4/3/2014
Lecture 6
Application of Thermodynamics
in Phase Diagrams
A. K. M. B. Rashid
Professor, Department of MME
BUET, Dhaka
Today’s Topics





The phase diagrams and its applications
The structure of phase diagrams
Construction of unary phase diagram
The free energy – composition (G - X) diagrams
Problem solving
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What are Phase Diagrams?
 A phase is a homogeneous system of which the intensive properties are
uniform or, at most, vary continuously throughout the system.
 The study of phase transformations, like fusion, vaporisation, sublimation,
and allotropic transformations is extremely important in materials science.
 Phase diagrams are the primary thinking tool in materials science which
provide the basis for predicting or interpreting the changes in internal
structure of a material that accompany its processing or subsequent
service.
(a) Phase diagram for
pure metal copper
(b) Phase diagram for unary
ceramic compound SiO2
Fig. 7.1
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(c) Binary phase diagram of
Pb-Sn metallic system
(d) Binary phase diagram of
SiO2-Al2O3 ceramic system
Fig. 7.1
(e) Ternary phase diagram of CaO-SiO2-Al2O3 ceramic system
Fig. 7.1
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 Phase diagram is a map that shows
the domains of stability of phases
and their combinations.
 A point in this diagram represents a
state of the system and lies within
a specific domain on the map.
 Reading the phase diagram will then tell you, at that state,
when it comes to equilibrium,
1.
2.
3.
what phases are present,
the state of those phases, and
the relative quantities of each phase.
Equilibrium in Multi-component System
 The limits of stability of phases are indicated
by the lines, called phase boundaries, on
these diagrams and they are defined by the
conditions under which pairs of phases may
coexist at equilibrium.
P
 Analogy similar to the unary heterogeneous
systems can be applied to determine the
conditions for equilibrium for multi-component,
multi-phase systems.
T
 For a system containing C component and P phases, the change in
entropy for any process would be
P
dS’sys
C
1 dU’a + Pa dV’a – 1
=
mka dnka
a
a
a
T
T
T
a=1
k=1


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 Conditions for equilibrium for multi-component, multi-phase systems:
Ta = Tb = Tg =...... = TP
Pa = Pb = Pg =...... = PP
m1a = m1b = m1g = …. = m1P
m2a = m2b = m2g = …. = m2P
. . . . . . . . . . . . .
. . . . . . . . . . . . .
mCa = mC b = mC g = …. = mC P
The Gibbs Phase Rule
 A crucial aspect in phase diagrams is the number of variables that need to
be specified in order to determine a thermodynamic state, indicated by its
degrees of freedom.
 It is the smallest number of intensive variables, such as temperature,
pressure, concentrations, and so on, that the experimenter can vary
independently without changing the number of phase at equilibrium.
 This can be calculated from the difference in the number of total number
of variables (both dependent and independent) required to identify the
system and the number of available relations corresponding to the
conditions for equilibrium in the system.
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 The state of any phase is completely determined by its pressure,
temperature and composition.
 If a phase contains C number of components, then (C -1) variables must
be specified for the compositions, and 2 variables for P and T.
 This makes a total of (C +1) variables for each phase.
 So, for a system containing P phases, the total number of variables
m = P (C +1)
 However, if the system is in equilibrium, all of these variables are not
independent: they are related by the conditions for equilibrium.
 Equations on the equality of the
temperatures and pressures yield
2(P -1) relations
Ta = Tb = Tg =...... = TP
Pa = Pb = Pg =...... = PP
m1a = m1b = m1g = …. = m1P
 While the equality of the chemical
potentials of each species for the
P phases yield C (P -1) additional
relations.
m2a = m2b = m2g = …. = m2P
. . . . . . . . . . . .
mCa = mC b = mC g = …. = mC P
 So the total number of relations is
n = 2 (P -1) + C (P -1)
n = (C + 2) (P – 1)
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 Accordingly, for a system with C components and P phases,
the number of degrees of freedom (i.e. the number of independent
variables) is:
F = number of total variables (m) – number of relations (n)
F = P (C +1) – (C +2)(P -1)
F = C–P +2
 This equation is known as the Gibbs phase rule.
Example : One-component system
 In a single-phase region, two variables need to
be specified
(e.g., T and P).
 In a two-phase region, one variable needs to
be specified
F = C–P +2
P
L
S
Other variables change automatically in order to maintain
the existing two-phase equilibria.
triple
point
G
If T and P both are changed, the two-phase equilibria
would be disturbed by disappearing one of the two phases.
T
 In a three-phase region, no variables need to be specified
There is only one point (the triple point) which occurs at a determined temperature
and pressure (zero degree of freedom).
