Statistical Thermodynamics: Molecules to Machines
Statistical Thermodynamics: Molecules to MachinesVenkat
Viswanathan May 5, 2015Module 4:Non-interacting systems:two- level
systems and ideal gasesLearning Objectives: Discuss the statistical
mechanics of non-interacting particles/molecules Introduce
distinguishability and indistinguishability of molecules Analyze
the statistical thermodynamics of a system of 2-state parti- cles
Develop the statistical thermodynamics of a diatomic moleculeKey
Concepts:Non-interacting particles, distinguishable versus
indistinguishable, single- particle partition function, 2-state
model, diatomic molecule, quantum- mode partition function, de
Broglie wavelength, symmetry factor, elec- tronic degeneracy, ideal
gas equation of state, equipartition theorem.Non-interacting
particlesDistinguishable particlesConsider a system of 2
distinguishable particles, meaning that they can be uniquely
identified according to some criterion (e.g. fixed spatial position
of particles on a surface or in space). The two particles contain
energy levels sA (with i = 1, 2, ..., a) and sB (with i = 1, 2,
..., b). In theiicanonical ensemble, the partition function Q is
given by:ab.AB .Q = . exp .
E . = . . expk Ti=1 j=1
i + j kBT. aA
. b
.B .= . exp .i=1
i .kBT
. expj=1
jkBT
= q
AqBwhere qA and qB are the single-particle partition functions
for the two particles.Generally, for a system with N
non-interacting, distinguishable par- ticles, each with a
single-particle partition function q, the partition func- tion is
given by Q = qN .Indistinguishable particlesMany systems are
composed of particles that cannot be distinguished from each other;
for example, particles in the gas phase cannot be dis- tinguished
from each other.For a system of N non-interacting,
indistinguishable particles, the partition function summation
contains instances where particles exist in different states but
cannot be distinguished. On one extreme, if all particles exist in
different states, the over-counting factor is N ! (number of ways
of reassigning N particle labels). On the opposite extreme, if all
particles are in the same state, the indistinguishability and the
degeneracy are the same, so there is no over-counting factor.At
finite temperature, the particles are more likely to exist in a di-
versity of states that are over-counted by the factor N !. Thus to
correct for the indistinguishability at finite temperature, the
partition functionis given by Q = 1 qN .Two-state, non-interacting
particlesConsider a collection of non-interacting, distinguishable
particles that can exist in two states: the ground state (with zero
energy) and the excited state (with energy 0). We will analyze this
system using two of the ensembles derived in the previously to show
equivalence.Microcanonical EnsembleFor a system of N particles, we
consider a state with a fixed energy E = 0n, where n is the number
of particles in the excited state. The degeneracy in the number of
states with this fixed energy E is given by: =N !n! (N n)!which is
the number of permutations of n in N (N choose n).Therefore, the
entropy of the system is given by:
(1)S = kB log
.N !n!(N
.n)!
(2)To approximate the entropy for large number of particles N
(ther- modynamic limit), we use Stirlings approximation: The
logarithm of N ! is first written as log N ! = .N
log i The summation is then converted to an integral to give:N
N. log n n=11
dn log n = N log N N + 1 N log N N(3) Stirlings approximation is
written as log N ! N log N N . The entropy is now written as:S = kB
log
.N !n! (N
.n)! kB [N log N N n log n + n (N n) log(N n) + (N n)]= kBN [(1
x) log(1 x) + x log x](4)where x = n/N = E/ (0N ) is the fraction
in the excited state.Using thermodynamic relations, the temperature
is found to be:1. S .=TE
V ,N
. x .=E
V ,N
. S .
x
V ,N
= kBN0N
[log(1 x) log x] x.thus= exp
0 .
