θ, φ
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Rovibrational energy levels of diatomic molecules Rigid rotator
Rigid rotator
Rotational energy levels:
EJ =~2
2IJ(J + 1)
Rotational quantum number: J = 0, 1, 2, . . .Degeneracy of the energy level:
gJ = 2J + 1
Energy spacing:
EJ+1 − EJ =~2
I(J + 1) = hν, J = 0, 1, 2, . . .
Rotational frequency:ν = 2B (J + 1)
Rotational constant:
B =h
8π2I
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Rovibrational energy levels of diatomic molecules Rigid rotator
Limits of validity of rigid rotor and harmonic oscillatorassumption for the nitrogen molecule
Leroy’s rovibrational potential can be used for the ground electronicstate of the nitrogen molecule
Potential curve of N2 for J=0,20,40,. . .
The 9390 (v,J) rovibrational energy levels for N2 are obtainedtogether by solving the Schrodinger equation
These energy levels are either bound (below the dissociation energy)or quasi-bound (above the centrifugal barrier)
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Rovibrational energy levels of diatomic molecules Rigid rotator
Limits of validity of rigid rotor and harmonic oscillatorassumption for the nitrogen molecule
Rotational energy spacing
0 50 100 150 200 250 300Rotational quantum number J [ - ]
0.0
2.5x10-21
5.0x10-21
7.5x10-21
10.0x10-21
12.5x10-21
15.0x10-21
E(v
,J+
1) -
E(v
,J)
[ J
]
12v = 0
Vibrational energy spacing (J=0)
0 10 20 30 40 50 60Vibrational quantum number v [ - ]
0
1x10-20
2x10-20
3x10-20
4x10-20
5x10-20
E(v
+1,
0) -
E(v
,0)
[ J
]
Non rigid rotor: E (J + 1, v)− E (J, v) is not ∝ JNon harmonic oscillator: E (v + 1, 0)− E (v , 0) 6= constant
[paper AIAA 2009-3837]
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 42 / 80
Statistical physics
Outline
1 The Dawn of the quantum theory
2 The Schrodinger equation and a particle in a box
3 Rovibrational energy levels of diatomic molecules
4 Statistical physicsPure perfect gasMixture of inert perfect gasesMixture of reacting perfect gases
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 42 / 80
Statistical physics Pure perfect gas
Microscopic description of a nitrogen molecule gas
Potential energy diagram for N2 [Gilmore, 1964]
Translational energy spacing
ET2,1,1 − ET
1,1,1 = 3h2
8mN2a2 =
kBθTN2
, with θTN2= 10−15 K
(a = 1cm)
Internal energy spacing
RotationERJ=1 − ER
J=0 = ~2
IN2= 2kBθ
RN2
,
with θRN2= 2.88 K
VibrationEVv=1 − EV
v=0 = hνN2 = kBθVN2
,
with θVN2= 3393 K
ElectronicEEN2(A) − EE
N2(X) = kBθEN2
, with
θEN2= 72 293 K
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 43 / 80
Statistical physics Pure perfect gas
Gas composed of independent particles (perfect gas) thatare identical
Let us assume that the total energy of level I for a molecule is givenby its translational, rotational, vibrational and electroniccontributions: εI = εTI + εRI + εVI + εEI , I = 1, 2, . . .
Consider the following system of independent and identical particlesNumber of molecules:
∑I NI = N
Total energy:∑
I NI εI = EVolume: V
A macrostate NI is a population distribution N1,N2, . . . over theenergy levels
At some given instant, the molecules are distributed over the energy levels in adistinct way.In the next instant, due to molecular collisions, the populations of some levels maychange, creating a different set of NI ’s, and hence a different macrostate.
