Statistical Thermodynamics Exercise 11 HS Exercise 11

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1 Exercise 11 Release: on-line Return: your assistant Discussion: your tutorial room Macromolecules (or polymers) are large molecules consisting of smaller subunits (referred to as monomers) In addition to biopolymers such as DNA or polysaccharides, synthetic polymers were studied extensively for diverse applications (ie packaging, membranes, ) The statistical treatment of macromolecules with the aim to predict their properties will be at the heart of this exercise Due to their size and complexity methods developed in earlier chapters are not applicable anymore Especially their complexity led to the development of a broad range of models that describe their physics For simplicity, we will use a framework which is analogous to the model of the ideal gas The basic assumption made in the model of an ideal polymer chain is the absence of volume interactions The chain is made up of immaterial elements connected to its nearest neighbours The interactions with elements of the same or other macromolecules and with solvent molecules will be neglected The main differences in the treatment of an ideal polymer chain are the type of bonding and the flexibility of the polymer chain Even though this model is lacking in accuracy and generality, it will provide us with useful concepts Problem 1: Ideal chain models The freely jointed chain model Figure 1-1: One conformation of a flexible polymer 1 The freely jointed chain model neglects the fact that bond lengths vary All bonds are of the same length l = r i It is assumed that there are no correlations of the orientation between the bond vectors r i As illustrated in Fig (1-1) r i is the vector connecting the two neighbouring elements A i and A i 1, R is the vector connecting the first and the last element of a chain and θ ij is the angle between the bond vectors r i and r j 1 Polymer Physics, Michael Rubinstein and Ralph H Colby, 23, p 51 page 1 of 5

2 a) What is the average end-to-end vector R and the mean-square end-to-end vector R 2 in the freely rotating chain model Hint: Note that r i r j = cos θ ij l 2 b) In a real chain there will be forces acting on the individual elements of the chain which lead to correlations between the bond vectors Compute an expression for the mean-square end-to-end distance in terms of Flory s characteristic ratio c) Correlations can be thought of originating from hindrances Which polymer will have the larger Flory s characteristic ratio? (a) Polystyrene (b) Polypropylene Figure 1-2: Comparison of two polymers d) Calculate the root-mean-square end-to-end distance for polypropylene with M = 1 7 g mol 1 in an ideal conformation with C = 59 Compare the end-to-end distance with the contour length of this polymer Hint: For l you should use the main-chain bond length Use the bond length of a sp 3 hybridized C-C bond Kuhn monomers and Kuhn lengths In the framework of ideal polymers some properties are independent of the chemical structure For a simple treatment of all ideal polymers, one might define an equivalent freely jointed chain This newly defined chain has the same mean-square end-to-end distance R 2 and the same maximum end-to-end distance R max as the actual polymer Using this concept, a complex chemical structure simplifies to a straight chain of N Kuhn momomers with an effective length b (Kuhn length) e) Calculate the Kuhn length b of polypropylene How many monomers make up one Kuhn monomer? Calculate the molecular weight M of the Kuhn monomer Hint: For the calculation of R max use the all-trans conformation with a bond angle of θ = 68 as outlined in Fig 1-3 The freely rotating chain model As mentioned above, in contrast to the freely jointed chain, real chains will have preferential bond angles θ i Also due to steric hindrances, elements of the chain will have preferential orientations ψ i relative to neighbouring elements In the freely rotating chain model the page 2 of 5

3 Statistical Thermodynamics Exercise 11 HS 215 Figure 1-3: Illustration of Rmax 2 variations in potential U (ψi ) are neglected All torsional angles ψi are therefore equally likely f) Show that the mean-square end-to-end distance and the Flory s characteristic ratio C of a freely rotating chain are given as hr2 i = nl2 and C = 1 + cos θ 1 cos θ (11) 1 + cos θ 1 cos θ (12) Hint: The correlations transmitted down the chain are given as h~ri ~rj i = l2 (cos θ) j i (13) Also note that (cos θ) j i decays rapidly j i (cos θ) j i = exp [ j i ln (cos θ)] = exp, sp (14) with a persistence segment sp = 1 ln (cos θ) (15) g) Calculate C for a bond angle of θ = 68 Polymers usually have values of C that vary between 4 and 1 What is the reason for this large deviation from the calculated value? 2 Polymer Physics, Michael Rubinstein and Ralph H Colby, 23, p 5 page 3 of 5

4 The Kratky-Porod model The Kratky-Porod model (also known as the worm-like chain model) can be considered as the limiting case of the freely rotating chain model In this model, the bond lengths l and the bond angle θ go to zero It is especially useful when describing stiff polymers such as DNA h) Calculate the Kuhn length b in terms of the persistence length l p Hint: The persistence length is given as l p = s p l Use a series expansion at small angles θ for cos θ i) The mean-square end-to-end distance of the worm-like chains is given as ( ( R 2 = 2l p R max 2lp 2 1 exp R )) max (16) l p Discuss the limiting cases where the contour length is much smaller and much larger than the persistence length Problem 2: Dependence of chain elongation on force If a force f is applied to a freely jointed chain along the z coordinate, internal energy is given by U = f R z, (21) where f is a constant force acting on one end of the chain and its equal and opposite acting on the other end of the chain R z corresponds to the end-to-end vector in the z-direction Due to the applied force, the different conformations are no longer equally likely Their probability distribution can be expressed by a Boltzmann distribution a) Show that the partition function of an elongated chain can be simplified to [ 4πkB T Z(T, f, N) = fb sinh ( fb )] N (22) Hint: As discussed above the end-to-end vector R is made up of the N individual bond vectors r i Fig (1-1) corresponds to one possibility of the conformations to be accounted for with an energy U conf For the derivation of the partition function one has to sum over all possible conformations The orientation of the bond vector r i can be fully described by two angles θ i and ψ i in the spherical coordinate system In principle we would therefore have to sum over all possible orientations of the bond vectors θ i and ψ i Since the bond vectors are continuously distributed, the sum of all possible conformations of a freely jointed chain therefore corresponds to an integrals over all possible θ i and ψ i Z = conformations exp ( U conf ) = 2π 2π π π exp ( U conf ) N sin θ i dθ i dψ i (23) The z component of the end-to-end vector can be calculated by the angle θ i between R and the z axis N R z = b cos θ i (24) i=1 i=1 page 4 of 5

5 b) Give an expression for the Gibbs free energy Hint: Note that in the framework of ideal chains there are no volume dependencies c) The average end-to-end distance R is the negative of derivative of the Gibbs free energy with respect to the force f Derive an expression for R d) Consider an ideal chain with N Kuhn monomers of length b The chain is carrying a positive charge +e at one end and a negative charge -e at the other end What will be its average end-to-end distance R z at room temperature in an electric field E = 1 5 V cm 1 acting along the z axis The chain has N = 1 4 Kuhn monomers of length b = 6 Å Give a ratio of R z to the maximum end-to-end distance R max Hint: Ignore direct Coulomb interactions between the charges page 5 of 5

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