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5.60 Thermodynamics & Kinetics
Spring 2008

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

vacuum

V2gas

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 Calculate S = k ln Ω in each state Molecular degeneracy g = V/v
System degeneracy Ω = gN/N! = (V/v)N/N!
For expansion from volume V1 to V2,
V1 V2

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 V2 V1

N

Should look familiar! And ΔG = ΔH − TΔS = −nRTln

Entropy change is positive, free energy change is negative, as we expect.

5.60 Spring 2008

Lecture #26-27

page 2

Ideal gas mixture

NA VA

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 S1 = kln ΩA + kln ΩB = kln ΩAΩB

(V v ) (VB v ) = kln A
NA

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 finalstate entropy is
S2

(V v ) = kln Ω = kln

N

NA !NB !
N

ΔS = S2 − S1

(V v ) = kln

NA !NB !

(V v ) (VB v ) − kln A
NA

NB

NA !

NB !

( V v )( ) = kln N N (VA v ) (VB v )
NA +NB
A

B

= kln

V NA V NB N VA A 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 N N ΔS = kln NA NB = −kln XA A − kln XB B = −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 Ω = 10 = 5!/2!3!

How many distinct configurations Ω?

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: Degeneracy g:

0 1

εint

εint 3

εint

We’ve redefined the zero of energy as theground state energy. “Configurational” molecular partition function is
qconf = =

∑e states

i
εi

−εi,conf kT

= e 0 kT + e

−εint kT

+ e −εint

kT

+ e −εint

kT

∑ ge energy levels
εi

−εi kT

= e 0 kT + 3e

−εint kT

= 1 + 3e

−εint kT

For a solution of noninteracting polymer molecules,
N Qconf = qconf = 1 + 3e

(

−εint kT

)

N

We can...