tangent with the two phases, a mixture of two phases with compositions given by the tangent points has a lower energy, the tangent line, than either of the two phases in equilibrium. Therefore, these two tangent points determine the composition range of a two-phase region at this temperature in Figure 1.16. For compositions to the left of the tangent point 1 on the α phase, the α phase has the lowest Gibbs energy and so this tangent point determines the maximum composition of single phase alpha on the phase diagram in Figure 1.16.
At T3, Figure 1.18c, two tangent lines can be drawn: one intersecting the alpha and liquid phases at points 1 and 2, and one intersecting the liquid and beta phases at points 3 and 4. As at T2, G* is lower for the two-phase mixtures of alpha + liquid (points 1 and 2) and liquid + beta (points 3 and 4) between these common tangent points on the tangent lines. Again, the extensions of the tangent lines to the vertical axes give the activities (actually RT ln a) of A and B for the compositions of the three phases at the tangent points. Again, to the left of the α + liquid, two-phase region (point 1) the single phase α has the lowest energy and is the stable phase. Similarly, for compositions to the right of point 4, tangent to the beta phase, beta has the lowest energy and is the stable phase.
At T4, a single tangent line intersects all three phases at points 1, 2, and 3. This is the invariant eutectic temperature where the three phases are in equilibrium so all three must have the same G and G*A
B*
where the tangent line intersects the left and right vertical axes, respectively. To the left of point 1, alpha has the lowest Gibbs energy and is the stable phase while to the right of point 3, beta has the lowest energy and is the stable phase.
Finally, at T5, both solid phases are lower in energy than the liquid as given by the tangent line between them intersecting at points 1 and 2. Between points 1 and 2, the tangent has the lowest energy so these two points give the limits of the two-phase α + β region. To the left of point 1, α has the lowest energy and is the stable phase while to the right of point 2, β has the lowest Gibbs energy and is the stable phase.
Conversely, the phase diagram shown in Figure 1.16 could be constructed from the Gibbs energy- composition diagrams in Figure 1.18. This example demonstrates the very close relationship between Gibbs energy-composition diagrams and phase diagrams. It is possible to create either from the other.
Understanding this relationship is invaluable in being able to use phase diagrams to initiate develop- ment of kinetic models for the rates of reactions and phase transitions.
in his England, the unit of heat was the British thermal unit, the amount of heat necessary to raise 1 lb of water 1°F! Today, the energy unit is a joule, so the goal is to convert Joule’s measurement of the energy equivalent of heat into joules and compare that to a calorie. His best measurement of the equivalence of work and heat was 722.55 ft-lb/BTU. There are 2.2046 lb/kg. Therefore, carrying out the conversion step by step:
1 1 12 2 54 1
100 0 3048
ft ft in ft cm in
cm m m
= ×
(
.)
× .(
.)
×( )
= .1 1 1
2 2046 0 4525 lb lb
lb kg kg
= × =
. . .
Therefore,
1ft lb⋅ =0 3048. m×0 4525. kg×9 8. m s2=1 35164. J
where 9.8 m/s2 is the acceleration of gravity necessary to change mass into a force. So 772.55 ft-lb are 772 55. ft lb⋅ ×1 35164. J ft lb⋅ =1 04421. J×103
is the amount of heat to raise the temperature of 1 lb of water 1°F. Therefore, the number of joules required to heat 1 g of water 1°C is
1 04421 10 2 2046 1
1000
9
5 4 144
. × 3
(
J lb F⋅°)
× .(
lb kg)
×(
g kg)
×(
° °F C)
= .(
JJ g C⋅°)
=1 calorieSince the amount of heat necessary to raise 1 g of water depends on the starting temperature, the IUPAC has determined, based on more recent measurements, that the value of the thermodynamic calorie is
1 calorie = 4.184J.
So James Prescott Joule’s measurement in his laboratory, almost 150 years ago, was off by only about 1% from the currently accepted value.
