Title: Lecture 12: Phase diagrams
1Lecture 12 Phase diagrams
- PHYS 430/603 material
- Laszlo Takacs
- UMBC Department of Physics
2Phases and the phase rule
- Phase Uniform agglomeration of material - no
boundaries inside - Gibbs phase rule f n - p 2
- f degrees of freedom, i.e. number of freely
selectable thermodynamic parameters. Typically
temperature, pressure, concentrations. - n number of chemical components.
- p number of phases in equilibrium
- When phases are in equilibrium, atoms of any
component can be transferred between phases.
Equilibrium requires equal chemical potentials,
mathematically equations that reduce the number
of freely selectable parameters.
3The phase diagram of UF6 as a function of p and
T n 1, thus f 3 - p
4Pressure-temperature phase diagram of single
component systems sulfur and silicon dioxide.n
1, thus f 3 - p
5 The T-p phase diagram of FeUsually processes
are carried out at atmospheric pressure, thus we
will usually neglect the pressure dependence.
Extreme high pressure does cause phase changes.
6- n 2, thus f 4 - p
- The phase diagram of a two-component system is
usually shown for a fixed - ambient - pressure.
The remaining variables are temperature and the
concentration of either component. Notice - allotropic phases (Ti),
- solid solutions,
- compound phases,
- two-phase regions,
- transformations,
- etc.
7The lever rule on the amount of each phase in a
two-phase region
- Sample is 1 mole total, contains
- (1 - c) mole of component A
- c mole of component B
- We can also consider the sample as
- m? mole of phase ? with c? plus
- m? mole of phase ? with c?
- Total of component A in the sample
- 1 - c m? (1 - c?) m? (1 - c?)
- Total of component B in the sample
- c m? c? m? c?
- The sum of these two equations gives
- 1 m? m? as it should be.
- From the last two equations we get
- m? (c? - c) / (c? - c?)
- m? (c - c?) / (c? - c?)
- m? / m? (c? - c) / (c - c?)
?
?
c? c c?
c cB
A
B
Notice that c? is not the same as cA, even if
phase ? is a solid solution based on A.
8What determines the phase diagram?
- Equilibrium is given by the minimum of the Gibbs
potential, G. If more than one phase are
possible, the phase with the lowest G is the
equilibrium state. - G H - TS At low temperature enthalpy dominates.
- Entropy becomes increasingly important at
higher T. - Ideal solution Two chemically similar components
- G H - T (Sv SM) GM - TSM
- Assume
- GM changes linearly from A to B.
- Configurational entropy SM -Nk c ln c (1-c)
ln (1-c) - (same idea as with vacancies)
- The slope of SM goes to /- 8 when c --gt 0 or 1
9G(T,c) of an ideal solution
- GM is a simple weighted average of the Gibbs
potential of the components and ?Sm is the
configurational entropy. - Notice that the entropy term becomes more
important with increasing temperature, increasing
the curvature of the G(c) curve.
10- Comparing G for the ideal solution and the
similar G for the liquid phase allows the
construction of a hypothetical phase diagram. - Notice that S is lower for low, L is lower for
high temperature. - Notice the changing curvature.
- In intermediate states, neither L nor S gives the
smallest possible G. The state is a mixture of
the two common tangent gives the concentrations,
the lever rule gives the relative amounts.
11Why the common tangent?
- For an average concentration of c, we have x
fraction with c1 and (1 - x) with c2. - c x c1 (1 - x) c2, thus
- x (c2 - c) / (c2 - c1) 1 - x (c - c1) / (c2
- c1) - The Gibbs potential for the mixture
- G G1 x G2 (1 - x)
- G1 (c2 - c) / (c2 - c1) G2 (c - c1) / (c2
- c1) - This is a linear equation that describes a
straight line between the end points. We get the
lowest G when finding the lowest straight line
still having end points on the G(c) curve - that
is the common tangent.
G1
G2
c1 c c2
12The Ag-Au system is close to an ideal solution.
13Regular solutions
- Beside mixing entropy, differences in nearest
neighbor bond energy provide a variation of the
enthalpy term HAA, HBB, HAB. - Suppose N atoms (A and B together), z nearest
neighbors, c is concentration of B. - NAA 1/2 N (1-c) z(1-c) 1/2 (total A)
(mean A neighbor) - NBB 1/2 N c z c
- NAB N z c (1-c)
- Hm NAAHAA NBBHBB NABHAB 1/2 Nz (1-c)HAA
cHBB 2c(1-c)H0 - where H0 HAB - 1/2(HAA HBB)
- The last term in Hm has a maximum for H0 gt 0,
competes with the minimum caused by
configurational entropy. The slope of entropy is
/- 8 at the borders, the slope of Hm is finite. - Can result in limited solubility.
- G Hm - TSm
14- Gibbs free energy for a regular solution
- H0 lt 0 makes Gm sharply peaked - ordering,
compound formation - H0 gt 0 can result in two minima. For c1 lt c lt c2,
the lowest energy state is a mixture - phase
separation, limited mutual solubility. - If HAA HBB and c ltlt 1, the solubility limit
from dG/dc 0 is - c1 exp(-zH0/kT)