Material Science Test 2 Study Guide

Material Science Test 2 Study Guide - Material Science Test...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Material Science Test 2 Study Guide Describe what is a glass transition temperature For any amorphous solid , it is the critical temperature that separates rubber behavior from glassy behavior In a volume vs. temperature graph identify the liquid, glass, semi- crystalline and crystalline regions. Also identify T g and T m . Look at graphs: Chapter 6, pg. 187 T g =glass transition temperature Below T g , molecular motion is extremely slow, above it we have molecular motion T m =melting point Specific volume (y-axis) vs. Temperature (x-axis) Liquid to crystalline solid transformation Zero slope at T m (discontinuous change in volume occurs at the melting temperature: Will not have a T g ) Liquid to glass transformation The slope changes at T g Between T g and T m the new slope makes a supercooled liquid Semi-crystalline The slope will change at T g and at T m At T m the slope changes in a discontinuous change in volume Difference between crystals and non-crystals Non-crystals (amorphous)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Short range order Corrosion resistant No grains No X-ray diffraction peaks Transparent Crystalline Short and Long Range Order Will corrode Grain boundaries (polymorphic) X ray diffraction peaks Nontransparent (scatters light) Define what is a polymer and how polymers are classified based on their backbone and based on their bond Polymer Organic-based macromolecules made of a “carbon backbone” to which several side groups are attached Two important properties of polymers and in general of any amorphous material No grain boundaries Transparent (do not scatter light) Classification of polymers Based on the nature of their backbone Vinyl’s Rubbers
Background image of page 2
Polyesters Polyamides Others Based on bonding nature Thermoplastics Have primary and secondary bonds Thermosets Primary bonds Determine the degree of polymerization, average molecular weight and polydispersity of a polymer (CH 2 -CH 2 ) n n=degree of polymerization Average molecular weight Carbon=12g/mol H=1g/mol 2(12)+4(1)=28g/mol Polydispersity of a polymer P=1 (all the molecules have the same molecular weight) Example Calculate polydispersity of polypropylene which has 2 types of chains: %60 of the chains have n=5000, %40 of the chains have n=10000 Step 1 Calculate the molecular weight of repeating unit
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3(12)+6(1)=42g/mol Step 2 Evaluate for each chain the molecular weight Chain 1 – M 1 =(42g/mol)(5000)=210000g/mol Chain 2 – M 2 =(42g/mol)(10000)=420000g/mol Step 3 (M n =Number average of molecular weight) Calculate M n =∑(N i M i )/ ∑(N i ) N i =number of molecules M i = with molecular weight M n =((%60)(210000)+(%40)(420000))/ (%60+%40)= 294000g/mol
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 15

Material Science Test 2 Study Guide - Material Science Test...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online