Different states can be associated to the same energy level εI when itis degenerate. The occupancy of an energy level with degeneracy aicompletely defines a microstate for a given macrostate.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 44 / 80
Statistical physics Pure perfect gas
Macrostate and microstate
Let us examine some possible macrostates for∑
I NI = 5 particleswith total energy
∑I NI εI = 20 [au] for a system with 8 energy levels
Index I 1 2 3 4 5 6 7 8energy εI [au] 0 1 2 3 5 7 9 11degeneracy ai 2 1 3 2 4 1 3 2
macrostate 1: NI 0 1 1 0 2 1 0 0macrostate 2: NI 1 0 2 0 0 1 1 0macrostate 3: NI 1 1 0 1 1 0 0 1
Example of three possible microstates for macrostate 1
X
I = 3I = 2 I = 5 I = 6
X XX X X
X X
XX
X
X
X X
microstate 1
microstate 3 (bosons)
microstate 2
X
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 45 / 80
Statistical physics Pure perfect gas
Exercise
Consider particles weakly interacting in a box (cube of volumeV = a3). Introducing the notation n2 = n2
x + n2y + n2
z , withnx , ny , nz = 1, 2, . . ., their translational energy is given byεT = h2/(8mV 2/3)n2
Show that the number of states with an energy less than ε isΓ = 4
3πV /h3(8mε)3/2
Compute the number of nitrogen molecules N0 in a volume of 1 cm3 attemperature T = 0 C and pressure p = 1 atm.Compute Γ assuming that ε = 10 < ε >, where < ε >= 3
2 kBT is theaverage kinetic energy of these particles.Show that in this case the great majority of states are emptyΓ/N0 1.
As consequence, it is possible to group the translational energy levels of similar energiesand to divide the energy scale into successive regions containing a range of energy valuessuch that the following inequality holds ai NI
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 46 / 80
Statistical physics Pure perfect gas
Counting the number of microstatesTwo types of particles
Molecules and atoms with an odd number of elementary particles arecalled fermions and obey a Fermi-Dirac statistics.Molecules and atoms with an even number of elementary particles arecalled bosons and obey a Bose-Einstein statistics.
Quantum mechanical property: indistinguishability of particles
The number of microstates W in a given macrostate N1,N2, . . .follows Pauli’s exclusion principle
For fermions, only one molecule may be in any given degenerate state at anyinstant (C(ai ,NI ) combination of ai states taken NI times without repetition)
W (NI ) =∏I
ai !
(ai − NI )!NI !
For bosons, the number of molecules that can be in any given degenerate state atany instant is unlimited (C(ai + NI − 1,NI ) combination of ai states taken NI timeswith repetition)
W (NI ) =∏I
(NI + ai − 1)!
(ai − 1)!NI !Magin (AERO 0033–1) Aerothermodynamics 2018-2019 47 / 80
Statistical physics Pure perfect gas
The most probable macrostate
The total number of microstates of the system is defined as
Ω =∑
over all sets of NIsuch that
∑I NI=N
and∑
I NI εI =E
W (NI )
The most probable macrostate NI is that macrostate which has themaximum number of microstates. It is derived to be
NI =ai
eαeβεI ± 1
where quantities α and β are Lagrange multipliers associated with theconstraints (+1 for fermions and -1 for bosons)
That constitutes the thermodynamic equilibrium of the system.For N large, it can be shown that only the largest term in the summakes any effective contribution to Ω and one has:
Ω = Wmax
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 48 / 80
Statistical physics Pure perfect gas
The most probable macrostate
ProofFor mathematical convenience, one assumes that NI 1 and ai 1,and for fermions, that ai NI
The finiteness of the total energy requires that NI → 0 as εI →∞, however the
high-lying energy levels for which NI is small contribute little to the value of lnW .
For lower levels, we recall that suitable values of NI and ai can be formally obtained
by grouping the translational energy levels of similar energies and dividing the
energy scale into successive regions containing a range of energy values.