A.2 S
tirlinG’
SA
PProximAtion A.2.1 Plausibility Argument Stirling’s approximation isln
( )
N! ≅NlnN N− (A.1)which is made plausible from the following very non rigorous approach:
ln N N ln ln ln N n
N
! ln
( )
= ln(
1 2 3 4× × × ×)
= 1 + 2 + =∑
1
(A.2)
and approximating the sum with an integral (here N and n are just integers and dummy variables)
ln ln ln
ln ln
n n dn n n n
N N N N N N
N N N
≅ = −
= − + ≅ −
∑ ∫
11 1
1
(A.3)
ln( !)N ≅NlnN N− (A.4) Substituting the integral for the sum certainly seems valid if N is large. The integral is evaluated by integration by parts where the value of the integral at the lower limit is just 1 which can be ignored compared to any reasonably large value of N.
A.2.2 How Good Is Stirling’s Approximation?
For example, even for N = 100, N! = 9.3 × 10157 so ln N! = 363.7 while N ln N − N = 100 × ln 100 − 100 = 360.5 or a difference of about 0.9%. Therefore, for the large numbers involved in sol- ids, N≅1022cc−1 Stirling’s approximation is certainly good enough!
A.3 S
tAtiSticAlA
PProAcHtotHee
ntroPyofm
ixinGA.3.1 Molar Entropy of Mixing of Two Crystalline Elements
The entropy of mixing of an ideal solution can be modeled from Boltzmann’s equation for the entropy S k= BlnW
where W, in this case, is the number of ways a total number of N atoms consisting of a atoms, Na, and Nb b atoms can be distributed among N lattice sites, where N N= a+Nb. Therefore,
W N
N Na b
= !
! ! (A.5)
is the number of ways that the a and b atoms can be arranged over the N total sites. Taking the natu- ral logarithm of both sides of Equation A.5
lnW=ln ! lnN − Na! ln− Nb!.
And with the application of Stirling’s approximation this becomes,
ln ln ln ln
ln ln ln
W N N N N N N N N N
W N N
N N N
N
a a a b b b
a a
b b
= − − + − +
= −
−
and if N = NA, Avogadro’s number, one mole of atoms, then
lnW N N ln ln
N
N
N N N
N
N
A a N
A
a A
A b
A
= − b
−
so the entropy becomes,
S k N X X X X
S R X X X X
mix id B A a a a a
mix id a a a a
,
,
ln ln
ln ln
= −
(
+)
= −
(
+)
(A.6)where Xa and Xb are the mole fractions of a and b, respectively. This is the same as Equation 1.47 in the text obtained through a strictly thermodynamic path. For a 50–50 solution, Smix id, =
−Rln .0 5= −5 762. J mol K⋅ .
A.3.2 Molar Entropy of Mixing of Two Compounds
NiO and MgO form almost an ideal solid solution. Therefore, the ideal molar entropy of mixing of these two compounds is also given by Equation A.6 since the two cations are distributed over a
total of NA lattice sites. However, what is not always appreciated is that the molar entropy of mixing is different if the two compounds are Al2O3 and Cr2O3 that also form an almost ideal solid solution.
In this case, the total number of sites that the aluminum and chromium ions are distributed over is not NA but 2NA since there are two cations for each mole of the two constituent compounds.
Therefore, now 2NA=NAl+NCr and
W N
N N
A Al Cr
=
(
2)
!! ! (A.7)
and taking the natural logarithm of both sides of the equation lnW=ln
(
2NA!)
−ln(
NAl!)
−ln(
NCr!)
and applying Stirling’s approximation
lnW≅2NAln
(
2NA)
−2NA−NAllnNAl+NAl−NCrlnNCr+NCrlnW≅(NAl+NCr) ln
(
2NA)
−NAllnNAl−NCrlnNCrlnW N ln N n
N N l N
Al Al N
A
Cr Al
A
≅ −
−
2 2
lnW N N ln ln
N X N N
N X
Al A
A
Al O Cr A
A
≅ − Cr O
−
2
2
2
2 3 2 2 3
lnW≅ −2NA
(
XAl O2 3lnXAl O2 3+XCr O2 3lnXCr O2 3)
. Therefore, the molar entropy of mixing isSmix id, = −2R X
(
Al O2 3lnXAl O2 3+XCr O2 3lnXCr O2 3)
(A.8) just twice that for the molar entropy of mixing for a solution of elements. So, for a 50–50 solid or liquid solution, Smix id, = −2Rln .0 5= −11 526. J mol K⋅ . Obviously, if the solution consisted of Al2O3 and Cr2S3—but not very likely—then there are 5 NA sites over which the cations and anions can be distributed and the molar entropy would beSmix id, = −5R X