this problem is a bit more difficult than the other problems. the density and associated percent…

this problem is a bit more difficult than the other problems. the density and associated percent crystallinity for two polytetrafluoroethylene materials follows: ρ(g/cm³) crystallinity (%) 2.10 64.0 2.30 77.0 (a) compute the density of totally crystalline polytetrafluoroethylene. ρc select 2.17 2.77 1.87 1.57 g/cm³ (b) comp select g/cm³ (c) determine the percent crystallinity of a specimen having a density of 2.20 g/cm³. % crystallinity = select

this problem is a bit more difficult than the other problems. the density and associated percent crystallinity for two polytetrafluoroethylene materials follows: ρ(g/cm³) crystallinity (%) 2.10 64.0 2.30 77.0 (a) compute the density of totally crystalline polytetrafluoroethylene. ρc select 2.17 2.77 1.87 1.57 g/cm³ (b) comp select g/cm³ (c) determine the percent crystallinity of a specimen having a density of 2.20 g/cm³. % crystallinity = select

Answer

Explanation:

Step1: Recall the density - crystallinity relationship formula

The formula for percent crystallinity ($%C$) is $%C=\frac{\rho_c(\rho - \rho_a)}{\rho(\rho_c-\rho_a)}\times100$, where $\rho$ is the density of the semi - crystalline polymer, $\rho_c$ is the density of the totally crystalline polymer, and $\rho_a$ is the density of the totally amorphous polymer. Let $\rho_1 = 2.10\ g/cm^3$, $%C_1=64.0%$, $\rho_2 = 2.30\ g/cm^3$, and $%C_2 = 77.0%$.

Step2: Substitute the first set of values into the formula

$0.64=\frac{\rho_c(2.10 - \rho_a)}{2.10(\rho_c-\rho_a)}$ $0.64\times2.10(\rho_c-\rho_a)=\rho_c(2.10 - \rho_a)$ $1.344\rho_c-1.344\rho_a=2.10\rho_c-\rho_c\rho_a$

Step3: Substitute the second set of values into the formula

$0.77=\frac{\rho_c(2.30 - \rho_a)}{2.30(\rho_c-\rho_a)}$ $0.77\times2.30(\rho_c-\rho_a)=\rho_c(2.30 - \rho_a)$ $1.771\rho_c-1.771\rho_a=2.30\rho_c-\rho_c\rho_a$

Step4: Solve the system of equations for $\rho_c$ and $\rho_a$

From $1.344\rho_c-1.344\rho_a=2.10\rho_c-\rho_c\rho_a$ and $1.771\rho_c-1.771\rho_a=2.30\rho_c-\rho_c\rho_a$, we can subtract the first equation from the second: $(1.771\rho_c-1.771\rho_a)-(1.344\rho_c - 1.344\rho_a)=(2.30\rho_c-\rho_c\rho_a)-(2.10\rho_c-\rho_c\rho_a)$ $0.427\rho_c - 0.427\rho_a=0.20\rho_c$ $0.427\rho_c-0.20\rho_c=0.427\rho_a$ $0.227\rho_c=0.427\rho_a$ $\rho_a=\frac{0.227}{0.427}\rho_c$. Substitute $\rho_a$ into the first equation $0.64=\frac{\rho_c(2.10 - \frac{0.227}{0.427}\rho_c)}{2.10(\rho_c-\frac{0.227}{0.427}\rho_c)}$ After simplification and solving, we get $\rho_c = 2.77\ g/cm^3$ and $\rho_a = 1.50\ g/cm^3$.

Step5: Calculate the percent crystallinity for $\rho = 2.20\ g/cm^3$

Using the formula $%C=\frac{\rho_c(\rho - \rho_a)}{\rho(\rho_c-\rho_a)}\times100$, substitute $\rho_c = 2.77\ g/cm^3$, $\rho_a = 1.50\ g/cm^3$, and $\rho = 2.20\ g/cm^3$ $%C=\frac{2.77(2.20 - 1.50)}{2.20(2.77 - 1.50)}\times100$ $=\frac{2.77\times0.70}{2.20\times1.27}\times100$ $=\frac{1.939}{2.794}\times100\approx69.4%$

Answer:

(a) 2.77 g/cm³ (b) 1.50 g/cm³ (c) 69.4%