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5.60 Thermodynamics & Kinetics

Spring 2008

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5.60 Spring 2008 Lecture #28 page 1

MODEL SYSTEMS
Starting with QM energy levels for molecular translation, rotation, & vibration,
solve for q and Q, & all the thermodynamics, for these degrees of freedom.
The results are the fundamentals of molecular statistical mechanics.

We’ll derive the results for a classical model that maps onto QM vibrations.
Then we’ll compare to results (given, not derived) for translation and rotation.

Double-stranded polymer model

Each monomer in one strand interacts with a monomer in the other strand.

Interaction energy for each monomer pair is –ε0.

The strands can “unzip” from one end, rupturing the interactions of the end

monomers, then the next ones, then the next, and so on. Each ruptured

interaction raises the energy by ε0. The three lowest-energy states and the

energy levels are illustrated below.

0 ε0 2ε0 μ

Configurational
ª
energy levels εconf
μ nε0
ª ª
9ε0
8ε0
ª ª ª
7ε0
6ε0
ª ª ª μ
5ε0
4ε0
ª ª ª
3ε0
2ε0
ª ª ª
ε0
0
ª ª ª

Nondegenerate evenly spaced levels, separated by energy ε0: ε = nε0 (n = integer)

5.60 Spring 2008 Lecture #28 page 2

For a very long polymer, there is a large number of levels. Then we can extend
the sum over states in qconf to infinity because the highest energies are much
bigger than kT anyway (so the corresponding terms in the sum are negligible).

∞
qconf = ∑ e −εn kT
≈ ∑ e −nε0 kT
= 1 + e −ε0 kT
+ e −2 ε 0 kT
+ e −3 ε 0 kT
+"
n n

1
( ) + (e )
2 3
kT kT −ε0 kT kT
= 1 + e −ε0 + e −ε0 + " ≡ 1 + x + x2 + x3 + " = where x ≡ e −ε0
1−x
1
qconf = kT
1 − e −ε0

So qconf takes a very simple closed form. Everything else follows.

N
⎛ 1 ⎞
Qconf = ( qconf )
N
=⎜ kT ⎟
⎝1 − e 0
−ε
⎠
⎛ 1 ⎞
Aconf = −kT ln Qconf = −NkT ln qconf = −NkT ln ⎜
⎝1 − e
−ε0 kT ⎟
⎠
= NkTln 1 − e −ε0 kT ( )
⎛ ∂A ⎞
μconf = ⎜ conf ⎟ = kTln 1 − e −ε0 kT
⎝ ∂N ⎠T,V
(
Aconf scales with N, μconf = Aconf/N)
dln qconf 1 −ε0 kT ⎛ ε 0 ⎞ e −ε0 kT 1
Uconf = NkT 2 = −NkT 2 −e ⎜ 2⎟ = Nε (
= Nε 0 ε kT )
dT 1−e −ε0 kT
( ⎝ kT ⎠
0
1−e −ε0 kT
) e 0 −1 ( ) ( )
CV conf =
dUconf
= Nε0
−e ε0 kT − ε0 kT ( 2
) = Nk ⎛ ε ⎞0
2
e ε0 kT

⎜ kT ⎟
dT
( ) (e )

2 2

e ε0 kT − 1 ⎝ ⎠ ε0 kT
−1
Aconf Uconf kT ⎤
Sconf = −
T
+
T
⎡
= Nk ⎢ −ln 1 − e −ε0
⎣
( kT
) + eε 0
ε0 kT
− 1 ⎥⎦

Particularly important are Uconf and CVconf and their high-T and low-T limits. Both
quantities scale with N, so we have them per molecule too.

Low-T limit: Uconf = 0, CVconf = 0. As we’ve seen before, at low T all the molecules
are in the ground state, and a slight increase in T leaves them there, so the
system energy does not increase.

High-T limit: lim Uconf = Nε0 1

= NkT , lim CV conf = Nk
T→∞ (1 + ε0 kT − 1) T→∞
5.60 Spring 2008 Lecture #28 page 3

Once kT exceeds the energy spacing, then further increase in T increases the
occupation of higher levels, but the amount of energy increase with T doesn’t
change any further: Uconf ∂ T, CVconf is T-independent in the high-T limit.

