statistical thermodynamics. objectives of the...
TRANSCRIPT
Statistical Thermodynamics. Objectives of the Theory
• Creating a bridge between theory of the microworld (theory of individualmolecules and their interactions) and theory of macroscopic phenomena
• Explanation (quantitative) of the properties of macroscopic systems (e.g.thermodynamic functions) using the knowledge of the properties of indi-vidual molecules (obtained from molecular spectroscopy or quantum chem-istry)
• Providing rigorous definitions of thermodynamic quantities and derivationsof the laws of thermodynamics from the laws of quantum mechanics
• Obtaining information on the properties of single molecules and their in-teraction from the knowledge of macroscopic (bulk) properties of matter(mainly of historical significance)
Example
Phenomenological thermodynamics (Boyle, Carnot, Joule, Clausius) :
Free energy (Helmholtz): F (T, V )=E−TS.
dF = −S dT − p dV⇓
S = −(∂F
∂T
)V
p = −(∂F
∂V
)T
⇓(∂S
∂V
)T
=
(∂p
∂T
)V
The functions S=S(T, V ) i p=p(T, V ) for any substance.
Statistical thermodynamics (Maxwell, Boltzmann, Einstein, Smoluchowski):
Free energy: F=−kT lnQ, where Q=ΣI e−EI/kT (statistical sum)
In the case of the gas of N atoms of mass m we have:
p(T, V ) =NkT
V, S(T, V ) = Nk ln
(2πmkT )3/2V
h3N+
5
2Nk
Statistical Thermodynamics. Subject of Research
Using the universal constants (such as k, h, c, e, me) and ,,material parame-ters” specific for the molecules of the considered substance, such as:
• masses and spins of atomic nuclei (http://www.nist.gov)
• molecular bond lengths
• angles between the bonds
• force constants
• electronic excitation energies
• intermolecular potentials
the formalism of statistical thermodynamics allows us to predict:
• thermodynamic functions (entropy, free enthalpy, heat capacity, etc.)
• equilibrium constants
• equation of state
• rates of chemical reactions
• electric and magnetic properties of molecules i
• temperatures and heats of phase transitions
• parameters characterizing critical phenomena
Statistical Thermodynamics. Subfields of Theory
Statistical thermodynamics, and generally statistical mechanics is a very largefield in exact sciences. It can be divided into:
− classical statistical thermodynamics
− quantum statistical thermodynamics
or into:
− statistical thermodynamics of equilibrium states
− statistical thermodynamics of irreversible processes
With regards to calculation techniques we have a different division:
− theories using analytical
− computer simulation methods (Monte Carlo, molecular dynamics).
Recommended literature:
1. F. Reif Fizyka statystyczna, PWN, 1973. Chapters. 3, 4, 6. (available in English)
2. H. Buchowski Elementy termodynamiki statystycznej, WNT, 1998. Rozdz. 1, 3.
3. R. Ho lyst, A. Poniewierski, A. Ciach Termodynamika, WNT, 2005, Rozdz. 13, 16, 17, 18.
Specific properties of macroscopic systems
Macroscopic systems exhibit three important properties (features) distinguish-ing them from microscopic systems:
1. In macroscopic systems occur irreversible processes leading to equilibriumstates in which the properties of the system do not depend on time (andthere are no mass flows).
2. The equilibrium states are uniquely specified by a very small number of pa-rameters (for one-component systems only three parameters are sufficient,e.g. T, V, N).
3. Properties of macroscopic systems are in general random variables withsmall relative fluctuations. Usually these fluctuations decrease with thesize of the system as 1/
√N .
Relative fluctuation δ(X) of random variable X is defined as :
δ(X) =
√σ2(X)
〈X〉=σ(X)
〈X〉where 〈X〉 denotes the average value of the variable X, and σ2(X) is thevariance of this variable (the square of the standard deviation σ):
σ2(X) = 〈(X − 〈X〉)2〉 = 〈X2〉 − 〈X〉2
Property 1 contradicts the Poincare theorem, which states the any confineddynamical system of finite energy returns arbitrarily close to its initial state.
