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Subject
Chemistry
Date
Feb 20, 2024
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MS&E 3030 Prelim – Closed Book
Prelim #2 – November 7, 2022
Solutions
1.
(20 pts) Cadmium (Cd) and zinc (Zn) form a simple eutectic phase diagram with moderate solubility of Zn
in solid Cd (up to 4.35 at.% at the eutectic). Both have relative low melting and boiling points as given in
the table below.
Neither the solid nor liquid solutions can be modeled by ideal mixing.
From literature, phase change values for Cd and Zn are:
T
melt
(K)
∆
h
m
(J/mol)
T
boil
(K)
∆
h
v
(J/mol)
Cadmium
594 K
6,210
1,040
99,870
Zinc
693 K
7,320
1,180
115,000
A Cd-Zn alloy is fully equilibrated at 550 K in the two-phase
L
+(Cd) region; the solid has a composition
Cd
0
.
97
Zn
0
.
03
and the liquid has a composition Cd
0
.
88
Zn
0
.
12
. The ambient is nominally pure N
2
, but partial
pressures of Cd and Zn are measured and found to be
2
.
98
×
10
−
5
bar and
3
.
0
×
10
−
7
bar respectively (as
shown in the schematic below). The phase diagram is included for your edification (unneeded).
For Zn, the saturated vapor pressure over the liquid is given approximately by
ln
h
p
o
Zn(
l
)
(bar)
i
=
−
13
,
831
T
+ 11
.
72
The databases do not have an equivalent expression for Cd.
(a)
(3 pts) Using the Clausius-Clapeyron relationship and the data above, obtain a comparable expression
for the equilibrium vapor pressure of Cd as a function of temperature.
Given we need to have ln
p
= 0 at the boiling point and the Clausius-Clapeyron slope, we can just
integrate from
T
b
to
T
ln
h
p
o
Cd(
l
)
(
T
)
i
=
−
Z
T
T
b
∆
h
vap
R
d
1
T
=
−
∆
h
vap
RT
+
∆
h
vap
RT
b
=
−
12
,
012
T
+ 11
.
55
(b)
(3 pts) Estimate the free energy of melting for Cd and for Zn at this temperature (550 K).
Just based on one-term Taylor series expansion of ∆
g
m
from
T
m
as given in the formula sheet or derive
∆
g
m
[Cd]
=
∆
h
m
1
−
T
T
m
= (6210 J/mol)
1
−
550
594
= 460 J/mol
∆
g
m
[Zn]
=
∆
h
m
1
−
T
T
m
= (7320 J/mol)
1
−
550
693
= 1
,
510 J/mol
1
(c)
(12 pts) Determine the activities of Cd and Zn in the two solutions for the various reference states
listed below. Give final answers to 3 significant digits.
Part of this should was a very nice give away. The solid and liquid alloys are in complete equilibrium
and hence they must have the exact same activities.
The activity with respect to the gas standard
state is just the partial pressure and so can just be copied from the figure. For the liquid, we have
the equilibrium vapor pressure of the pure standard states and hence just use
a
(
l
)
Cd
=
p
Cd
/p
o
Cd(
l
)
. The
solids need to use the free energy of melting to convert the activities.
At 550 K, the equilibrium vapor pressures and the activities with respect to the standard liquid state
are
ln
h
p
o
Cd(
l
)
i
=
−
12
,
012
T
+ 11
.
55 =
−
12
,
012
550
+ 11
.
55 =
−
10
.
29
p
o
Cd(
l
)
=
3
.
397
×
10
−
5
bar
a
(
l
)
Cd
=
p
Cd
p
o
Cd(
l
)
=
2
.
98
×
10
−
5
3
.
397
×
10
−
5
= 0
.
8772
ln
h
p
o
Zn(
l
)
i
=
−
13
,
831
T
+ 11
.
72 =
−
13
,
831
550
+ 11
.
72 =
−
13
.
427
p
o
Zn(
l
)
=
1
.
4744
×
10
−
6
bar
a
(
l
)
Zn
=
p
Zn
p
o
Zn(
l
)
=
3
.
0
×
10
−
7
1
.
4744
×
10
−
6
= 0
.
2034
For the solid state, need just the relationship between activities so both point to the same chemical
potential.