Self-Assessment Question 7.1
What is the degree of freedom of the liquid phase of Fe-C-Si alloys?
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The Structure of Phase Diagrams
 Representation of structure of phase diagram using three types of
co-ordinate systems:
1.
2.
3.
All the axes are thermodynamic potentials (T, P or m)
One axis is a potential and the others are not
All axes are not potentials.
(a) Both axes are potential
(b) One axis is potential
(c) Neither axis is potential
Additional Benefits of
Using Non-Potential Axes
 For diagrams constructed using all-potentialaxes are simple and easy to understand.
However, they do not give information about the
relative amounts of the phases presents in two- or
three-phase equilibria.
 For diagrams constructed using onepotential-axis, two-phase equilibria are
presented as areas.
 Similarly, three-phase equilibria are presented
as area when diagrams are constructed using
all non-potential-axis.
This makes easy to calculate the relative amount of
co-existing phases by constructing tie lines and using
lever rules.
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Number of Axes Required
 Using Gibbs Phase Rule, the single phase regions of a component
have the highest number of degrees of freedom:
F = C-P +2 = C+1
That is, single-phase regions require the largest number of variables, i.e.,
(C+1) for their specification (e.g., 2 for unary, 3 for binary, 4 for ternary etc.).
 The graphical space in which the phase diagram is constructed
must have (C+1) independent co-ordinates.
Unary systems is two-dimensional [in (T, P) space]
Binary systems is three-dimensional [ in (T, P, a2) or (T, P, X2) space]
Ternary systems is four-dimensional [in (T, P, X2, X3) space]
 Because of its limitation in
presentation, most multicomponent phase diagrams are
represented as sections.
This is obtained by fixing a value for
one (for binary systems) or two (for
ternary systems) of the independent
variables.
Fig. 7.3:
A binary phase diagram plotted on
(a) thermodynamic potentials (P, T, a2) space.
(b) a more complex (P, T, X2) space.
(c) a section taken at constant P.
(d) the familiar (T, X2) space.
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Construction of Unary Phase Diagram
 Generally plotted using (T, P) co-ordinate
systems.
P
L
S
 The lines, indicating two-phase equilibria,
appeared on the diagram have the
general form of equation P = P (T).
G
T
 During construction of the diagram:

Chemical potential surface of each phase is produced first.

The curve of intersection of any two chemical potential surface is then traced
down to (P, T) space to obtain the relationship P = P (T).
The chemical potential
expressed earlier in terms
of internal energy
The Chemical Potential
m = U’
n S’, V’
 Using Gibbs free energy
dG = - SdT + VdP + mdn
m =
the chemical potential can
more suitably be defined as
G’
n
T, P
 For unary system, G = nG:
m =
G’
n
=
T, P
(nG)
n
= G
T, P
 Then the dependence of the chemical potential upon temperature and
pressure in a unary system can be written as
dm = dG = - S dT + V dP
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 If a phase is taken through an arbitrary change in state,
dma = - Sa dTa + Va dPa
m
ma = ma (P, T)
 This can be integrated over the limits
of T and P to yield the chemical
potential surface of a phase with the
function ma = ma (T, P)
 A chemical potential surface for
b phase represented by the
function mb = mb (T, P) can also
be constructed in similar way.
T
P
 When the curves of both phases are
superimposed into a single plot, the two
surfaces intersect along a space curve AB.
m
B
At any point on that space curve, the
temperatures, pressures, and chemical
potentials of the two phases are identical.
T a = Tb
Pa = Pb
ma = mb
A
As we have seen earlier that, these three
conditions must be met precisely in order for
a and b phases to coexist in equilibrium.
 Thus the curve of intersection of the two chemical
potential surfaces, AB, is the locus of points for which
the a and b phases are in equilibrium.


B’
A’
 When the a = b equilibrium curve AB traced down to the P-T space,
the line for P = P (T) relation, indicated by the curve A’B’, is obtained.
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The Clapeyron Equation
 We have stated earlier that, for unary systems, the two phase equilibria
curves in (P, T) space may each be described mathematically by the
function P = P (T).
 The Clapeyron equation is a differential form of this equation.
 For any pair of coexisting phases in the unary system, integration of the
Clapeyron equation yields a mathematical expression for the corresponding
phase boundary on the phase diagram.
Repeated application to all the pairs of phases that may exist in the system yields all
possible two-phase domains. Intersections of the two-phase curves produce triple point
where three phases coexist.
 Thus, Clapeyron equation is the only relation required for calculating a
unary phase diagram.
