Learn how to estimate number-average molecular weight from colligative properties and viscosity-average molecular weight from viscometer measurements. Classify polymers based on crystallinity and solve problems involving osmotic pressure and falling time data analysis.
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Presentation Transcript
(Molecular-weight Estimation from Experimental Data) At the end of this lecture, you should be able to: - Estimate number-average molecular weight from the colligative properties (those properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution). - Estimate viscosity-average molecular weight from a viscometer measurements. - Classify polymers based on the degree of crystallinity . Mustansiriyah University/ Materials Eng. Dept. Dr Ahmed A. AYASH March 2020
QUESTION 1: Given the following data of osmotic pressure (?) of polycarbonate solution, what is the number-average molecular weight of this polymer? (T = 25 C; density = 1.0 g/cm3.) ?/? (pa L/g) 0.255 0.26 0.27 0.27 0.29 0.295 C (g/L) 0.001 0.009 0.015 0.019 0.023 0.029 https://www.protec-arisawa.com/reverse- osmosis-the-power-to-dominate-the-ocean/
SOLUTION: The relation between the osmotic pressure , concentration and the number-average molecular weight is as follows (please this equation is given in the M.wt lecture): (1)
Now, return to Eq. 1: (1) Comparing this equation with the general form of a line equation ,it is easy to note that the termRT Mn Therefore, one can conclude that the value of the term Mn = 8314*298/0.2093 ( or Mn = 11,837,419 g/mole). represents the intercept value the red circled area in the graph . RT Mn is 0.2093 pa cm3/g and hence L Pa K mol NOTE: the gas constant (R ) is 8314 L Pa K 1 mol 1
QUESTION 2: Given the following data of the falling time of polystyrene-cyclohexane solution in a viscometer, what is the viscosity-average molecular weight of this polymer? (T = 35 C) t0=3 4 7 9 10 12 14 ??????? ???? (???) C (g/L) 0 0.001 0.009 0.015 0.019 0.023 0.029 SOLUTION: The viscosity average molecular weight is correlated with the intrinsic viscosity through Mark-Howink equation as follows:
The intrinsic viscosity in the above equation can be found from the reduced viscosity When the polymer solution concentration approaching zero (i.e. very dilute solution) : This can be found graphically by plotting the reduced viscosity against the concentration using the given data : Falling time (sec) t0 = 3 8 14 22 32 44 56 C (g/L) 0 0.1 0.2 0.3 0.4 0.5 0.6 0 16.67 18.33 21.11 24.17 27.33 29.44 ?red NOTE: this equation is given in the M.wt Lecture.
The intrinsic viscosity from the plot is 13.4 and Mark-Howink constants are 0.08 and 0.5 for K and a , respectively. Now,, applying Mark-Howink equation gives: 13.4 = 0.08 (Mv)0.5 Mv = 28000 g/mol. H.W: Repeat QUESTION 2 using the value of the inherent viscosity where:
QUESTION 3 Classify polystyrene whether it is crystalline , semi-crystillane or amorphous, if it has a density of 0.75 g/cm3 and the density of totally amorphous material is 0.65 g/cm3. Furthermore, the unit cell volume equals to 88 X10-23 cm3/unit cell and four styrene repeating units are contained within each unit cell. Solution: To find whether the given polystyrene is crystalline , semi-crystillane or amorphous, one may need to calculate the degree of crystallinity: To use the above equation , one may need to calculate the density of totally crystalline polystyrene using the following relation: ?? ? = ?? ?? (1)
Where n represents the number of repeat unit within the unit cell (for polystyrene n = 4), and A is the repeat unit molecular weight, which for polystyrene is C8H8: A= 8 (12.01)+8(1.008)= 104.14 g/gmol Vc is unit cell volume and equals to 88X10-23 cm3/unit cell, NA is Avogadro number (6.022 X 1023 repeat unit/mol). Now, substituting the above parameters in Eq.1, one can find the density of totally crystalline polystyrene: ? (4 ?????? ????? )(104.14 (88?10 23??3 ???) ??= )= 0.786 g/mol ????)(6.022?1023?????? ???? ??? Now, one can use the given information and the calculated density value of totally crystalline polystyrene to find the degree of crystallinity as follows: Degree of Crystallinity=0.786(0.75 0.65) 0.75 0.786 0.65 0.77 or 77% So, polystyrene can be classified as a crystalline polymer since the degree of crystallinity is more than 50%.
