Two amorphous linear polymers have high number densities of pendant groups. Assu
ID: 998631 • Letter: T
Question
Two amorphous linear polymers have high number densities of pendant groups. Assume that the pendant groups of polymer "A" are C_4 groups (i.e. each pendant group has 4 C-C bonds); the pendant groups of polymer "B" are C_16 groups (i.e. each pendant group has 16 C-C bonds). A C_4 group can be described as -CH_2-CH_2-CH_2-CH_3; a C_16 group is of a similar structure but four times longer. Every thing else of the two polymers, e g. the degree of polymerization, is the same. Which one tends to have a lower glass transition temperature, 7g? Why? Which one tends to have a lower melting point, T_m? Why? Clearly state all your assumptions. A constitutive equation is the governing equation that relates stress, sigma(t), to strain,epsilon(t), where t is time. For instance, for a linear elastic material, the constitutive equation is the Hooke's law: sigma(t) = E. epsilon(t), where E is the Young's modulus. Derive the constitutive equation of the two-parameter Voigt model, in which a spring and a dashpot are placed in parallel with each other. Denote the spring constant as k and the dashpot viscosity as eta In a constant-stress-rate (CSR) test, a tensile stress is applied and it is increased at a constant rate, R_sigma, i.e.sigma(t) = Rat. If a polymer can be characterized by the two-parameter Voigt model, derive its strain,epsilon(t), for the CSR test.Explanation / Answer
QUESTION 3
a) If the polymer chain has high mobility the Tg is low. The C16 group is bulkier than the C4 group and the polymer chain with the C16 group has more free volume, which enables it to move freely. Therefore, the polymer with the C16 pendant group has the lower Tg.
b) As the bulkiness of the pendant group increases, Tm decreases. Therefore, the polymer with the C16 pendant group has the lowe Tm.