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 .

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rejkjavik rejkjavik · Answer 1 · 2019-10-10T08:50:25+02:00

Answer:

A) [tex]\rho_c = 1.4223 g/cm^3[/tex]

B) 66.6%

Explanation:

A) expression for crystallanity is given as

[tex]C =\frac{\rho_c(\rho_S -\rho_A)}{\rho_S (\rho_c -\rho_A)}[/tex] ........1

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

[tex]\rho_s[/tex] = density of specimen

C = 0.743 [given]

[tex]0.743 =\frac{\rho_c(1.408 -\rho_A)}{1.408 (\rho_c -\rho_A)}[/tex]

[tex]1.2427\rho_c - 1.2427\rho_A - 1.408\rho_c + \rho_c \rho_A = 0[/tex]

[tex]-0.1653\rho_c - 1.2427\rho_A = - \rho_c\rho_A[/tex] ........2

SUBSTITUTE[tex] \rho_s[/tex] value of second specimen in 1 equation we get

[tex]-0.923 \rho_c - 0.419\rho_A = -\rho_c \rho_A[/tex] .......3

ON COMPARING 2 AND 3rd equation we get

[tex]\rho_c = 1.4223 g/cm^3[/tex]

[tex]\rho_A = 1.3085 g/m^3[/tex]

b)

[tex]C =\frac{\rho_c(\rho_S -\rho_A)}{\rho_S (\rho_c -\rho_A)}[/tex]

substitute value to ger crystallanity

[tex]C =\frac{1.422 (1.382 -1.3085)}{1.382 (1.422 -1.3085)}[/tex]

C = 66.6 %