Solution Exercise 11 HS 2015

Statistical Thermodynamics Solution Exercise 11 HS 2015

Solution Exercise 11

Problem 1: Ideal chain models a) The end-to-end vector of a polymer chain is deﬁned as

n ~ X Rn = ~ri . (1.1) i=1

Since there are no correlations between the individual ~ri all chain orientations will be equally likely. The mean end-to-end vector is therefore equal to zero. ~ hRni = 0 . (1.2)

the mean-square end-to-end distance, however, provides a more useful description of the system. With * n   n + 2 ~ 2 ~ ~ X X hR i = hRni = hRn · Rni =  ~ri ·  ~rj , (1.3) i=1 j=1 and with the hint given one gets

n n 2 2 X X hR i = l hcos θiji . (1.4) i=1 j=1 Again we use the assumption that there are no correlations between the orientations of the bond vectors. For all i 6= j this will lead to hcos θiji = 0, however for i = j (one and the same bond) θii will always be equal to 0° which leads to hcos θiii = 1. With this Eq. (1.4) then simpliﬁes to hR2i = nl2 . (1.5) b) When considering correlations between the individual bond vectors the hcos θiji are not necessarily equal to zero or one anymore. In general one might assume, however, that for bonds that are inﬁnitely far apart from each other there won’t be any correlations, i.e

lim hcos θiji = 0 . (1.6) |i−j|→∞

Without any further knowledge about the polymer we cannot write an analytic expression for the mean-square end-to-end distance apart from the most general form in which the correlations between the ith element and all the other elements in the chains is denoted as n 0 X Ci ≡ hcos θiji . (1.7) j=1

With these considerations Eq. (1.4) reduces to

n n n 2 2 X X 2 X 0 2 hR i = l hcos θiji = l Ci = Cnnl , (1.8) i=1 j=1 i=1

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where the Flory’s characteristic ratio Cn is the average correlation over all bonds

n 1 X C = C0 (1.9) n n i i=1

It is interesting to note that for all polymers Cn > 1 and as n → ∞ Cn approaches a ﬁnite value C∞. For long polymers Eq. (1.8) can then be rewritten as 2 ∼ 2 hR i = C∞nl . (1.10) c) Neglecting some exceptions such as polyethylene (C∞ = 7.4) a more sterically hindered polymer has a larger C∞. Polypropylene has a characteristic ratio at inﬁnite chain length of C∞ = 5.9, whereas for polystyrene C∞ = 9.5.

Physical Properties of Polymers Handbook, James E. Mark, 2007, p.445-452 is a good refer- ence. d) Polypropylene has a repeat unit of (C3H6). One repeat unit therefore has a molar mass −1 of M(C3H6) = 42.08 g mol . Since we are interested in the number of backbone bonds n, we will have to ﬁnd the average molar mass of the repeat unit per backbone bond. In polypropylene there are two backbone bonds. Therefore, we get an average molar mass per −1 backbone bond of Mb = 21.04 g mol . From this it follows that the chain consists of 107 g mol−1 n = = 48 · 104 (1.11) 21.04 g mol−1

backbone bonds. The bond length can be assumed to be l = 1.54 A. The mean-square end-to-end distance therefore is

2 hR2i = 5.9 · 48 · 104 · (1.54 A) 2 = 6 720 000 A . (1.12)

This corresponds to a root-mean-square end-to-end distance of

p 2 Rrms = hR i = 2600 A ≈ 0.3 µm (1.13) e) From Eq. (8.44) in the lecture script we get an expression for the Kuhn length b hR2i C nl2 b = = ∞ . (1.14) Rmax Rmax For a chain in the all-trans conformation (see Fig. (1-3) on the exercise sheet)

Rmax = nl cos (θ/2) . (1.15) Eq. (1.14) can then be simpliﬁed to C nl2 C l b = ∞ = ∞ . (1.16) nl cos (θ/2) cos (θ/2)