µ
=
G
o
A
(
s
)
+
RT
ln
a
(
s
)
A
=
G
o
A
(
l
)
+
RT
ln
a
(
l
)
A
RT
ln
"
a
(
s
)
A
a
(
l
)
A
#
=
G
o
A
(
l
)
−
G
o
A
(
s
)
= ∆
g
m
A
a
(
s
)
A
=
a
(
l
)
A
exp
∆
g
m
RT
a
(
s
)
Cd
=
a
(
l
)
Cd
exp
∆
g
m
[Cd]
RT
= (0
.
8772) exp
460
(8
.
3144)(550)
= 0
.
9700
a
(
s
)
Zn
=
a
(
l
)
Zn
exp
∆
g
m
[Zn]
RT
= (0
.
2034) exp
1510
(8
.
3144)(550)
= 0
.
2830
a
(
s
)
Cd
a
(
l
)
Cd
a
(
v
)
Cd
a
(
s
)
Zn
a
(
l
)
Zn
a
(
v
)
Zn
Solid
Cd
0
.
97
Zn
0
.
03
(
s
)
0
.
9700
0
.
8772
2
.
98
×
10
−
5
0
.
2830
0
.
2034
3
.
0
×
10
−
7
Liquid
Cd
0
.
88
Zn
0
.
12
(
l
)
0
.
9700
0
.
8772
2
.
98
×
10
−
5
0
.
2830
0
.
2034
3
.
0
×
10
−
7
For some of the numbers, one could also start from taking Roult’s law for the solid alloy and conclude
a
(
s
)
Cd
= 0
.
97, which is consistent with the numbers starting from the partial pressures.
(d)
(2 pts) Based on these results, determine the free energy of mixing of Cd and Zn into a solid solution
at 550 K.
Another plug in from the tables above recognizing that we want the solid solution (since both compo-
nents would be solid at 550 K)
∆
g
mix
=
RT
[
X
A
ln
a
A
+
X
B
ln
a
B
]
=
(8
.
3144)(550) [(0
.
97)(ln 0
.
97) + (0
.
03)(ln 0
.
2830)]
=
−
308
.
3 J/mole
2
2.
(12 pts) The phase diagram for Mo and Pt is shown below. Dashed lines indicate uncertainty in the precise
location of the lines, but assume they are correct for this problem. Identify the nature (name) of the phase
transformations that occur at the indicated temperatures.
(1 pt) 2623
o
C
melting
(1 pt) 2080
o
C
eutectic
(1 pt) 1880
o
C
peritectoid
(1 pt) 1780
o
C
peritectoid
(1 pt) 1280
o
C
eutectoid
(1 pt) 2175
o
C
congruent melt
(1 pt) 2020
o
C
peritectic
(1 pt) 1600
o
C
eutectoid
(1 pt) 1475
o
C
eutectoid
(1 pt) 1300
o
C
peritectoid
(1 pt) 1800
o
C
order/disorder transformation
(congruent) polymorphic
(1 pt) 1769
o
C
melting
The transformation at 1800
o
C is the one twist on this diagram. It isn’t a binary invarient, but might be
referred to in many ways. It looks similar to a congruent melt, but involves only solid phases, so “congruent
polymorphic” transformation might be reasonable.
It is also just from something that is labelled as the
MoPt
2
compound to a fully disordered (Pt) alloy, so is also an order/disorder transformation.
Multiple
answers accepted.
3.
(15 pts) To purify a copper rich Mn-Cu alloy, it is loaded into a low pressure furnace held isothermally at
1800 K. A flowing mixture of 1%
H
2
S(
g
)
(by volume) in
H
2
(
g
)
gas is passed over the sample at a
total
pressure of 0.1 bar
, and the system is allowed to come to complete equilibrium. While
H
2
(
g
)
and
H
2
S(
g
)
will create some small partial pressure of
S
2
(
g
)
, the partial pressure will be small and can be mostly ignored.
The Mn reacts with hydrogen sulfide to form a skin of pure
MnS(
l
)
while the copper remains stable and
no copper sulfides form. After reaching full equilibrium, the sample was analyzed and the composition was
Mn
0
.
005
Cu
0
.
995
. This composition is well within the Henrien and Raoultian limits for the components.
