Question

The density and associated percent crystallinity for two poly(ethylene terephthalate) materials are as follows: rho (g/cm3 ) Crystallinity (%) 1.408 74.3 1.343 31.2 (a) Compute the densities of totally crystalline and totally amorphous poly(ethylene terephthalate). (b) Determine the percent crystallinity of a specimen having a density of 1.382 g/cm3 .

Answer 1

A)
`\rho_c = 1.4223 g/cm^3`

B) 66.6%

Step-by-step explanation:

A) expression for crystallanity is given as

`C =(\rho_c(\rho_S -\rho_A))/(\rho_S (\rho_c -\rho_A))` ........1

where, \rho_c = density of crystalline, \rho_a = density of amorphous material,

`\rho_s` = density of specimen

C = 0.743 [given]

`0.743 =(\rho_c(1.408 -\rho_A))/(1.408 (\rho_c -\rho_A))`

`1.2427\rho_c - 1.2427\rho_A - 1.408\rho_c + \rho_c \rho_A = 0`

`-0.1653\rho_c - 1.2427\rho_A = - \rho_c\rho_A` ........2

SUBSTITUTE
`\rho_s` value of second specimen in 1 equation we get

`-0.923 \rho_c - 0.419\rho_A = -\rho_c \rho_A` .......3

ON COMPARING 2 AND 3rd equation we get

`\rho_c = 1.4223 g/cm^3`

`\rho_A = 1.3085 g/m^3`

b)

`C =(\rho_c(\rho_S -\rho_A))/(\rho_S (\rho_c -\rho_A))`

substitute value to ger crystallanity

`C =(1.422 (1.382 -1.3085))/(1.382 (1.422 -1.3085))`

C = 66.6 %