Untangling Complex Systems, page 20
systems?
• What is the entropy production?
• How many regimes do we distinguish in non-equilibrium thermodynamics?
• Make examples of irreversible processes in the linear regime.
• What are the rules governing the relationships among forces and flows in linear regime?
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Untangling Complex Systems
• How does an out-of-equilibrium system in the linear regime evolve?
• Explain the Theorem of Minimum Entropy Production.
• What is the relationship between the chemical flow and the chemical force for an elemen-
tary reaction?
• What does the Glansdorff-Prigogine stability criterion tell?
• Present the linear stability analysis for a chemical system wherein a reaction with two
variables occurs.
3.10 KEY WORDS
Entropy production; Flow and Force; Symmetry principle and Onsager reciprocal relations; Linear
stability analysis; Attractors and repellers.
3.11 HINTS FOR FURTHER READING
A landmark paper about the non-equilibrium thermodynamics is the Nobel lecture given by
Prigogine in December 8, 1977 and published in Science in 1978.
In this book, I use MATLAB for the numerical solution of differential equations and for plotting
the results. Alternatively, there exists a software package called SageMath, which is free and can
be downloaded from the website: http://www.sagemath.org/. SageMath is similar to the scientific computing environment called Python.
Finally, in October 2017, there was a special issue of Chaos: An Interdisciplinary Journal
of Nonlinear Science celebrating the 100th birthday of Ilya Prigogine: It is volume 27, issue 10.
3.12 EXERCISES
3.1. Imagine having a solution of tryptophan (whose molecular formula is C H N O )
11
12
2
2
dissolved in water and contained inside a cuvette, at a temperature of 300 K. The height of
the solution is 3 cm. Determine the ratio of the concentration of tryptophan between the
top and the bottom of the solution, knowing that tryptophan molecules feel the terrestrial
gravitational field ( g = 9.8 m/sec2). How much is the same ratio when we consider a solu-
tion of Human Serum Albumin (its molecular mass weight is 67,000 gmol−1) in water?
3.2. The cellular membrane is made of a phospholipid bilayer and acts as a barrier that pre-
vents the intracellular fluid from mixing with the extracellular fluid. These two solutions
have different concentrations of their ions and, hence, different electrical potentials. In
particular, the inside solution is at a more negative potential than the external solution. If
the concentration of K+ is 5 mmol/L outside the cell, and 140 mmol/L inside the cell, how
much is the potassium transmembrane potential?
3.3. Imagine that a system hosts a spontaneous chemical reaction. The dependence of Gibbs
free energy of the system ( G) and the entropy of the universe ( S ) on the extent of the
univ
reaction ξ are shown in Figure 3.19. Predict the shape of the first and second derivatives of G and S
with respect to
univ
ξ.
3.4. Determine the physical dimensions of the thermal conductivity k by making a dimensional
analysis of Fourier’s Law (equation [3.32])
3.5. Determine the physical dimensions of viscosity η by using equations [3.38] and [3.39].
3.6. Considering Fick’s Law (equation [3.40]), which are the physical dimensions of the diffu-
sion coefficient D ?
k
3.7. Estimate the ratio between the mobility and the diffusion coefficient for sodium cation at
298 K in the following two situations: (1) in the presence of a gravitational field; (2) in the
presence of an electric field.
Out-of-Equilibrium Thermodynamics
83
G
Suniv
equilibrium
equilibrium
ξ
ξ
FIGURE 3.19 Trend of G and S as function of the extent of reaction.
univ
Tx?
Th
Tc
L 1
L 2
FIGURE 3.20 Heat conduction between two reservoirs and across two blocks.
3.8. Let us imagine having two blocks of two different materials, 1 and 2, with two different
thermal conductivities ( k and , respectively), two different lengths ( and ), and the
1
k 2
L 1
L 2
same section ( Ar). Block 1 is in physical contact with a hot reservoir (being at T ) on one
h
side and with block 2 on the other side. Block 2 is in physical contact with block 1 on one
side and with a cold reservoir (being at T ) on the other side (see Figure 3.20). Determine c
the temperature T at the surface in conjunction between 1 and 2, when the entire system
x
reaches the stationary state.
3.9. Suppose that the function describing the trend of a body temperature along the spatial coor-
x
dinate x is initially (at time t = 0) T =
+
−
273 10e 5 (see Figure 3.21). The body has a length
of 10 in the x unit. Imagine fixing the values of T at the extremes of the body at the following
values: T(0) = 283.00 K and T(10) = 274.35 K. How does T evolve in time along x? Which is the function describing T( x) at the stationary state, for waiting for an infinite amount of time?
