The density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows: \begin{tabular}{cc} \hline\(\rho\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) & crystallinity \((\%)\) \\\ \hline \(2.144\) & \(51.3\) \\ \(2.215\) & \(74.2\) \\ \hline \end{tabular} (a) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene. (b) Determine the percent crystallinity of a specimen having a density of \(2.26 \mathrm{~g} / \mathrm{cm}^{3}\).

Firstly, we will determine the densities of totally crystalline (\(\rho_c\)) and totally amorphous (\(\rho_a\)) PTFE materials, by setting up two equations based on the given data: 1. \(2.144 = \rho_c (0.513) + \rho_a (1-0.513)\) 2. \(2.215 = \rho_c (0.742) + \rho_a (1-0.742)\) Now we will solve these simultaneous equations to find \(\rho_c\) and \(\rho_a\).

To solve the equations, we can first solve Equation 1 for \(\rho_a\): \(\rho_a = \frac{2.144 - \rho_c (0.513)}{1 - 0.513}\) Now, we can substitute this expression for \(\rho_a\) into Equation 2 and solve for \(\rho_c\): \(2.215 = \rho_c (0.742) + \frac{2.144 - \rho_c (0.513)}{1 - 0.742}(1-0.742)\) After solving this equation, we obtain \(\rho_c \approx 2.28 \, \mathrm{g}/\mathrm{cm}^3\) Now, let's use the value of \(\rho_c\) to find the density of the amorphous phase, \(\rho_a\): \(\rho_a = \frac{2.144 - 2.28(0.513)}{1 - 0.513} \approx 1.98 \, \mathrm{g}/\mathrm{cm}^3\) So, the density of totally crystalline PTFE material is approximately \(2.28 \, \mathrm{g}/\mathrm{cm}^3\) and the density of totally amorphous PTFE material is approximately \(1.98 \, \mathrm{g}/\mathrm{cm}^3\).

Now, we have to determine the percent crystallinity of a specimen with a density of \(2.26\, \mathrm{g}/\mathrm{cm}^3\). We can use the density equation \(\rho = \rho_c x_c + \rho_a x_a\) and solve for the percent crystallinity \(x_c\). For the given density \(\rho\), we have: \(2.26 = 2.28x_c + 1.98(1-x_c)\) Now, we must solve this equation for \(x_c\).

To solve for \(x_c\), we first rearrange the equation: \(x_c = \frac{2.26 - 1.98}{2.28 - 1.98}\) After solving, we obtain \(x_c \approx 0.7\) Now, convert the fraction to a percentage: \(\%\) crystallinity \(= 0.7 \times 100 \approx 70\%\) Thus, the percent crystallinity of the specimen with a density of \(2.26 \, \mathrm{g}/\mathrm{cm}^3\) is approximately 70%.