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
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
A theory is more impressive the greater the simplicity of its
premises, the more different of things it relates, and the more
extended is its area of applicability. Therefore, the deep impression
which classical thermodynamics made upon me. It is the only
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
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 ,
! 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
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
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
Question from last time:
How does equilibrium differ from a steady-state?
1
Initial State
Insulator
Fix @
T2
Conductor
T ( X ,0) = TO ;
Fix @
T1
0< X < L
Temperature (T)
T2
TO
T1
Distance (X)
5-24
2-24
Transient State
Insulator
Fix @
T2
Conductor
T = f (
Enthalpy & Entropy of
Transformation
At equilibrium -
=
H TS = H TS
Thus,
H / = T / S /
S / L
H / L
=
TM
and
and
G = G
H H = T (S S )
/
H
S / = /
T
L /V
H
S L /V =
TV
Notes from R.T. DeHoff, Thermodynamics in Materials Science (McGraw-Hill,
H(T)-H(T
Reading - beginning Chapter 5
1
4.6 Compute the U when 12 liters of Ar at 273 K
and 1 atm are compressed to 6 liters with final
pressure = 10 atm. (a.) Find U=U(P,V) & integrate.
U = U ( P ,V )
dU = MdP + NdV
dU = MdP + N (VdT VdP )
dU = NVdT + (M NV )dP
o
1. The absolute value of the entropy at 298K of Al is SAl
= 28.3 J/(mole K), of O2 is SOo 2
o
= 205.03 J/(mole K) and of SAl
= 51.1 J/(mole K). A stoichiometrically balanced
2 O3
statement of the reaction is written
xAl + yO2 = zAl2 O3
where x, y and z
Definition of Symbols & Terms
10/26/14
2-33
Application of the Extremum Condition
with Constraints to Entropy for a
Unary, Two-Phase System
Notes from R.T. DeHoff, Thermodynamics in Materials Science (McGraw-Hill, 1993)
5-38
2-38
Extensive Properties, Z
M
The Thermodynamic State Function: Enthalpy
For an simple pure fluid, consider the physical meaning of the -PV term alone;
Question: What are the units of PV ? What are the units of P?
Imagine that there is a completely homogeneous system that has been arb
Sketches of the surfaces representing
the chemical potential as a function of
temperature and pressure for the solid
phase (a), the liquid phase (b) and superimposed in (c)
2-3
Plot G vs T at Constant P
Integrate dG at constant pressure from TO to T.
d =
Entropy a Maximum at Equilibrium.
Entropy production is always positive.
In an isolated system with no entropy transfer,
the change in entropy is limited to entropy
production.
In an isolated system, the entropy can only
increase.
As the system evolves to
MSE 321 : Homework 1
Due : Monday, October 12, 2015
1. Do problem 2.5 in the text.
2. Do problem 2.6 in the text.
3. The molar Gibbs free energy of mixing (free energy per mole of solution) in a
binary solution is defined as:
Gmixing
= RT [XA ln XA + XB l
MSE 321 : Homework 3
Due : Wednesday, October 28, 2015
1. Please do problem 4.1 in the text
2. Please do problem 4.3 in the text
3. Please do problem 4.4 in the text
4. Please do problem 4.7 in the text
5. Please do problem 4.9 in the text
MSE 321 : Homework 4
Due : Friday, November 6, 2015
1. One mole of an ideal gas is expanded adiabatically and reversibly from 300 K
to 560 K. Compute the change in Helmholtz free energy. Hint: Consider the
variables F, S, and T.
2. Please do problem 4.13
MSE 321 : Homework 5
Due : Friday, November 20, 2015
1. Please do problem 7.1 in the text.
2. Please do problem 7.2 in the text.
3. Please do problem 7.5 in the text.
4. Please do problem 7.6 in the text.
MSE 321 : Homework 2
Due : Monday, October 19, 2015
1. Please do problem 3.2 in the text
2. Please do problem 3.7 in the text
3. Please do problem 3.8 in the text
4. Please do problem 3.9 in the text
5. Please do problem 3.11 in the text
6. Please do prob