Viscosity and polymers - Lab Report Example

AND CALCULATIONS EXPERIMENT 17: Viscosity of Liquids Question Assume that ∆P and L are constants. The viscosity is therefore a function of r only. Integrating equation 2 with respect to t we get; V= Rearranging to make the subject, we get Absolute error = Relative error = Absolute error/ actual = Therefore, the relative is fourfold the radius of the viscometer. Question 2Values obtained from this formulaDensity = mass/volume Mass = density x volume Viscosity = density x area/timeSince a similar viscometer was used, the cross-sectional area is constant and therefore the viscosity formula reduces to density/time.

Table 1: Methanol + Water% methanolAverage time(sec)Density(gcm-3ViscosityGcm-1s-1Volume of Methanol (ml)Volume of water (ml)053.03750.9770.0180102085.9250.93920.0112840109.35750.90140.0084660102.150.86360.008648083.92750.82580.018210049.8850.7880.016100% Toluene Average Time(sec)DensityViscosityVolume of TolueneVolume of Xylene084.34750.8570.010102084.3470.85780.01284081.68750.85860.01466080.6050.85940.01648079.0450.86020.018210078.530.8610.01100Data Analysis From the results above, viscosity decreases with increase in methanol content.

For Toluene and Xylene, viscosity is constant. Viscosity reduces with increase in intermolecular forces and vice versa. If the resulting solution forms stronger intermolecular forces as in methanol and water the viscosity is lowered. For weak intermolecular forces, as in toluene and Xylene mixture remains constant, i.e. equal to the viscosity of the individual liquids.Question 3Viscosity is inversely proportional to the intermolecular forces in a solution. Viscosity of water is higher than that of methanol.

Water molecules combine with methanol molecules to form stronger intermolecular forces which results to a decrease in viscosity.EXPERIMENT 32: The Monomer Linkage Properties and Molecular Weight of Poly (vinyl alcohol)Question 1Mv/Mn = [(1+a)ᴦ(1+a)]1/a = SFrom gamma function tables for a=0.5, ᴦ(1+a) = ᴦ1.5 = 0.886227 For a = 0.6’ ᴦ(1+a) = ᴦ1.6 = 0.893515 S = [1+0.6(0.893515)]1/0.6 = 1.81Question 2Decrease in molecular weight =1/ Mn’ – 1/Mno Where Mn’ is the number average molecular weight of polymer in the sample after cleavage.

Mno is the number average molecular weight of the polymer in sample before cleavage.The molecular weight of the monomer unit = Mo.Change in the average molecular weight is expressed as a ratio ƒ. Ƒ = Simplifying Ƒ = (Mo/ Mn’ ) – (Mo/ Mno)Multiplying both sides by MvMv Ƒ = [(Mo/ Mn’ ) – (Mo/ Mno)] MvBut Mv/Mn = S From equation 17.Therefore, Mv Ƒ = MoS’ - Moso Ƒ = (MoS’ - Moso)/ MvFactorizing, Ƒ = MoS (1/ Mv’ – 1/Mvo)Question 3An experiment which measure the direct force generated during polymerization of the protein or biopolymer molecules using an optical trap can be conducted.

Question 4For PVOH, whose structure is spherical the assumption is appropriate. The assumption yields poor results in non-spherical non-polar molecular structures with longer hydrocarbon chain.Question 5The water molecule in vapor form enters the PVOH molecule to minimize the solvolytic associations. The molecule consequently expands and swells out. In aqueous solutions, the polymer knots up and avoid interaction with water molecules. This is as a result of the temperature difference between the two media.Question 6