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Physics 7A – Lecture 7 Winter 2008

Physics 7A – Lecture 7 Winter 2008. Prof. Robin D. Erbacher 343 Phy/Geo Bldg erbacher@physics.ucdavis.edu. Announcements. No Quiz today, Quiz 5 is next week. Join this Class Session with your PRS clicker! Final Exam review schedule is posted, 3 days full!

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Physics 7A – Lecture 7 Winter 2008

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  1. Physics 7A – Lecture 7Winter 2008 Prof. Robin D. Erbacher 343 Phy/Geo Bldg erbacher@physics.ucdavis.edu

  2. Announcements • No Quiz today, Quiz 5 is next week. • Join this Class Session with your PRS clicker! • Final Exam review schedule is posted, 3 days full! • Website has moved: power outages fried the mac: • http://www.physics.ucdavis.edu/physics7/ • Check Physics 7 website frequently for updates. • Turn off cell phones and pagers during lecture.

  3. ReviewMultiple- Atom Systems:Particle Model of Ebond ,Particle Model of Ethermal

  4. Particle Model of Ebondand Ethermal Example: H2O What is Ebond in terms of KE and PE of individual atom (atom pair)? What is Ethermal in terms of KE and PE of individual atom (atom pair)?

  5. Multiple Atom Systems: Ebond • Typically every pair of atoms interacts • Magnitude of Ebond for a substance is the amount • of energy required to break apart “all” the bonds • i.e. we define Ebond = 0 when all the atoms are separated • We treat bonds as “broken” or “formed”. Bond • energy e(per bond) exists as long as the bond exists. • The bond energy of a large substance comes • from adding all the potential energies of particles • at their equilibrium positions. • Ebond = ∑all pairs(PEpair-wise)

  6. 4.4 A 6.6 A 11 A 2.2 A 8.8 A ~ 0 J ~ 0 J Atom-atom potential for each atom Separation (10-10m) What is the change in bond energy (∆Ebond) by removing the red atom? Bond energy -8 x 10-21J -0.5 x 10-21J ~ 0 J

  7. Atom-atom potential for each atom Separation (10-10m) What is the bond energy Ebond for the entire molecule? -8 x 10-21J -8.5 x 10-21J -8.5 x 10-21J -8.5 x 10-21J -8.5 x 10-21J Ebond = -42 x 10-21 Joules

  8. Atom-atom potential for each atom Separation (10-10m) What is the bond energy Ebond for the entire molecule? =5 bonds. Energy required to break a single pair of atoms apart: +8x10-21 J Ebond ≈ -40 x 10-21 Joules Approximation! (doesn’t include NN neighbors)

  9. Recap: Particle Model of Ebond • Ebond for a substance: amount of energy required • to break apart “all” the bonds (magnitude only) • i.e. we define Ebond = 0 when all the atoms are separated • The bond energy of a substance comes from • adding all the potential energies of particles at their • equilibrium positions. Ebond = ∑all pairs(PEpair-wise) • A useful approximation of the above relation is: • Ebond~ - (total number of nearest neighbor pairs) x () • Ebond of the system is negative, determined by: • 1) the depth of the pair-wise potential well e(positive) • 2) the number of nearest-neighbors.

  10. A B Clicker: Etotal, Ebond, Ethermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is Etot greater? a) Situation A has a greater Etot b) Situation B has a greater Etot c) Both have the same Etot d) Impossible to tell

  11. A B Clicker: Etotal, Ebond, Ethermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is Ebond greater? a) Situation A has a greater Ebond b) Situation B has a greater Ebond c) Both have the same Ebond d) Impossible to tell

  12. A B Clicker: Etotal, Ebond, Ethermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is Ebond greater? a) Situation A has a greater Ebond b) Situation B has a greater Ebond c) Both have the same Ebond d) Impossible to tell We did not break a bond - Ebond did not change!