Irreversible process: decompression of the gas of 200 atoms into the vacuum
Density fluctuations, experiment
Density fluctuations, theory
As an example we compute the fluctuation of the number of particles in asmall volume element v in a larger container of volume V .If we had only one particle then:
Pn=1 = v/V ≡ p Pn=0 = 1− v/V ≡ 1− p Pn>1 = 0,
where Pn=k denotes the probability of finding k particles in volume v. Tus,
〈n〉 = 〈n2〉 = 0 · (1− p) + 1 · p = p
σ2(n) = 〈n2〉 − 〈n〉2 = p− p2 = p(1− p)
If we have N particles then Pn=k, 〈n〉 and σ2(n) can be obtained from theBernoulli distribution but there is not need for that. It is sufficient to defineN independent random variables ni, where ni is the number of particles withthe number (label) i in the volume v:
n =N∑i=1
niThen
〈ni〉 = p, σ2(ni) = p(1− p), 〈n〉 = Np, σ2(n) = Np(1− p)
and finally
δ(n) =
√σ2(n)
〈n〉=
√1− pNp
=
√1− p〈n〉
Postulates and three most important probability distributions
Definitions:
Macrostate: is defined as the macroscopic state of the system specified
by a small number of parameters need to defined it
Micorstate: is defined as a specific quantum state of the system
(in quantum mechanics) or a small cell in the phase space
(in classical machanics)
Postulates:
Postulate 1: In a isolated macroscopic system spontaneous processes occursuch that the number of possible microstates increases
Postulate 2: If an isolated system (of fixed energy) is in a state of equilibriumthen
all microstates of this energy are equally probable
Distributions (Ensembles) :
1. Microcanonical. For isolated system. Fixed E, V,N.
2. Canonical (Gibbs). For thermostatic system. Fixed T, V,N .
3. Grand canonical (Gibbs). For open systems. Fixed µ, T, V , where µ de-notes the chemical potential.
Microcanonical distribution
Definition: The number of states Ω(E, V,N) is the number of microstatesof the system of volume V , particle number N and constant energy(contained in the interval from E to E+δE, δE=10−30J).
Definition: Statistical (quantum) definition of the temperature
1kT =
(∂ ln Ω∂E
)V,N
Definition: Empirical definition of the temperature (Tt.p=273.16 K)
T =1
klimp→0
pV
N
Definition: S Statistical (quantum) definition of entropy (Boltzmann, Planck)
S = k ln Ω
Definicja: Statistical (quantum) definition of pressure
p = kT
(∂ ln Ω
∂V
)E,N
Canonical distribution (Gibbs)
A thermostatic system with temperature T does not have a fixed energy.
Such a system can be in a quantum state i of energy Ei with probability:
Pi = 1Q e−Ei/kT
Q =∑i e−Ei/kT
where Q=Q(T, V,N) is the statistical sum. Knowing the statistical sum we
can easily obtain all thermodynamic functions of the system, for instance:
E = kT 2
(∂ lnQ
∂T
)V
S = k lnQ+E
T
p = kT
(∂ lnQ
∂V
)T
F = −kT lnQ
Canonical distribution, c.d.
If we neglect the interaction between molecules and if the temperature T isnot too low then Q has a particularly simple form:
Q =qN
N !
where N is the number of molecules in the system, and q is the statistical sumfor a single molecule, called also the molecular partition function
q =∑i
e−εi/kT
In the definition of the partition function q the summation i goes over allquantum states ψi with energy εi of a single moleculeFor rigid molecules q is approximately equal to the product
q = qtr qrot qvib qel qnucl
of partition functions corresponding to various degrees of freedom
− translational qtr
− rotational qrot
− vibrational qvib
− electronic qel
− nuclear qnucl
Canonical distribution. Calculation of the partition function qtr
The factorization of Q is possible when the energy εi of the ith quantum statecan be represented as a sum of the energy of translation, rotation, vibration,etc.
εi = εtrn1n2n3
+ εrotJKM + εvib
v1...vf+ εel
l + εnuclλ
The function qtr qrot qvib, etc. are defined formally in the same way as as q,for instance the translational partition function qtr has the form:
qtr =∞∑
n1n2n3
exp(−εtrn1n2n3
/kT ),
where the energies of the translational excitations are given by the formulafor the energeis of a particle in the box of volume V = L3:
εtrn1n2n3
=h2
8mL2(n2
1 + n22 + n2
3)
The summation over n1, n2, n3 can be factorized and replaced by integration:
qtr =∞∑
n1n2n3
exp
(−n2
1 + n22 + n2
3
a2
)=
[ ∞∑n
exp
(−n2
a2
)]3
=
[∫ ∞0
exp
(−x2
a2
)dx
]3
=π3/2a3
8
where a=√
8mkT L/h.
Replacing summation by integration
∞∑n=1
exp(−n2/a2) ≈∫ ∞
0
exp(−x2/a2)dx =
√πa
2when a >> 1
Canonical distribution. Calculation of the partition function qtr
After the substitution a=√
8mkT L/h we find
qtr =(2πmkT )3/2V
h3
or
qtr =V
λ3B
, where λB =h
√2πmkT
is the so called thermal de Broglie wave length.
