Thermodynamic data
A tutorial course
Session 2: unary data (part 2)
Alan Dinsdale
“Thermochemistry of Materials” SRC
Review
• The aim was to give you some experience in
handling data
– Conversion of data from one format to another
– Calculation of unary phase diagram
– Estimation of lattice stabilities
Heat capacity data for KCl
Deltaf H
S298
-436684.1
a
82.555
bT
c T^2
50.47661 0.005924377
Upper Temperature
Limit
Enthalpy change
dT^(-2)
7.49668E-06
-144173.9
700
0
143.5698
-0.1680399
9.9657E-05
-8217836
1044
26283.9
73.59656
0
0
0
2000
0
• Convert these data to G-Hser
• Hint : Start with the low temperature range. Use the equations for Cp
given in slide 11 and equate them to the coefficients in the above
table. Then include ΔfH298 and S298
• Then go to the second temperature range. Adjust the first and second
coefficient to make sure that H and S are the same as line 1 at the
temperature limit
• Then go to the third range but this time include also the enthalpy and
temperature of transition
The answers
A
BT
C T ln(T)
d T^2
e T^3
Upper
temperature
limit
f T^(-1)
-4.525468E+05 2.584275E+02 -5.047661E+01 -2.962189E-03 -1.249447E-06 7.208695E+04
700
-4.971616E+05 8.704243E+02 -1.435698E+02 8.401995E-02 -1.660950E-05 4.108918E+06
1044
-4.437308E+05 4.062578E+02 -7.359656E+01
2000
0
0
0
Phase diagram for KCl
• Calculate the phase diagram for KCl between 300 to 1800 K, 0 to 3
GPa
• We have three phases: HALITE, LIQUID and CSCL
• We can take the HALITE phase as the reference phase. The data for
the other phases will be relative to that and will take the form
A+BT+CP
• HALITE:
G=0
• LIQUID:
G = 26284-25.17624521*T+3.69441e-6*P
• CSCL:
G = 3207.841581-0.078831938*T-1.592096e-6*P
• what value do you calculate for the triple point when all three
phases are in equilibrium ?
• Hint: The boundary between HALITE and the LIQUID and CSCL
phases is given by setting the appropriate Gibbs energy expression
to zero and then expressing T as a function of P
Lattice stability data for bct Ga
• Bct_a5 is the phase labelled as II on the next slide
• Derive lattice stability data for this bct_a5 phase
• Hint: extrapolate the phase boundaries between
phase II and liquid, and phase II and phase I
(ortho_Ga). These temperatures can be marked
on the next diagram and a straight line can drawn
between them to show how the Gibbs energy of
this bct_a5 phase varies with temperature
Phase diagram for Ga
Gibbs energy differences for Ga
This session will be concerned with:
• Unary data (part II)
– Magnetic model
– Pressure dependence
– Lattice stabilities
– Unstable phases
Magnetics
• Model we generally use for magnetic contribution to
thermodynamic properties from Hillert and Jarl based
on work by Inden
𝐺𝑚𝑎𝑔 = 𝑅 𝑇 ln 𝐵0 + 1 𝑔 𝜏
• Where 𝜏 = 𝑇 𝑇 ∗
• T* the critical temperature
– TC, the Curie temperature for ferromagnetic materials
– TN, the Neel temperature for antiferromagnetic materials
• B0 is the average magnetic moment per atom
Magnetic materials
• Bcc Fe (TC = 1043 K) – therefore all ferritic
steels
• Fcc Ni (TC = 633 K) – therefore most austenitic
steels and superalloys
• Co (hcp, fcc), Mn (cbcc, fcc, bcc), Dy (hcp), Gd
(hcp), Fe3O4
𝐺𝑚𝑎𝑔 = 𝑅 𝑇 ln 𝐵0 + 1 𝑔 𝜏
79𝜏 −1 474 1
𝑔 𝜏 =1−
+
−1
140 𝑝 497 𝑝
𝜏3
𝜏9
𝜏 15
+
+
6 135 600
𝜏 −5 𝜏 −15 𝜏 −25
𝑔 𝜏 =−
+
+
/𝐷