Consider a  b Equilibrium
 Change in chemical potential of a phase when taken through any arbitrary
change in its state
dma = Va dPa – Sa dTa
 Similarly, for any arbitrary change state of b phase, the change in
chemical potential
dmb = Vb dPb – Sb dTb
 For equilibrium, dma = dmb, dPa = dPb, and dTa = dTb, so that
Va dP – Sa dT = Vb dP – Sb dT
dP/dT = (Sb – Sa)/(Vb – Va)
dP/dT = S/V
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dP/dT = S/V
 In experiments, change in S is not measure directly.
Calorimetric measurements at constant pressure (i.e., QP) results values for
heat of transformations (HF, HV, etc.).
Since, for a b equilibrium, G = 0 = H – TS, we have S = H/T.
 Thus, expression for Clapeyron equation becomes
dP
H
=
dT
T V
Check units:
LHS:
dP/dT = atm/deg
RHS:
H = cal/mol,
V = cc/mol
H/TV = cal/cc-deg
1 atm = 41.293 cal/cc
Applications of Clapeyron Equation
dP
H
=
dT
T V
1.
Predicting the effects of P on Tt (e.g., Tm, Tb)
describing the relationship between P and T at which a and b phases
can exist in equilibrium
2.
Calculating precisely Ht (e.g., HF, HG)
knowing the derivative (dP/dT), and V of the transformation
3.
Obtaining an equation for a given equilibrium, P = P(T)
knowing any reference point to integrate dP/dT
(e.g., normal melting point, normal boiling point)
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Integration of Clapeyron Equation
dP
= H
dT
T V
 Using appropriate reference points, the Clapeyron equation can be
integrated for a two-phase equilibrium to yield the phase boundary
P = P(T) for the equilibrium.
 While integrating, the dependency of H and V on P and T must,
however, be considered first.
 To obtain H = H (P, T) for a  b equilibrium:
d (H) = dHb - dHa
 Now, for the function H = H (P, T):
dH = CPdT + V(1 – Ta) dP
 For all practical purposes, dependency of H on P can be ignored.
Thus, dHP = CPdT, and
d (H)  CP dT
(7.21)
where CP = CPb – CPa = a + bT + cT-2
 Eq (7.21) is often known as the Kirchhoff’s equation.
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 To obtain V = V (P, T) for a  b equilibrium:
d (V) = dVb - dVa
 If b is an ideal gas phase and a is either a solid or
liquid phase, then Vb >>Va , and
V = VG – Va  VG = RT/P
At STP, molar volume of:
Gas = 22400 cc
Solid/Liquid ≈ 10 cc
(for ideal gas)
 If both b and a phases are condensed phases (solid or liquid),
For an approximate calculation (where P < a few tens of atm),
V can be treated as constant.
For a precise calculation, dependency of V on P and T must be considered
according to the familiar formula
V = V (P,T):
dV = Va dT - Vb dP
The Clausius - Clapeyron Equation
 Consider the equilibrium between the condensed phase and its vapour
phase, i.e., (a  G) equilibrium
 If CPG = CPa, then H is independent of T. Then
dP / dT = H / TV
dP / dT  HG / TVG = PHG / RT2
dP / P = (HG / RT2) dT
d ln P =
HG
dT
T2
(7.25)
 Eq. (7.25) is known as the Calusius - Clapeyron equation.
This is an approximate equation, derived by assuming that
1. HG is constant
2. VG >>Va so that V  VG.
3. Vapour phase behaves ideally.
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Integration of Clausius - Clapeyron Equation
d ln P = (HG / RT2) dT
 Integrating between the limits (P1, T1) and (P2, T2):
ln
P2
P1
= –
HG
R
 Integrating indefinitely:
ln
1
1
–
T2
T1
ln P
ln P = – (HG / RT) + C
P = A exp (– HG / RT)
C = ln A
Slope = - H/R
where A = ln C
1/T
The Condensed Phase Equilibria
 The Clausius – Clapeyron equation is applied only when there is a gas
phase involved in the equilibra.
Example: S  G equilirbium, L  G equilirbium, etc.
 For all other equilibria involving condensed phases only (e.g., a  b
equilirbium, S  L equilirbium, etc), the Clausius-Clapeyron equation
cannot be applied.
 In such cases, the original Clapeyron equation should be used.
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dP
dT
=
S
V
 For a precise calculation, T and P dependency of S and V must be considered
while integrating the Clapeyron equation.
 An approximate calculation of the phase boundaries between two
condensed phases can be made by ignoring T and P dependency of H and V.
 In such cases, considering S and V as constant, integration of the
Clapeyron equation becomes straightforward:
P2 - P1 =
S
V
( T2 - T1 )
P2 - P1 =
H
V
ln (T2 / T1)
H = latent heat
of fusion, etc.
