GROUP (7)
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4.5 The Effect of Temperature
 As the temperature increase, the upper levels become
relativity more populated, and this corresponds to an increase
in the average energy of the particles, (i.e. an increase in the
value of U/n), which for fixed value of V and n, U will
increase. Also as T increases the value of β decreases and
the shape of the exponential distribution changes will be as
shown in figure (4.3).
 As the macrostate of the system is fixed by fixing the values
of U, V, and n, then T as a dependent variable is fixed too.
 When the number of particles in the system is very large, then
the number of arrangements within the most probable
distribution , Ωmax ,is the only term which makes a significant
contribution to total number of arrangements, Ωtotal ,which
the system may have ; that is Ωmax is significantly larger than
the sum of all other arrangements ; hence :
Ωtotal ≈Ωmax
thus:
ln Ωtotal = ln Ωmax
i=r
= n ln n − n − ∑ (ni ln ni − ni )
i=o
i=r
i=r
= n ln n − n − ∑ ni ln ni + ∑ ni
i=o
2
i=o
Figure (4.3): The influence of temperature on the most
probable distribution of particles among
energy levels in a closed system of
constant volume.
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i=r
= n ln n − ∑ ni ln ni
i=o
i=r
n − ∈i
n − ∈i
= n ln n − ∑
e KT ln
e KT
p
p
i=o
i=r
n
∈i − ∈i
= n ln n − ∑ (ln n − ln p −
)e KT
p
KT
i=o
i=r
∈
n
− i
= n ln n − (ln n − ln p) ∑ e KT
p
i=o
i=r
∈
n
− i
+
∑ ∈i e KT
PKT
i=o
n
= n ln n − (ln n − ln p)p
p
i=r
∈
n
− i
+
∑ ∈i e KT
PKT
i=o
i=r
∈
n
− i
= n ln n − n ln n + n ln p +
∑ ∈i e KT
PKT
i=o
𝐢=𝐫
∈
𝐧
− 𝐢
= 𝐧 𝐥𝐧 𝐩 +
∑ ∈𝐢 𝐞 𝐊𝐓
𝐏𝐊𝐓
𝐢=𝟎
but:
∈
U = ∑ ni ∈i = ∑
n − i
∈i e KT
p
4
=
n
p
∈
− i
∑ ∈i e
KT
therefore:
i=r
∈i
UP
−
= ∑ ∈i e KT
n
i=o
thus:
ln Ω = n ln P +
n UP
U
= n ln P +
PKT n
KT
or:
𝐔 = 𝐊𝐓 𝐥𝐧 Ω − 𝐧𝐊𝐓 𝐥𝐧 𝐏
4.6 Thermal Equilibrium within a System and the
Boltzmann Equation
 Consider a particles system at constant V in thermal
equilibrium at temperature T with a heat reservoir, thus n and
∈i ′s of the particles system are constant; therefore P is
constant. For a small exchange of energy between the
particles system and the heat reservoir, we have:
 Applying the First Law of Thermodynamics gives:
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Since the particles system is at constant volume, δ w = 0,
i.e.
 Applying the Second Law of Thermodynamics gives:
 As both S and Ω are state functions, the above equation can
be written as:
𝑺 = 𝐤 𝐥𝐧 Ω
This equation is known as Boltzmann’s Equation.
 The previous equation is thus the required quantitative
relationship between the entropy of a system and its "degree
of mixed-up-ness" given as Ω and defined as: the number of
ways which the available energy of the particles system is
shared among the particles of the system.
 Thus, the equilibrium state of the system is that state at which
S is maximum at the considered fixed volume of U, V and n;
this is based on the following "the most probable state of the
system is that in which Ω is maximum at the considered U, V,
and n of the system.
 Therefore, the Boltzmann’s Equation provides
a physical quality to entropy.
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4.7 Heat Flow and the Production of Entropy
 Consider two closed systems A and B. Let the energy of A to
be UA and the number of complexions of A to be ΩA ;
similarly, let the energy of B to be UB and the number of
complexions of B to be ΩB. If the two bodies are brought in
contact with each other , the product ΩA ΩB will generally not
have its maximum possible value and heat will flow either
from A to B or from B to A to maximize the product ΩA ΩB . If
the heat flow from A to B, this means that T(A) > T(B), also
this means that the increase in the number of ΩB due to this
heat exchange is larger than the decrease in the value of ΩB .
