PHYSICS POLYMER - Spring 2007
EMA 6165 - EXAM 2
STUDENT UFID#:_________
___________
(UF ID No.)
INSTRUCTIONS: Read the entire exam carefully. Answer every question to the
best of your ability, based upon the available information. State
all assumptions used to answer/solve each exam question.
SECTION I: Polymer Structures/Properties.
State the polymerization mechanism, i.e., condensation, addition, ring opening.
State the relative Tg and/or Tm for each polymer (24pts)
Polymer
1. poly(propylene)
2. poly(arylene
ethylene imide)
3. Nylon 6,6
4. poly(phenyl
terepthalamide)
(Kevlar)
5. poly(ethylene
terephthlate)
6. poly(bisphenol A
carbonate) (Lexan)
Polymerization
Mechanism (2pt)
Tg
(1pt)
Tm
(1pt)
POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
Concepts. Answer the following statements with TRUE or FALSE. If the statement is
false, state the reason(s). If you believe a statement can be made TRUE by a simple
qualifier, then state it clearly. (Normally only a single word change is required.) (12 pts)
7.
Spinodal decomposition represents nucleation and growth kinetics of phase segregation.
8.
The osmotic pressure of the dissolved polymer chain scales inversely with the
hydrodynamic volume of the chain.
9.
The second virial coefficient equals zero at a chi value of 3/2.
10. The chi parameter is temperature independent.
11. The chain radius of gyration increased with evaporation of the solvent from the SEBS
triblock copolymer spread on a surface (Wang et al, Langmuir 2007 hmwk assignment)
12. The restorative force of an ideal elastomer scales with the extension (length) of the
polymer (Helmhotz)
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POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
II.
Graphical Relationships: You must provide the appropriate equations for answering each
question. You must label axes appropriately and provide appropriate units and orders of magnitude.
STATE ALL ASSUMPTIONS required. SHOW ALL Equations required. It is preferred that the
relationships are linearized.
13. Chi parameter versus molar mass (state assumptions and show all equations)
(4pts)
14. Chi parameter versus chain concentration (state assumptions clearly)
(4pts)
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POLYMER PHYSICS - Spring 2007
15. Chain characteristic ratio versus chain molar mass
EMA 6165 - EXAM 2
(4pts)
16. Sketch the Number average mean square end to end distance versus the Weight (4pts)
average mean square end to end distance of chains in polymer fractions separated
by GPC (Assume a PDI = 2.0)
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POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
SECTION II: Definitions/Concepts:
Give a BRIEF description for each of the following terms in the SPACE provided. 16 pts,
Use equations or words or figures.
17. UCST:
18. Chemical potential:
19. Spinodal decomposition:
20. Chain expansion factor:
21. Cohesive Energy Density:
22. SEBS:
23. Scattering vector (Hashimoto Paper):
24. Binodal Decomposition :
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POLYMER PHYSICS - Spring 2007
25.
SECTION III: Engineering Problems
CLEARLY state any and all assumptions applied developing your the solution.
Given the following information, calculate the influence of hindered rotation on the
chain dimensions for the polymer given in problem 10. (10 pts)
Polymer: Poly(ethylene-blockpropylene)
Mn = 420 kg/mol
cis
trans
- gauche
-eclipse
+ gauche
+eclipse
**
EMA 6165 - EXAM 2
Rotational Angle
0/360
180
90
60
270
240
Probability
for
Ethylene
0.13
0.2
0.26
0.02
0.34
0.05
16 pts
Probability
for
Propylene
0.18
0.15
0.22
0.07
0.28
0.1
The values provided are for purposes of this exam only and not to be construed as accurate.
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POLYMER PHYSICS - Spring 2007
26.
EMA 6165 - EXAM 2
Review the article by Brown and Russell (Macromolecules 1996, 29, 798-800) and the
one by Zhang et al (J. Applied Polymer Science, Vol. 96, 961-965 (2005).
Compare the calculated Me (entanglement molar mass) of a linear poly(styrene-altermethyl methacrylate) versus that a four armed, star polymer of the same composition
and molar mass. You must review both papers to determine the effect of the stars on
the radius of gyration for the polymer. State all assumptions clearly. This is due in my
secretarys (317 office Materials Engineering) by 5 PM on Friday, April 6, 2007 or as
an electronic submission at the same time.
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POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
HONOR CODE: We, the members of the University of Florida community, pledge to hold ourselves and our peers to
the highest standards of honesty and integrity.
