Untangling Complex Systems, page 16
∇µ
µ
µ
k =
∇ nk − k
S ∇ T +
∇ P [3.91]
∂ nk
P
T, P
∂ T, nk
If we assume that our system is in mechanical equilibrium, ∇ P = 0. Introducing [3.91] in [3.88],
after rearranging the term − S
2
1
k∇ T in SkT ∇ ( ), we obtain:
T
1
1 µ
p* = J
k
µ
u −
Jk( k + TSk) ∇
J
−
∂
∇
∑
∑
n
T
k
T
k [3.92]
n
∂
k
k
k
T, P
The term (µ
( H . Therefore, from equation
nk )
k +
k
TS ) = Hk is the partial molar enthalpy, that is ∂∂ P, T
[3.92], it derives that a heat flow that includes a matter flow is given by
Jq = Ju −
JkHk
∑ [3.93]
k
If we express the entropy production per unit volume as function of J that is called the reduced heat
q
flow, we have
1
1 ∂µ
p* = J
k
q∇
J
−
k
∑ ∇ n [3.94]
T
T n
k
∂
k
k
T, P
Let us imagine having a two-component system. From the Gibbs-Duhem relation [3.70], it derives that
µ
∂µ
n
1
2
(∇µ ) + (∇µ ) =
∂
∇ 2 =
1
1
n 2
2
n 1
n
n
0 [3.95]
,
,
∇ +
1
n
P T
P T
n
2
∂
,
∂
1
n
T P
2 T, P
Out-of-Equilibrium Thermodynamics
63
Moreover, in the absence of volume flow, we may write, analogously to equation [3.69], the follow-
ing equation:
0 = J V +
1 1
J 2 2
V [3.96]
Introducing equations [3.95] and [3.96] into [3.94], the definition of entropy production becomes
1
µ
1
1 1 ∂ 1
*
J
n V
p = Jq∇
1
n
−
+
∇ [3.97]
T T
1
n 2 V 2 n
∂ 1 T, P
We are in the presence of two independent forces (∇( 1 and − 1
n V
µ
n or
T (1 + 1 1
1
2
n 2
V )( ∂∂ 1
n )
∇
T )
1
T, P
− 1
n V
µ
n and two independent flows ( J and J or J ).
T (1 + 2 2
2
1
n 1
V )( ∂∂ 2
n )
∇ 2
T, P
)
q
1
2
The phenomenological linear laws are:
1
1
n V
µ
1 1 ∂ 1
Jq =
Lqq∇
L
− q 1
1+
n
∇
T
T
1
n 2 V 2 n
∂ 1
T, P
[3.98]
1
1
n V ∂µ
J
1 1
1
∇
1 =
L q 1
L
−
+
11
1
n
∇
T
T
n
1
∂
2 V 2
n 1 T, P
The term L
L
1
1 q
q∇ ( ) = −
∇
1
represents the Soret effect. It is also expressed as − C 1 DS∇ T, where D
2
T
T
T
S
is the coefficient of thermal diffusion. The ratio
D
c
S
S =
[3.99]
D 1
(where D is the ordinary diffusion coefficient for species 1) is called the Soret coefficient, that has
1
the dimension of [ T]−1. A thermal gradient applied to a solution contained in a closed vessel gives
rise to a concentration gradient. The value of the concentration gradient can be obtained from the
second equation in [3.98], setting J
1 = 0:
D
∇ C 1 = − C S
1
∇ T = − C c
1 S∇ T [3.100]
D 1
The Soret coefficient is generally small, of the order of 10−3 or 10−2 K−1.
The cross term − L 1
n V
1
µ
q 1
1
1 1
in first equation of [3.98] represents the Dufour
T ( + n 2 V 2 )( ∂
n
∂ n 1 )
∇
T, P
1
effect, that is a heat flow generated by a concentration gradient. The Dufour effect is expressed as
− C
1
1
µ
1 D ∇
n V
D n 1 = − Lq 1
1
1 1
where D is the Dufour coefficient. The Onsager reciprocal
T ( + n 2 V 2 )( ∂
n
∂ n 1 )
∇
T, P
1
D
relation L
imposes that
q 1 = L 1 q
D
µ
1 1
D
n V ∂ 1
= T 1+
[3.101]
D
S
n 2 V 2 n
∂ 1 T, P
Equation [3.101] has been verified in ideal gas mixtures and liquid isotope solutions (Würger 2014).
3.4 EVOLUTION OF OUT-OF-EQUILIBRIUM SYSTEMS IN LINEAR REGIME
A system is maintained out-of-equilibrium when at least one force is not null. If we fix the value of
the force, how does the system evolve when it works in linear regime?
We are now considering two examples: heat conduction and diffusion. We will notice that in the
linear regime, the system evolves to a stationary state that has time-independent properties.
64
Untangling Complex Systems
3.4.1 The case of heaT conducTion
Suppose to have a system of length L and rectangular section Ar (see Figure 3.13), which is in physical contact with two heat reservoirs at its ends. One reservoir is at high temperature ( T ), and the
h
other is colder, at T . If we assume that the temperature gradient is not large, the heat will flow from
c
the hot to the cold bath by conduction along the x-axis.