Inserting the bond angle θ = 68°, the characteristic ratio C∞ = 5.9 and the bond length l = 1.54 A one gets 5.9 · 1.54 A b = = 11 A. (1.17) cos (34°)

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The length of the Kuhn monomer is therefore roughly 11 A. This is larger than the size of a single repeat unit of the polymer. To get a feeling of how many repeat units are within a Kuhn monomer one could derive an expression for the ratio of the backbone bonds n to Kuhn monomers N. Also from the lecture script (Eq. (8.43)) and with Eq. (1.15) of this solution we get 2 2 2 2 2 Rmax n l cos (θ/2) n cos (θ/2) N = 2 = 2 = . (1.18) C∞nl C∞nl C∞ This gives a ratio of n C C 5.9 = ∞ = ∞ = = 8.6 . (1.19) N cos2 (θ/2) cos2 (34°) 0.6873 Since there are 2 backbone bonds in each repeat unit, there are roughly 4 repeat units in each Kuhn monomer!

Using the same considerations as above, one has to consider the average molar mass per backbone bond Mb of the original polymer to get to the molar mass of one Kuhn monomer. −1 The molar mass of a repeat unit in polypropylene is M(CH2CH(CH3)) = 42.08 g mol . Since there are two backbone bonds per repeat unit, the average molar mass per backbone −1 bond Mb is equal to 21.04 g mol . Since the total mass of the polymer has to be conserved

Mpolymer = NM0 = nMb . (1.20) Therefore n M = M = 8.6 · 21.04 g mol−1 = 181 g mol−1 . (1.21) 0 N b The molecular weight can then be calculated by deviding by Avogadro’s constant NA. In the equivalent freely rotating chain model chemical information is contained within the Kuhn monomer size, making a comparison between polymers easy. f) By using Eq. (8.34) from the lecture script and Eq. (1.3) from the exercise sheet one gets

n n n  i−1 n  2 X X X X 2 X hR i = h~ri · ~rji =  h~ri · ~rji + h~ri i + h~ri · ~rji i=1 j=1 i=1 j=1 j=i+1

n n  i−1 n  X 2 2 X X i−j X j−i = h~ri i + l  (cos θ) + (cos θ)  (1.22) i=1 i=1 j=1 j=i+1

n  i−1 n−i  2 2 X X k X k = nl + l  cos θ + cos θ . i=1 k=1 k=1 In the ﬁrst line of the expression we made a simple expansion of the sum over j. This enables us to get all terms of i = j out of the summation. From line two to line three a change in variables was undertaken. The new variable of summation k = |j − i| gives the distance in linking units from the position at which i = j. With Eq. (1.4) of the exercise sheet we are able write the summation over k as an inﬁnite sum. We are making use of the fact, that cos θ|j−i| is a rapidly decaying function.

n ∞ ∞ X X X hR2i = nl2 + 2l2 cosk θ = nl2 + 2nl2 cosk θ . (1.23) i=1 k=1 k=1

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After factorizing out one cos θ,

∞ X hR2i = nl2 + 2nl2 cos θ cosk θ (1.24) k=0 Eq. (1.23) can be rewritten using the following relation 1 = 1 + x + x2 + x3 + ... , (1.25) 1 − x as cos θ 1 + cos θ hR2i = nl2 + 2nl2 = nl2 . Q.E.D. (1.26) 1 − cos θ 1 − cos θ When comparing Eq. (1.26) with Eq. (1.10) one also gets

1 + cos θ C = . (1.27) ∞ 1 − cos θ g) For a chain with a bond angle of θ = 68°

C∞ ≈ 2 . (1.28)

When comparing this value to the characteristic ratio C∞ of real polymers one sees, that the correlations are underestimated by a factor of 2 to 5. Better agreements are achieved when taking hindrance of the bond torsion into account. h) When using the series expansion for cos θ at small angles

