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$ KS 3.012 PS 9 THERMODYNAMICS SOLUTIONS
Issued: 11.24.04
Due: 12.03.04
3.012
Fall 2004
THERMODYNAMICS
1. Phase separation of a regular solution. Given below is thermodynamic data for an A-B
solid solution that exhibits regular solution behavior, with a miscibility gap in the solid state.
You have a sample with composition XB = 0.15 which is cooled from very high temperatures
to 250 K.
J
mole
J
µB,o = −1,000
mole
J
Ω = 7,000
mole
µA,o = −2,000
As a reminder, you can derive the molar free energy of solution for the regular solution from
the molar free energy of mixing and the molar free energy of the heterogeneous A-B mixture:
€
ΔG mix,rs = G RS − G heter
G RS = ΔG mix,rs + G heter = ΩX A X B + RT [ X A ln X A + X B ln X B ] + µA,o X A + µB ,o X B
a. What is the critical temperature for this system?
The €
critical temperature marks the first point on the phase diagram as temperature is lowered where
phase separation occurs (at the top center of the miscibility gap). The critical point is identified by
finding the point at the center of the composition axis (XB = 0.5) where the second derivative of the
free energy goes to zero (i.e., at the onset of the ‘humped’ shape in the free energy):
 ∂ 2ΔG mix,RS 

 X = 0.5 = 0
2
 ∂X B  B
Plugging in the expression for the regular solution free energy of mixing and solving for T, we arrive
at the expression given in class for the critical temperature:
€
Tcrit =
7000
Ω
= 420.9K
=
2R 2(8.3144)
b. What are the locations (XB values) of the spinodals and the phase boundaries
(boundaries between the miscilibility gap and pure α1, α2 phases) for this regular
solution at€the given temperature?
3.012 PS 9
1 of 9
12/6/04
The phase boundaries and spinodals are determined by examining the free energy vs. composition
curve. At a temperature below the criticial point (within the miscibility gap), the phase boundaries
are set by the ends of the common tangent that runs between the ‘humps’ of the free energy curve,
while the spinodals are set by the saddlepoints on the plot where the free energy’s curvature
changes sign. These points are illustrated on the diagram below.
∂G
=0
∂X B
€ G
∂ 2G
<0
∂X B2
∂ 2G
=0
∂X B2
SPINODALS
€
Phase boundaries
(BINODALS)
€
€
∂G
=0
∂X B
XB
∂ 2G
>0
∂X B2
Mathematically, the phase boundaries are set by the points where the first derivative of the free
energy of the solution is zero. We can derive the analytic solution to this situation:
€
∂
∂G RS
Ω(1− X B ) X B + µA ,o (1− X B ) + µB ,o X B + RT{(1− X B ) ln(1− X B ) + X B ln X B } = 0
=
∂X B ∂X B
[
]
We write the free energy above completely in terms of XB to calculate the partial derivative. Taking
the derivative, we can simplify the expression:
€
XB
∂G RS
= 0 at the phase boundaries
= Ω − 2ΩX B − µA,o + µB ,o + RT ln
∂X B
(1− X B )
The spinodal points will be identified by looking at the second derivative of the free energy:
€
 1
∂ 2G RS
1 
=
−2Ω
+
RT
+

 = 0 at the spinodal points
∂X B2
 X B (1− X B ) 
It is instructive to look at a plot of all 3 expressions: the molar free energy of the solution, its first
derivative, and its second derivative vs. XB at the given temperature:
€
3.012 PS 9
2 of 9
12/6/04
molar free energy of the regular solution
2000
1000
0
0
0.2
0.4
0.6
0.8
1
-1000
-2000
-3000
-4000
-5000
XB
slope
3000
2500
2000
1500
1000
500
0
-500 0
0.2
0.4
0.6
0.8
1
-1000
-1500
-2000
-2500
XB
curvature
50000
40000
30000
20000
10000
0
0
0.2
0.4
0.6
0.8
1
-10000
XB
3.012 PS 9
3 of 9
12/6/04
We can note several characteristics from the curves: The free energy curve displays the expected
‘W’ shape for a regular solution below the critical temperature. The slope curve crosses zero at 3
points as XB varies from 0 to 1: once at each of the free energy minima, and once at the local
maximum in the hump between the two valleys of the ‘W’. The curvature passes through zero at two
locations as expected, and is symmetric about XB = 0.5. Now, to answer the posed question, the
approximate phase boundary and spinodal locations from these graphs are:
Phase boundaries:
Spinoodals:
XB = 0.025 and XB = 0.9
XB = 0.18 and XB = 0.82
c. At the given temperature and composition, will this system phase separate by
nucleation and growth or by spinodal decomposition? Show why.
The mechanism of phase separation- nucleation and growth vs. spinodal decomposition- for binary
systems cooled a temperature within the phase boundaries is determined by the location of the
system on the composition/temperature diagram. If the composition falls inside the phase
boundaries but outside the spinodal boundaries, nucleation and growth occurs; inside the spinodals,
spinodal decomposition is the mode of phase separation. A the given composition, XB = 0.15, we
are between the left-hand phase boundary and the left-hand spinodal, indicating that nucleation and
growth will occur.
T = T2
α
G
T = 250 K
spinodals
0
3.012 PS 9
XB = 0.15
4 of 9
XB
1
12/6/04
2. Two-state quantum systems. Suppose you have a collection of N non-interacting
molecules that are indistinguishable (and identical). Each molecule resides in one of two
microstates: a ground state with energy 0, and an excited state with energy ε0.
a. What is the molecular partition function for this system?
The partition function is a sum of Boltzmann factors for each of the possible molecular states. There
are only two states for each molecule, with energies 0 and ε0, respectively:
all molecular states
q=
∑
e
−
εi
kT
=e
−
0
kT
+e
−
ε0
kT
= 1+ e
−
ε0
kT
i=1
b. What is the partition function of the entire system of N molecules?
€
Because the given state is composed of non-interacting molecules, the system partition function is
obtained by simply as a product of N single-molecule partition functions. In addition, because the
molecules are indistinguishable, we correct for over-counting total system states that are
indistinguishable by dividing by N!:
N
ε

