Lecture06

Lecture06 - Physics 7A-2 (C/D) Professor Chertok Fall, 2008...

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Unformatted text preview: Physics 7A-2 (C/D) Professor Chertok Fall, 2008 Lecture 6 Original presentations copyright M. Chertok 2008. All rights reserved. Questions? Outline for today Office hours? Wrap up of DLMs 9 & 10 from last week What’s ahead this week Quiz 6 / Reading assignment M. Chertok, Physics 7A 3 Office hours I’m getting an average of 4 students at my office hours M. Chertok, Physics 7A 4 DLM09 3.1.4 Follow up of DLM08 FNTs: Relate particle spacing to “non-interaction” of ideal gas molecules. Use PE graph Two atoms and r=1.2σ. What is total energy if: a) atoms at rest, or, b) atoms have total KE 0.1ε Use total energy plus fact that KE≥0 to plot KE for atom-atom potential Plot Etot and KE for various Etot How do bound particles behave? Unbound? Numerical: how much energy need to break bond? Connect macroscopic and microscopic energy systems boiling water example M. Chertok, Physics 7A 5 PE for 2-atom system e en ship Potential Energy (in units of well depth, ") Pair-wise or Atom-Atom Potential (as modeled with the Lennard-Jones Potential) 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Pair-wise Potential (as modeled with the Lennard-Jones Potential) hat . tion um rticle -1.2 distance between centers of atoms, r (in units of particle diameter, ! ) M. Chertok, Physics 7A 6 DLM09, (ii) 3.2.1 - Particle model of Ebond Geometrical picture of little spheres packing closely. Each touching neighbor pair is one “bond” What if you have 100 atoms? NA atoms? You can’t count all those bonds M. Chertok, Physics 7A 7 Close packing in solids,liquids n-n’s: 6 in same plane 3 below 3 above TOTAL = 12 M. Chertok, Physics 7A 8 Double counting? So, in a large sample each particle has 12 n-n’s Why “large” important? In a large sample of N atoms, is total # n-n bonds = 12N? Why or why not? total # n-n bonds = 6•N Do not use this for small N! M. Chertok, Physics 7A 9 Clicker questions 6 atoms: The total number of bonds is 6*6 = 36 a) True, b) False 2 moles of water: The total number of bonds is a) NA b) 6NA c) 12NA M. Chertok, Physics 7A 10 Particle Model of Ebond Model #5 Nearest neighbors Nearest neighbor pair (n-n) # of nearest neighbors Non-nearest neighbor pairs Total # of n-n pairs Bond energy Ebond = ∑all pairs (PEpair-wise) Ebond ≈ -(total # of n-n pairs) × ε = -6Nε (Why approximately??) |ΔEbond| ≈ |ΔmΔH| at a phase change M. Chertok, Physics 7A 11 DLM10 3.2.2 FNTs from DLM09 Bond energy for cluster of 6 atoms Rule of thumb: if you can count the # n-n pairs, then do so Compare Ebond predictions from model to empirical values for molecules 3.2.3, 3.2.4 Particle Model of Thermal Energy M. Chertok, Physics 7A 12 modes (= degrees of freedom) (plus in and out of the page) 3 springs, so 3 KE modes + 3 PE modes = 6 total modes M. Chertok, Physics 7A 13 Particle Model of Ethermal Remember: thermal energy is due to microscopic random motion For solids, liquids: atoms are “stuck” in place, but oscillate in how many different directions? More or less motion as T increases? T is the indicator for thermal energy For gases, what is the motion? Translational, in 3 directions Equipartition: count the ways an atom can store energy These are called “modes” M. Chertok, Physics 7A 14 Equipartition Ethermal/mode ~ T Actual formula is Ethermal/mode = ½kBT kB is Boltzmann’s constant kB = 1.38 x 10-23 J/K Most of upcoming DLs is practicing counting #modes This allows us to determine the total Ethermal M. Chertok, Physics 7A 15 Example: thrown baseball A major league pitcher can throw a baseball really fast M(baseball) ~ 5 oz = 0.145 kg v(baseball) = 100 MPH = 44 m/s How do the mechanical and thermal energies compare? KE = ½mv2 = ½(0.145kg)(44m/s)2 = 140.4 J PE = mgh = (0.145kg)(10m/s2)(1m) = 1.5 J M. Chertok, Physics 7A 16 Example: thrown baseball Thermal energy #modes? Solid →3PE + 3KE = 6 If a baseball weighs 145g and it’s mostly carbon (molar mass 12 g), then the ball is about 12 moles Eth = (#modes/atom)(12•NA)(½kBT) PV=nRT = NkBT, So nR=NkB, and R=NAkB = 8.3 J/mol•K Eth = (6)(½)(12•R)(T) = (3)(12)(8.3J/mol•K) (300K)=90kJ. Roughly 600 times greater than Emech! M. Chertok, Physics 7A 17 Result from A.I.M. A.I.M. result: connection between