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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
Lecture #26-27
page 1
STATISTICAL THERMODYNAMICS
Calculation of macroscopic thermodynamic results
Entropically driven examples:
Free expansion of a gas
V1
gas
V2
gas
vacuum
Lattice model for ideal gas translation:
Molecular volume v, Total volume V
All molecular positions have equal energy εtrans = 0
All system microstates have equal energy Etrans = 0
V1
Calculate S = k ln Ω in each state
V2
Molecular degeneracy g = V/v System degeneracy Ω = gN/N! = (V/v)N/N! For expansion from volume V1 to V2, N
Ω
(V v ) N!
ΔS = kln Ω2 − kln Ω1 = kln 2 = kln 2 N
Ω1
(V1 v ) N!
ΔS = Nkln
V2
V
= nRln 2
V1
V1
Should look familiar! And ΔG = ΔH − TΔS = −nRTln
V2
V1
Entropy change is positive, free energy change is negative, as we expect.
5.60 Spring 2008
Lecture #26-27
NA
VA
Ideal gas mixture
page 2
NB
VB
N = N A + NB V = VA + VB Assume same initial (p,T) for A & B ⇒ same (p,T) for mixture
Assume equal molecular volumes & lattice cell sizes.
Then initially
(V v ) (VB v )
= kln A
NA
S1 = kln ΩA + kln ΩB = kln ΩAΩB
NB
NA !
NB !
After mixing: Count how many ways to distribute NA
molecules of A and NB molecules of B among the (V/v)
lattice sites
As before, the number of ways to distribute N molecules among (V/v) sites is (V/v)N. To correct for indistinguishability, divide by NA!NB! So the final state entropy is (V v )
= kln Ω = kln
N
S2
NA !NB !
(V v )
= kln
N
ΔS = S2 − S1
NA !NB !
(V v ) (VB v )
− kln A
NA
NA !
NB
NB !
(
)
V v)
(
= kln
N
N
(VA v ) (VB v )
NA +NB
A
B
= kln
V NA V NB
VANA VBNB
Since the initial pressures are the same, the initial volumes must be in the ratio
of the number of molecules, i.e. VA/V = NA/N = XA and VB/V = XB, so
V NA V NB
ΔS = kln NA NB = −kln XANA − kln XBNB = −Nk ( XA ln XA + XB ln XB )
VA VB
(> 0)
With a simple microscopic model we can derive the macroscopic entropy change!
5.60 Spring 2008
Lecture #26-27
page 3
Ideal liquid mixture
Lattice model is different from gas because all the cells are occupied. Then in
the pure liquid there is no disorder at all!
SA = kln ΩA = kln1 = 0
+
SB = kln ΩB = kln1 = 0
Mixture: N molecules for N sites. First molecule has N choices, second (N – 1), etc. # ways to put the molecules into sites = N! Correct for overcounting by dividing by NA!NB!
ΔSmix = Smix − ( SA + SB ) = Smix = kln Ωmix = kln
N!
NA !NB !
Stirling’s approximation lnN! ≈ NlnN – N
⇒
(
ΔSmix = Nkln N − Nk − NAklnNA − NAk + NBkln NB − NBk
= ( NA + NB ) kln N − NAkln NA − NBkln NB = NAkln
= −Nk ( XA ln XA + XB ln XB )
)
N
N
+ NBkln
NA
NB
Real liquid has additional states - positional disorder, molecular rotation, etc. –
but these occur in both the pure and mixed liquids, so ΔSmix is dominated by the
disorder in molecular positions that the lattice model describes reasonably well.
***********************************************************************
Combinatorics:
Simple example
Mix 2 molecules A
+ 3 molecules B
How many distinct
configurations Ω?
Ω = 10 = 5!/2!3!
5.60 Spring 2008
Lecture #26-27
page 4
Energy & entropy changes
We saw one example earlier, with 4-segment polymers.
Molecular state:
Energy ε:
0
Degeneracy g:
1
εint
εint
εint
3
We’ve redefined the zero of energy as the ground state energy.