Applying Stirling’s formula ln a! = a ln a− a for very large values of a,
one gets, lnW =∑
I
[NI ln
(aiNI± 1)± ai ln
(1± NI
ai
)], where the
upper sign stands for bosons and -1 for fermionsThe constrained optimization problem is:∑
I
[ln(
aiNI± 1)]
dNI = 0,∑
I dNI = 0,∑
I εIdNI = 0
Introducing Lagrange multipliers, the thesis is obtained by cancelingthe terms between brackets in the relation∑
I
[ln(
aiNI± 1)− α− βεI
]dNI = 0
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 49 / 80
Statistical physics Pure perfect gas
The limiting case: Boltzmann distribution
The values of α and β can be obtained from the constraints∑I NI = N and
∑I NI εI = E
We investigate the limiting case for which |eαeβεI | 1, the physicalmeaning of this assumption will be discussed later. The Boltzmanndistribution is derived as NI = aie
−αe−βεI .Notice that this result is based on indistinguishability of particles as opposed to a purely
classical derivation based on Boltzmann statistics.
Introducing the partition function Q =∑
I aie−βεI , quantity α is
given by α = exp(Q/N) and the following equation is obtained fromthe conservation of particles
NI
N=
1
Qaie−βεI
The energy can be obtained from the relation
E =N
Q
∑I
aiεI e−βεI
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 50 / 80
Statistical physics Pure perfect gas
Boltzmann’s relationUsing the sparse population property ai NI shown in a previousexercise and the formula ln(1 + x) ∼ x for |x | 1, one obtainslnW =
∑I NI (ln ai
NI+ 1). The number of microstates is given by
ln Ω = lnWmax = N(ln QN + 1) + βE
All possible microstates of a system corresponding to given values of N, E , and the εI ’sare a priori equally probable. Since Ω = Wmax , in the Boltzmann limit, this system spendmost of its time in the macrostate corresponding to the Boltzmann distribution.
Boltzmann postulated a functionalrelationship between the entropy of asystem and the number of microstates
S = kB ln Ω
= kBN(ln QN + 1) + kBβE
where kB is Boltzmann’s constant
The entropy is additive: S1 + S2 =kB ln Ω1 + kB ln Ω2 = kB ln(Ω1Ω2)
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 51 / 80
Statistical physics Pure perfect gas
Link with Gibbs’ relation
Writing Gibbs’ relation for a single-component gas
dS =1
TdE +
p
TdV − µ
TdN
Using the definition of temperature ( ∂S∂E )V ,N = 1T , and the relations
β = β(E ,N) andQ = Q[V , β(E ,N)], one obtains
β =1
kBT
The energy can be conveniently be derived from the partition functionas follows
E = NkBT2
(∂ lnQ
∂T
)V ,N
The pressure is given by the relation
p = T (∂S
∂V)T ,N − (
∂E
∂V)T ,N = T (
∂S
∂V)T ,N = NkBT (
∂ lnQ
∂V)T ,N
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 52 / 80
Statistical physics Pure perfect gas
Exercise
In the Boltzmann limit, let us examine the number of ways W inwhich a distribution can occur for values NI = N∗I + ∆NI around theBoltzmann distribution N∗I .
Assuming that |∆NI | N∗I , expand ln(N∗I + ∆NI ) neglecting thirdorder terms.Show that the number of microstate satisfies the following relation
ln WWmax
∼ − 12
∑I
(∆NI
N∗I
)2
N∗I
Consider 1 cm3 of nitrogen gas at 1atm pressure and 0 C. Supposethat the average deviation |∆NI |/N∗I is 0.1%, compute the value of theratio W /Wmax .
We note that the right-hand side is negative showing that the extremum is a maximum.
In a system with a large number of particles, even a small deviation in the distribution
numbers from the Boltzmann values N∗I leads to an enormous reduction in the number of
microstates corresponding to a given macrostate.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 53 / 80
Statistical physics Pure perfect gas
Exercise
Show that the Lagrange multiplier is given in the Boltzmann limit bythe following expression
α = ln
[VN
(2πmkBT
h2
)3/2]
+ ln(Q int
).