 The gradient of P = P (T) line (i.e. S/V) for S  L equilibrium for most
substances is positive, which means that V for the transformation is
positive (since S is always positive), i.e., the volume expands.
The only exception is water, for which the slope is negative, since ice
contracts when it becomes water.
P
S
P
L
WATER
ICE
G
STEAM
T
T
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Trouton’s rule
The entropy of vapourisation of most elements is constant.
SG =
HG
Tb
 21 cal/deg-mol
Richards’ rule
The entropy of fusion of most elements is constant.
SF =
HF
TF
 9 J/mol-K
The Triple Point
 At the triple point, S  L, L  G, and S  G equilibrium lines meet.
At the triple point, Ga = Gb = Gg
P – T diagram for pure iron
 For most elements and compounds, the triple point pressure is well below
the atmospheric pressure.
The only exception is CO2, for which Ptp = 0.006 atm.
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 The triple point of S, L and G phases can be determined if any two of the
three equilibrium (S  L, L  G, S  G) lines are known.
 Using the equations for L  G and S  G equilibrium lines, the triple
point temperature and pressure is calculated as follows:
PG = AG exp (– HG / RT)
PS = AS exp (– HS / RT)
 At the triple point (Ptp, Ttp), these two lines intersect. Thus
Ptp = AG exp (– HG / RTtp) = AS exp (– HS / RTtp)
Ttp =
HS – HG
R ln (AS/AG)
;
Ptp = AS exp
HG
HG – HS
Free Energy – Composition Diagrams
 The most useful tool to obtain connections between the phase
diagrams and their underlying thermodynamics principles.
 The fundamental principles of G-X diagrams are:
For a system of definite composition at constant P and T,
 The stable phase has the lowest free energy, G.
 The free energy is the same for coexisting phases.
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G-X Diagrams for Ideal Solutions
 The molar free energy of ideal binary solutions:
GS = G1 + GM
where,
GM,
(1)
the molar free energy of mixing of component 1 and 2, is:
GM = HM – TSM = RT (X1 ln X1 + X2 ln X2)
(2)
and G1, the molar free energy of unmixed solution is:
G1 = X1 G01 + X2 G02 = G01 + (G02 – G01)X2
(3)
 Combining eq.(1) and eq.(3), we get,
GS = G01 + (G02 – G01) X2 + GM
(4)
HM = 0
GM = RT (X1 ln X1 + X2 ln X2)
Low T
 The GM – X plot shows that:
GM
High T
 The curve is symmetrical at X2=0.5 and
has a vertical slope at X1=1 and X2=1.
-TSM = GM
 The curve has a minimum value of
–RT ln2 at X2=0.5. This magnitude
increases linearly with T.
1
X2
G01 + (G02 – G01)X2
GS
2
G02
GS = G01 + (G02 – G01) X2 + GM
 Since, throughout the curve, GM is –ve
and GS is less than G1, components 1
and 2 prefer to form a solution.
G01
1
GM
X2
2
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 Thus, for a given phase of the solution, G-X diagram is a plot of the
molar Gibbs free energy of mixing, GM, versus the mole fraction of
component B, XB, at a fixed P and T.
 Each existing phase in a system has its own G-X curve.
 The competition for domains of stability of the phases and interactions
that produce two and three phase fields that separate them can be
visualised by comparing the G-X curves for all the phases in the system.
 For such a comparison of free energies of mixing to be valid,
it is absolutely essential that the energies of each component in all
the phases be referred to the same reference state.
G-X Diagrams for Non-ideal Solutions
 The form of G-X curve for non-ideal or real solution is given by the equation
GM = RT (X1 ln a1 + X2 ln a2)
GM = Gex + RT (X1 lnX1 + X2 lnX2)
where the excess free energy of
mixing, Gex, can be positive or
negative depending on the type
of deviation from ideality.
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 To produce G-X curves for non-ideal binary solutions of components
A and B, the following steps can be followed:
 Let the stable forms of pure A and B at the
given T and P be a (fcc) and b (bcc),
respectively.
GM
 The molar free energies of fcc A and bcc B
are shown as point a and b in the figure.
c
d
 To draw the free energy curve of the fcc a
phase, convert the stable bcc arrangement of
B atoms into an unstable fcc arrangement.
This requires an increase in free energy, bc.
Ga
GM
a
b
e
 Construct the free energy curve for the a
phase now be by mixing fcc A and fcc B as
shown in the figure. The distance de will give
GM for such solution of composition XB.
A
XB
B
 A similar procedure produces the molar free energy curve for the
b phase.