3
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H
2
(
g
) +
1
2
S
2
(
g
)
⇒
H
2
S(
g
)
∆
G
o
(
T
) =
−
90
,
290 + 49
.
39
T
Mn(
l
) +
1
2
S
2
(
g
)
⇒
MnS(
l
)
∆
G
o
(
T
) =
−
262
,
600 + 64
.
4
T
Cu(
l
) +
1
2
S
2
(
g
)
⇒
CuS(
l
)
∆
G
o
(
T
) =
−
53
,
600 + 66
.
5
T
(a)
(12 pts) Determine the activities of Mn, Cu, MnS, H
2
, and H
2
S in the system at equilibrium; be sure
to indicate the appropriate standard state for each activity.
The only reaction of interest is the Mn reacting with H
2
S to form MnS and H
2
.
Mn(
l
) + H
2
S(
g
)
=
⇒
Mn(
l
) + H
2
(
g
) +
1
2
S
2
(
g
)
∆
G
o
(
T
) = 90
,
290
−
49
.
39
T
=
⇒
MnS(
l
) + H
2
(
g
)
∆
G
o
(
T
) =
−
262
,
600 + 64
.
4
T
∆
G
o
(
T
) =
−
172
,
310 + 15
.
01
T
We evaluate ∆
G
o
at 1800 K to get the activity constant
K
A
∆
G
o
(1800 K)
=
−
145
,
292 J/K-mol
K
A
= 16
,
452
Now, just need the activities from the formula. The relative fractions H
2
and H
2
S are given, which
become partial pressures and hence activities with the total pressure of 0.1 bar. The MnS is pure so
its activity is 1.0, leaving only the activity of the Mn is unknown. We have
a
(
g
)
H
2
= 0
.
099,
a
(
g
)
H
2
S
= 0
.
001,
a
(
l
)
MnS
= 1
.
0 giving
K
A
= 16
,
452
=
h
a
(
l
)
MnS
i h
a
(
g
)
H
2
i
h
a
(
l
)
Mn
i h
a
(
g
)
H
2
S
i
=
(1
.
0)(0
.
099)
a
(
l
)
Mn
(0
.
001)
a
(
l
)
Mn
=
(1
.
0)(0
.
099)
K
A
(0
.
001)
=
(1
.
0)(0
.
099)
(16
,
452)(0
.
001)
=
0
.
006018
The only one left is the activity of the copper. We just use Raoult’s law to get
a
(
l
)
Cu
=
X
Cu
= 0
.
995.
Putting it all together, we have
a
(
l
)
Mn
a
(
l
)
MnS
a
(
g
)
H
2
a
(
g
)
H
2
S
a
(
l
)
Cu
0
.
006
1
.
0
0
.
099
0
.
001
0
.
995
(b)
(2 pts) From your results, estimate Henry’s constant for Mn in Cu.
From above,
k
Mn
=
a
(
l
)
Mn
/X
Mn
= 1
.
2.
(c)
(1 pts) Based on your value of Henry’s constant, is mixing of Mn and Cu exothermic or endothermic,
and
very
briefly how did you determine this?
As we have positive deviation from ideality (
k
Mn
>
1), the interactions must be endothermic.
4.
(10 pts) Using the Ellingham diagram on the next page, draw appropriate lines to answer the following
questions.
You must draw the correct lines on the diagram to obtain credit. Identify each with
the corresponding section letter (a,b,c,. . . ).
(a)
(2 pts) Josephine will be annealing a Ti sample in a nearly pure H
2
ambient at 1200
o
C. She is con-
cerned that any water vapor in the gas might cause oxidation of the Ti.
What is the
maximum
concentration
of H
2
O that she can tolerate in the H
2
gas to avoid oxidation of the Ti at 1200
o
C?
The ratio H
2
/H
2
O is 2
×
10
6
which means there must be less than 5
×
10
−
7
H
2
O concentration (5 ppm).
(b)
(2 pts) What is the approximate standard state enthalpy of formation for
ZrO
2
(
s
)
(with units)?
-1100 kJ/mol
(c)
(2 pts) Kwame needs to anneal an NiO (oxide) in a CO/CO
2
ambient and not have the oxide decompose.