The internal partition function Q int is not small. This equation shows a posteriori thatthe condition |eαeβεI | 1 for the Boltzmann limit reduces to
V
N
(2πmkBT
h2
)3/2
1.
This requirement would be violated and the gas degenerate, for example, when theparticle mass is very small and the number density is sufficiently large, as in the case of anelectron gas in metals. It is also violated for normal molecular values of the mass m andnumber density N/V when the temperature T is very low. In this situation the particlestend to crowd together in the lower energy states.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 54 / 80
Statistical physics Mixture of inert perfect gases
Mixture of inert perfect gases
Consider a mixture of species i ∈ S = H + e (heavy species + freeelectrons), with H = Ha + Hp (atoms + polyatomic species)
The total number of particles of each species Ni is kept constantindependently: it is assumed that species are inert and do not react(frozen gas)
The energy level I of species i is εi ,IThe system is assumed to be constrained as follows
Number of particles of species i :∑
I Ni,I = Ni
Total energy:∑
i∈S
∑I Ni,I (εi,I + mih
Fi ) = E
Volume: V
where hFi is the formation enthalpy
The total number of particles is N =∑
i∈S Ni
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 55 / 80
Statistical physics Mixture of inert perfect gases
Energy of chemical bonds
Formation enthalpy
Energy is released in the gas or absorbed by the gas when chemicalreactions occurA common level from which all the energies are measured can beestablished by means of a formation enthalpy hFiNotice that assigning a formation enthalpy to species is an arbitraryconvention, solely a difference of energy can be measured in a chemicalreactor (1st law of thermodynamics)This formation enthalpy can be fixed for instance at 0 K
Formation entropy
There is no need for a formation entropyThe entropy of a perfect crystal at 0 K is exactly equal to zero (3rd lawof thermodynamics)Notice that the semi-classical expression of the entropy presented inthis introduction does not satisfy this property
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 56 / 80
Statistical physics Mixture of inert perfect gases
Boltzmann distribution for a mixture of perfect gases
The most probable macrostate of gas mixture can be obtained in theBoltzmann limiting case following a method similar to the onedeveloped for the single species gas
Ni ,I = ai ,I e−αi−βmih
Fi e−βεi,I
Boltzmann distribution for equilibrium population Ni ,I of energy levelsI of species i Ni ,I
Ni=
1
Qiai ,I exp(
−εi ,IkBT
)
Energy level I of species i : εi,IDegeneracy of level εi,I of species i : ai,IPartition function of species i : Qi =
∑I ai,I exp(
−εi,IkBT
)
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 57 / 80
Statistical physics Mixture of inert perfect gases
Evaluation of the thermodynamic properties in terms of
the partition function Qi =∑
I ai ,I exp(−εi,IkBT
)
For a pure gas of Ni particles of species i in equilibrium in a volumeV , the energy per unit mass is given by
ei =kB
miT 2
(∂ lnQi
∂T
)V
+ hFi
The partial pressure is given by pi = NikBT (∂ lnQi∂V )T
The enthalpy per unit mass is expressed as
hi = ei +piρi
The entropy per unit mass is given by
si =kB
mi
[T
(∂ lnQi
∂T
)V
+ 1
]+
kB
miln
Qi
Ni
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 58 / 80
Statistical physics Mixture of inert perfect gases
Translational energy of a particle in a box
Translational partition function for species i
QTi (V ,T ) ∼
∫ ∞0
exp[
−h2P
8mia2kBT(n2
x + n2y + n2
z)]
dnxdnydnz
= V(
2πmikBTh2
P
) 32
(Hint:∫∞
0 e−αx2dx =
√π
4α , α > 0)
Translational thermodynamic properties for species i ∈ S
eTi (T ) = 32
kBTmi
pi = NiV kBT = nikBT (perfect gas)
hTi (T ) = eTi (T ) + kBTmi
= 52
kBTmi
sTi (pi ,T ) =hTi (T )
T + kBmi
ln
[kBTpi
(2πmikBT
h2P
) 32
]Magin (AERO 0033–1) Aerothermodynamics 2018-2019 59 / 80
Statistical physics Mixture of inert perfect gases
Rotational energy of rigid rotorRotational partition function for diatomic molecule i ∈ Hp
QRdiai (T ) ∼
∫ ∞0
(2J + 1) exp(− θRi
T J(J + 1))
dJ
= − TθRi
[exp
(− θRi
T J(J + 1))]∞
0= T
θRi
θRi : characteristic rotational temperature
Polyatomic molecule
QRi (T ) = 1
σi
(TθRi
)L/2
σi : steric factor: symmetry and indistinguishability propertiesσi = 1 for heteronuclear diatomic molecules (CO, NO, NO+, . . .)σi = 2 for CO2 and homonuclear diatomic molecules (N2, O2, N+
2 , . . .)