  13. A B Clicker: Etotal, Ebond, Ethermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is Ethermal greater? a) Situation A has a greater Ethermal b) Situation B has a greater Ethermal c) Both have the same Ethermal d) Impossible to tell

  14. A B Clicker: Etotal, Ebond, Ethermal KE A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is Ethermal greater? a) Situation A has a greater Ethermal b) Situation B has a greater Ethermal c) Both have the same Ethermal d) Impossible to tell KE We increased Ethermal by putting more energy into the system

  15. Initial Final Clicker:Individual Atoms Okay. Now let us look at the same problem from the perspective of the KE and PE of individual atoms. Which situation is correct in going from initial to final states?

  16. Initial Final Clicker:Individual Atoms Okay. Now let us look at the same problem from the perspective of the KE and PE of individual atoms. Which situation is correct in going from initial to final states?

  17. Particle Model of Ethermal • Ethermal is the energy associated with the random • microscopic motions and vibrations of the particles. • We increased Ethermal by putting more energy into the • system. • KE and PE keep changing into one another as the • atoms vibrate, just like in the mass-spring system, so we • cannot make meaningful statements about • instantaneousKE and PE. • We can make statements about average KE and PE. • Increasing Ethermal increases both KEaverage and PEaverage .

  18. Particle Model of Ethermal The energy associated with the random motion of particles is split between PEoscillation and KE .

  19. Mass on a Spring Energy PE Etot KE position As we increase Etot we increase PEavg and KEavg PEavg = KEavg = Etot/2

  20. Particle Model of Ebondand Ethermal • The energy associated with the random • motion of particles is split between PEoscillation, KE. • For particles in liquids and solids, let’s not forget • the part that corresponds to Ebond of the system. • Ebond of the system is determined by both the • depth of the pair-wise potential well and the • number of nearest neighbors (# NN) .

  21. Particle Model of Ebondand Ethermal • Ebond of the system is determined by both the • depth of the pair-wise potential well and the # NN. • Solids/Liquids: • KEall atoms = (1/2)Ethermal • PEall atoms = PEbond+ PEoscillation • = Ebond (PEbond )+ (1/2)Ethermal (PEoscillation) • KEall atoms + PEall atoms = Ethermal + Ebond • Gases (monoatomic): • The gas phase has no springs: • no PEoscillation or PEbond

  22. Summary of Ebondand Ethermal • Potential Energy contributes to both Ebond and • Ethermal. • Kinetic Energy contributes to Ethermal, but • not Ebond. • Number of bonds is important for Ebond, • not really for Ethermal.

  23. Equipartition of Energy

  24. Intro to Equipartition of Energy • If the atoms in the molecule do not move too far, the forces between them can be modeled as if there were springs between the atoms. • Each particle in a solid or liquid oscillates in 3 dimensions about its equilibrium positions as determined by its single-particle potential.

  25. Intro to Equipartition of Energy • Another way of stating: Each particle has six “ways” to store the energy associated with its random thermal motion. • We call this “way” for a system to have thermal energy a “mode”.

  26. ? ? Question What is Temperature in terms of Ethermal? Answer: Temperature IS Thermal Energy!

  27. But Wait a Minute… [Energy] = [Joule] [Temperature] = [Kelvin] Answer revised: Temperature is proportional to Thermal Energy Ethermal. The constant of proportionality is kB: Boltzmann’s Constant kB = 1.38  10-23Joule per degree Kelvin

  28. Equipartition of Energy Gas Liquids and Solids To be precise, energy associated with the component of motions/vibrations (“modes”) in any particular direction is (1/2)kBT : Ethermal per mode = (1/2) kBT a.k.a. Equipartition of Energy

  29. What are Modes? • Modes :Ways each particle has of storing energy. • Ex. Mass-spring has one KE mode • and one PE mode.