− 0 
kT
1+ e 
qN 

Q=
=
N!
N!
c. What is the internal energy of this system as the temperature approaches infinity?
The internal energy is obtained€from the ensemble average energy. The most convenient route to
obtain the ensemble average energy is through the direct relationship between U and the system
partition function:
 
−ε o / kT N 
1+
e
(
)  = kT 2 ∂ N ln 1+ e−ε o / kT − ln N!
∂
∂
lnQ
= kT 2 ln
U =< E >= kT 2
(
)

∂T
N!
∂T  
∂T

 
 εo  −ε o / kT
 2 e
 kT 
2
U = kT N
(1+ e−ε o / kT )
[
]
As the temperature approaches infinity, the expression reduces to:
€
 εo 
 2
ε
 kT 
2
U = kT N
=N o
2
(1+ 1)
d. Using the molecular partition function, determine what temperature gives a probability
for a molecule€
to reside in the excited state equal to 1/2. (Hint: The answer will seem
3.012 PS 9
5 of 9
12/6/04
unusual- it is a unique property of a two-state system. But you should readily be able
to solve for T!)
We are asked to look at a molecular state probability:
pε o =
1
2
This is the probability that a given molecule is in the excited state with energy ε = εo:
€
−
εo
−
εo
1 e kT
e kT
pε o = =
=
ε

2
q
− o 
1+ e kT 


−
εo
kT
e
1
=
ε

− o 
2
1+ e kT 


We can rearrange this expression to solve for T:
€
1
1
= ε

2  kTo
e + 1


ε
 o
1
e kT + 1 = 1

2
e
εo
kT
=1
For the last equation to be true, T must be infinity. This bizarre result is a characteristic of two-state
quantum systems, and matches the prediction for the internal energy derived in part (c).
€
3.012 PS 9
6 of 9
12/6/04
3. Predicting molecular collapse of a polymer chain at cold temperatures[DJI1]. Let’s
determine the thermodynamic properties of a 6-repeat unit polymer chain, modeled as a set
of 6 connected beads. The links between each bead (covalent bonds in the real polymer) are
flexible and allow the chain to reside in numerous possible microstates, which are
enumerated in the figure below. Within these microstates, the internal energy of the chain is
greater when there are no bead-bead contacts (contacts between beads which are not
covalently linked to one another). The microstates have 3 different possible energies, based
on the number of bead-bead contacts present: states with no contacts have an energy 2εo,
states with 1 bead-bead contact have an energy εo, and states with 2 bead-bead contacts
have an energy 0. As shown in the figure below, there are 4 states with 2 contacts, 11 states
with 1 contact, and 21 states with zero contacts. Use this simple model to answer the
following questions:
a. What is the partition function for this 6-unit polymer?
The molecular partition function for the polymer sums over the states; we can group together the
Boltzmann factors for states with the same energy:
 −εo 
 − 2ε o 
 −εo 
 − 2ε o 
 −0
q = 4 e kT  + 11 e kT  + 21e kT  = 4 + 11e kT  + 21e kT 










b. What is the probability of the polymer being in a state with energy 0, εo, or 2 εo? (Hint:
you want to know the probability of being in any of the states with the energy of
€ interest, not just the probability of being in one specific state of that energy).
We know the probability of being in one specific microstate is given by the ratio of the Boltzmann
factor over the partition function:
pj =
e
−
εj
kT
q
Now, to get the total probability for finding the polymer in any of the different conformations we must
sum the probabilities for each microstate that has the same energy:
€
3.012 PS 9
7 of 9
12/6/04
 −0 
 e kT  4
pε = 0 = 4 p0 = 4
=
 q  q


 −εo 
 e kT 
pε =ε o = 11p0 = 11
 q 


 − 2ε o 
 e kT 
pε = 2ε o = 21p0 = 21
 q 


c. Calculate the internal energy of the polymer chain as a function of temperature.
We again make use of the relationship
between the ensemble average energy and the partition
€
function:
 
 −εo 
 − 2ε o 
∂ lnQ
2 ∂
kT
ln 4 + 11e  + 21 e kT 
U =< E >= kT
= kT
∂T
∂T  




2
ε
2ε

− o
− o
kT
kT
11
ε
e
42
ε
e
 o
o
+
2

kT 2
∴U = kT 2  kT
q




ε


− o 
εo
−

11+ 42e kT 
kT 
=
ε
e
o



q





d.
€ Calculate the value of the critical temperature Tc, where the polymer is 50% likely to
be in one of the fully collapsed (2-contact) states. You do not need to solve the final
equation- simply reduce the required relationship as much as possible (a
graphical/numerical solution is required to obtain the exact value of Tc).
The probability of finding the system in one of the fully collapsed states is:
pone collapsed state =
e
−
0
kT
q
=
1
q
There are 4 different fully collapsed states, and we want to know the temperature where the total
probability is 50%:
€
3.012 PS 9
8 of 9
12/6/04
pcollapsed = 4 pone collapsed state =
4
4
=
ε
2ε
o
q 
−
− o
kT
kT
4
+
11e
+
21e




1
4
=
εo
2ε

2
−
− o
 4 + 11e kT + 21e kT 


The last equality must be satisfied at T = Tcrit. To find the critical temperature, we would solve for T
in the last equality, which could be done numerically.
€
3.012 PS 9
9 of 9
12/6/04