Print Name: ___________________ (Last) ___________________ (First)
Signature:___________________________________________________________
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POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
Rubber Elasticity:
= NV kT ( 2 )
=
=
Thermodynamics of Mixing
2
RT
2 MC
1
( 2 )
MC Mo
RT
MC
H m = N 1 2 1, 2 k T
1
3
( )
2
(
0 = N v kT x x 3y 2
x
0x =
0x =
E 1 / 2 E 1 / 2
H m = Vm 1 + 2 (1 2 )
V
V1
2
RT
Mc
1/ 3 x
)
1
x 2
RT
Mc
1
x 2
x
f * 2
=
N V RT( 2 )
*
f
1
2
G mix = RT ln (1 2 ) + 1 2 + 1 2
x
1, 2 = B1 +
C
E
S
T
f =
l
l
T ,V
T ,V
f
E
f =
+ T
T
l
T ,V
l ,V
2 (1 2 ) C
2
1
1
=
1+
1
2
X2
RT
V1
Nk
S = c [
2
2
( x + y 2 + z 2 3 ) + (ln( x y z )) ]
2
C
x (t ) =
2
C
1
2
ln(1 2 ) + 1 x 2 + 12
t
a
1
c = 1
2
T
Crystallization & Kinetics
4d N kT
1
P = 0 v3 * 7
R0
1
2G
1
=
+ 1 + 2
22
(1 2 )C X
2
=
f * 2
ri 2
=
N RT 2 ( 2 )
r0
f* v
1
=
C
V1
( 1 2 )2
RT
1 1 = R V u
T
T o H V 1
f
f
f 1
T
f
= h
Tm o
Energy
f
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POLYMER PHYSICS - Spring 2007
EMA 6165 - EXAM 2
v = 2b l + 2v f ab e
General Thermodynamics
GI = bo ao nsi
Q=
G = Go expQ * RT *exp K * RT
I
D
gI
G = b S g aN
I T
0.5
n 1
Tm
T
n
n 1
)
Physical Aging Behavior/Viscoelasticity
(B
f o )(T To )
( f o f ) + (T To )
(B f o )(T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) ( f o f ) + (T To )
8.86 (T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) 101.6 + (T To )
ln1 (t1 , T1 ) 17.4 (T To )
=
ln 2 (t 2 , T2 ) 51.6 + (T To )
= Ae
f = f o + T To
f
1 V
V P l ,T
Chain Dimensions/Molecular Weight
(t ) = 1 exp( kt
Vo
B
V
f
1 V
V T P ,l
P=
1 2g
SK =
=
Nk i
ln At =
=
05
.
Gn
C Tm
=
kTC TC TC
m1 mo 1
mo s
N M
N
Mn =
i
i
i
2
Kc
1
1 16 2 r
sin 2 + 2 A2 c
=
+
2
R M W M W 3
2
6
n
p( r , n ) = 2 l 2
3
n
p ( x , y , z , n) = 2 l 2
3
2
g
r
3
2
( )
4 r 2 e
(
2
e
3 x 2 + y 2 + z 2
2 nl 2
3 r 2
2 nl 2
dr
)
dxdydz
ro2
nl 2
=
=
6
6
r 2 = nl 2
r2
3
= nl
2
1 + cos(180 )
1 cos(180 )
1 + cos (180 ) 1 + cos
1 cos (180 ) 1 cos
a
= KM V
[ ] = 2 .5 4
3
2
R eo
M
3
2
M
1
2
3
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POLYMER PHYSICS - Spring 2007
r2
[ ] =
M
3
2
M
1
2
EMA 6165 - EXAM 2
ln1 (t1 , T1 ) 17.4 (T To )
=
ln 2 (t 2 , T2 ) 51.6 + (T To )
3
[ ] =
[ o ]
= Ae
5 3 = 2Cm 1 1 M 2
T
1
De =
f = f o + T To
f
=
K
M
(t ) o e i
=
G=
(t ) o
*
= n
=
Vo
B
V
f
P R 4
8LQ
= A e E / RT
= A eV/V
0
Gf
Go
= 1 + 2.5 + 14.1 2
(t ) = o cos t
w = 0.5 max min
Viscoelasticity and Mechanics
EYoung 's = 2 G Shear (1 + ) = 3 B (1 2 )
ln At =
(B
f o )(T To )
( f o f ) + (T To )
(B f o )(T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) ( f o f ) + (T To )
8.86 (T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) 101.6 + (T To )
Fluid Dynamics (Rheology)
R 4 P
Q=
8 L
Wh 2 h
P
Q =
2 (s + 2 ) 2m * L
QP =
S
Dh 3 sin 2 P
*
12
L
Dh 2 sin 2
QP =
2 (s + 2 )
h
P
2m * L
S
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POLYMER PHYSICS - Spring 2007
2
E h
2
&
m = bulk D 2 tan D B D S
4
sin
Q D = 1/2 2 D 2 Nh cos sin
F2
P=
2
9 2 r1 r 2
2
P=
8LQ
R4
F2
P=
2
9 2 r1 r 2
2
P=
8LQ
R4
t
F0 =
x2
Bt = B
Re =
h0
= B R DR
hf
DR =
BR =
X=
vf
v0
Rf
R0
Z
R0
yield = 0.028 ETensile
F=
R0 fz
Q
E = 2 (1 2 ) B
DV
R0 p
P=
Q
3
Q =V A
(n 1 )
= (T )
Miscellaneous
E = E f v f + E m vm
vf
v
1
=
+m
E
Ef
Em
k
=
Cp
4 h2
t cooling =
Blow Molding
E = 2 (1 ) G
2
p
= m (T )&
EMA 6165 - EXAM 2
8 (Tm T w )
ln 2
(T D TW )
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