The density of totally crystalline nylon 6,6 at room temperature is \(1.213 \mathrm{~g} / \mathrm{cm}^{3} .\) Also, at room temperature the unit cell for this material is triclinic with the following lattice parameters: $$ \begin{array}{ll} If the volume of a triclinic unit cell, is a function of these lattice parameters as $$ V_{\mathrm{tri}}=a b c \sqrt{1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma+}{2 \cos \alpha \cos \beta \cos \gamma} $$ determine the number of repeat units per unit cell. a=0.497 \mathrm{~nm} & \alpha=48.4^{\circ} \\ b=0.547 \mathrm{~nm} & \beta=76.6^{\circ} \\ c=1.729 \mathrm{~nm} & \gamma=62.5^{\circ} \end{array} $$

Short Answer

Expert verified
Answer: There are approximately 26 repeat units per unit cell of totally crystalline nylon 6,6 at room temperature.

Step by step solution

01

Calculate the volume of the triclinic unit cell

First, we need to find the volume of the triclinic unit cell using the given formula and lattice parameters: $$ V_{\mathrm{tri}}=a b c \sqrt{1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma} $$ Plug in the given values for the lattice parameters: $$ V_{\mathrm{tri}}=(0.497)(0.547)(1.729) \sqrt{1-\cos ^{2} 48.4^{\circ}-\cos ^{2} 76.6^{\circ}-\cos ^{2} 62.5^{\circ} +2 \cos 48.4^{\circ} \cos 76.6^{\circ} \cos 62.5^{\circ}} $$ Compute the volume: $$ V_{\mathrm{tri}} \approx 0.474 \mathrm{~nm}^{3} $$
02

Determine the volume of a single nylon 6,6 repeat unit

To find the volume of a single nylon 6,6 repeat unit, we will use the given density and the molar mass of nylon 6,6. The molar mass of nylon 6,6 is \(226.32 \mathrm{~g/mol}\) (12 carbon atoms, 22 hydrogen atoms, 2 nitrogen atoms, and 2 oxygen atoms). First, convert the density to \(\mathrm{g/nm}^3\): $$ 1.213 \frac{\mathrm{g}}{\mathrm{cm}^3} \times \frac{1 \mathrm{~nm}^3}{1\times10^{-21} \mathrm{~cm}^3} = 1.213 \times 10^3 \frac{\mathrm{g}}{\mathrm{~nm}^3} $$ Now determine the volume of one nylon 6,6 repeat unit: $$ \frac{1 \mathrm{mol}}{N_A} \times \frac{226.32 \mathrm{~g}}{1 \mathrm{mol}} \times \frac{1 \mathrm{~nm}^3}{1.213 \times 10^3 \mathrm{~g}} \approx 1.839 \times 10^{-2} \mathrm{~nm}^3 $$ Where \(N_A\) is Avogadro's number.
03

Calculate the number of repeat units per unit cell

Now, we can find the number of repeat units per unit cell by dividing the total unit cell volume by the volume of a single repeat unit: $$ \frac{0.474 \mathrm{~nm}^{3}}{1.839\times10^{-2} \mathrm{~nm}^3} \approx 25.79 $$ However, the number of repeat units must be an integer. Therefore, we round the result to the nearest whole number: $$ \mathrm{Number~of~repeat~units} \approx 26 $$ Hence, there are approximately 26 repeat units per unit cell of totally crystalline nylon 6,6 at room temperature.

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Most popular questions from this chapter

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 \(325 \mathrm{~K}\) ?

Consider the diffusion of oxygen through a low(- density polyethylene (LDPE) sheet \(15 \mathrm{~mm}\) thick. The pressures of oxygen at the two faces are 2000 \(\mathrm{kPa}\) and \(150 \mathrm{kPa}\), which are maintained constant. Assuming conditions of steady state, what is the diffusion flux [in \(\left.\left(\mathrm{cm}^{3} \mathrm{STP}\right) / \mathrm{cm}^{2} \cdot \mathrm{s}\right]\) at \(298 \mathrm{~K}\) ?

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