  30. Equipartition of Energy: Restated Gas Liquids and Solids Equipartition of Energy Restated In thermal equilibrium, Ethermal is shared equally among all the “active” modes available to the particle. In other words,each “active” mode has the same amount of energy given by : Ethermal per mode = (1/2) kBT

  31. Modes of an atom in solid/liquid Every atom can move in three directions 3 KEtranslational modes

  32. Modes of an atom in solid/liquid Every atom can move in three directions 3 KEtranslational modes Plus 3 potential energy modes along three directions 3 PEmodes Total number of modes is 3PE + 3KE = 6 Ethermal = 6(1/2)kBT

  33. Modes of an atom in a Monoatomic Gas Every atom can move in three directions 3 KEtranslational modes 0 PEmodes (no bonds) Gas: No bonds, i.e. no springs • Total number of modes is 3KE = 3: • KEavg = 3(1/2)kBT= Ethermal • PEavg = 0

  34. 2 KErotational modes Modes of a Molecule in a Diatomic Gas 3 KEtranslational modes 2 vibrational modes (1 KE, 1PE) (associated with atom-atom interaction within the molecule)

  35. 2 KErotational modes Modes of a Molecule in a Diatomic Gas 3 KEtranslational modes 2 vibrational modes (1 KE, 1PE) (associated with atom-atom interaction within the molecule) • Total number of modes is 6KE + 1PE = 7 • Ethermal = 7(1/2)kBT • Sometimes (at lower temperatures), however, not all the modes are “active”. (Freezing out of modes)

  36. Equipartition of Energy: Ethermal per Mode = ½ kBT

  37. Equipartition of Energy: Ethermal per Mode = ½ kBT When energy is added to a system, what does it mean to have more places (modes) to store it in?

  38. Clicker Question: Energy for Temperature Does it take more or less energy to raise the temperature of a diatomic gas than a monatomic gas? More Less Equal I don’t even know how to start guessing

  39. diatomic(no vibrations) monatomic Real Heat Capacity Measurements… All measurements at 25 Cunless listed otherwise (100 C) (500 C)

  40. Answer: Energy for Temperature Does it take more or less energy to raise the temperature of a diatomic gas than a monatomic gas? More C = ∆Ethermal / ∆T ∆ Ethermal per molecule= number of active modes  (1/2)kB∆T ∆ Ethermal per N atoms= number of active modes  (1/2)kB∆T  N

  41. Measuring Heat Capacity Thermometer Thermometer Stirring rod Stirring rod Substance of interest Substance of interest Heating element Heating element

  42. Specific Heat of Solids (0C) All measurements at 25 Cunless listed otherwise (-100 C)

  43. Molar Specific Heat of Solids All measurements at 25 Cunless listed otherwise (0C) (-100 C)

  44. 6 modes Molar Specific Heat of Solids All measurements at 25 Cunless listed otherwise kBNA = R = 8.31 J/moleK : Gas constant or Ideal gas constant kB(Boltzmann constant) = 1.38 x 10-23 Joule/Kelvin NA(Avogadro’s number)= 6.02 x 1023 (0C) (-100 C)

  45. Specific Heat of Liquids All measurements at 25 Cunless listed otherwise

  46. Molar Specific Heat of Liquids All measurements at 25 Cunless listed otherwise 6 modes

  47. Molar Specific Heat of Liquids For polyatomic substances , the values of molar specific heat of liquids are greater than the values for solids. Limitation of our model… For monatomic substances, the value of molar specific heat of liquids is similar to the values for solids. Our model works well here! All measurements at 25 Cunless listed otherwise 6 modes

  48. Specific Heat for Gases (500 C) (100 C) All measurements at 25 Cunless listed otherwise

  49. diatomic(no vibrations) monatomic Molar Specific Heat for Gases Oops! What’s going on??! All measurements at 25 Cunless listed otherwise (500 C) (100 C)

  50. Discrepancy Explained… Closed box of gas Open box of gas CV measurement CP measurement Open box: Some heat goes into pushing air out of the way Closed box: all heat goes into the gas’s energy

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