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

..

..