10
315 1500
/𝐷
𝜏≤1
𝜏>1
518 11692 1
𝐷=
+
−1
1125 15975 𝑝
• p can be thought of as the fraction of the magnetic
enthalpy absorbed above the critical temperature
– Usually taken to be 0.4 for bcc_a2 and 0.28 for other common
phases
Entropy contribution
𝑆𝑚𝑎𝑔 = −𝑅 ln 𝐵0 + 1 𝑓 𝜏
474 1
𝑓 𝜏 =1−
−1
497 𝑝
2𝜏 3 2𝜏 9 2𝜏 15
+
+
3
27
75
2𝜏 −5 2𝜏 −15 2𝜏 −25
𝑓 𝜏 =
+
+
/𝐷
5
45
125
/𝐷
𝜏≤1
τ>1
Enthalpy contribution
𝐻𝑚𝑎𝑔 = 𝑅 𝑇 ln 𝐵0 + 1 ℎ 𝜏
where
79𝜏 −1 474 1
ℎ 𝜏 = −
+
−1
140𝑝 497 𝑝
𝜏 3 𝜏 9 𝜏 15
+
+
2 15 40
𝜏 −5 𝜏 −15 𝜏 −25
ℎ 𝜏 =−
+
+
/𝐷
2
21
60
/𝐷
𝜏>1
𝜏≤1
Heat capacity contribution
𝐶𝑝𝑚𝑎𝑔 = 𝑅 ln 𝐵0 + 1 𝑐 𝜏
where
474 1
𝑐 𝜏 =
−1
497 𝑝
9
15
2𝜏
2𝜏
2𝜏 3 +
+
3
5
−15
−25
2𝜏
2𝜏
𝑐 𝜏 = 2𝜏 −5 +
+
/𝐷
3
5
/𝐷 𝜏 ≤ 1
𝜏>1
Heat capacity of bcc Fe
Gibbs energy differences for Fe
Enthalpy differences for Fe: ref fcc
Pressure dependence
• Our basic Equation of State
𝐺 =𝐻 −𝑇𝑆+𝑃𝑉
• The PV term can be quite complicated as the
volume will vary both with temperature and
pressure
• For relatively modest pressures the volume
can be represented by
𝑎1 𝑇 2 𝑎2 𝑇 3
𝑉 = 𝐴 1 + 𝑎0 𝑇 +
+
+ 𝑎3 𝑇 −1
2
3
• And the compressibility by
𝜅 = 𝐾0 + 𝐾1 𝑇 + 𝐾2 𝑇 2
• One way of combining these is in terms of the so
called Murnaghan model
𝐺𝑝𝑟𝑒𝑠
𝑎1 𝑇 2 𝑎2 𝑇 3
𝐴 𝑒𝑥𝑝 𝑎0 𝑇 + 2 + 3 + 𝑎3 𝑇 −1
=
𝐾0 + 𝐾1 𝑇 + 𝐾2 𝑇 2 𝑛 − 1
(1 + 𝑛𝑃 𝐾0 + 𝐾1 𝑇 + 𝐾2 𝑇 2 )1−1/𝑛 − 1
T-P phase diagram for Fe
Lattice stabilities: Difference in Gibbs
energies between phases
• Often we are interested in the thermodynamic
properties of phases outside the region where
they are stable
– Eutectics – we would like to have data for liquid
Ag and Cu at 780°C – well below the melting
points of the elements
– Solubility – in order to model the Fe-Cr phase
diagram we need data for Cr in austenite (fcc). The
only stable solid phase of Cr is bcc. Therefore we
need to derive data for fcc Cr
Extrapolation of P-T phase diagrams
Extrapolation of binary phase diagrams
Cr-Ni
Cr-Ni thermodynamic properties
Trends in entropies of fusion: fcc
Trends in entropies of fusion: hcp
Trends in entropies of fusion: bcc
Trends across transition metal series
Trends across transition metal series
Trends across transition metal series
Trends down a series in the periodic table
Use of ab initio calculations
• The present set of element data were derived
largely from experimental data. Ab initio
calculations were not thought to be sufficiently
accurate in 1991
• Now we have much more confidence in ab initio
calculations. They can be used for calculation of
enthalpy differences between phases and
entropy of phases at 25°C
• However ab initio has also told us that some
phases of elements are mechanically unstable eg.
Fcc Cr, fcc W, even bcc Ti at low temperatures
• This calls into question the very basis behind
lattice stabilities
Next session
• This will be concerned mainly with the liquid phase
• Exercise:
• The data for fcc and liquid Ag are given by:
-7209.512 + 118.202013 T - 23.8463314 T ln(T) - 1.790585E-3 T2 - 0.398587E-6 T3 - 12011 T-1
-3587.111 + 180.964656 T - 33.472 T ln(T)
(fcc)
(Liquid)
• Calculate the melting point, the heat capacity and the
difference in Gibbs energy and entropy between the two
phases between 50 K and 5000 K assuming that it is valid
to extrapolate the data. Can you explain the odd
behaviour ? What should the curves look like ?