If the gas mixture is 1% CO and 99% CO
2
, what is the maximum temperature to which he can heat
his NiO sample without having it decompose to Ni?
900
o
C
4
(d)
(2 pts) Bright annealing of copper occurs when the oxygen partial pressure is reduced sufficiently low
that the oxide is unstable and decomposes, leaving only pure clean copper on the surface. If we heat the
copper to just below its melting point, how low does the
p
O
2
need to be to perform this bright annealing?
5
×
10
−
6
bar
(e)
(2 pts) To control the partial pressure of oxygen during annealing in a gas ambient, the carrier gas is
often first passed over a reactive metal (typically Ti sponge) to remove most of the oxygen (forming
TiO
2
). Gollum plans to anneal pure Zr at 1100
o
C; what is the maximum allowed temperature for the
Ti sponge to ensure no ZrO
2
is formed?
920
o
C
5.
(15 pts) McGyver is feeling pretty good now, having reset his boss’s Pb block. But it required that him to
melt essentially all the ice in the ice/water tank and completely drain the batteries. Still trying to save his
penguins, he looks for a way to refreeze the water tank and recharge the facility batteries so he can watch the
latest new episodes on Netflix. Sitting in a back corner of the lab, he finds an unused tank of liquid nitrogen
(LN2 - which is at its boiling point of 77 K). Unfortunately, there are only 19.6 kg (700 moles) of LN2 left
in the tank.
5
While he could just use LN2 to just freeze the water, McGyver knows that is irreversible and he can do
better. He remembers numbers for N
2
and H
2
O of potential use:
c
(
s
)
P
c
(
l
)
P
c
(
v
)
P
∆
h
m
∆
h
v
H
2
O
42
J/K-mol
75
J/K-mol
33
.
6
J/K-mol
6
,
000
J/mol
40
,
650
J/mol
N
2
-
28
J/K-mol
29
J/K-mol
720
J/mol
5
,
600
J/mol
Problem Statement:
There is a 90 kg tank (5000 moles) of water currently at 273 K and a 12 MJ
(
1
.
2
×
10
7
J) battery that is currently empty. McGyver has 19.6 kg (700 moles) of liquid nitrogen at 77 K
and the ambient is at 310 K. Does McGyver have enough liquid nitrogen to freeze the water in the tank and
fully recharge the battery?
(a)
(4 pts) Sketch the initial and final states (1 pt), explicitly identify a system (1 pt), and indicate/label
all thermodynamic heats and works (2 pt).
One possible system definition is shown below. Another equally reasonable system would have the water
turning to ice also inside the system with no ∆
Q
ice
; the choice outside is based on the recognition that
the energy needed to freeze the ice will be transferred under isothermal conditions (273 K) and hence
is easy to handle as a thermodynamic heat across the boundary. While it is possible to analyze the
system with the liquid nitrogen outside of the chosen system, it is more difficult as the thermodynamic
heat then transferred to the system would be coming in at varying temperatures.
(b)
(2 pts) Write an explicit mathematical statement that you will answer/use in order to determine if
McGyver has a chance to succeed.
There are three potential questions that can be used to determine if McGyver can be successful.
Entropy is critical in each.
i. Calculate ∆
S
irr
for the process as described, freezing the water and fully charging the battery.
Question: If ∆
S
irr
>
0, then McGyver can succeed.
ii. Let ∆
S
irr
= 0 (thermodynamic best).
Calculate the maximum amount of work that could be
transferred into the battery. Question: If ∆
W
′
>
12 MJ, then McGyver can succeed.
6
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iii. Let ∆
S
irr
= 0 (thermodynamic best).
Calculate the maximum amount of thermodynamic heat
that can be extracted from the water tank.
Question:
If ∆
Q
ice
>
30 MJ, then McGyver can
succeed.
The analysis below follows the first path, though values for the others are given as well.
(c)
(5 pts method) Solve the problem based on the question you posed in part (b) and the system you defined
in part (a). Your solution must be consistent with those parts of the problem.
To succeed, we need ∆
Q
ice
= (5000 mol)(6000 J/mol)) = 30
,
000
,
000 J of thermodynamic heat trans-
ferred into the system (so ∆
Q
ice
>
0).