 In the following figure, G-X diagrams for both a and b phases are
constructed in a single plot for comparison.
c
d
GM
a
b
Gb
Ga
A
XB
B
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Comparing Free Energy of Solutions
with that of its Unmixed Components
G-X curves “convexing downwards”
 Consider mixing of two separate solutions
of A and B, marked by the points a and b.
 The free energy of unmixed solutions is
given by the point c.
GM
 The free energy of mixed homogeneous
solution will be represented by the point d,
which is lower than that of point c.
d
 Thus, the resultant single-phase solution is
stable relative to any two unmixed portions.
 This is true for any single phase region in
which the G-X curve is “convex downwards.”
b
c
a
A
XB
B
G-X curves “convexing upwards”
 The two separated solutions represented
by point p and q have a free energy
corresponding to point y, which is lower
than that for the single homogeneous
solution represented by point x.
 The configuration with the lowest free
energy for this composition is obviously the
point z on the line mn, where two separate
solutions of compositions m and n are in
equilibrium with each other.
x
GM
q
y
p
n
z
m
A
XB
B
 Thus, the stable system with compositions
 from XB=0 to XB=m are composed of a single solution
 from XB=m to XB=n consists of a mixture of two solutions of compositions
m and n
 from XB=n to XB=1 the stable system is again composed of a single solution.
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 Single phase solutions from XB=m to XB=n
are metastable with respect to the unmixed two-phase system.
x
GM
 This situation is typical for systems
exhibiting a miscibility gap in the phase
diagram and is associated with sufficiently
great positive deviation from the ideal
behaviour.
q
y
p
n
z
m
B
XB
A
System Consisting Two or More Separate Phases
 Similar kind of miscibility may occur.

The A-rich alloys will have the lowest free energy as a homogeneous
a phase and B-rich alloys as b phase.

For alloys with compositions near the cross-over in the G curves, the
total free energy can be minimised by the atoms separating into two
phases.
G0b
Ga
G0a
G1a
G1b
G0a
Gb
G1
a1
A
G0b
Ge
mAa = mAb
be
ae
B
mBa = mBb
Ge
b1
X0
Geb
a
A
X0
B
Construction of a tangent line common to two (G-X) curves for a pair of phases
identifies the compositions of those two phases that coexist in equilibrium at the
indicated temperature and pressure.
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Examples of Some Common G-X Diagrams
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Problem Solving
7.13 The vapour pressure of liquid zinc as a function of temperature is
given as:
log P (mm Hg) = – 6620/T – 1.255 logT + 12.34.
Calculate the heat of vaporisation of zinc at its boiling point 907 C. If
heat of sublimation of zinc at the boiling temperature is 30 kcal/mol,
what will be the heat of fusion of zinc at its boiling temperature?
7.15 Mercury boils at 375 C with HG=14130 cal/mol. The heat capacity of
liquid mercury is 6.61 cal/mol-K; that of mercury vapour, 4.97 cal/mol-K.
What is the vapour pressure over liquid mercury at 25 C? At 100 C?
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7.18
Below the triple point (-56.2 C) the vapour pressure of solid CO2 is
given as
ln P (atm) = –3116 / T + 16.01.
The molar heat of melting of is 8330 J.
Calculate the vapour pressure exerted by liquid CO2 at 25 C, and
illustrate why solid CO2, sitting on the laboratory bench, evaporates
rather than melts.
7.19
The triple point of iodine I2 occurs at 112.9 C and 11.57 kPa. The heat
of fusion at the triple point is 15.27 kJ/mol, and the following vapour
pressure data are available for solid iodine:
Vapour pressure, kPa
Temperature, C
2.67
84.7
5.33
97.5
8.00
105.4
Estimate the normal boiling point of molecular iodine.
7.22
Estimate the change in the equilibrium melting point of copper
caused by a change of pressure of 10 kbar.
The molar volume of copper is 8.0x10–6 m3 for the liquid, and
7.6x10–6 m3 for the solid phase. The latent heat of fusion of copper
is 13.05 kJ/mol. The melting point is 1085 C.
7.25
Carbon has two allotropes, graphite and diamond. At 25 C and 1 atm,
graphite is the stable form.
Calculate the pressure that must be applied to graphite at 25 C in
order to bring about its transformation to diamond.
Given data:
S298 (graphite) = 5.73, S298 (diamond) = 2.43 J/mol/K,
H298 (graphite) – H298 (diamond) = -1900 J/mol,
@1 atm and 25 C, D(graphite) = 2.22 and D(diamond) = 3.515 g/cc
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4/3/2014
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Lecture 7
Thermodynamics of Reactive Systems
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