Hillert M. Phase Equilibria, Phase Diagrams and Phase Transformations: Their Thermodynamic Basis
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3
Systems with variable composition
3.1 Chemical potential
In Chapter 1 we were concerned mainly with closed systems but have also considered
the addition of more matter through terms with dN .Without explicitly stating it, it was
presumed that the properties of the system were not affected by this addition. That would
hold for a one-component system and also if the added matter has the same composition as
the initial system. We shall now take changes of composition into account by generalizing
G
m
dN to µ
i
dN
i
where the subscript i identifies different components.
dU = T dS − PdV +µ
i
dN
i
− Ddξ. (3.1)
The µ
i
quantity is a potential just like T, −P and −D.Itwas first introduced by Gibbs [3]
and is called chemical potential. Its close relation to G
m
will be explained in Section 4.1.
µ
i
and N
i
are conjugate variables and the terms µ
i
dN
i
may thus be included in Y
a
dX
a
in
the generalized form of the combined law, introduced in Section 1.9.For any component
j of the system the chemical potential may be defined from Eq. (3.1)as
µ
j
= (∂U/∂ N
j
)
S,V ,N
k
,ξ
. (3.2)
The subscript ‘N
k
’ indicates that all N
i
are kept constant except for N
j
.Atequilibrium
with respect to the internal process, where ξ is a dependent variable, we have
µ
j
= (∂U/∂ N
j
)
S,V ,N
k
. (3.3)
The summation µ
i
dN
i
is taken over all components in a chosen set of independent
components. In chemical thermodynamics one often takes the summation over all molec-
ular species but then one must also define a set of independent reactions. That procedure
is less general and will be avoided in the present text.
From the new, more general form of the combined law introduced above, it is evident
that the number of external variables which can be varied independently of each other
under equilibrium conditions is c + 2, if there are c independent components. In general,
µ
j
is a function of S, V, N
i
and ξ and we can write
µ
j
= µ
j
(S, V , N
i
,ξ), (3.4)
or under conditions of internal equilibrium
µ
j
= µ
j
(S, V , N
i
). (3.5)
46 Systems with variable composition
This is another equation of state but it is not a fundamental equation and, thus, it does
not contain all the properties.
When considering systems with variable composition it is useful to define many new
quantities. A large part of the present chapter is devoted to discussions of such quantities.
Exercise 3.1
Show how µ
j
can be evaluated from a derivative of S instead of U.
Hint
Use the entropy scheme.
Solution
−dS = (−1/T )dU + (−P/T )dV + (µ
i
/T )dN
i
− (D/T )dξ yields:
(∂ S/∂ N
j
)
U,V,N
k
,ξ
=−µ
j
/T .
3.2 Molar and integral quantities
Let us consider a homogeneous system at equilibrium and define a part of it as a subsys-
tem. The size of the subsystem may be expressed by the value of any extensive variable.
The most natural way may be to use the content of matter because from the experimental
point of view it is easier to control the content of matter than the volume or entropy. One
usually uses the total content of matter, N, defined by
N = N
i
. (3.6)
Sometimes we shall use the content of a particular component, N
j
, instead of the total
content of matter, N.
As a measure of the content of matter Gibbs used the mass, but today it is more
common to use the number of atoms or species. We shall use the latter method but it
should be emphasized that it is often necessary to specify what species are considered,
which Gibbs did not have to do. On the other hand, thermodynamic models of special
kinds of substances are often based upon considerations of atoms and it is then convenient
to interpret N and N
i
as the number of atoms (or groups of atoms). The number is usually
expressed in units of moles, i.e. approximately 6 ×10
23
pieces (Avogadro’s number, N
A
).
The volume V is proportional to N in a homogeneous system and we may define a
new quantity, the molar volume
V
m
= V/N . (3.7)
This quantity has a defined value at each point of a system. It is thus an intensive
variable like T , −P and µ
i
.However, its properties are quite different, a fact which
becomes evident if we consider a system consisting of more than one phase, i.e. regions
3.2 Molar and integral quantities 47
exhibiting different properties. In each homogeneous phase V
m
has a different, constant
value but T , −P and µ
i
must have the same value in the whole system at equilibrium
(with one exception which we shall deal with later). This is the property of a potential as
noted in Section 1.10 and we may conclude that V
m
is not a potential. It is very important
to distinguish between two kinds of intensive variables, potentials and molar quantities.