Entropy & probability distributions

Low-T limit: Sconf = 0 = klnΩconf since only the ground state is occupied.
High-T limit:
⎡ ε0 kT ⎤
lim Sconf = Nk ⎢ −ln (1 − (1 − ε 0 kT ) ) + ⎥ = Nk ⎡−ln
⎣ ( ε0 kT + 1)⎤⎦
T→∞
⎣ 1 + ε 0 kT − 1 ⎦
= Nkln (kT ε0 ) = kln (kT ε0 )
N

Note high-T limits for q and Q:

N
kT ⎛ kT ⎞
lim qconf = 1 = lim Qconf lim qconf = , lim Qconf = ⎜ ⎟
T →0 T→0 T→∞ ε 0 T→∞ ⎝ ε0 ⎠
q is a measure of how many states the molecule has thermal access to.

For kT >> εo, it’s just the ratio kT/εo

If kT = 10εo then molecules have thermal access to ~ 10 states.

Boltzmann distribution Pi(εi) gives probabilities for each state:

Pi(εi)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 xε0

Molecular energy εi

Most likely molecular energy ε is 0 (for nondegenerate levels)

Wide range of molecular levels may be occupied
Average molecular energy <ε> >> 0

System energy U = N<ε> >> 0

Individual molecular energies vary widely, but system energy does not

How come?
∑ e ∑
−Ei kT
Recall Q = = ΩE e −Ei kT
i
system system

states i energies Ei

5.60 Spring 2008 Lecture #28 page 4

e −Ei kT ΩE e −Ei kT ΩE e −Ei kT

Also recall Pi = ⇒ PE = i
= i

∑e
−Ei kT i
∑e
−Ei kT Q
i i

Measurement of macroscopic system energy always yields the same result

⇒ P(E) ≈ 1 for that system energy!

System degeneracy Ω(Ei) increases sharply as system energy Ei increases.

e.g. Ω(0) = 1; Ω(ε0) = N; Ω(2ε0) = N(N – 1)/2 + N ≈ N2/2; etc.

This weights probability in favor of higher system energy.

Boltzmann factor decreases as system energy increases.

This weights probability in favor of lower system energy.

Average is a balance between these factors. Probability is very sharply peaked!

How much does the system energy fluctuate?

Molecular average energy = <ε>, molecular standard deviation σ ≈ <ε>

System energy = N x molecular average energy = N<ε>

System standard deviation = N σ ≈ N <ε>

N ε N 1012
Relative system energy variation = = ≈ 24 = 10 −12
N ε N 10
Fluctuations are immeasurably small for a macroscopic system!

System entropy S = −k ∑ pi ln pi
for system at constant T
i

But we can approximate S = kln Ω where Ω(E) is the degeneracy for the most
probable level. This is OK because the range of system energies is very small.

Vibrational partition function & thermodynamics

The double-stranded polymer model used here gives the same energies as
quantum mechanical vibrational modes of molecules and materials.

Classical vibration: E = ½mv2 + ½kx2 = K.E. + P.E., where m is mass, v is velocity, k

is force constant (for this section only, normally it’s the Boltzmann constant),
and x is displacement.
1 k
Natural resonance frequency ν 0 =
2π m
Vibrational amplitude & energy can take on any value, continuously.
5.60 Spring 2008 Lecture #28 page 5

QM vibrational states: nondegenerate, spaced by equal amounts. Spacing is

1 k
ε 0 = hν 0 = h h ª Planck’s constant
2π m

We’ve already done this problem! We can define the zero of vibrational energy
as the lowest vibrational level, and we get identical results.

∞
qvib = ∑ e −εn kT
≈ ∑ e −nε0 kT
= 1 + e −ε0 kT + e −2ε0 kT
+ e −3 ε 0 kT
+"
n n

1
= 1 + e −ε0 kT
(
+ e −ε0 ) + (e
kT 2 −ε0 kT 3
) + " ≡ 1 + x + x2 + x3 + " =
1−x
where x ≡ e −ε0 kT

1
qvib = kT
1 − e −ε0

⎛ 1 ⎞
Qvib = ( qvib )
N
=⎜ kT ⎟
⎝1 − e 0
−ε
⎠
⎛ 1 ⎞
Avib = −kTln Qvib = −NkTln qvib = −NkTln ⎜
⎝ 1 − e −ε0 kT ⎟
⎠

