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Chapter 14: Problem 10

Using the definitions for total chain molecule (9 length \(L\) (Equation 14.11) and average chain endto-end distance \(r\) (Equation 14.12), determine the following for a linear polytetrafluoroethylene: (a) the number-average molecular weight for \(L=2000 \mathrm{~nm}\) (b) the number-average molecular weight for \(r=15 \mathrm{~nm}\)

Short Answer

Expert verified
Question: Determine the number-average molecular weight (\(M_n\)) for each of the following cases: (A) a polymer chain with a total chain length of 2000 nm, and (B) a polymer with an average chain end-to-end distance of 15 nm. Answer: (A) For the polymer chain with a total chain length of 2000 nm, the number of subunits (N) is 10,000, and the number-average molecular weight (\(M_n\)) is 1,000,000 g/mol. (B) For the polymer with an average chain end-to-end distance of 15 nm, the number of subunits (N) is 5625, and the number-average molecular weight (\(M_n\)) is 562,500 g/mol.

Step by step solution

01

(Part A: Determine the number of subunits with \(L=2000 \mathrm{~nm}\))

Given the total chain length, \(L = 2000 \mathrm{~nm}\), we will use Equation 14.11 to determine the number of subunits, \(N\). First, we will convert the chain length to a more manageable unit (Å) to match the length of each subunit \(b\): 1 nm = 10 Å Thus, \(L = 2000 * 10 = 20000 \mathrm{~Å}\). Assuming the length of each subunit \(b = 2 \mathrm{~Å}\), we can now calculate the number of subunits using Equation 14.11: \(N = \frac{L}{b} = \frac{20000 \mathrm{~Å}}{2 \mathrm{~Å}}\)
02

(Part A: Calculate number-average molecular weight with N subunits)

With the calculated \(N\), find the number-average molecular weight (\(M_n\)) by multiplying the number of subunits, \(N\), by the molecular weight of one tetrafluoroethylene subunit: \(M_n = N * (100 \frac{\mathrm{g}}{\mathrm{mol}})\)
03

(Part B: Determine the number of subunits with \(r=15 \mathrm{~nm}\))

Given the average chain end-to-end distance, \(r = 15 \mathrm{~nm}\), we will calculate the number of subunits, \(N\), using Equation 14.12: \(r = \sqrt{N} * b\) Convert the end-to-end distance to Å: \(r = 15 * 10 = 150 \mathrm{~Å}\) Using the same subunit length, \(b = 2 \mathrm{~Å}\), the equation becomes: \(150 \mathrm{~Å} = \sqrt{N} * 2 \mathrm{~Å}\)
04

(Part B: Calculate number-average molecular weight with N subunits)

With the calculated \(N\), find the number-average molecular weight (\(M_n\)) by multiplying the number of subunits, \(N\), by the molecular weight of one tetrafluoroethylene subunit: \(M_n = N * (100 \frac{\mathrm{g}}{\mathrm{mol}})\)

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Most popular questions from this chapter

A random poly(styrene-butadiene) copoly\(\Theta\) mer has a number-average molecular weight of \(350,000 \mathrm{~g} /\) mol and a degree of polymerization of 5000. Compute the fraction of styrene and butadiene repeat units in this copolymer.

Carbon dioxide diffuses through a high-density polyethylene (HDPE) sheet \(50 \mathrm{~mm}\) thick at a rate of \(2.2 \times 10^{-8}\left(\mathrm{~cm}^{3} \mathrm{STP}\right) / \mathrm{cm}^{2} \cdot \mathrm{s}\) at \(325 \mathrm{~K}\). The pressures of carbon dioxide at the two faces are \(4000 \mathrm{kPa}\) and \(2500 \mathrm{kPa}\), which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at \(325 \mathrm{~K}\) ?

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(a) Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of \(250,000 \mathrm{~g} / \mathrm{mol}\) and a degree of polymerization of 4640 . (b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?
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