One should try to avoid using the words ‘intensive variable’ and when using them one
should specify what kind one is considering.
In the same way we may define the molar quantity for any extensive property obeying
the law of additivity, e.g. the molar content of component i. Usually, it is denoted by x
i
and is called mole fraction
x
i
= N
i
/N. (3.8)
However, it is sometimes essential to stress its close relation to other molar quantities.
‘Molar content’ is thus preferable.
The molar quantities have been defined for a homogeneous system or for a homo-
geneous part of a system. The definition may very well be extended to the whole of a
system with more than one phase but such a molar quantity is not strictly an intensive
quantity and may be regarded as an average of an intensive quantity.
Let us return to a homogeneous system at equilibrium, i.e. with D = 0, and define
avery small subsystem enclosed inside an imaginary wall. We shall let the subsystem
grow in size by expanding the wall but without making any real changes in the system,
i.e. without changing P, T or composition. For this process we have
dS = S
m
dN (3.9)
dV = V
m
dN (3.10)
dN
i
= x
i
dN. (3.11)
However, dξ = 0 because no internal process is going on. We may thus evaluate the
change in U as follows
dU = T dS − PdV +µ
i
dN
i
= (TS
m
− PV
m
+ µ
i
x
i
)dN. (3.12)
The value of the expression in parentheses is constant. By integrating over the expan-
sion we obtain
U = (TS
m
− PV
m
+ µ
i
x
i
)N = TS− PV + µ
i
N
i
(3.13)
U/N = U
m
= TS
m
− PV
m
+ µ
i
x
i
. (3.14)
Using the definition of Gibbs energy in Section 1.11 we get
G ≡ U − TS+ PV = µ
i
N
i
. (3.15)
G
m
,which was introduced as a notation for H
m
− TS
m
in Eq. (1.48), is thus identical to
the molar Gibbs energy,
G
m
≡ U
m
− TS
m
+ PV
m
= H
m
− TS
m
. (3.16)
48 Systems with variable composition
In Section 1.11 we did not recognize changes in composition. For that case Eqs (3.15)
and (3.16)would yield
G = µ
i
N
i
(3.17)
G
m
= µ
i
x
i
. (3.18)
It may be noted that for a pure substance x
i
= 1 and G
m
will thus be equal to the chemical
potential of the substance.
Extensive quantities like U are sometimes called integral quantities in order to be
distinguished from molar quantities. When defining N as the mass, as Gibbs did, one
usually calls the quantities, obtained by dividing with N, specific instead of molar. Most
of the thermodynamic relations are valid independent of how N is defined.
In many cases it is convenient to consider one mole of formula units or groups of
atoms and the molar quantities are then defined by dividing with the number of formula
units or groups of atoms, expressed as moles, i.e. units of approximately 6 × 10
23
.
It can be easily shown that all the relations between integral quantities also apply to
molar quantities, e.g.
S
m
=
1
N
S =
−1
N
∂G
∂T
P,N
i
=
−1
N
∂ NG
m
∂T
P,N
i
=−
∂G
m
∂T
P,N
i
=−
∂G
m
∂T
P,x
i
.
(3.19)
because all x
i
are constant if all N
i
are kept constant.
Exercise 3.2
Cuprous oxide has a density of 6000 kg/m
3
.Give its molar volume in two different ways
and also its specific volume.
Hint
The atomic mass is for Cu 63.546 and for O 15.9994.
Solution
The mass of one mole of the formula unit Cu
2
Ois143.09 g or 0.14309 kg and the
molar volume is thus 0.14309/6000 = 24 × 10
−6
m
3
per mole of Cu
2
Oor8× 10
−6
m
3
per mole of atoms, i.e. moles of Cu
0.67
O
0.33
. The specific volume is 1/6000 = 167 ×
10
−6
m
3
/kg, whether one considers Cu
2
OorCu
0.67
O
0.33
.