L = 2 for linear molecules, L = 3 otherwise
Rotational thermodynamic properties
eRi (T ) = hRi (T ) = LkBT2mi
sRi (T ) =hRi (T )
T + kBmi
[L2 ln
(TθRi
)+ ln
(1σi
)]Magin (AERO 0033–1) Aerothermodynamics 2018-2019 60 / 80
Statistical physics Mixture of inert perfect gases
Vibrational energy of harmonic oscillator
Vibrational modes of CO2
Number of vibrationalmodes for a moleculecomposed of N atoms
3N− 3− L
Diatomic molecule:3× 2− 3− 2 = 1CO2 molecule (linear):3× 3− 3− 2 = 4
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 61 / 80
Statistical physics Mixture of inert perfect gases
Vibrational energy of harmonic oscillator
Vibrational partition function for normal mode m of molecule i ∈ Hp
Shifting the zero-point energy to zero, one obtains
QVim(T ) =
∞∑v=0
exp(−θVimv
T
)=∞∑v=0
[exp(
−θVimT )
]v= 1
1−exp(−θVimT )
(Hint:∑n
v=0 xv = 1−xn+1
1−x →∑∞
v=0 xv = 1
1−x for |x | < 1)
θVim: characteristic vibrational temperature
Vibrational thermodynamic properties
eVi (T ) = hVi (T ) =kB
mi
∑m
θVim
exp(θVimT )− 1
sVi (T ) =hVi (T )
T− kB
mi
∑m
ln[1− exp(−θVimT )], i ∈ Hp
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 62 / 80
Statistical physics Mixture of inert perfect gases
Vibrational energy of harmonic oscillator
Low temperature limit
limT→ 0 K
eVi (T ) = 0
High temperature limit
limT→ ∞ K
eVi (T )
T=
kB
mi
∑m
θVim “ 00 ”
=kB
mi
∑m
θVim limT→ ∞ K
−1T 2
θVim exp(θVimT )−1
T 2
=kB
mi(3N− 3− L)
Example: diatomic molecule
eVi (T ) ∼ kB
miT forT → ∞ K
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 63 / 80
Statistical physics Mixture of inert perfect gases
Electronic energy of an atom / molecule
Electronic partition function for species i ∈ H
QEi (T ) =
∑n
gEin exp(
−θEinT
)
Cut-off criterion: n = 1, 2, . . . , nmax with finite value of nmax to avoiddivergence of QE
i (T ) when nmax →∞Electronic thermodynamic properties
eEi (T ) = hEi (T ) =kB
mi
∑ng
Einθ
Ein exp(
−θEinT )
QEi
sEi (T ) =hEi (T )
T+
kB
milnQE
i
Spectroscopic data for electronic energy level nθEin: characteristic electronic temperaturegEin : degeneracy
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 64 / 80
Statistical physics Mixture of inert perfect gases
Species thermodynamic properties
Species enthalpies for species i are obtained by summing over theenergy modes
Atoms: hi (T ) = hTi (T ) + hEi (T ) + hFi , i ∈ Ha
Molecules: hi (T ) = hTi (T ) + hEi (T ) + hRi (T ) + hVi (T ) + hFi , i ∈ Hp
Electrons: he(T ) = hTe (T ) + hFe
Species energies are obtained in a similar fashion