θ2 cos θ ∼= 1 − , (1.29) 2 and a series expansion ln(1 − x) ∼= −x + ... (1.30)

the persistence segment sp can be approximated as

∼ 1 ∼ 2 sp = − = . (1.31) θ2 θ2 ln 1 − 2

The Flory characteristic ratio (Eq. (1.8)) might similarly be rewritten in the limit of a worm-like chain as 2 2 − θ 1 + cos θ ∼ 2 ∼ 4 C∞ = = = . (1.32) 1 − cos θ θ2 θ2 2 The expression for the Kuhn length b in Eq. (1.16) then gives

C 4 b = l ∞ ∼= l = 2l , (1.33) cos (θ/2) θ2 p

when using the hint given that the persistence length lp is

lp = spl . (1.34)

page 4 of 6 Statistical Thermodynamics Solution Exercise 11 HS 2015 i) In the limit of Rmax lmax Eq. (1.6) on the exercise sheet simpliﬁes to

2 hR i ≈ 2lpRmax = bRmax , (1.35)

which corresponds to the ideal chain limit. For Rmax lp we get

2 2 hR i = Rmax , (1.36) where we used the series expansion 1 exp(−x) ∼= 1 − x + x2 + ... , (1.37) 2 with ! R x = max . (1.38) lp This essentially corresponds to a rod-like behaviour of the polymer. From the persistence length one can therefore extract informations on the rigidity of a polymer chain.

Problem 2: Dependence of a chain elongation force a) The statistical weights of the possible chain conformations under an elongation force are represented by Boltzmann factors exp (−U/kBT ). The possible orientations of the chain may be represented in spherical coordinates. With Eq. (2.1), (2.2) and (2.3) of the exercise sheet one then gets for a force acting along the z axis   Z Z N N X fRz fb X Y Z(T, f, N) = exp = ... exp cos θ (sin θ dθ dψ ) . (2.1) k T k T i i i i B B i=1 i=1

Since all N integrals are identical, Eq. (2.1) can be written as

" #N Z 2π Z π fb Z(T, f, N) = sin θi exp cos θi dθidψi . (2.2) 0 0 kBT

Integration then yields

" #N Z π fb Z(T, f, N) = 2π sin θi exp cos θi dθi 0 kBT  N " # 2πkBT fb fb =  exp − exp −  (2.3) fb kBT kBT " #N 4πk T fb = B sinh . fb kBT

The integration over θi was performed after the following substitution

u = cos θi . (2.4)

page 5 of 6 Statistical Thermodynamics Solution Exercise 11 HS 2015 b) Since there are no volume dependencies in the framework of an ideal chain the Gibbs free energy G can be expressed as (see Eq. (5.25), p. 44, lecture notes)

∂ ln Z G(T, f, N) = −kBT ln Z(T, f, N) + kBT = −kBT ln Z(T, f, N) . (2.5) ∂ ln V T,N

When inserting Eq. (2.3) we get the desired expression for the Gibbs free energy

 !  fb fb G(T, f, N) = −kBTN ln 4π sinh − ln  . (2.6) kBT kBT c) As mentioned on the exercise sheet, the average end-to-end distances of a stretched chain can be calculated by taking the derivative of the Gibbs free energy with respect to the force f. " # ∂G fb k T hRi = − = bN coth − B (2.7) ∂f kBT fb Note that for a force that is only acting along the z-axis

hRi = hRzi , (2.8)

since the contributions to the mean end-to-end distance from x and y remain 0! d) A force on a charged particle in an electric ﬁeld is calculated by

f = qE . (2.9)

For our case the force acting on either end of the chain is therefore equal to

f = 1.602 × 10−12 N . (2.10)

With Eq. (2.7) we get a mean end-to-end distance of

hRi = 0.46 µm . (2.11)

The maximum end-to-end distance is calculated to be

4 Rmax = Nb = 10 · 6 A = 60 000 A = 6 µm . (2.12)

This leads to a ratio of hRi 0.46 µm = ≈ 0.1 . (2.13) Rmax 6 µm The chain is therefore stretched by roughly 10 %. Even though there is a strong electric ﬁeld acting, the chain will still be mainly coiled up.

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