 When an amount of heat δq is transferred from A to B at
constant total energy , then :
thus:
δq 1
1
d ln ΩA ΩB = d ln ΩA + d ln ΩB = ( − )
K TB TA
 Of course, the flow will cease when ΩA ΩB will reach its
maximum value, i.e. when d ln ΩA ΩB = 0 , and the
condition for that is TA = TB; and that condition at thermal
equilibrium will prevail between the two bodies.
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 Thus, in the microscopic analysis, an irreversible process is
one which takes the system from a less probable to the most
probable state; while in the corresponding macroscopic
analysis an irreversible process takes the system from a nonequilibrium to the equilibrium state.
 Therefore, what is considered in Classical Thermodynamics
to be an impossible process turns out as result of microstate
examination to be an improbable process.
4.8 Configurational Entropy and Thermal
Entropy
 Consider two crystals at the same temperature and pressure,
one containing atoms of the element A and the other
containing atoms of element B. When the two crystals are
placed in contact with one another, the spontaneous process
which occurs is the diffusion of A atoms into the crystal B
lattice sites and diffusion of the B atoms into the crystal A
lattice.
 As this is a spontaneous process, the entropy of the system
will increase and it might be predicted that equilibrium will be
reached (i.e. the entropy of the system will reach a maximum
value) when the concentration gradients in the system have
been eliminated (this is similar to the case of heat flow under
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temperature gradient, and this flow will be ceased and the
thermal equilibrium state will be reached when the
temperature gradient in the system is eliminated).
 Following the discussion of Denbigh, consider a crystal
consisting of four atoms of A placed in contact with a crystal
containing four atoms of B as shown in figure (4.4).
 The mixing process in this system can be written as
at constant
U ,V and n
(A + B)unmixed →
(A + B)mixed
 Since the number of ways of arrangements is given by the
relation :
thus:
 Now , for the initial state where the arrangement in crystal A
is 4 : 0 and the arrangement in crystal B is 0 : 4 , thus :
 when one atom of A is interchanged with one atom of B
across x y ; thus :
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Figure (4.4): representation of a crystal of A in
contact with a crystal of B.
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 when two atoms are interchanged , thus :
 when three atoms are interchanged , thus :
 when the four atoms are interchanged , thus :
 Thus:
∑ Ω = 1+ 16 + 36 + 16 + 1 = 70
which equals Ω4+4
 The arrangement 2 : 2, thus, is the most probable, 36 / 70,
therefore it corresponds to the equilibrium state .This
arrangement, as expected, corresponds to the elimination of
the concentration gradient.
 Defining the configurational entropy ,(S conf ), as the
entropy component which is caused from the number of
distinguishable ways in which the particles of the system can
be mixed over position in space ; and the thermal entropy ,
(Sth ) as that component of the entropy which arises from the
number of ways in which the energy of the system can be
shared among the particles , thus :
S total = S th + S conf
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 For the previous example:
∆S conf = S conf (2) - S conf (1)
= K lnΩconf (2) – K lnΩconf (1)
Ωconf (2)
= k ln
Ωconf (1)
 In general:
S tot = S th + S conf
= K ln Ωth + K ln Ω conf
= K ln Ωth Ω conf
Ωth (2) Ωconf (2)
∆Stotal = k ln
Ωth (1) Ωconf (1)
 Thus, for two closed systems placed in thermal contact or for
two chemically identical open systems placed in thermal
contact:
Ωconf (2) = Ωconf (1)
thus:
Ωth (2)
∆Stotal = k ln
= ∆Sth
Ωth (1)
 Similarly , if particles of A are mixed with particles of B and
if this mixing process doesn’t effect the distribution of the
particles among the energy levels , i.e. Ωth (2) = Ωth (1) ,
thus :
Ωconf (2)
∆Stotal = k ln
= ∆Sconf
Ωconf (1)
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Such a situation corresponds to “ideal mixing” and requires
that the quantization of energy within the two initial crystals is
identical; such a situation is the expectation rather than the
rules.
 In general, when two or more pure elements are mixed at
constant U, V and n, Ω 𝑡ℎ (2) ≠ Ω 𝑡ℎ (1) complete spatial
randomization of the constituent particles does not occur. In
such case, either clustering of like particles (indicating
difficulty in mixing) or ordering (indicating a tendency towards
compound formation) occurs.
 In all cases, however, the equilibrium state of the mixed
system is that which, at constant U, V and n, maximizes the
product
(Ω th Ω conf )2
(Ω th Ω conf ) 1
.
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