Chapter 16, Problem 16.14P

Interpretation Introduction

(a)

Interpretation:

The number of polyethylene chains in the given rope is to be calculated.

Concept Introduction:

The molecular weight of a polymer is not fixed. It has a certain range under which its molecular weight falls. For a linear polymer, degree of polymerization represents the average length of the polymer. Degree of polymerization is defined as:

  $\begin{array}{c}\text{Degree of polymerization}=\frac{\text{average molecular weight of polymer}}{\text{molecular weight of repeating unit}}\\ n=\frac{{\left(\text{MW}\right)}_{\text{av}\text{.}}}{{\left(\text{MW}\right)}_{\text{repeating unit}}}\end{array}$   ....... (1)

The formula to calculate number of chains in the given mass of polymer is:

  $\text{Number of chains}=\frac{m\times N_A}{{\left(\text{MW}\right)}_{\text{av}\text{.}}}$   ....... (2)

Here, $m$ is the mass of the polymer, $N_A$ is the Avogadro's number, and ${\left(\text{MW}\right)}_{\text{av}\text{.}}$ is the average molecular mass of the polymer.

Answer to Problem 16.14P

The number of polyethylene chains in the given rope is $3.487\times {10}^{21}\text{ chains}$.

Explanation of Solution

Given information:

A rope made up of polyethylene weighs $0.25\text{ lb}$ per foot. One chain of this polymer contains $7000$ repeat units.

The length of the rope is $10\text{ ft}$.

The repeating unit of polyethylene is $-\left[{\text{C}}_2{\text{H}}_4\right]-$. Its molecular weight is:

  $\begin{array}{c}{\left(\text{MW}\right)}_{-\left[{\text{C}}_2{\text{H}}_4\right]-}=\left(2\times 12\right)+\left(4\times 1\right)\\ =28\text{ g/mol}\end{array}$

Now, use equation (1) and calculate the average molecular weight of polytetrafluoroethylene as:

  $\begin{array}{c}n=\frac{{\left(\text{MW}\right)}_{\text{av}\text{.}}}{{\left(\text{MW}\right)}_{-\left[{\text{C}}_2{\text{H}}_4\right]-}}\\ 7000=\frac{{\left(\text{MW}\right)}_{\text{av}\text{.}}}{28\text{ g/mol}}\\ {\left(\text{MW}\right)}_{\text{av}\text{.}}=196000\text{ g/mol}\end{array}$

Calculate the mass of $10\text{ ft}$ polyethylene rope as:

  $\begin{array}{c}m=\left(0.25\frac{\text{lb}}{\text{ft}}\times \frac{454\text{ g}}{\text{lb}}\right)\times \left(10\text{ ft}\right)\\ =1135\text{ g}\end{array}$

Calculate the number of chains in the given rope of polyethylene using equation (2) as:

  $\begin{array}{c}\text{Number of chains}=\frac{m\times N_A}{{\left(\text{MW}\right)}_{\text{av}\text{.}}}\\ =\frac{\left(1135\text{ g}\right)\left(6.023\times {10}^{23}\text{ chains/mol}\right)}{196000\text{ g/mol}}\\ =3.487\times {10}^{21}\text{ chains}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The total length of chains in the given rope is to be calculated.

Concept Introduction:

The formula to calculate the length $\left(L\right)$ of 1 chain of a polymer is:

  $L=n\times L_{\text{repeat unit}}+\left(n-1\right)\times L_{\text{C-C}}$   ....... (3)

Here, $n$ is the number of repeating units in the polymer chain, $L_{\text{repeat unit}}$ is the length of one repeating unit, and $L_{\text{C-C}}$ is the length between two carbon atoms in the chain.

Answer to Problem 16.14P

The total length of chains in the given rope is $9.637\times {10}^{12}\text{ km}$.

Explanation of Solution

Given information:

A rope made up of polyethylene weighs $0.25\text{ lb}$ per foot. One chain of this polymer contains $7000$ repeat units.

The length of the rope is $10\text{ ft}$ .The length between each carbon atom is approximately $0.15\text{ nm}$ and length of each repeat unit is $0.24495\text{ nm}$.

From the given data:

  $\begin{array}{c}n=7000\\ L_{\text{repeat unit}}=0.24495\text{ nm}\\ L_{\text{C-C}}=0.15\text{ nm}\end{array}$

Use equation (3) to calculate the length of $1$ polymer chain as:

  $L=n\times L_{\text{repeat unit}}+\left(n-1\right)\times L_{\text{C-C}}$

Substituting the values:

  $\begin{array}{c}=7000\times \left(0.24495\text{ nm}\right)+\left(7000-1\right)\times \left(0.15\text{ nm}\right)\\ =2764.5\text{ nm}\end{array}$

Now, calculate the total length of the chains in the given polymer rope as:

  $\begin{array}{c}{L}_{Total}=\left(3.487\times {10}^{21}\text{ chains}\right)\times \left(2764.5\text{ nm}\right)\\ =9.637\times {10}^{24}\text{ nm}\\ =9.637\times {10}^{12}\text{ km}\end{array}$