Polymers- Rubber Elasticity Theory

Question 1

Elastic properties of rubber

Answer 1

Extremely high extensibility, up to x10, generated by low mechanical stress.
Complete recovery after mechanical deformation.
These are both due to deformation-induced changes in entropy

Question 2

What is the origin of polymer elasticity at molecular level due to?

Answer 2

The fact that the molecules prefer disordered, coiled conformations. This is entropy driven

Question 3

Formula for change in internal energy of a system

Answer 3

dU=dQ-dW
dQ is heat absorbed by the system
dW is the work done by the system

Question 4

Formula for work done by the system

Answer 4

dW=-fdl
f is tensile force
dl is change in length of elastomer

Question 5

Deriving formula for tensile force in terms of energy changes for a reversible process (entropic elasticity formula)

Answer 5

For reversible process: dQ=TdS 
Using previous formulae: fdl=dU-TdS 
Divide through by dl
f=(δU/δl)T,V - T(δS/δl)T,V
Where δs are curly ds. T,V is subscript for constant T and V

Question 6

Which terms are dominant for metals, ceramics and elastomers in entropic elasticity formula?

Answer 6

For metals and ceramics the internal energy term is dominant but for elastomers the change in entropy gives the largest contribution to the force. So can be approximated as:
f=-T(δS/δl)T,V

Question 7

How do pressure, volume, internal energy and entropy of ideal gases change when they are compressed?

Answer 7

Pressure increases
Volume decreases
Internal energy constant
Entropy decreases

Question 8

Difference between natural rubber and vulcanised natural rubber

Answer 8

Natural rubber just has lots of the polymer chains mixed around together.
Vulcanised has some sulfur crosslinks between the rubber molecules.
Repeating unit [CH2-C(CH3)=CH-CH2]n (has wings)

Question 9

Structural requirement for a polymer to be elastic

Answer 9

Amorphous
Above Tg
Lightly cross-linked

Question 10

Ideal rubber

Answer 10

Consists of flexible cross-linked polymer chains undergoing violent liquid-like motions.
The chains are joined to the network at both ends (sub-molecules).
The chains or sub-molecules have equal contour length

Question 11

Freely jointed chain model

Answer 11

One rubber molecule (the chain) has lots of zigzags (bonds). Two bonds have an angle between them and form two equal length sides of a triangle. These have length l. The length of the open side is rf (f sub). Over the whole molecule:
rf^2=nl^2
n is number of bonds of length l

Question 12

Statistical problem of where the two chain ends of a rubber molecule will be

Answer 12

If one end is the origin, find the probability that the other chain end lies in a certain volume, dV, whose centre is a distance r from the origin.

Question 13

How does the number of conformations that a single chain can take up depend on end-to-end separation?

Answer 13

Number of conformations possible decreases as end-to-end separation increases (eventually only a straight line possible)

Question 14

Gaussian chain model

Answer 14

P(x,y,z)dxdydz=(exp(-(x2+y2+z2)/ρ2))/(rt(π)ρ)^3 dxdydz
Where =3/2 ρ2
Graph of P(x,y,z) vs r starts high, curves down to diagonal then starts to level off near x axis

Question 15

Entropy of a Gaussian chain

Answer 15

s=s0+klnP(x,y,x), k is Boltzmann
Sub in formula for Gaussian chain model
s=s0-k[3ln(rt(π)ρ)+(x2+y2+z2)/ρ2]
So entropy goes down with increasing end-to-end distance as the number of conformations it can adopt decreases. This is the molecular origin of rubber elasticity

Question 16

Elastic behaviour of Gaussian chain in different directions formulae for tensile force

Answer 16

fx=-T(δs/δx)=2kTx/ρ2
fy= 2kTy/ρ2
fz= 2kTz/ρ2
f bar = (2kT/ρ2)r bar
So Gaussian chain is linearly elastic and the force constant is linear in T

Question 17

How can elastic properties of molecular network in a rubber be represented?

Answer 17

By interconnected linearly elastic springs

Question 18

How do extension ratios work?

Answer 18

When something stretched there is no change in volume so
Xi,Yi,Zi=XYZ (initial dimensions to new dimensions)
Extension ratios: λx=X/Xi, λy=Y/Yi, etc
λxλyλz=1
This applies to each sub-molecule

Question 19

Work done by chain during deformation

Answer 19

Integrate fx from Xi to λxXi with respect to x (capitals actually LC).
wx=(kT/ρ2)(λx^2-1)xi^2
Same for wy and wz just replace x

Question 20

Total work done by external force

Answer 20

W=Σ(wx+wy+wz) from1 to v (number of chains (sub-molecules))

Question 21

Sum of one component of work from 1 to v

Answer 21

Σwx=(kT/ρ2)(λx^2-1)Σxi^2
Σwx= (kT/ρ2)(λx^2-1)vx^2 i (i sub outside arrows)
x^2 i is same for y and z so equals r^2 i/3
Where i is mean square end-to-end distance of chains in the unreformed state. For Gaussian chain:
r^2 0=3/2 ρ^2 (arrows)

Question 22

Formula for shear modulus

Answer 22

G=NkT x (r^2 i/r^2 0)
Where N is v/V (number of sub-molecules per unit volume)
The r^2 terms have arrows around them and sub i or 0.
k is Boltzmann constant
T is temperature
r^2 in arrows sub i is mean square end-to end distance of the chains in the undeformed state
r^2 in arrows sub 0 is mean square end-to-end distance of the chains for a Gaussian state

Question 23

Nominal stress for simple stretching along z direction

Answer 23

σn=f/XiYi=G(λ-1/λ2)

Question 24

True stress for simple stretching along z direction

Answer 24

σt=f/XY=G(λ2-1/λ)

Question 25

How well do theoretical results for relationship between stress and extension ratio compare to experimental results?

Answer 25

At low values of λ (0-1.5, theory matches well with curve up with decreasing gradient. Les than 1 has negative stress and above 1 positive stress. Theory then largely a straight diagonal line as λ increases whereas experimental goes below and then crosses as it goes to exponential increase.

Question 26

Reasons for discrepancies at large extension ratios

Answer 26

The theory neglects network defects, chain entanglements, etc.
The chains are no longer Gaussian at high extensions.
At high extensions vulcanised natural rubber crystallises

Question 27

Mechanical hysteresis graph

Answer 27

Stress vs %extension
Loading curve always above unloading curve apart from the joins at both ends. This is for a strain-crystallising elastomer