“Configurational” molecular partition function is
qconf =
∑e
states
−εi,conf kT
= e 0 kT + e
−εint kT
kT
+ e −εint
+ e −εint
kT
i
=
∑ ge
energy levels
εi
−εi kT
= e 0 kT + 3e
−εint kT
= 1 + 3e
−εint kT
εi
For a solution of noninteracting polymer molecules,
(
N
Qconf = qconf
= 1 + 3e
−εint kT
)
N
We can determine the thermodynamic properties:
(
Aconf = −kT ln Qconf = −NkT ln 1 + 3e
Uconf
(
−εint kT
) = −NkT ln (1 + 3e
⎛ ∂ln 1 + 3e −βεint
⎛ ∂ ln Qconf ⎞
= −⎜
⎟ = −N ⎜⎜
∂β
∂β
⎝
⎠V,N
⎝
) ⎟⎞
⎟
⎠V,N
−βεint
)
−βε
3ε e int
= N int −βε
1 + 3e int
Energy scales with N: molecules are not interacting with each other so total
energy is just a sum of individual molecule energies.
5.60 Spring 2008
Lecture #26-27
page 5
Average energy per molecule is
εconf =
Uconf 3εint e −βεint
=
−βε
N
1 + 3e int
∑ ε e −βε
But we also know ε = ∑ εiPi =
i
q
i
Sconf
i
i
=
−βε
0 + 3εint e int
−βε
1 + 3e int
- same result
−βε
Aconf Uconf
1 ⎛ ∂ln Qconf ⎞
N 3εint e int
−βεint
=−
+
= kln Qconf − ⎜
+
⎟ = Nkln 1 + 3e
T
T
T ⎝ ∂β ⎠V,N
T 1 +
3e −βε
int
(
)
Also scales with N – sum over individual molecule entropy contributions
Average molecular configurational entropy is
(
sconf = kln 1 + 3e
−βεint
)
+
−βε
1 3εint e int
T 1 + 3e −βεint
In high-T (low-β) limit, it’s kln(4) as expected. In low-T limit, it’s kln(1) = 0.
⎛ ∂ln Q ⎞
−βε
⎛ ∂A ⎞
μconf = ⎜
= −kT ln 1 + 3e int
⎟ = −kT ⎜
⎟
⎝ ∂N ⎠T,V
⎝ ∂N ⎠T,V
(
)
Chemical potential is just A per molecule, and A scales with N so it’s just A/N.
CVconf
−βε
1 ⎛ ∂Uconf ⎞
N ∂ ⎛ 3εint e int ⎞
⎛ ∂Uconf ⎞
=⎜
⎟ = − kT 2 ⎜ ∂β ⎟ = − kT 2 ∂β ⎜
−βεint ⎟
⎝ ∂T ⎠V,N
⎝
⎠V,N
⎝ 1 + 3e ⎠
(
3ε N 1 + 3e
= − int
2
kT
−βεint
) ( −ε
)−e
(1 + 3e )
int
e
−βεint
−βεint
−βεint
( −3ε
int
e
−βεint
)
2
Scales with N, so we can think of a configurational heat capacity per molecule.
Complicated function, but its limits are understandable:
5.60 Spring 2008
Lecture #26-27
page 6
CVconf → 0 as T → 0
At low T, all molecules are in the lowest state. If
kT increases infinitesimally, all the molecules are
still in the lowest state! So the configurational
energy Uconf doesn’t change!
εint
kT
0
CVconf → 0 as T → ∞
At high T, the molecules are equally distributed
among all the states. If kT increases, they are
still equally distributed among all the states! So
Uconf doesn’t change.
Low-T
limit
kT
0
εint
High-T
limit
The low-T limit CV → 0 is common to almost every degree of freedom since
ultimately a temperature is reached at which only the lowest level is occupied.
The high-T limit CV → 0 is characteristic of systems or degrees of freedom
with a finite number of states, i.e. a maximum possible energy. In that case,
ultimately a temperature is reached at which the equilibrium distribution is ~
equal probability of all the levels being occupied. This is the case for molecular
configurations as discussed here and for spin states of nuclei and electrons.