HW5 Solutions - ChemE 178: Homework #5 SOLUTIONS Due...

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ChemE 178: Homework #5 SOLUTIONS Due Friday, October 17 1.) Consider a freely-jointed chain with a distance between joints of 0.6 nm. a.) Compute the root mean square end-to-end distance of such a chain that has 10000 links. For a FREELY JOINTED CHAIN: where n, the number of links is 10,000 and l, the distance between joints, is 0.6nm: r 2 1 2 = n 1 2 l = (10 ,000) 1 2 *0 .6 nm = 60 nm b.) Compute the radius of gyration of the chain in part a.) For a linear Gaussian chain, radius of gyration and mean squared end-to-end distance are related by: R g 2 = 1 6 r 2 R g 2 1 2 = 1 6 r 2 1 2 = 1 6 *60 nm = 24.5 nm c.) Suppose that the links in the chain can be completed aligned in a straight line, by applying a force to the end of the chain, for example. What is the end-to-end distance of this (stretched or aligned) chain? If we align the chain in a straight line, the end-to-end distance is the same as the countour length of the chain. Calculate the contour length: nm 85 . 899 , 4 2 5 . 109 sin * nm 6 . 0 * 000 , 10 2 sin * l * n L = = β = Note: It is not possible to align C-C-C bonds at a 180 degree angle, so the sin (beta/2) term takes care of the tetrahedral symmetry of carbon. 2.)(similar to Y&L 3.5): For the linear molecule of polyethylene with molecular weight 119,980 g/mol calculate a.) The contour length From Y&L 3.5 we know l=0.154nm. We must find n, the number of links. For polyethylene, the molecular weight of the repeat unit (C 2 H 4 ) is M o =12*2+1*4=28 g/mol. Neglecting the end groups, the degree of polymerization is: x n = MW M o = 119980 g / mol 28 g / mol = 4285 2 2 nl r =
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This note was uploaded on 12/09/2011 for the course CHEM ENG 178 taught by Professor Segalman during the Fall '08 term at University of California, Berkeley.

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HW5 Solutions - ChemE 178: Homework #5 SOLUTIONS Due...

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