Determining Fraction of Cross-Linked Units in Isoprene

Abstract.  The objective of the present work is to demonstrate an experimental methodology to determine the fraction of cross-linked units in isoprene. For this purpose isoprene was stretched out at a constant elongation. For each elongation, the experimental effort ranged of temperature for the region 303-344 K to evaluate the force. Then calculated effective Young’s modulus, so can be obtained the average of cross-link fraction between polymer molecules. It is shown that once the temperature was dropped, the weight was decreased. The effect of higher temperature had on the elasticity of the isoprene was amplified under more weight. Therefore, the change of temperature may be due to effect to stress, which has affected the elastic property of elastomer. Then crosslink fraction can be evaluated which is 0.01.

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Introduction

Elastomeric materials are applied widely to demanding applications on account of their inherent reversible deformation behavior[1]. The chains of molecules in elastomer have a natural elasticity: Elastomer will be stretched after acquiring the force, once remove the force elastomer spring back to the original length. The molecular chains of an elastomer principally perform as springs which are called crosslinks. Furthermore, these crosslinks and the elastomers will break if obtain too much force.

When informing the tensile strength of materials loaded, it is common to consider forces in term of the stress (the force by the cross-sectional area):

fa =
FA
      (1)

where fa is the stress, the stress is attained by using the measured force (F) and the original material area[2]. A statistical mechanical treatment of elasticity, based on a model which assumes that the polymer chains are freely jointed and that the distance between the ends of a given polymer chain is characterised by a random distribution, can be written algebraically as:
fa=ρRTzMσ–1σ2=YTσ–1σ2
    (2)

Here M is a molar mass of monomer that has relative with a density of the polymer (ρ), z is the average number of monomer units between crosslinks and σ is the fractional elongation which evaluated from σ = L/L0. Calculate an effective Young’s modulus by plotting fa against T(σ – (1/σ2). Thus z can be attained and then use this equation Fcl =
nN
=
1z
to attain the fraction of cross-links[3].

Experimental

1. Turned on the balance and unscrewed the set-screws with the shaft at the bottom of the sample for balance. Measured the sample length. Turned on the tap for compressed air. Set the air flow between 9 and 10 with the bench tap.

2. Set the temperature controller to a temperature between 70 and 80 °C. Wait a few minutes until the temperature was stable. Tared the balance. Expanded the sample approximately 10-20% by pulled the connecting shaft down and then screwed the set-screw to the shaft. Took a note from the digital thermometer when the measurement was stable.

3. Reduced the temperature between 5 to 10 °C. Noted the results when the measurement was stable. Continued this process until the temperature was between 20 and 30 °C. Unscrewed the set-screws and then expanded the sample again approximately 10-20%. Performed sets of measurements in which the temperature was risen to between 70 and 80 °C.

4. Repeated the above to get about 5 sets. Recorded the width and thickness of the sample. After the experiment was finished, turned off the heater controller, air flow and balance.

Results and discussion

The elastomer sample is isoprene (2-methylbutadiene) which has thickness 0.09 mm, width 6.6 mm, original length 23.5 mm and molar mass 68.12 g/mol. In this experiment assume the density of the elastomer is 0.98 g/cm3.

Table 1.  Length stretched 25 mm by connecting rod, σ = 1.06

Temperature (K)

Weight (kg)

fa (Pa)

343.05

-0.0045

74318.18

332.85

-0.00407

67216.67

323.15

-0.00371

61271.21

313.05

-0.00332

54830.30

303.05

-0.00294

48554.55

Table 2. Length stretched 28 mm by connecting rod, σ = 1.19

Temperature (K)

Weight (kg)

fa (Pa)

303.05

-0.00946

156233.33

313.05

-0.01014

167463.64

322.95

-0.0108

178363.64

333.25

-0.01135

187446.97

343.35

-0.01194

197190.91

Table 3. Length stretched 32 mm by connecting rod, σ = 1.36

Temperature (K)

Weight (kg)

fa (Pa)

343.05

-0.02278

376215.15

333.05

-0.0217

358378.79

322.85

-0.02082

343845.45

313.15

-0.01978

326669.70

303.15

-0.01884

311145.45

Table 4. Length stretched 38 mm by connecting rod, σ = 1.62

Temperature (K)

Weight (kg)

fa (Pa)

303.05

-0.02801

462589.39

312.75

-0.02834

468039.39

323.05

-0.02947

486701.52

333.25

-0.03041

502225.76

343.65

-0.03132

517254.55

Table 5. Length stretched 42 mm by connecting rod, σ = 1.79

Temperature (K)

Weight (kg)

fa (Pa)

344.15

-0.0387

639136.36

334.55

-0.03711

612877.27

322.85

-0.03545

585462.12

312.35

-0.03397

561019.70

303.25

-0.03269

539880.30

Plot fa against T
σ–1σ2
via equation 2,
fa=ρRTzMσ–1σ2=YTσ–1σ2

 