3.3 More about characteristic state functions
We have seen that
U = TS− PV + µ
i
N
i
(3.20)
3.3 More about characteristic state functions 49
in a multicomponent system. It is evident that we get the following relation for the Gibbs
energy
G = U − TS+ PV = µ
i
N
i
. (3.21)
The relations between the characteristic state functions can be summarized as follows
µ
i
N
i
= G = H − TS = U + PV − TS = F + PV (3.22)
µ
i
x
i
= G
m
= H
m
− TS
m
= U
m
+ PV
m
− TS
m
= F
m
+ PV
m
. (3.23)
In Section 1.11 we discussed various forms of the combined law obtained by changing
from S to T and from V to P in the set of independent variables. We can now generalize
them as follows:
dU = T dS − PdV +µ
i
dN
i
− Ddξ (3.24)
d(U − TS) = dF =−SdT − PdV +µ
i
dN
i
− Ddξ (3.25)
d(U + PV) = dH = T dS + V dP + µ
i
dN
i
− Ddξ (3.26)
d(U − TS+ PV) = dG =−SdT + V dP + µ
i
dN
i
− Ddξ. (3.27)
It is evident that the chemical potentials for a substance can be evaluated from any
one of these characteristic state functions if it is given in terms of its natural variables
µ
j
= (∂U/∂ N
j
)
S,V ,N
k
,ξ
= (∂ F/∂ N
j
)
T,V,N
k
,ξ
= (∂ H/∂ N
j
)
S,P,N
k
,ξ
= (∂G/∂ N
j
)
T,P,N
k
,ξ
.
(3.28)
We may consider ξ as a dependent variable under equilibrium conditions but, in view
of Section 2.2, that fact does not change the value of a partial derivative. We could thus
omit ξ and write
µ
j
= (∂U/∂ N
j
)
S,V ,N
k
= (∂ F/∂ N
j
)
T,V,N
k
= (∂ H/∂ N
j
)
S,P,N
k
= (∂G/∂ N
j
)
T,P,N
k
.
(3.29)
The remaining extensive variables, N
i
, can also be replaced by their conjugate potentials,
µ
i
, and we can get four new forms of the combined law
d(U − µ
i
N
i
) = d(TS− PV) = T dS − PdV − N
i
dµ
i
− Ddξ (3.30)
d(U − TS− µ
i
N
i
) = d(−PV) =−SdT − PdV − N
i
dµ
i
− Ddξ (3.31)
d(U − PV − µ
i
N
i
) = d(TS) = T dS + V dP − N
i
dµ
i
− Ddξ (3.32)
d(U − TS+ PV − µ
i
N
i
) = 0 =−SdT + V dP − N
i
dµ
i
− Ddξ. (3.33)
The first three forms define new characteristic state functions. The fourth form
is unique because it defines a function which is identically equal to zero since
U − TS− PV = µ
i
N
i
.For reversible conditions, D = 0, or in the absence of internal
processes, dξ = 0, it yields a direct relation between the c + 2 potentials, the so-called
Gibbs–Duhem relation. Consequently, one of the potentials is no longer an independent
variable.
SdT − V dP + N
i
dµ
i
= 0. (3.34)
50 Systems with variable composition
This relation is often given in terms of molar quantities
S
m
dT − V
m
dP + x
i
dµ
i
= 0 (3.35)
Forapure substance x
i
= 1 and the chemical potential is identical to the molar Gibbs
energy, G
m
,asnoted after Eq. (3.18). The Gibbs–Duhem relation thus simplifies to
S
m
dT − V
m
dP + dG
m
= 0. (3.36)
Equation (3.31)gives a characteristic state function which is equal to (−PV) and is
particularly interesting in statistical thermodynamics. This characteristic state function
is sometimes denoted by and is called ‘grand potential’. It can be evaluated from the
so-called grand partition function, ,
=−kT ln . (3.37)
The grand partition function is defined for a so-called grand canonical ensemble for which
T, V and µ
i
are the independent variables and = (T, V,µ
i
). It is sometimes useful
in calculations of equilibrium states because it may yield relatively simple relationships.