282
x
− 5
280
T = 273 + 10e
T 278
276
274
x
FIGURE 3.21 A trend of T as a function of the spatial coordinate x.
84
Untangling Complex Systems
3.10. Imagine having sodium ions at the bottom of a container full of water and having area Ar
and height h (along the z-axis). To describe the diffusion of the sodium ions from the bot-
tom to the top of the container, it is necessary to solve the diffusion equation [3.114].
1. Verify that the following Gaussian function
2
n
z
−
C ( z t ) =
0
e 4 Dt
,
(
Ar)(π Dt /
)1 2
is the solution of equation [3.114] for a system having n moles of Na+ and being
0
D the
diffusion coefficient of sodium ions.
2. Determine the expression that defines the average distance covered by sodium ions as
a function of time.
3. Calculate the one-dimensional average distance covered by sodium ions after 1 min-
ute, 1 hour, and 1 day, knowing that D = 1.33·10−5 cm2/sec in water at 298 K.
3.11. Determine the equation that defines the one-dimensional mean square distance traveled by
a molecule through diffusion. Then, calculate the one-dimensional mean square distance
covered by sodium ions, having D = 1.33·10−5 cm2/sec in water at 298 K, and a protein,
having D = 100 μm2/sec in water at 298 K, after 1 minute, 1 hour and 1 day.
3.12. Demonstrate the validity of Minimum Entropy Production Theorem in the case of heat
conduction for a system such as that described in Section 3.4.1.
3.13. Investigate the stability of the stationary states for the following reaction, by using the
graphical and the linear stability methods.
k
1 →
X + Y ←
2 X
k
−1
Assume that the reaction is performed in a CSTR where the only Y is injected and sucked
out. The concentration of Y introduced into the reactor [ Y] is so large that it can be con-
0
ceived constant. Let τ
be its residence time into the CSTR.
0 = 1/ k 0
3.14. Apply the linear stability analysis to the chemical reaction described by equation [3.168].
3.15. The linear stability analysis to predict the course of a love story. 23 True love is the noblest and most exciting feeling a human can ever experience in his/her life. However, it may
become a source of pain as well because it is not mutual and because it can fade. We may
experience love at first sight, or love may take time to grow. Sometimes the first impression
is crucial, sometimes not. Love is the subject of many stories in the literature: the most
well-known love story is that of Romeo and Juliette. Other notorious stories are those of
Orpheus and Eurydice, Paris and Helen, Tristan and Isolde, Eloise and Abelard, or Dante
and Beatrice. Everyone has lived or is living a love story, either concrete or virtual. We
all know that a lover asks himself/herself if the counterpart likes him/her; if his/her feel-
ing and actions fit well with those of his/her partner or not; if the relationship with his/
her partner will be everlasting or destined to fail. We may try to answer these queries
through a model that looks like a chemical mechanism. In fact, such a model describes a
love relation between a male and a female by an ensemble of sentimental transformations
conceived as they were chemical reactions. Let us get started by indicating the two main
characters of the relation: a male ( P) and a female ( M). P and M stand for Pierre and Marie Curie: this couple is a prototype of a successful love story in the scientific environment.
Marie was a hard-working student at the Sorbonne University of Paris in the early 1890s.
23 This is a form of agent-based modeling that we will know in Chapter 13.
Out-of-Equilibrium Thermodynamics
85
She was spending every spare hour reading in the library or the laboratory. The industri-
ous student caught the attention of Pierre Curie, director of one of the laboratories where
Marie was working. Let us indicate the feeling of Pierre for Marie with F and with I
P
M
Pierre’s thought about Marie. F is a variable that can assume positive or negative values.