One can now calculate the translational statistical sum Qtr = (qtr)N/N !, the
translational part of the internal energy Etr, heat capacity CtrV , entropy Str:
Calculation of Etr is particularly simple. It is enough to notice thatQtr∼T 3N/2.
We find :Etr =
3
2NkT Ctr
V =
(∂E
∂T
)V
=3
2Nk
Str =Nk ln(2πmkT )3/2V
h3N+
5
2Nk = Nk ln
v
λ3B
+5
2Nk,
where v=V/N . This is the Sackur-Tetrede equation - applicable when v> λ3B
.
Canonical distribution. Rotation of heteronuclear diatomics
For linear molecules the energies of rotational excitations are given by:
εrotJ =
~2
2IJ(J + 1)
where I is the moment of inertia of a molecule (I=µR2 for diatomics)The rotational partition function is thus given by the formula
qrot =∞∑J=0
J∑M=−J
e−εrotJ /kT =
∞∑J=0
(2J + 1) e−J(J+1)ΘrotT ≈
T
γΘrot
valid for T Θrot where Θrot is the characteristic temperature jest of rotations
Θrot =~2
2IkThis large-T formula is applicable also to homonuclear molecules with γ = 2(for heteronculear ones γ = 1). For T ≤ Θrot the series converges very quickly.Most often T>>Θrot since Θrot is of the order 1 K or less. For H2, Θrot=85 K.
Erot =NkT CrotV =
(∂E
∂T
)V
=Nk
Srot = Nk lnT
γΘrot
+Nk,
Rotational heat capacity. Heteronuclear molecule
Hear capacity of para (1), orto (2), and 1:3 para-orto
mixture (3) for hydrogen
Para hydrogen Snucl=0, J = 0, 2, 4, 6, . . .Orto hydrogen Snucl=1, J = 1, 3, 5, 7, . . .
Canonical distribution. Molecular vibration of a diatomic
For diatomic molecules the vibrational energy is given by the formula:
εvib = nhν
The vibrational partition function takes then the from:
qvib =∞∑n=0
e−nhνkT =
1
1− e−Θv/T,
where Θvib is the characteristic temperature of vibration
Θv =hν
k
Temperatures Θvib are high: Θvib≈6000 K for H2, Θvib≈3000 K for N2 or CO.
For the vibrational contributions to the internal energy and heat capacity weget:
Evib =Nhν
ehν/kT − 1⇒ Cvib
V = Nk
(hν
kT
)2 ehν/kT
(ehν/kT − 1)2
Vibrational heat capacity for a diatomic molecule
or for one normal mode of a polyatomic molecule
Canonical distribution. Rotations of polyatomic molecules
For a spherical, symmetric, or asymmetric top one can derive for high T :
qrot =
√π
γ
√T 3
ΘAΘBΘC
,
where ΘA, ΘB, i ΘC are characteristic temperatures
ΘA =~2
2IAk, ΘB =
~2
2IBk, ΘC =
~2
2ICk
This formula is valid only whenT ΘA, T ΘB, and T ΘC.IA, IB and IC are moments of inertia relative to principal axes of a moleculeand γ is the symmetry number equal to the number of permutations of iden-tical nuclei which can be effected by a rotation: γ=n for the Cnv, γ=2n forthe Dnh, γ=12 for the Td, γ=24 for the Oh symmetry group.
Since Qrot=(qrot)N the thermodynamic functions are:
Erot =3
2NkT Crot
V =3
2Nk
Srot = Nk ln
[√π
γ
√T 3
ΘxΘyΘz
]+
3
2Nk,
Canonical distribution. Vibrations of polyatomic molecules
For a polyatomic molecule with f normal modes of vibrations with frequenciesνj the vibration energy is the sum of energies of all modes. Therefore, qvib
must be a product:
qvib =
f∏j=1
1
1− e−hνj/kT
The internal energy and heat capacity per one molecule are sums of contri-butions from each normal mode:
Evib =
f∑j=1
hνj
ehνj/kT − 1
A crystal consisting of N atoms can be treated as a one big molecule withf=3N normal modes. If we assume that all these modes have the same fre-quency equal νE then
Evib = 3NhνE
ehνE/kT − 1CvibV = 3Nk
(hνE
kT
)2 ehνE/kT
(ehνE/kT − 1)2
This is Einstein’s theory of the heat capacity of crystals - quite good forintermediate and high temperatures but poor for low temperatures.