Figure 1. Sample graph between fa and T (σ – (1/σ2)

Can be obtained Y from a slope in figure 1, so  Y = 1264.9

Then calculated z from equation 2, so   z = 94.56

And calculated the fraction of cross-links (Fcl):

     Fcl =
1z
= 0.011

As noted in the results sections, once the temperature was dropped, the weight was decreased as shown in table 1, 3 and 5. In addition, higher temperatures also affect more weight of the elastomer because the molecules are more excited and want to be in a random state as given in table 2 and 4. Therefore, the change of temperature may be due to effect to stress, which has affected the elastic property of elastomer[4]. Moreover, during recorded the values shown that there was a fluctuation of the weight because the force is trying to pull the elastomer back to its equilibrium length[5]. By looking at figure 1, according to calculate the equation, the crosslink fraction is 0.01 that means approximately 100 molecules of isoprene have one crosslink in the polymer.

Conclusions

The applied experimental methodology for determining the fraction of cross-linked units in isoprene (2-methylbutadiene). Isoprene was stretched out at a constant elongation. The results also showed that once the temperature was dropped, the weight was decreased. The effect of higher temperature had on the elasticity of the isoprene was amplified under more weight. When isoprene was heated, the molecules stretched out, making them more excited and able to be in a random state. Therefore, the change of temperature may be due to effect to stress, which has affected the elastic property of elastomer. Constitutive equations are evaluated the crosslink fraction which is 0.01 that means approximately 100 molecules of isoprene will have one crosslink in the polymer. The balance led to some of the sources of error. After tested 3 sets of temperature, the weight values have fluctuated. Hence, the teacher assistant moved the balance to reduce this error. However, there was still fluctuation weight error because the force is trying to pull the elastomer back to its equilibrium length. If have the chance to conduct this experiment again, an experiment should test more range of temperature such as 273 – 373 K to observe other property changes.

References

1   Polymerization, S. et al. (2016) ‘Communications Independent Control of Elastomer Properties through Stereocontrolled Synthesis’, pp. 13076–13080.

2   Roylance, D. (2008) ‘Mechanical properties of materials’. pp. 1-128.

3   Technologies, A. and Composites, P. (2020) ‘POLYMER’. pp. 1-43.

4   Urry, D. W., Harris, D. and Prasad, U. (1984) ‘B 1984 Academic Press, Inc.’, 125(3), pp. 1082–1088.

5   Engineering, P. and Collection, S. P. (1994) ‘The effect of annealing on stretched trans-1 , 4-polyisoprene networks’. pp. 290-300.

Appendices

Appendix A: Calculation crosslinks fraction.

The elastomer sample is isoprene (2-methylbutadiene) which has thickness 0.09 mm, width 6.6 mm, original length 23.5 mm and molar mass 68.12 g/mol. In this experiment assume the density of the elastomer is 0.98 g/cm3.

The calculation example; Temperature 343.05 K, weight -4.5g and stretched length 25 mm Calculate Force (F) from                                           F = mg

So      F = (-0.0045 kg) x (-9.81 m/s2)

      F = 0.044145 N

      F ≈ 0.044 N

Then, calculate Force per unit area (fa): fa =
FA

      fa =
0.044145 N(0.00009m) X (0.006m) 

      fa = 74318.18182 Pa

      fa ≈ 74318.18 Pa

And      σ =
LL0

      σ =
25 mm23.5mm

      σ = 1.063829787

      σ ≈ 1.06

Plot fa against T
σ–1σ2
via equation 2 
fa=ρRTzMσ–1σ2=YTσ–1σ2

Figure 2. Sample graph between fa and T(σ – (1/σ2)

Can obtain Y from slope in figure 2, so   Y = 1264.9

Then calculate z from equation 2, so   z = 94.55952109

       z ≈ 94.56

And calculate the fraction of cross-links (Fcl):

     Fcl =
1z
= 0.01057535 ≈ 0.011

NB. Density of the elastomer (
ρ
)  = 980 kg/m3

 Molar mass (M)  = 0.06812 kg/mol

 Ideal gas constant (R)  = 8.314 m3•atm/mol•K

[1]Polymerization, S. et al. (2016) ‘Communications Independent Control of Elastomer Properties through Stereocontrolled Synthesis’, pp. 13076–13080.

[2] Roylance, D. (2008) ‘Mechanical properties of materials’.

[3] Technologies, A. and Composites, P. (2020) ‘POLYMER’.

[4] Urry, D. W., Harris, D. and Prasad, U. (1984) ‘B 1984 Academic Press, Inc.’, 125(3), pp. 1082–1088.

[5] Engineering, P. and Collection, S. P. (1994) ‘The effect of annealing on stretched trans-1 , 4-polyisoprene networks’.