The fact that it applies under constant values of µ
i
,which may be difficult to control
experimentally, does not limit its usefulness in such calculations.
In this connection it may be mentioned that the ordinary partition function Z is defined
for an ordinary canonical ensemble for which T, V and N
i
are the independent variables.
It can be used to evaluate the Helmholtz energy
F =−kT ln Z. (3.38)
Furthermore, for a microcanonical ensemble one keeps U, V and N
i
constant and can
evaluate S(U, V , N
i
).
The remaining two new forms of the combined law and their characteristic state
functions have not found much direct use. However, in the next section they will
prove useful in some thermodynamic derivations. It should finally be emphasized that a
large number of additional forms may be derived by selecting some of the N
i
and some
of the µ
i
as independent variables. We shall discuss one such example in Section 14.5.
Exercise 3.3
Prove the well-known equality (∂ G/∂ N
A
)
T,P,N
j
= (∂ F/∂ N
A
)
T,V,N
j
from Eq. (3.29)by
changing variables using Jacobians.
Hint
T and all N
j
are not to be changed. Simplify the notation by omitting them from the
subscripts. Change from N
A
, P to N
A
, V . Then express all derivatives of G in terms of
F using P =−(∂ F/∂V )
T,N
i
=−F
V
and G = F + PV = F − VF
V
.
3.4 Additivity of extensive quantities. Free energy and exergy 51
Solution
∂G
∂ N
A
P
=
∂G/∂ N
A
∂G/∂V
∂ P/∂ N
A
∂ P/∂V
∂ N
A
/∂ N
A
∂ N
A
/∂V
∂ P/∂ N
A
∂ P/∂V
= (∂ G/∂ N
A
)
V
− (∂G/∂ V )
N
A
(∂ P/∂ N
A
)
V
/(∂ P/∂V )
N
A
But, (∂G/∂ N
A
)
V
= F
N
A
− VF
VN
A
;(∂G/∂V )
N
A
= F
V
− F
V
− VF
VV
=−VF
VV
;
(∂ P/∂ N
A
)
V
=−F
VN
A
;(∂ P/∂V )
N
A
=−F
VV
. Inserting these we get (∂G/∂ N
A
)
P
=
(F
N
A
− VF
VN
A
− (−VF
VV
)(−F
VN
A
)/(−F
VV
) = F
N
A
= (∂ F/∂ N
A
)
V
.
3.4 Additivity of extensive quantities. Free energy and exergy
The extensive quantities that were primarily defined in Chapter 1 are additive with no
restrictions. The values of V, U , S and N of a composite system are always equal to the
sum of the values for the subsystems. This is also true for the contents of all components
N
i
.InEqs (3.31) and (3.32)wedefined quantities that are equal to −PV and TS. They are
also extensive but it is evident that they are additive only if the potentials P and T have the
same values in the subsystems. The same is true for F, H and G because they are defined
by subtracting TS or adding PV .Asaconsequence, it was mentioned in Section 1.12
that one cannot compare Gibbs energy values for states at different temperatures.
The problem can sometimes be solved by accepting that T and P in the definition
G = U − TS+ PV are the values in the surroundings. Then one can add the Gibbs
energy for two subsystems that are kept at different T and P,
G
1
+ G
2
= U
1
(T
1
, P
1
) − TS
1
(T
1
, P
1
) + PV
1
(T
1
, P
1
) + U
2
(T
2
, P
2
)
− TS
2
(T
2
, P
2
) + PV
2
(T
2
, P
2
) = U
1
(T
1
, P
1
) + U
2
(T
2
, P
2
) − T [S
1
(T
1
, P
1
)
+ S
2
(T
2
, P
2
)] + P[V
1
(T
1
, P
1
) + V
2
(T
2
, P
2
)] = U
1+2
− TS
1+2
+ PV
1+2
= G
1+2
.