Debye’s theory of heat capacity of crystals
The assumption that all frequencies in the crystals are identical is too drastic.In reality there exists a distribution of frequencies given by the funciton g(ν)such that
g(ν)dν = the number of normal modes with frequencies between ν and ν+dν
The sum over all normal modes can then be given by the integral:
Evib =
∫ νmax
0
hν g(ν)dν
exp(hν/kT )− 1
Debye assumed that
g(ν) = 4πV
(1
cl+
2
ct
)ν2 = Aν2 for ν < νmax
where cl i ct are the longitudinal and transverse sound velocities in the crystal.Then:
Evib = A
∫ νmax
0
hν3dν
exp(hν/kT )− 1= A
k4T 4
h3
∫ ΘD/T
0
x3 dx
ex − 1
where ΘD = hνmax/h is the so called Debye temperature. When T ΘD
then:Evib ∼ T 4 oraz CV ∼ T 3.
When νmax=∞ (photons) we get the Planck distribution and the Stefan law.
Spectral function g(ν) in a crystal
Comparison of Einstein’s and Debye’s theories
Comparison of Debye theory with experiment
Spectrum of relict radiation from a big-bang. T=2.73KComparison with Planck’s distribution
Canonical distribution. Electronic excitations
Electronic partition function is computed directly from its definition preform-ing the summation over relevant electronic states.
qel =∑i
e−εeli /kT
Usually it is sufficient to include only the lowest electronic states resultingfrom the spin-orbit coupling. For example, for the NO molecule the groundelectronic state 2Π splits into two states 2Π3/2 i 2Π1/2 differing by the excitationenergy equal to ∆εel/k=178 K. For NO qel has then the form:
qel = 2 + 2e−Θel/T
where Θel =178 K. Such partition function gives at T=178 K a characteris-tic maximum on the heat capacity as a function of temperature, known asSchottky anomaly.If the nuclei have a spin different from zero (e.g. the nitrogen or deuterium)then to the product of all partition functions we have to include the factor ofthe nuclear partition function
qnucl =∏j
(2sj + 1),
where sj is the spin of the jth nucleus.
Canonical distribution. Application to chemical equilibrium
The equilibrium constant for the chemical reaction
nA A + nB B nC C + nD D
in the gas phase is defined as follows:
Kp(T ) =pnCC p
nDD
pnAA p
nBB
,
where X is the partial pressure of substance X, X=A, B, C, D, and nX isit stoichiometric coefficient. Kp(T ) is the equilibrium constant in terms ofpressures. It is useful to consider also the equilibrium constant in termsof particle numbers KN(T ), defined analogously as Kp(T ), but with partialpressures replaced by numbers of molecules NX of substances:
KN(T ) =N
nCC N
nDD
NnAA N
nBB
,
Using the equation p = NkT/V connecting partial pressures with particlenumbers it is easy to show that these two equilibrium constants are related asfollows
Kp(T ) =
(kT
V
)nC+nD−nA−nB
KN(T ).
Canonical distribution. Application to chemical equilibrium
Using the canonical (Gibbs) distribution one an derive the following, basicformula for the equilibrium constant KN(T ):
KN(T ) =qnCC q
nDD
qnAA q
nBB
,
where qX is the partition function of molecule X defined formally the sameway as before
q =∑i
e−εi/kT
but with molecular energy levels εi measured relative to the energy of free,separated atoms (rather than relative to the energy of the ground state of amolecule. The energy of separated atoms is higher than E0 and differs fromE0 by the atomization energy D0 (energy of dissociation into atoms). Thus
εi = εi −D0
and, in consequence,q = eD0/kTq.
Canonical distribution. Application to chemical equilibrium
When we use the usual partition funcions qX computed relative to the energyof the ground state then the equilibrium constant KN(T ) is expressed as:
KN(T ) = e∆D0/kTqnCC q
nDD
qnAA q
nBB
.
where
∆D0 = nCDC0 + nDD
D0 − nAD
A0 − nBD
B0
and DX0 is the atomization energy of the molecule X. Since DX
0 is equal to theenergy of atoms minus the energy of the ground state EX
0 and since the totalnumber of atoms does not change during the chemical reaction we have
∆D0 = −∆E0
where∆E0 = nCE
C0 + nDE
D0 − nAE
A0 − nBE
B0
is the reaction energy at zero temperature (∆E0 < 0 for exothermic reactions).
The obtained expression for KN(T ) can be computed using spectroscopic dataand gives usually more accurate results than measurement of KN(T ).