1 / 24

Chapter 4

Chapter 4. Effect of Chemical Forces on Physical Properties. Melting Points Thermal Expansion Stiffness or Young’s moduli Theoretical Strengths Surface Energy . Melting Points. What does the ∆ refer to?

laurel
Download Presentation

Chapter 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 4 Effect of Chemical Forces on Physical Properties Melting Points Thermal Expansion Stiffness or Young’s moduli Theoretical Strengths Surface Energy

  2. Melting Points What does the ∆ refer to? ∆Sf for most crystalline solids, from solid Ar to MgO and Ta with very high MPs are ≈ 10-12 J/mol K

  3. Properties From Bonding: Tm Energy r r o r Energy smaller Tm Equilibrium distance larger Tm r o r Eo = “bond energy” • Bond length, r • Melting Temperature, Tm • Bond energy, Eo Tm is larger if Eo is larger.

  4. What Determines MP? Bond Energy i) Most important z1 and z2: higher z1 and z2 results in?? ii) Second is ro - larger ro result in ???? b)Covalent character of ionic bond If CCIB results in uneven bonding, the melting point is reduced….

  5. Examples Which has higher MP: NaCl or MgO? NaCl or NaBr? Why? MgO or Al2O3? Why?

  6. Thermal Expansion

  7. Properties From Bonding: length, L o unheated, T 1  L heated, T 2 r o Energy Equilibrium distance r Larger  Eo Eo smaller  • Coefficient of thermal expansion,  coeff. thermal expansion  L  = ( T - T ) 2 1 L o •  ~ asymmetry at ro  is larger if Ebond is smaller.

  8. Generalizations: for ceramics typically < metals is a function of T. Thus you need to specify T range c) Atomic packing can be crucial More open structures - like those of covalent ceramics - have lower ’s e.g. SiC and Si3N4 Quartz and fused silica. Atoms vibrate in the “open spaces” d) Thermal expansion anisotropy can also be very important

  9. This animation will not play on your computer

  10. Young’s Moduli What is relationship between energy and force? F = dE/dr Many problems in science and engineering can be solved by minimizing energy or equating forces. You should get the same answer at the end.

  11. ro ro ro2 Stress = Y x strain where Y is a measure of stiffness. Strong bonds are stiff … Y = slope of F vs. r curve at ro ro

  12. Exercise: Plot corresponding E vs. r plots of the two bonds shown in this figure. Which one will be stiffer?

  13. This animation will probably not play in your computer

  14. Theoretical Bond Strengths So = ?? You can easily show that rf = 1.25 ro

  15. rf = 1.25 ro If you actually find the second derivative of E vs. r, i.e. solve it exactly (see Problem 4.2) you will get:

  16. This animation will not play in your computer

  17. Surface Energy

  18. Calculation of Surface Energies Ns = number of broken bonds per unit area. Ebond = bond energy CN = coordination number CNplane = coordination number in plane NA = number of atoms in plane A = area of plane.

  19. Summary Strong bonds result in solids with high melting points that are also stiff and posses high theoretical strengths and low thermal expansion coefficients. They also result in solids with high surface energies. Large bond energy large Tm large E small 

More Related