= or x =1 x
kBT
1 + The Helmholtz free energy per particle f = F /N is given
by:f = e Ts = 0x + kBT
.1log1 +
.1.1 +
+
log 1 +
. ..1 + = kBT 1 + log + kBT log(1 + ) + kBT 1 + log .. 0 ..= kBT
log
1 + exp
kBT
= kBT log q(5)where we define q = 1 + exp . 0 ., which will
become the single-particle canonical partition function.Canonical
ensembleThe canonical partition has a fixed temperature T and
number of par- ticles N , thus the energy states fluctuate
according to the Boltzmann distribution found in the previously.
For this system, the partition func- tion Q is written as:Q = . .
... . exp .
0m1
0m2 ...
0mN .m1 m2mN. 1
kBT kBT.N
kBTN=. exp .m=0
0m .kBT
..= 1 + exp
0 ..kBT
= qNwhere we again define the single-particle partition function
q.An alternative method to derive the partition function is to
extend the microcanonical solution to the canonical solution, such
that:Q = . (E) exp .
E .N N !
=exp
0 .n
1Nnk TE.. 0 ..N
n=0 N
n!(N n)!
kBT= 1 + exp
= qkBTwhich makes use of the binomial expansion from Module
3.The Helmholtz free energy is found to be:.. 0 ..F = kBT log Q =
NkBT log q = NkBT log
1 + exp
kBTwhich is in agreement with the microcanonical approach.Using
Q as a generating function, the average energy is given by: (E)e
==
1 . log Q .
. log q .=
= 0 exp(0)
Figure 1:Thermodynamic func-NN
V ,N
V ,N
1 + exp(0)
tions for the 2-state model.The first figure shows F /(NkBT )
(solidwhere = 1/(kBT ). As 0 (T ), e 0/2 (entropy-driven mix of
states at largeT ). As (T 0), e 0 (ground-state energy at zero T ).
The entropy is given by:
curve), E/(NkBT ) (dashed curve), and S/(NkB ) (dashed-dotted
curve) versus kBT /0; the dashed curve gives the asymptotic value
for entropy S kB log 2 at large T . The figure be- low shows the
energy E/(N0 ) versus kBT /0.s = S
= e f = k
exp(0)
log[1 + exp(
)]NTB
0 1 + exp(0 + kB0 As 0 (T ), s kB log 2 (entropy-driven mix of
states at large T ). As (T 0), s 0 (zero entropy at zero T ).Ideal
gas of diatomic moleculesConsider N non-interacting, diatomic
molecules in a fixed volume V and temperature T (canonical
ensemble). The molecules in the gas phase are indistinguishable,
thus the partition function is given by Q = 1 N N !
, where q is a single-molecule partition function. In Module 1,
weanalyzed the quantum mechanics of a diatomic molecule, resulting
in translational, rotational, vibrational, and electronic quantum
mechani- cal modes (Fig. 2).Quantum mechanics predicts the
energetic contributions associated with:h2
Figure 2: Three quantum mechani- cal modes that are relevant to
the thermodynamic behavior of a diatomic molecule. Quantum
mechanics associ- ated with electronic states are not de-
picted.n=
.n + n + n .translational modes8ML2xyzEtransErot
222 k2l=2R2 l (l + 1)rotational modes.1 .Evib
j=j + 2
kvibrational modeswhere M = m1 + m2, = m1m2 , = . k , R is the
atomic separation,m1 +m2and k is the spring stiffness.Owing to the
fact that these quantum mechanical modes are decou- pled, the
single-particle partition function is separable into quantum- mode
partition functions, such that:q = qtransqrotqvibqelec(6) where
qtrans is the sum over translational modes, qtot is the sum over
rotational modes, etc.We find solutions for each of these
quantum-mode partition func- tions:Translational partition
functionThe translational partition function is written as. . .
exp
2.n2 + n2 + n2(7)nx =1 ny =1 nz =1
8kBTML2xyz2Define the translational temperature trans =h,
resulting theBtranslational partition function:.