= NkTln 1 − e
−ε0 kT ( )
⎛ ∂A ⎞
μ vib = ⎜ vib ⎟ = kT ln 1 − e −ε0 kT

⎝ ∂N ⎠T,V
( )
Avib scales with N, μvib = Avib/N

2 dln qvib 1 −ε0 kT ⎛ ε 0 ⎞ e −ε0 kT 1

Uvib = NkT = −NkT 2
−e ⎜ 2 ⎟ = Nε 0 ( )
= Nε 0 ε kT
dT 1−e −ε0 kT
( ⎝ kT ⎠ ) 1−e −ε0 kT
e 0
−1 ( ) ( )
CV vib =
dUvib
= Nε0
−e ε0 kT − ε0 kT ( 2
) = Nk ⎛ ε ⎞
0
2
e0 ε kT

⎜ kT ⎟
ε kT
dT
( ) ( )
2 2

e ε0 kT − 1 ⎝ ⎠ e0 −1
Avib Uvib kT ⎤
Svib = −
T
+
T
⎡
= Nk ⎢ − ln 1 − e −ε0
⎣
( kT
) + eε 0
ε0 kT
− 1 ⎦⎥

Results are important for molecular & material vibrations.

Vibrational energy & heat capacity results & limiting values:

5.60 Spring 2008 Lecture #28 page 6

Low-T limit: Uvib = 0 (= N(½ε0) with the zero as usually defined), CVvib = 0.

High-T limit: lim Uvib = NkT , lim CV vib = Nk

T→∞ T→∞

Molecular vibrational frequencies ~

1000-3000 cm-1. kT at 300 K ~ 200 cm-1.
⇒ most molecules in ground vibrational
states at room T (low-T limit). CVvib
3R
Crystal lattice acoustic vibrational
frequencies ~ 30 cm-1 ⇒ most crystals
are in the high-T limit. For N atoms in
an atomic crystal, there are 3N
vibrational modes, so at room T, CV =
3Nk = 3nR. This was used to determine kT ε 0
molecular weights!

No one could explain why CV → 0 at low T until Einstein suggested in 1905 that
if energy was quantized, not continuous, then kT can be much lower than the
first excited state energy. (Not possible if energy is continuous.)

Molecular translation & rotation, classical equipartition of energy

Results are derived in statistical mechanics course 5.62 (and in your text).
One key result: for each degree of freedom (3 translational, 2 or 3 rotational),
high-T limit for energy is <ε> = ½kT & for heat capacity is Cv = ½k.

<εtrans> = ½kT x 3 = 3/2 kT

<εrot> = ½kT x 2 = kT (linear) or ½kT x 3 = 3/2 kT (nonlinear)

<εvib> = kT per vibrational mode

This is the classical equipartition of energy. Why does it come about?

Each degree of freedom has kinetic energy given classically by ½mv2. (½Iω2 for

rotation where I = moment of inertia and ω = angular velocity.)

Vibrational degrees of freedom: kinetic energy ½mv2 & potential energy ½kx2.

All these “squared” energy terms can be written in the form ay2.

5.60 Spring 2008 Lecture #28 page 7

The average molecular energy for any of these degrees of freedom is given by
∞ ∞
<ε> = ε = ∑ εie −εi kT
∑e −εi kT

i
i

But if the levels are spaced close together relative to kT, then we can convert
the sums into integrals. If we treat the energy classically then it’s just

∞ 2
kT ∞
ay2e −ay kT ∫ x2e −x dx
2

ε =
∫−∞
dy
= −∞
where x2 = ay2/kT

∞ − ay2 kT ∞
−x2
∫ e
dy ∫ e dx
−∞ −∞

Integrate numerator by parts

∫ ( )
∞ ∞
x2e −x dx = [x ≡ u, xe −x ≡ dv, v = ( −1/2 ) e −x ]+ABC
2 2 2 2

∫−∞ −∞
x xe −x dx
1 1 ∞ −x2 1 ∞
e dx = ∫ e −x dx
2 2
= − xe −x ∫
∞
−∞ +
2 2 −∞ 2 −∞

⇒ <ε> = ½kT

½kT energy per kinetic and potential energy degree of freedom in high-T limit