(3.39)
because U, S and V are all additive. It should thus be realized that one could add Gibbs
energies of two subsystems at different T or P from tabulated values only by first breaking
them down into U, S and V values. Then one must decide on the relevant T and P for the
whole system. It would normally be T and P of the surroundings which appears natural if
only one subsystem has contact with the surroundings and the other one is an inclusion.
If each subsystem has its own surroundings, then there is probably no good reason to try
to compare or add their G values.
A particularly interesting case is found when a homogeneous system has different T or
P than the surroundings. One may then be interested in predicting the maximum amount
of work that can be extracted from the system when it moves towards equilibrium with
the surroundings. Since the surroundings are always regarded as a homogeneous, infinite
reservoir, its T and P are constant and a process under constant T and P should of course
be treated with the Gibbs energy function. Equation (3.27)gives, if there is no exchange
of matter,
dG =−SdT + V dP + µ
i
dN
i
− Ddξ =−Ddξ. (3.40)
52 Systems with variable composition
Foraspontaneous process Ddξ is positive and Ddξ/T is equal to the internal production
of entropy. If there were a mechanism by which one could extract another kind of work
than through a volume change, PdV, then it should have been included in W in the first
definition of the first law, Eqs (1.1) and (1.2), and it would have been considered all
through the derivations and appear in Eq. (3.40). Of course, extracted work should be
given with a minus sign. Equation (3.40)would thus have been modified
dG =−dW
extr
− Ddξ. (3.41)
Forareversible process one obtains
dW
extr
=−dG; W
extr
= G(initial) − G(final). (3.42)
This gives the maximum work that can be extracted. It is clear that G(final) is evaluated
for T and P of the surroundings. It is also evident that G(initial) must be given as
G(initial) = U(T
in
, P
in
) − TS(T
in
, P
in
) + PV(T
in
, P
in
), (3.43)
because any two extensive quantities can only be compared if the law of additivity applies.
The quantity W
extr
can be regarded as the part of the energy of the initial system
that is free to be transformed into useful work. That is why Gibbs energy was initially
called Gibbs free energy. If the surroundings are instead a reservoir of constant T and
V then one should repeat the derivation starting from the Helmholtz energy and that is
the reason why it was initially called Helmholtz free energy. Often one extracts work by
allowing the system to react with a chemical compound in the surroundings, usually O
2
used for burning a fuel. In that case, the appropriate free energy function would be found
by considering a reservoir with constant T, P and µ
O
,
d = d(G − N
O
µ
O
) =−SdT + V dP + µ
j
dN
j
− N
O
dµ
O
− dW
extr
− Ddξ.
(3.44)
Forareversible process under constant P, T, N
j
and µ
O
,wefind
W
extr
= G(initial) − G(final) − µ
O
[N
O
(initial) − N
O
(final)], (3.45)
where N
O
(final) is the total content of O after the system has received enough O
2
from
the surroundings to burn the fuel.
It is evident that, what has here been called free energy, must be defined in different
ways depending on the surroundings or on how the system reacts with the surroundings.
In mechanical engineering it is often called exergy.
3.5 Various forms of the combined law
In Section 3.3 the discussion was based on the energy scheme, which starts from the
combined law in the form
dU = T dS − PdV +µ
i
dN
i
− Ddξ, (3.46)
where all the independent variables are extensive ones. It defines the following set of
conjugate pairs of variables (T, S), (−P, V ) and (µ
i
, N
i
). However, there are many more
3.5 Various forms of the combined law 53
possibilities to express the combined law in terms of only extensive quantities as inde-
pendent variables. Using the new characteristic state functions, obtained in Section 3.3,
we can change variables in the combined law. For example, let us replace S by
(TS− PV)/T + PV/T , obtaining
dS = d[(TS− PV)/T ] + (P/T )dV + V d(P/T ). (3.47)
By inserting this expression we get
dU = T d[(TS− PV)/T ] + PdV + TVd(P/T ) − PdV +µ
i
dN
i
− Ddξ
= T d[(TS− PV)/T ] + TVd(P/T ) + µ
i
dN
i
− Ddξ. (3.48)
By subtracting (P/ T ) · TV (which is equal to PV) from U,wecan form a new charac-
teristic state function with only extensive quantities as independent variables
d(U − PV) = d[U − (P/T ) · TV]
= T d[(TS− PV)/T ] − (P/T )d(TV) +µ
i
dN
i
− Ddξ. (3.49)
This form of the combined law defines a new set of conjugate pairs, {T, [(TS−
PV)/T ]}, (−P/ T, TV) and (µ
i
, N
i
).