And we can set ∆
W
′
= 12
,
000
,
000 J to fully recharge the
battery (have to transfer energy out of the system so again ∆
W
′
>
0). Now, straightforward analysis
based on transfers across the boundary specified and analysis of the changes within the system.
From the first and second law perspective across the boundary,
∆
H
=
∆
Q
A
+ ∆
Q
ice
−
∆
W
′
= ∆
Q
A
+ 18
,
000
,
000 J
∆
S
=
∆
Q
A
310 K
+
∆
Q
ice
273 K
+ ∆
S
irr
=
109
,
890 J/K-mol +
∆
Q
A
310 K
+ ∆
S
irr
Inside the system, the engine can be ignored (∆
H
= ∆
S
= 0) and we only have changes in the enthalpy
and entropy of boiling the LN2 and heating it to 310 K.
∆
H
=
n
LN2
∆
h
vap
+
Z
310
77
c
N
2
(
l
)
P
dT
=
(700 mol) [(5600 J/mol) + (29 J/K-mol)(310
−
77 K)]
=
8
,
649
,
900 J
∆
S
=
n
LN2
"
∆
h
vap
77 K
+
Z
310
77
c
N
2
(
l
)
P
T
dT
#
=
(700 mol)
5600 J/mol
77 K
+ (29 J/K-mol) ln
310 K
77 K
=
79
,
182 J/K-mol
Now just a matter of equating. From ∆
H
, we determine ∆
Q
A
which then can be substituted into the
∆
S
expressions to determine ∆
S
irr
.
∆
H
=
8
,
649
,
900 J = ∆
Q
A
+ 18
,
000
,
000 J
∆
Q
A
=
−
9
,
350
,
100 J
(not sure I would have guessed the sign corrrectly)
∆
S
=
79
,
182 J/K-mol = 109
,
890 J/K-mol +
∆
Q
A
310 K
+ ∆
S
irr
∆
S
irr
=
(79
,
182 J/K-mol)
−
(109
,
890 J/K-mol)
−
−
9
,
350
,
100 J
310 K
=
−
546
.
4 J/K-mol
(d)
(4 pts - final numerical answer correct and answers question) Use your results to quantitatively answer
the question in part (b) and determine if McGyver can succeed.
Question:
If ∆
S
irr
>
0, then McGyver can succeed.
∆
S
irr
is not positive and hence, in this case,
McGyver needs to find another alternative or accept either the battery not being full charged or the
water not completely converted to ice.
The other solutions, based on the best possible thermodynamically, would give the maximum ∆
W
′
=
11
.
83 MJ which is just slightly less than the required 12 MJ. Alternatively, to fully charge the battery,
only able to extract 28.75 MJ from the water tank (need 30 MJ).
6.
(10 pts) Karlos has an interesting system; it is based on a calorimeter with a piston allowing the volume to be
readily modified. For this system, reasonable control parameters are the volume
V
and the entropy
S
, and one
7
would like to track the pressure as a function of these two parameters (entropy can be controlled by controlling
the thermodynamic heat in and out of the calorimeter under reversible conditions since
dS
=
δQ/T
). But
this means finding
dP
(
V, S
)
which, as he has taken thermodynamics, poses no significant challenge.
dP
=
dP
(
V, S
) =
f
(
. . .
)
dV
+
g
(
. . .
)
dS
where
f
(
. . .
)
and
g
(
. . .
)
are functions of
T, S, P, V, α, κ
T
,
C
V
, and/or
C
p
.
Determine explicitly
either
the function
f
(
. . .
)
and
g
(
. . .
)
(focus on only one . . . both are non-trivial).
Write the final answer at the bottom of the page. Note that your answer must be supported by appropri-
ate thermodynamic reduction work below. You need only consider
P dV
work and may assume reversible
processes.
Fundamentally, we have
dP
=
dP
(
V, S
) =
∂P
∂V
S
dV
+
∂P
∂S
V
dS
And we just need to thermodynamically reduce these derivatives to physical measurables.
There are many many ways, some simple, some more challenging. Below are fairly short solutions but may
not always be obvious to follow this trajectory.