Species entropy for species i are given by
Atoms: si (pi ,T ) = sTi (pi ,T ) + sEi (T ), i ∈ Ha
Molecules: si (pi ,T ) = sTi (pi ,T ) + sEi (T ) + sRi (T ) + sVi (T ), i ∈ Hp
Electrons: se(pi ,T ) = sTe (pi ,T ) + kB
meln 2
⇒ The spin contribution is added to calculate the free electron entropy. Itcan be formally treated as a degeneracy
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 65 / 80
Statistical physics Mixture of inert perfect gases
Thermodynamic properties
Mixture energies and enthalpies are obtained by summing over thespecies properties weighted by the mass fractions
e =nS∑i=1
yiei (T ) and h =nS∑i=1
yihi (T )
These quantities are linked by the expression e = h − p/ρ
The contribution of the entropy of mixing is added to evaluate themixture entropy
s(p,T ) =∑j∈S
yjsj(pj ,T )
=∑j∈S
yjsj(p,T ) +kB
ρ
∑j∈S
nj ln(1/xj)
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 66 / 80
Statistical physics Mixture of inert perfect gases
Thermodynamic properties: asymptotic values
Equipartition theorem: energy per unit mass for fully excited energymode δ of species i
eδi (T ) = 12
kT
mi= 1
2
RT
Mi= 1
2RiT
When the mode δ is not excited then its energy vanishes
eδi (T ) = 0
Species energy for species i obtained by summing over the modes
ei (T ) =∑δ
eδi (T )
Atomic species with translational modes excited: ei (T ) = 32
kBTmi
Molecular species with translational and rotational modes excited: ei (T ) = 52
kBTmi
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 67 / 80
Statistical physics Mixture of inert perfect gases
Exercise1 Compute the specific heat ratio for:
A monoatomic gas such as He,Air at ambient temperature.
Molecular mass data: MN = 14× 10−3kg /mol, MO = 16× 10−3kg /mol, and
MHe = 4× 10−3kg /mol.
2 Explain why your voice changes when you breath helium.
The speed of a sound wave through helium will then be much higher. So by inhaling
helium and using it as the source of the perceived sound, the frequency of the voice
changes but neither the pitch, since your vocal chords are vibrating at the same speed as
when you are using air, nor the configuration of the vocal tract. So while the base
frequency of the chords remains the same, the frequency of the sound heard by others is
increased.
3 Give bounds for the specific heat of a nitrogen gas at the exit of thenozzle of the Longshot facility assuming that the nitrogen moleculesare not dissociated and their vibrational energy mode partially excited.
Some high-temperature effects in nitrogen can be simulated by means of cold helium.