. trans..qtrans =
. expn2Tn=1The translational temperature is generally very
small. For example,O2 in a box with L = 1cm results in trans = 1.5
1015K. Thistemperature represents the temperature where the
translational modes become excited.For any appreciable temperature
(T trans) we can approximatethe summation by an integral:.
. trans 2..3
. T
.3/2
. 2MkBT .
3/2qtrans
dneT0
==V4transh2where we have used V = L3, and eax2 dx = 1 , .02a
We can define the de Broglie wavelength = .h2B
to arrive atthe final result qtrans = V .Rotational partition
functionFor each l-index value, there are (2l + 1) m-index values
(degeneracy). Therefore, the rotational partition function
is:.2.qrot = .(2l + 1) expl=0
k
2k T R2 l(l + 1)
(8)We define the rotational temperature rot = 2k k
, resulting inqrot = .
.
.
BR2l=0(2l + 1) exp T l(l + 1) . The rotational temperature
isgenerally small (though larger than trans), e.g. the rotational
temper-ature for O2 is rot = 2.08K.For any appreciable temperature
(T rot), the rotational partitionfunction is approximated by the
integral:qrot
dl(2l + 1) exp0
. rot T
.l(l + 1) =
T SHAPE \* MERGEFORMAT
rot
(9)We must correct for rotational symmetries in the molecule,
such thatrot . The symmetry factor accounts for the number of
equiva- lent orientations ( = 1 for heteronuclear diatomic
molecules and = 2 for homonuclear diatomic molecules).Vibrational
partition functionThe vibrational partition function is written
as:. .1 ..qvib = . expj=0
k
kBT
j + 2
(10)We define the vibrational temperature vib = k , which
dictates when vibrational modes become excited. This temperature is
usually large; for example, the vibrational temperature for O2 is
qvib = 2274K.Using the property .
j1 , we have:1zqvib = evib /(2T ) . expj=0
. vib . T=
evib /(2T )1 evib /(T )
(11)This exact result for qvib is valid over the entire
temperature range, which is necessary since vib tends to be very
large. In comparison, the approximate solutions for qtrans and qrot
are valid at temperatures T trans, rot (note, trans, rot are very
small).Electronic partition functionThermal excitation of the
electronic states in a molecule leads to the electronic partition
function:qelec = g0 + g1e1 /kBT + g2e2 /kBT ...(12)where gi is the
electronic degeneracy of the ith state, and i are the electronic
excitation energy of the ith state relative to the ground state. We
define the electronic temperature elec = 1/kB , which is typi-cally
extremely large (elec 104 105K).For most applications with T elec,
the electronic partition func- tion is written as qelec g0. The
ground-state electronic degeneracy g0 must be found for the
molecule (or atom) of interest. Typically, theseare tabulated for
many simple molecules (for example, see page 207 ofDill and
Bromberg1). For O2, the ground-state degeneracy is g0 = 31 Ken A.
Dill and Sarina Bromberg.(see page 158 of Hill2).Statistical
thermodynamics of O2The partition function for O2 is given by:
Molecular Driving Forces: StatisticalThermodynamics in Chemistry
and Bi- ology. Garland Science, 2003. ISBN 08153205152 Terrell L.
Hill. An Introduction to Statistical Thermodynamics. Courier Dover
Publications, 1960. ISBN 0486652424Q = 1 qN =N !
1 .. VN !3
. . T SHAPE \* MERGEFORMAT
rot
. . evib /(2T )1 evib /(T )
..Ng0
(13)where = .h2B
101/2T 1/2, rot = 2.08K, = 2, vib =2274K, and g0 = 3 (valid for
trans, rot T elec).The Helmholtz free energy is found using F = kBT
log Q to be:F = kBT
.N log
. V .3
.+ N log (qrotqvibqelec) logN != kBTN
.log
. V.N 3
.+ log (qrotqvibqelec) + 1
(14)which exhibits the necessary property that F (T , V , N ) is
a homogeneous first-order function of V and N , which only occurs
when the indistin- guishability factor of 1 is included in Q.From
thermodynamics, we write the equation of state:. F . SHAPE \*
MERGEFORMAT
V
T ,N
= p =
NkBT V
pV = NkBT(15)which is the ideal gas equation of state.The heat
capacity is found to be:CV = T3
. S . SHAPE \* MERGEFORMAT
T
V ,N
= T
. 2F .T 22
V ,Nevib /T vib
= 2 NkB + NkB + NkB T 2 .1which is valid for trans, rot T elec.