We may instead replace V by [( PV – TS)/ P + TS/P] and after some manipulations
we obtain a new characteristic state function with only extensive variables as independent
variables
d(U + TS) = d[U + (T /P) · PS]
=−Pd[(PV − TS)/P] + (T /P)d(PS) +µ
i
dN
i
− Ddξ. (3.50)
which yields a new set of conjugate pairs.
We may also rearrange the terms in the combined law before introducing new func-
tions. The entropy scheme uses
− dS =−(1/T )dU − (P/T )dV + (µ
i
/T )dN
i
− (D/T )dξ. (3.51)
It immediately defines a new set of conjugate pairs and two more alternatives are obtained
by replacing U or V in the way demonstrated above. One may also rearrange the terms
in the combined law as follows
dV = (T /P)dS − (1/P)dU + (µ
i
/P)dN
i
− (D/P)dξ. (3.52)
This may be called the volume scheme and it yields three more alternatives. We have
thus obtained the sets of conjugate pairs of variables given in Table 3.1.Ineach pair the
potential is given first and between them one can formulate a Gibbs–Duhem relation. In
each case the characteristic state function for the extensive variables is given to the left.
We may also define a number of content schemes by the following arrangement of
terms, but they will probably have very limited use and will not be discussed further.
− dN
j
= (T /µ
j
)dS − (1/µ
j
)dU − (P/µ
j
)dV +
k
(µ
k
/µ
j
)dN
k
− (D/µ
j
)dξ.
(3.53)
54 Systems with variable composition
Table 3.1 Some sets of conjugate pairs of state variables
From the energy scheme
U: T, S −P, V µ
i
, N
i
U − PV: T,(S −PV/T) −P/T, TV µ
i
, N
i
U + TS: T/P, PS −P,(V −TS/P) µ
i
, N
i
From the entropy scheme
−S: −1/T, U −P/T, V (µ
i
/T ), N
i
−S − PV/T: −1/T, H − P, V/T (µ
i
/T ), N
i
−S − U/T: −P/T, H/P −1/P, PU/T (µ
i
/T ), N
i
From the volume scheme
V: T/P, S −1/P, U (µ
i
/P), N
i
V −U/P: −1/T, TU/P −T/P, F/T (µ
i
/P), N
i
V + TS/P: T, S/P −1/P, F (µ
i
/P), N
i
Exercise 3.4
Suppose one would like to consider U as an independent variable. What would be its
conjugate potential?
Solution
From the entropy scheme we find −1/ T and from the volume scheme −1/P. Evidently,
the choice depends on what other conjugate pairs one would like to consider at the same
time.
Exercise 3.5
In Sections 9.1 and 10.7 we will find that the two axes in a phase diagram should be
taken from the same set of conjugate pairs. Suppose one would like to use U, F or G as
one of the axes in a unary system. How should the other axis be chosen?
Solution
From the entropy scheme we find that U could be combined with −P/T (which may
not be very practical) or V.From the volume scheme we find that U could be combined
with T/P (which again may not be very practical) or S.Wefind F in the volume scheme
only, and it can be combined with T or S/P (which is not very practical). We do not find
G in any scheme except in the form of µ
i
for a unary system.
3.6 Calculation of equilibrium
In calculations of equilibrium it is often assumed that the temperature and composition
can be kept constant. Instead of using the basic condition of equilibrium, d
ip
S = 0, it is