KE,PE and Eth, Ebond a bit complicated. Starting at T=0K, and adding Eth, what happened? Clicker question: a) PE increased b) KE increased c) PE and KE increased d) neither PE nor KE increased M. Chertok, Physics 7A 18 Result from A.I.M. Starting at T=0K, and adding Eth, BOTH KE and PE increased KEtot = ½Eth PEtot = Ebond + ½Eth We get Ebond from the phase change -- all PE, but Eth shared equally by KE and PE due to spring model M. Chertok, Physics 7A 19 Today in DLM11 3.2.5 FNTs from DLM10 3.2.6 Particle model of thermal energy 3.2.7 (not in your packets) Analyzing energies with two types of atoms M. Chertok, Physics 7A 20 Your TA makes a bet... example 50 atoms Total energy = -20 x 10-21 J Previously found the Ebond = -100 x 10-21 J What are KE, PE? KE/atom? PE/atom? Since Etotal = Ebond + Ethermal, we know Ethermal = 80 x 10-21 J. Thus, KE=½Eth = 40 x 10-21 J and PE = Ebond + ½Eth = (-100 + 40) x 10-21 J = -60x10-21 J Finally, KE/atom = 8.0 x 10-22 J, PE/atom = -1.2x10-21 J M. Chertok, Physics 7A 21 Two types of atoms PE (J) (Note: here we use absolute scales for axes) distance (m) What can we say about the two substances? Should there be a 3rd PE curve? M. Chertok, Physics 7A 22 Next in DLM12 3.3.2 Using modes to interpret data 3.3.3 Follow up of FNTs from DLM11 {3.4.1 Microscopic basis of heat capacity 3.4.2 Making sense of heat capacity data } M. Chertok, Physics 7A 23 Using modes to interpret data compartment of your refrigerator? How do the algebraic relationships relate to the graph? Which parts? What is the relationship? These are the kinds of questions you need to be asking yourself and getting confident about. You want to practice using this representation enough so that it really does become a useful tool to make sense of thermal phenomena and to be comfortable using it to construct explanations for particular phenomena. Energy Added or Removed (at constant pressure) flip this about y=x line total energy } Temp Within one phase, what is ΔEtotal? What is slope? M. Chertok, Physics 7A 24 total energy Microscopic basis of heat cap. Temp Within one phase: Q = ΔEth Also: ΔEtot = ΔEth Slope: ΔEtot/ΔT = ΔEth/ΔT Thus, slope = Q/ΔT This means the slope is the heat capacity! M. Chertok, Physics 7A 25 total energy Microscopic basis of heat cap. Temp Assume 1 mole: slope = cvm (If the volume changes, then some work is done by the heat, and the result changes = cmp) cvm = ΔEth/ΔT = ½R(#modes/atom) for 1 mole: Ethermal/mode = ½kBNAT We’ve connected specific heat to modes and equipartition! M. Chertok, Physics 7A 26 Compare with data You will examine cvm data for monoatomic gases diatomic gases triatomic gases solids By counting modes, you can predict (ok, verify) these Study graph on p. 55 Heat capacities drop at low temperature. Why? M. Chertok, Physics 7A 27 vaporization presented in Chapter 1, we would expect our models to provide us with the capability of explaining the heat capacity values, both at constant pressure and at constant volume for a large range of substances. Several of these data patterns are presented on this and the next page. Heat capacities vs. T CH4 NH3 Cvm R 9.5 8.5 7.5 6.5 5.5 4.5 3.5 N2 2.5 1.5 H2 Monatomic gases (He, Ar, Ne, etc.) 1000 1500 5 2 3 2 CO2 Cl2 7 2 This first graph shows the constant volume molar heat capacity of several gases from room temperature up to several thousand kelvin. The values of the heat capacities have been divided by the gas constant, R. There are several obvious trends. The monatomic gases have the lowest molar constant-volume heat capacity at 3/2 R and the values are independent of temperature. Diatomic gases seem to have higher values starting at about 5/2 R, while polyatomic gases have significantly larger values, but also a much more pronounced temperature dependence. These are some of the trends our models should enable us to provide explanations for. -y displacement from equilibrium Energy +y 300 500 2000 T [Kelvin] M. Chertok, Physics 7A 28 Reading / Quiz 6 Review Ch. 3; start on Ch. 4 Quiz 6 including FNTs from 11 covered in DLM12 M. Chertok, Physics 7A 29 ...
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This note was uploaded on 05/30/2009 for the course PHY 7A taught by Professor Pardini during the Fall '08 term at UC Davis.

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