Question

this problem is a bit more difficult than the other problems. the density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows:
ρ(g/cm³) crystallinity (%)
2.10 64.0
2.30 77.0
(a) compute the density of totally crystalline polytetrafluoroethylene. ρ_c = select g/cm³
(b) compute the density of totally amorphous polytetrafluoroethylene. ρ_a = 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 $X_c$ is $X_c=\frac{
ho_c(
ho-
ho_a)}{
ho(
ho_c -
ho_a)}\times100\%$, 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. We can rewrite it for two data points $(X_{c1},
ho_1)$ and $(X_{c2},
ho_2)$ as two equations:
$X_{c1}=\frac{
ho_c(
ho_1-
ho_a)}{
ho_1(
ho_c -
ho_a)}\times100\%$ and $X_{c2}=\frac{
ho_c(
ho_2-
ho_a)}{
ho_2(
ho_c -
ho_a)}\times100\%$. Let $X_{c1} = 64.0\%=0.64$, $
ho_1 = 2.10\ g/cm^3$, $X_{c2}=77.0\% = 0.77$, $
ho_2 = 2.30\ g/cm^3$.

Step2: Substitute the values into the equations

$0.64=\frac{
ho_c(2.10-
ho_a)}{2.10(
ho_c -
ho_a)}$ and $0.77=\frac{
ho_c(2.30-
ho_a)}{2.30(
ho_c -
ho_a)}$.
From $0.64=\frac{
ho_c(2.10-
ho_a)}{2.10(
ho_c -
ho_a)}$, we get $0.64\times2.10(
ho_c -
ho_a)=
ho_c(2.10-
ho_a)$, which simplifies to $1.344
ho_c-1.344
ho_a = 2.10
ho_c-
ho_c
ho_a$.
From $0.77=\frac{
ho_c(2.30-
ho_a)}{2.30(
ho_c -
ho_a)}$, we get $0.77\times2.30(
ho_c -
ho_a)=
ho_c(2.30-
ho_a)$, which simplifies to $1.771
ho_c-1.771
ho_a = 2.30
ho_c-
ho_c
ho_a$.
Subtract the first - derived equation from the second:
$(1.771
ho_c-1.771
ho_a-(1.344
ho_c - 1.344
ho_a))=(2.30
ho_c-
ho_c
ho_a-(2.10
ho_c-
ho_c
ho_a))$.
$1.771
ho_c-1.771
ho_a - 1.344
ho_c + 1.344
ho_a=2.30
ho_c-
ho_c
ho_a - 2.10
ho_c+
ho_c
ho_a$.
$0.427
ho_c - 0.427
ho_a=0.20
ho_c$.
$0.427
ho_c-0.20
ho_c = 0.427
ho_a$.
$0.227
ho_c = 0.427
ho_a$, so $
ho_a=\frac{0.227}{0.427}
ho_c$.

Step3: Substitute $

ho_a$ back into the first percent - crystallinity equation
Substitute $
ho_a=\frac{0.227}{0.427}
ho_c$ into $0.64=\frac{
ho_c(2.10-\frac{0.227}{0.427}
ho_c)}{2.10(
ho_c - \frac{0.227}{0.427}
ho_c)}$.
First, simplify the denominator: $
ho_c-\frac{0.227}{0.427}
ho_c=
ho_c(1 - \frac{0.227}{0.427})=
ho_c\frac{0.427 - 0.227}{0.427}=
ho_c\frac{0.2}{0.427}$.
The numerator is $
ho_c(2.10-\frac{0.227}{0.427}
ho_c)$.
$0.64=\frac{2.10-\frac{0.227}{0.427}
ho_c}{2.10\times\frac{0.2}{0.427}}$.
$0.64\times2.10\times\frac{0.2}{0.427}=2.10-\frac{0.227}{0.427}
ho_c$.
$0.64\times2.10\times\frac{0.2}{0.427}-2.10=-\frac{0.227}{0.427}
ho_c$.
$
ho_c\approx2.40\ g/cm^3$.

Step4: Calculate $

ho_a$
Substitute $
ho_c = 2.40\ g/cm^3$ into $
ho_a=\frac{0.227}{0.427}
ho_c$.
$
ho_a=\frac{0.227}{0.427}\times2.40\approx1.28\ g/cm^3$.

Step5: Calculate the percent crystallinity for $

ho = 2.20\ g/cm^3$
Use the formula $X_c=\frac{
ho_c(
ho-
ho_a)}{
ho(
ho_c -
ho_a)}\times100\%$.
Substitute $
ho_c = 2.40\ g/cm^3$, $
ho_a = 1.28\ g/cm^3$, and $
ho = 2.20\ g/cm^3$ into the formula:
$X_c=\frac{2.40\times(2.20 - 1.28)}{2.20\times(2.40 - 1.28)}\times100\%=\frac{2.40\times0.92}{2.20\times1.12}\times100\%=\frac{2.208}{2.464}\times100\%\approx89.6\%$.

Answer:

(a) $2.40$
(b) $1.28$
(c) $89.6$