start lecture 03
10/3/12
2-1
Predominance diagrams display domains of predominance of chemical compounds (in this example, copper compounds in an atmosphere containing sulfur and oxygen) as a function of chemistry of the gas atmosphere of the system.
Sket
Nucleation: an illustration of differences in scale
10/1/13
2-1
Last lecture
10/1/13
2-2
Predominance diagrams display domains of predominance of chemical compounds (in this
example, copper compounds in an atmosphere containing sulfur and oxygen) as a fu
MSE321 Thermodynamics in Materials
11/25/2013
Example Problem 7.1
Given the following expression for the internal energy of a system:
a. Calculate the three corresponding equations of state
b. Show that they are zero order homogeneous (T, P, are intensive
! nR $
V =#
&T
" P %
V
Ideal Gas
" V %
" nR %
$
' =$
'
# T &P # P &
! 1 $ ! V $
! P $ ! nR $
= # &#
& =#
&#
&
" V % " T %P
" nRT % " P %
V
V =
V
09/19/2001
( nRT ) P
-1
1
=
T
" V %
2
$
' = ( nRT ) P
# P &T
1 V
P nRT
=
=
2
P
V P T
nRT
1
V =
Chemical Potential of
Single Component
Phases
Chemical Potential - Open Systems
Combined 1st & 2nd law with extensive functions:
dU! = TdS! - PdV! + dn
By definition:
(
U /
=
Extensive enthalpy:
n
)
S,V
dH" = dU" + PdV" + V'dP
Substitute for dU/ in enthal
Chemical Potential of
Single Component
Phases
Chemical Potential - Open Systems
Combined 1st & 2nd law with extensive functions:
dU! = TdS! - PdV! + dn
By definition:
(
U /
=
Extensive enthalpy:
n
)
S,V
dH" = dU" + PdV" + V'dP
Substitute for dU/ in enthal
Characteristics of
Phase Diagrams
Experimentally determined PT phase diagram for pure bismuth (a) is
compared with the computed diagram in which the temperature and
pressure dependence on the parameters is neglected (b).
09/19/2001
7-2
, solid
crystal
+L
11/17/13
2-1
11/17/13
2-2
Sketches of the surfaces representing
the chenical potential as a function of
temperature and pressure for the solid
phase (a), the liquid phase (b) and superimposed in (c)
11/17/13
2-3
Superposition of chemical
potential surface
Find Conditions for Equilibrium
Constrained Extremum
Write a differential expression for the change in
entropy that the system may experience. Include
all possible independent variables.
Write differential equations that describe the
constraints for an
MSE 321 : Homework 1
Due : Monday, October 7, 2013
1. Do problem 2.5 in the text.
2. Do problem 2.6 in the text.
3. A binary regular solution is dened by the molar Gibbs free energy (free energy
per mole of solution):
Gm = XA Go + XB Go + RT [XA ln XA + X
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Problem Set 2 Solutions
1. Prove that the inexact differential, d w is equal to mechanical work, pdV:
i.e, d w = pdV
when the change in all other work i
Reading: Chapters: 1, 2, 3, 4, 5, 7 and 8.1 and 8.2 but excluding 8.2.1, and defini>ons of chemical poten>als in 8.4 Problems: All homework and exam problems plus examples covered in the reading.
Why is thermodynamics important? It applies everywhere alw
Volume Relations
to Temperature & Pressure
V = V (T, P )
V
V
dV =
dT +
dP
T P
P T
dV = VdT VdP
09/19/2001
Notes from R.T. DeHoff, Thermodynamics in Materials Science (McGraw-Hill, 1993)
4-1
Entropy Relations
to Temperature & Pressure
S = S (T ,
The Laws of Thermodynamics
0th Law - There is a temperature scale and energy (heat) flows
down hill on that scale.
1st Law - There exists a property of the universe, called its
energy, which cannot change no matter what processes occur.
Energy is conserve
2nd Law - There exists a property of the
universe, called its entropy, which always
changes in the same direction no matter
what processes occur. Entropy is not
conserved. Entropy always changes in one
direction (increases).
Entropy - a State Function
Ent
By definition: ST = SI + SII + SIII + SIV = 0
Experimental observation: ST = SI + SII + SIII = 0
Therefore: SIV = 0
for all materials the entropy is the same at 0K.
and the value is set at 0.
Moreover:
SII = ( SI + SIII )
The Third Law
There is a lower
Quasistatic/Reversible Processes
Recall that we showed that the work performed cannot be a state function
because:
cdw = c PdV 0
The differential work dw is sometimes called not a perfect differential because
of this property. It simply means that you nee
Some mathematics needed for thermodynamics
(All of which you have seen before in Math classes here)
The derivative of f(x) with respect to x is:
Rate of Change: an interpretation of a derivative is rate of change.
If f(x) represents a quantity at any x th
MSE 321 : Thermodynamics and Phase Equilibrium
General Course Information
Fall Quarter 2013
Lecture:
Time: MWF 9:30 - 10:20
Location: Mue 153
Sec AA (Tuesday) / Sec AB (Thursday):
Time: Tu 1:30 - 2:20
Location: Mue 154
Instructor : Professor Lucien N. Bru
In Chapter 4,
a unary, closed, homogeneous, non-reacting system
is considered.
In such a system in equilibrium with itself and its
surroundings, the state is completely specified by
two variables, for example P and T.
Then all other state properties follo
Last lecture .
10/3/13
2-1
We have discussed state variables that gives us information about a
body in an efficient way. Also there are:
Intensive Variables
P Pressure
T Temperature
Extensive Variables
N Number of Molecules/Atoms
V Volume
S Entropy
U Inte
J. Willard Gibbs
1839-1903,
The Architect of
modern equilibrium
thermodynamics.
10/31/13
2-1
Extremum in
Isolated System
Criterion for
Equilibrium
10/31/13
2-2
Extremum Principle for Entropy
For any real process that occurs within an isolated
system, the
Extensive Quantities of the State Functions: F/, G/, H/, S/, U/, V/
Using G/ as example:
G = G (T,P,n1,n 2 .n k .n c )
A differential form of G/:
and at constant T & P:
G / = dn k k=1 n k T ,P,n j k
c
(dG )
/
T ,P
8-1
Partial Molal Quantities of the Stat
G - T Diagram - , L, V
V
G (T) - GO (J/mole)
P (triple pt.)
L
Triple Point: a+L+V V T (K)
7-1
L
G - T Diagram - , L, V
V
G (T) - GO (J/mole)
P > P (triple pt.)
L T (K)
Melting Point: a+L
L
Boiling Point: L+V L V
7-2
G - T Diagram - , L, V
V
G (T) - GO (J
Quasistatic/Reversible Processes
Recall that we showed that the work performed cannot be a state function because:
cdw = - c PdV 0
The differential work dw is sometimes called "not a perfect differential" because of this property. It simply means that you
2nd Law - There exists a property of the universe, called its entropy, which always changes in the same direction no matter what processes occur. Entropy is not conserved. Entropy always changes in one direction (increases). 3rd Law - There is an absolute
In this lecture and in other lectures, some overheads include material adopted from the notes of Professor Craig Carter of the Department of Materials Science at MIT. Inclusion of this material supplements the material provided in the text, and provides y