evib
/T .2(16)The first contribution of 3NkB /2 is associated with
the 3N transla- tional degrees of freedom (kB /2 from each
direction), and the second term of NkB is associated with the 2N
rotational angles and (con- tributing kB /2 each). As T passes
through vib, the heat capacity goes from 5NkB /2 to 7NkB /2, thus a
single vibrational mode contributes kBT to the heat capacity. As
the temperature is raised, the individ-ual degrees of freedom get
turned on in order to further maximize the entropy, since more
active degrees of freedom (non-zero probability) con- tribute to
the entropy (Fig. 3).Equipartition theoremFor T trans, rot , the
heat capacity receives kB /2 for each degree of freedom, thus the
average energy per particle to be:
Figure 3: The heat capacity of a gas of diatomic molecules
versus temperature.() = (trans)(rot) = 3
kBT2+ 2
kBT(17)2which suggests that the average energy of a thermally
active degree of freedom contributes kBT /2 to the energy.Consider
a microscopic energy function that is quadratic, such that(x) =
cx2, where x is a generalized degree of freedom: Translational
modes and rotational modes both fit in this category,since: Etrans
=h2
n2 and Erot = k2
l(l + 1).n8mL2n2R2 The average energy is:.. (x) ..2
.cx2 .x (x) exp
kBT
x cx
exp
kBT() =
x exp
. (x) .=kBT
x exp
.cx2 .(18)kBT At large T , the sums can be approximated by the
integrals:()
dxcx2 exp . cx ...
(19) exp
cxkBT Noting the mathematical properties:C0 =
dxeax = 2
1/2 a1/2dC0
1/2C1 =
dxx2eax
=da = 2a3/2
(20)and setting a = c/(kBT ), we have:() = c
1/22 [c/(kBT )]3/2
[c/(kBT )]1/21/2
= kBT2
(21) The equipartition theorem states that the total energy will
parti- tion the amount kBT /2 into each thermally active degree of
freedom, provided the microscopic energies obey quadratic energy
functions.Consider a microscopic energy function that is linear,
such that (x) =cx, where x is a generalized degree of freedom:
Vibrational modes fit in this category, since E(vib) = .j + 1 .
k.j2 The average energy for this function gives:0 dxcx exp
cxkBT() =
0 exp
= kBT(22)cxkBT The equipartition theorem predicts that the
average energy per par- ticle for temperatures vib T elec is given
by:() = 3
kBT2+ 2
kBT2+ kBT(23)where 3 translational modes contribute kBT /2 each,
2 rotational modes contribute kBT /2 each, and 1 vibrational mode
contributes kBTThis picture is only valid at temperatures where the
modes are active. For example for T vib, the discrete vibrational
quantum modes are frozen down to the ground state energy level vib
= k/2 (independent of temperature).ReferencesKen A. Dill and Sarina
Bromberg. Molecular Driving Forces: Statistical Thermodynamics in
Chemistry and Biology. Garland Science, 2003. ISBN
0815320515.Terrell L. Hill. An Introduction to Statistical
Thermodynamics. Courier Dover Publications, 1960. ISBN
0486652424.B
N !
i=1
kBT
..
B
q
.
h..
8k ML2
3
n
2Mk T
3
B
2
rot
qrot = T
kBT
j=0
z =
2Mk T
= 4.410 mK
N !
.
.
kBT
2
2
2
..
..