The entropy production per unit length due to the conduction process will be given by the prod-
uct of the flow and the force:
*
d 1
Jq dT
p ( x) =
Jq
[3.102]
dx = −
T
T 2 dx
If we consider a portion of the system having an infinitesimal width along x-axis (see Figure 3.13), the net heat variation (according to the Conservation Law of Energy) is given by
dq
= J ( 0 )( )
( 0
)( )
q x
Ar − Jq x + dx Ar [3.103]
dt
where J ( )
(
)
q x
+
0 is the heat flow entering through Ar in x , whereas Jq x 0
dx is the heat flow getting
0
out through Ar in x
( + ) as a Taylor series and obtain
0 + dx. We can express Jq x 0
dx
dq
J
∂
q
J
∂
J
q
≈
( 0)( ) ( 0)
( )
q x
Ar − Jq x +
dx Ar = −
Ar dx [3.104]
dt
(
)
x
∂
0
x
∂
x
x 0
If we introduce the Fourier’s law [3.32] into equation [3.104], we achieve
dq
2 T
=
∂
k
[3.105]
2
( Ar)( dx)
dt
x
∂
x 0
If we employ the definition of the heat capacity at constant pressure ( Cp = dq/dT), finally we obtain
∂ T
k ( Ar)( dx) ∂2 T
=
[3.106]
dt
C
2
p
∂
x x 0
If the temperatures of the two reservoirs are kept constant, the system will evolve up to reach a
time-independent stationary state, wherein the derivative of T with respect to time becomes null.
The term ∂
( T is null when the second derivative of T with respect to x is null (see equation [3.106]).
dt )
The latter condition is true when T is a linear function of x. In this condition, i.e., at the stationary state, J is constant and, therefore,
q
x0 x0 + dx
Ar
Th
Tc
L
x
FIGURE 3.13 A system of length L is in physical contact with two reservoirs: one hot at T and the other cold h
at T . A portion of the system of section Ar and infinitesimal width ( dx) is shown.
c
Out-of-Equilibrium Thermodynamics
65
L
Tc
Jq dx = − k dT
∫
∫ [3.107]
0
Th
J
(
)
q L = − k Tc − Th [3.108]
From equation [3.108] it is evident that the higher the thermal gradient, the larger the conductivity,
the stronger the heat flow.
The total entropy production is obtained by integrating along the entire length L of the system
L
T
d
c
*
iS
Jq dT
1
1
1
P =
= −
dx
J
2
= −
q
dT J
dt
T
2
=
−
∫
∫
≥ 0 [3.109]
dx
T
q
Tc Th
0
Th
At the stationary state, the positive entropy production is counterbalanced by the entropy that is
exchanged with the environment:
deS
Jq Jq
diS
=
−
= −
[3.110]
dt
Th Tc
dt
At the stationary state, the total entropy of the system is constant: it is produced inside and released
into the environment.
TRY EXERCISES 3.8 AND 3.9
3.4.2 The case of diffusion
Suppose to have two reservoirs containing compound k at two different concentrations: C in res-
1 ,k
ervoir 1 and C in reservoir 2 (with
). The two reservoirs are kept in contact through a
2,
C
k
1, k > C 2, k
third vessel of length L along the x-axis and section Ar (see Figure 3.14a). Species k diffuses from x 0 x 0 + dx
Ar
C 1 ,k
C 2 ,k
L
(a)
x
Ck
Ck
(b)
x
(c)
x
FIGURE 3.14 (a) Diffusion of the species k into a vessel of length L and in between two reservoirs containing k at concentrations of C , and
, respectively. Two unstable profiles of (
1
C
C x): (b) convex (positive
,k
2 ,k
k
curvature) and (c) concave (negative curvature).
66
Untangling Complex Systems
reservoir 1 to 2. The variation of the number of k moles in a portion of the vessel having volume ( Ar) ( dx), according to the Conservation Law of Mass, is given by
∂ n
k
Jk
= J
dx ( Ar)
k ( x 0 )( Ar ) − Jk ( x 0 + dx )( Ar ) ≈ Jk ( x 0 )( Ar ) − Jk ( x 0 ) + ∂
[3.111]
∂
t
∂ x
x 0
∂ nk
Jk
≈ − ∂
Ar dx [3.112]
∂
(
)
t
∂ x x 0
It derives that
∂ Ck
Jk
≈ − ∂
[3.113]
∂
t
∂ x x 0
Introducing equation [3.40], known as the “first Fick’s law,” into equation [3.113], we achieve
∂ C
2
k
C
=
∂
D
k [3.114]
∂ t
k ∂ x 2
Equation [3.114] is termed as “diffusion equation” or “second Fick’s law.” It is formally equivalent to
equation [3.106] formulated for heat conduction. It reveals that the variation of C on time is directly
k
proportional to the curvature or concavity of the function C ( x). In particular, if C ( x) has a positive k
k
curvature, i.e., it is convex (sometimes referred to as concave up, meaning that the tangent line lies
below the graph of the function, see Figure 3.14b), C will grow until it annihilates the trough. If k
C ( x) has a negative curvature, i.e., it is concave (sometimes referred to as concave down, meaning
k
that the tangent line lies above the graph of the function, see Figure 3.14c), C will decrease up to k
level the hunch. Finally, when C is uniformly distributed or linearly distributed along x, it does not k
change over time.
If we fix the concentrations of species k into the two reservoirs at two different values, the system
will evolve towards a steady state, wherein the diffusion flow of k is constant over time, and the k
concentration depends linearly on x:
x
Ck ( x)
Jk dx = − Dk
dCk
∫
∫ [3.115]
0
C 1, k
J
C
k
( )
k x = C 1, k −
x [3.116]
Dk
The entropy production per unit length due to the diffusion process will be given by the product of
the flow and the force
µ
*
d
p ( x) = −
J
k
k
[3.117]
dx
T
At uniform T and for an ideal solution
µ
*
Jk d k