∂P
∂V
S
=
∂P
∂V
T
+
∂P
∂T
V
∂T
∂V
S
=
1
∂V
∂P
T
+
−
∂V
∂T
P
∂V
∂P
T
−
∂S
∂V
T
∂S
∂T
V
=
−
1
κ
T
V
−
α
κ
T
∂P
∂T
V
C
V
T
=
−
1
κ
T
V
−
α
2
T
C
V
κ
2
T
=
−
1
C
V
κ
T
V
C
V
+
α
2
V T
κ
T
=
−
C
P
C
V
1
κ
T
V
∂P
∂S
V
=
∂T
∂S
V
∂P
∂T
V
=
1
∂S
∂T
V
−
∂V
∂T
P
∂V
∂P
T
=
T
C
V
α
κ
T
=
αT
κ
T
C
V
7.
(16 pts) Basic concepts. For explanations, include both an English (verbal) definition followed by the explicit
mathematical definitions. A verbal definition that simply states the math is invalid (e.g. entropy is negative
the derivative of
G
with
T
at constant
P
). Provide just enough detail to convince me you understand.
(a)
(2 pts) Define the concept of a partial molar values, both in words and its mathematical form.
The incremental change in any extensive property per mole of the pure element added to the system
under isothermal and isobaric conditions, and in the limit that the addition does not significantly
change the composition. The partial molar enthapy of element A in an alloy including B and C atoms
would be mathematically defined as
H
A
=
∂H
∂n
A
T,P,n
B
,n
C
8
(b)
(2 pts) For a regular solution
∆
H
mix
A
= Ω
X
2
B
. What is the molar enthalpy of mixing (
∆
h
mix
) for an
A
0
.
3
B
0
.
7
alloy where
Ω = 10
kJ/mol?
For any partial molar expression, the molar is just a weighted sum of the partial molars:
∆
h
mix
=
X
A
∆
H
mix
A
+
X
B
∆
H
mix
B
=
Ω (0
.
3)(0
.
7)
2
+ (0
.
7)(0
.
3)
2
= 2 100 J/mol
(c)
(2 pts) Explain the concept of activity, and give its mathematical definition.
The activity is a generalized correction to move a material’s chemical potential from that of the chosen
standard state to the actual conditions in the system. Defined as
µ
A
=
G
A
=
G
o
A[
γ
]
(
T
) +
RT
ln
a
(
γ
)
A
(d)
(1 pts) Why is an activity meaningless unless it is linked to a specific standard state?
Without a starting point, the idea of onlyi providing the “change” to get to a known condition is
meaningless.
(e)
(2 pts) A condensed phase’s activity is approximately
a
(
γ
)
A
=
p
A
p
o
A
.
Experimentally, assume you can
quantitatively measure gas partial pressures in a system. Given a solid Cu
x
Ni
y
Li
z
sample at 1000 K,
sketch two specific systems that clearly indicate the meaning of
p
A
and
p
o
A
for determining
a
(
l
)
Li
.
(f)
(2 pts) Define Raoult and Henry’s Laws:
Raoult’s law states that as an alloy becomes nearly pure in a component, the activity of that component
will approach its molar fraction. Henry states the activity of a component in the limit it is at very low
concentrations will be proportional to the concentration with the proportionality constant known as
Henry’s constant. Mathematically,
Raoult’s Law:
lim
X
A
→
1
a
A
=
X
A
Henry’s Law:
lim
X
A
→
0
a
A
=
k
A
X
A
=
γ
o
A
X
A
(g)
(2 pts) Ge and Si are slightly endothermic in solid solutions (due to the small lattice constant differ-
ence). Sketch the activities (
a
(
s
)
Si
and
a
(
s
)
Ge
) of a binary Si
x
Ge
1
−
x
alloy as a function of
X
Ge
, indicating
the Henrien and Raoultian regimes. On a second graph, sketch the equivalent activity coefficient (
γ
(
s
)
Si
and
γ
(
s
)
Ge
) and mark Henry constants
k
Si
and
k
Ge
.
9
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(h)
(2 pts) Sketch
PT
and
PV
phase diagrams for a simple material, such as CO
2
or H
2
O, with only the
three standard phases (solid, . . . ). Indicate on the diagrams the triple point, the critical point, and the
phase fields for solid, liquid, vapor and gas.
(i)
(1 point) What is fugacity?
Fugacity is a measure of the chemical potential.
10
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