P
It will be positive when Pierre likes Marie, whereas it will be negative when Pierre dislikes
Marie. Physiologically, such feelings correspond to the release of excitatory (positive) or
inhibitory (negative) neurotransmitters. F can also be null if Pierre is indifferent towards
P
Marie and no neurotransmitter is released within his brain at the sight of Marie or thinking
about her. Similarly, the feeling of Marie for Pierre will be indicated by F and Marie’s
M
thought about Pierre with I . To describe the evolution of the relationship between Pierre
P
and Marie, we may use the following set of “sentimental” transformations:
F
k 1
P + I M →
n 1 FP + IM
P + F
k 2
M →
n 2 FP + FM
F
k 3
M + I P →
n 3 FM + IP
M + F
k 4
P →
n 4 FM + FP
The first and the third processes represent how F and F evolve spontaneously into the
P
M
brains of Pierre and Marie when they think about Marie and Pierre, respectively. The sec-
ond transformation describes how Pierre reacts when he gets acquainted with the feeling
of Marie for him, by interpreting her direct or indirect messages. Vice versa, the fourth
process describes how Marie reacts when she gets acquainted with Pierre’s feeling for
her, by interpreting his direct or indirect messages. The terms n , , , and look like
1 n 2 n 3
n 4
stoichiometric coefficients. They can be any, positive or negative, real numbers. For this
exercise, let us assume that n and may be 2, 1 or 0. If and are equal to 2, it means
1
n 3
n 1
n 3
that he/she likes her/him and their feelings of love grow autocatalytically; if n and are
1
n 3
1, it means that their sensations neither grow nor decrease; if they are 0, it means their feel-
ings fade. Note that Pierre’s thought about Marie ( I ) and Marie’s thought about Pierre ( I )
M
P
play as if they were catalysts of the first and third transformation, respectively. Their action
is positive when n
2, null when
1, and inhibitory when
0.
1 = n 3 =
n 1 = n 3 =
n 1 = n 3 =
If we look at the second and fourth processes, we notice that F , in the second, and F , in the
M
P
fourth, play as catalysts, as well. If n is equal to 1, it means that love messages sent by Marie
2
to Pierre work as efficient catalysts to increase his love for her. If n is equal to 1, it means that
4
love messages sent by Pierre to Marie work as effective catalysts to boost her love for him.
If
n or is equal to 0, it means that love messages sent by Marie/Pierre do not work at
2
n 4
all. Finally, when n or is equal to
2
n 4
−1, it means that love messages sent by Marie/Pierre
play as inhibitors of his/her love for her/him. In the following table, there is a summary of
all possible values and meanings of the four coefficients (Table 3.3).
Note that the values of products k
,
,
1 I
k
k
M
3 IP
2 P, and k 4 M define the time scale of the
dynamics of the love story. They may have day-1 as the unit if Pierre and Marie think
about and encounter each other, every day. In case Pierre and Marie meet only once a
week, k
,
,
1 I
k
k
M
3 IP
2 D, and k 4 B will have week−1 as units. Now, try to predict the dynam-
ics of the love story between Pierre and Marie. Pierre fell in love with Marie and wooed
her. On the other hand, it seems that Marie was a cautious lover. Can you predict what
happened? Write the ordinary differential equations for the four transformations listed
earlier, find the steady state solution, and use the linear stability analysis to predict the
course of the events. You are invited to imagine the evolution of the Pierre and Marie’s
relationship, but also you would like to predict the dynamics of your own love story and
other love affairs that you know.
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Untangling Complex Systems
TABLE 3.3
Values of the Stoichiometric Coefficients for the Love Story
Coefficients
2
1
0
−1
n ;
Positive
Null
Negative
1 n 3
n ;
Positive
Null
Negative
2 n 4
3.13 SOLUTIONS TO THE EXERCISES
3.1. When the system is at equilibrium the extended chemical potential at height h must be
1
equal to the extended chemical potential at height h :
2
µ
0
0
( )
( )
( )
( )
k h 1 = µ k + RTlnCk h 1 + PMk gh 1 = µ k h 2 = µ k + RTlnCk h 2 + PM k gh 2
If we rearrange the equation, we obtain:
C
PMk g
( )
k h
−
( h 2− h )
2
e RT
1
C ( ) =
k h 1
This ratio is equal to 0.999976 in the case of tryptophan, whereas it is 0.99213 in case
of the protein when h
is equal to 3 cm. This result underlines that the solutions
2 − h 1
are always more concentrated at the bottom than at the top due to the gravitational field.
The concentration gradient that originates inside the solution is the “equilibrium” condi-
tion in the presence of the gravitational field. It requires time to be reached. The solution
must decant. Finally, note that the higher the molecular weight of the solute, the more
pronounced is the concentration gradient.
3.2. At “equilibrium,” the electrochemical potential of the monovalent cations must be equiv-
alent inside and outside the cell:
µ ( )
0
( )
( )
0
( )
+
in = µ + + RTlnC + in + ψ
F
= µ + out = µ + + RTlnC + out + Fψ
K
K
K
in
K
K
K
out
It derives that
RT
C ( out)
ψ
k
in −ψ out =
ln
mV
F
C
= −86
( )
k in
3.3. The Gibbs free energy reaches a minimum at the equilibrium. On the left of the minimum,