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 68 / 80
Statistical physics Mixture of reacting perfect gases
Local Thermodynamic Equilibrium (LTE) composition
8-species carbon dioxide mixture: CO, CO2, O2, C, O, C+, O+, e−
⇒ Equilibrium composition derived from
Conservation of charge (quasi-neutrality)
Ne− -NC+ -NO+ =0
Conservation of the number of elements
NCO+2NCO2+ 2NO2+NO+NO+ =NO
NCO+NCO2+NC+NC+ =NC
Equilibrium conditions of a minimal reaction set
CO2 ↔ C + 2OCO ↔ C + OO2 ↔ 2OC+ + e− ↔ CO+ + e− ↔ O
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 69 / 80
Statistical physics Mixture of reacting perfect gases
Mixture of reacting perfect gases
Gas mixture of species i ∈ S = R∪ EIndependent chemical components (elements / charge) i ∈ EDependent chemical components i ∈ R
A convenient set of chemical reactions, sufficient to compute theequilibrium composition, is obtained by writing formation of thedependent components in terms of elements
Xi ∑j∈E
ν ijXj , i ∈ R,
Notice that these reactions do not correspond to elementary processes
Stoichiometric matrix ν ij , i ∈ R, j ∈ E , for 8-species carbon dioxide
j\i C O e− CO CO2 O2 C+ O+
C 1 0 0 1 1 0 1 0O 0 1 0 1 2 2 0 1e− 0 0 1 0 0 0 -1 -1
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 70 / 80
Statistical physics Mixture of reacting perfect gases
Boltzmann distribution
The system is assumed to be constrained as follows
Number of Independent chemical components j ∈ E :∑i∈S ν
ij
∑I Ni,I = Nj
Total energy:∑
i∈S
∑I Ni,I (εi,I + mih
Fi ) = E
Volume: V
The number of macrostates in the Boltzmann limiting case islnW =
∑i∈S
∑I Ni ,I (ln
ai,INi,I
+ 1)
Introducing Lagrange multipliers, the constrained optimizationproblem is written as:∑
i∈S
∑I
[ln
ai,INi,I−∑
j∈E νijαj − β(εi ,I + mih
Fi )]
dNi ,I = 0
The Boltzmann distribution is Ni ,I = ai ,I e−∑
j∈E νijαj−βmih
Fi e−βεi,I , for
species i ∈ S
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 71 / 80
Statistical physics Mixture of reacting perfect gases
Local Thermodynamic equilibrium
Summing of the energy levels, one obtains Ni = e−∑
j∈E νijαj−βmih
Fi Qi
with the partition function Qi =∑
I ai ,I e−βεi,I , for i ∈ S
⇒ The Boltzmann distribution is found to beNi,I
Ni= 1
Qiai ,I e
−βεi,I , forspecies i ∈ S
⇒ The equilibrium of reaction Xi ∑
j∈E νijXj , i ∈ R, is expressed by
means of a generalized Saha equation
Πk∈E
(NkQk
)ν ikNiQi
=e−
∑j,k∈E ν
ikν
kj αj−β
∑k∈E ν
ikmkh
Fk
e−∑
j∈E νijαj−βmih
Fi
= e−β(∑
k∈E νikmkh
Fk−mih
Fi )
since νkj = δjk for j , k ∈ EIt can be shown from the Gibbs’ equation that β = 1
kBT
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 72 / 80
Statistical physics Mixture of reacting perfect gases
Exercise
Show that the composition of a gas mixture satisfies van ’t Hoff’sequation ∑
j∈Eν ij ln xj − ln xi = lnKi , i ∈ R,
With the equilibrium constant
Ki (p,T ) = exp[(migi (p,T )−
∑j∈E ν
ijmjgj(p,T ))
kBT]
The species Gibbs free energy is defined by the relation
gi (p,T ) = hi (T )− Tsi (p,T )
= hFi −kBT
miln
Qi
N
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 73 / 80
Statistical physics Mixture of reacting perfect gases
A library for high enthalpy and plasma flows at VKI
Quantities relevant to engineering design for hypersonic flows
Heat flux & shear stress to the surface of a vehicleTheir prediction strongly relies on completeness and accuracy of thenumerical methods & physico-chemical models
Why a library for physico-chemical models?
Implementation common to several CFD codesNonequilibrium models, not satisfactory today, are regularly improvedBasic data are constantly updated(spectroscopic constants, cross-sections,. . .)
