Home assignments for the Course of Polymer Physical Chemistry
Section VI: Polymer Viscoelasticity and Section VII: Relaxation in Polymers
1. Estimate molecular weight between entanglement Me for a polymer if it has rubbery
plateau value GN=1 MPa at T=400K. Assume that the density of the polymer is 1g/cm3.
Me 
RT
GN
≈ 3320 [g/mol]
2. Estimate the viscosity of the PDMS melt with molecular weight M=106 g/mol and the
modulus plateau level GN=0.2 MPa at room T. Assume that relaxation time of a segment
at room T is 0=10-10 s, mass of a single segment is MS=250 g/moland the melt density is
=0.97 g/cm3. Using all these parameters, estimate:
- Me;
- viscosity and
- how extended in time is the rubbery plateau.
Me 
RT
2
GN
≈ 12,077 [g/mol]
2
M3
 ≈ 0.822*0.2*106*1.32*109*10-10 ≈ 21,700 [Pa*s]
2 0
12
12
M eM S
The rubbery plateau extends from the Rouse time of the chain with M~Me, a, till the reptation
time *.
M e2
 a  2  0 ≈ 2.3*10-7 [s]
MS

G N * 
GN
M3
 
 0 ≈ 0.132 [s]. So, the plateau extends for almost 6 orders in time.
M e M S2
*
3. Calculate the apparent activation energy V and 0 for the relaxation process shown below.
Assume that the relaxation process starts from the level (t)~0.5 and its temperature
variation follows Arrhenius temperature dependence =0exp(V/kT).
Relaxation function obtained for PPG (Mn=4000) by photon-correlation spectroscopy.
Temperatures from left to the right: 221K, 218K, 214K, 211K, 208K, 205K, 200K, 198K,
192K (data from [Bergman, et al. Phys.Rev.B 56, 11619 (1997)]).
First, you estimate  as the time at which the curves cross (t)~0.18. Using these data you plot
log vs 1/T and do a linear fit (see Fig.1 below).
The fit gives you a slope and an intercept at 1/T=0.
If you used decimal logarithm, then the intercept gives you ~ -65, i.e. 0~10-6510 s. The slope
is ~13,000, but you need to multiply it by the ln(10)~2.3. Using this slope multiplied by ln(10)
and by the gas constant R, you will get V~270 kJ/mol.
If you used natural logarithm, then the slope multiplied by the gas constant should give you
the same V, and 0 can be also recalculated from the intercept.
2
10
1
10
Equation
y = a + b*x
Weight
No Weighting
Residual Sum of
Squares
0.06287
Pearson's r
0.99849
0.99549
Adj. R-Square
Value
Tau
Intercept
Tau
Slope
Standard Error
-63.31138
2.40907
13041.96633
506.68322
0
 [s]
10
-1
10
-2
10
-3
10
-4
10
0.0045
0.0046
0.0047
0.0048
0.0049
0.0050
1/T [1/K]
Fig.1: Arrhenius plot of the relaxation time vs 1/T (symbols). The line shows the fit by a straight line
that provides estimates of the activation energy V (slope) and 0 (intercept).