Question

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

Explanation:

Step1: Recall the density - crystallinity relationship formula

The formula for percent crystallinity ($\%C$) is $\%C=\frac{
ho_c(
ho -
ho_a)}{
ho(
ho_c-
ho_a)}\times100$, where $
ho$ is the density of the semi - crystalline polymer, $
ho_c$ is the density of the totally crystalline polymer, and $
ho_a$ is the density of the totally amorphous polymer. Let $
ho_1 = 2.10\ g/cm^3$, $\%C_1=64.0\%$, $
ho_2 = 2.30\ g/cm^3$, and $\%C_2 = 77.0\%$.

Step2: Substitute the first set of values into the formula

$0.64=\frac{
ho_c(2.10 -
ho_a)}{2.10(
ho_c-
ho_a)}$
$0.64\times2.10(
ho_c-
ho_a)=
ho_c(2.10 -
ho_a)$
$1.344
ho_c-1.344
ho_a=2.10
ho_c-
ho_c
ho_a$

Step3: Substitute the second set of values into the formula

$0.77=\frac{
ho_c(2.30 -
ho_a)}{2.30(
ho_c-
ho_a)}$
$0.77\times2.30(
ho_c-
ho_a)=
ho_c(2.30 -
ho_a)$
$1.771
ho_c-1.771
ho_a=2.30
ho_c-
ho_c
ho_a$

Step4: Solve the system of equations for $

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

Step5: Calculate the percent crystallinity for $

ho = 2.20\ g/cm^3$
Using the formula $\%C=\frac{
ho_c(
ho -
ho_a)}{
ho(
ho_c-
ho_a)}\times100$, substitute $
ho_c = 2.77\ g/cm^3$, $
ho_a = 1.50\ g/cm^3$, and $
ho = 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%