Constraints for the library implementationHigh accuracy of the physical models
Laws of thermodynamics must be satisfiedValidation based on experimental data
Low computational costUser-friendly interface
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 74 / 80
Statistical physics Mixture of reacting perfect gases
MUTATION++ library
MUTATION++ library: MUlticomponent Transport AndThermodynamic properties / chemistry for IONized gases written inC++
⇒ Breeding of the following libraries: Pegase (VKI), Chemkin(Livermore), and EGlib (Ecole Polytechnique)
Modules1 Thermodynamic properties2 Chemistry3 Transport properties4 Energy / chemistry exchange terms
New features versus the fortran Mutation libraryFunctionality rebuilt into a modern, object oriented, extensible framework (J.B.Scoggins)Improved methodology when solving equilibrium compositionsEfficient calculation of costly model parameters using lookup tables defined a priorior during initialization based on error constraints
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 75 / 80
Statistical physics Mixture of reacting perfect gases
Definition of the mixture speciesEarth atmosphere: 79%N2 and 21%O2
Ablation of carbon phenolic composites
Mars atmosphere: 95.55%CO2, 2.7%N2, 1.6%Ar, and 0.15%O2
Titan atmosphere: 98.6%N2, 1.4%CH4
⇒ e.g. multitemperature models
11-species air: N2, NO, O2, N, O, N+2 , NO+, N+, O+
2 , O+, e−
8-species carbon dioxide: CO, CO2, O2, C, O, C+, O+, e−
⇒ e.g. electronic collisional-radiative (CR) models
17-species+7 pseudo-species: methane-nitrogen mixture+ CN(X), CN(A), CN(B), N2(X), N2(A), N2(B), N2(C)9-species+86 pseudo-species: air + N(1-46), O(1-40)
⇒ e.g. rovibrational CR models
1-species+9390 pseudo-species: N +9390 (v,J) levels of N2
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Statistical physics Mixture of reacting perfect gases
Chemical reactions: LTE composition
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
0.2
0.4
0.6
0.8
1.0
x i [ -
]
8-species carbon dioxide mixtureat 1 atm pressure
O
CO
CO2
O2
Ce
-
O+
C+
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
0.2
0.4
0.6
0.8
x i [ -
]
11-species air mixture at 1 atm pressure
O
NO
N2
O2
N
e-
O+
N+
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 77 / 80
Statistical physics Mixture of reacting perfect gases
Chemical reactions: LTE composition
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000Temperature [ K ]
0.0001
0.001
0.01
0.1
1
Spec
ies
mol
e fr
actio
n (L
TE
com
posi
tion)
Carbon phenolic at 0.1atm pressureC:H:O=0.2293:0.6606:0.1101 (Mutation library)
CO C
H
O
e-
O+
H+
C2H
2
C3
H2
C2
CH
C+
OH
C2H
CH4
C3H
CO2
OH
C2H
4
C6H
6
LTE composition
Magin (AERO 0033–1) Aerothermodynamics 2018-2019 78 / 80
Statistical physics Mixture of reacting perfect gases
Mixture enthalpy
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
25
50
75
100
125
h [
MJ
kg-1
]
11-species air mixture at 1 atm pressure
Total
TranslationalRotationalVibrational
Electronic
Form
atio
n
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
1
2
3
4
5
h [
MJ
kg-1
]
11-species air mixture at 1 atm pressure
Tot
alT
rans
latio
nal
RotationalVibrational
Electronic
Form
atio
nMagin (AERO 0033–1) Aerothermodynamics 2018-2019 79 / 80
Statistical physics Mixture of reacting perfect gases
Specific heat at constant pressure
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
5
10
15
20
c p [ k
J kg
-1 K
-1 ]
8-species carbon dioxide mixtureat 1 atm pressure
Frozen
Equilibrium
0 2500 5000 7500 10000 12500 15000
Temperature [ K ]
0
5
10
15
20
25
c p [ k
J kg
-1 K
-1 ]
11-species air mixtureat 1 atm pressure
Frozen
Equilibrium
Equilibrium specific heat: cp = (∂Th)p, frozen specific heat: c fp =∑
i yi (∂Thi )p
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