Physics 7A – Lecture 7 Winter 2009 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! Check Physics 7 website frequently for updates. Turn off cell phones during lecture.

Review Multiple- Atom Systems: Particle Model of Ebond , Particle Model of Ethermal

Particle Model of Ebond and 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)?

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)

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)

Continued: Particle Model of Ebond 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 nearest neighbor pairs per atom)/2 xN] 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.

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

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

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!

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

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 KE We increased Ethermal by putting more energy into the system

Clicker: Individual Atoms 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?

Clicker: Individual Atoms 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?

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 instantaneous KE and PE. We can make statements about average KE and PE. Increasing Ethermal increases both KEaverage and PEaverage .

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

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

Particle Model of Ebond and 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) .

Particle Model of Ebond and Ethermal Solids/Liquids: Ethermal = KE + PEoscillation: KEall atoms = (1/2)Ethermal PEall atoms = PEbond + PEoscillation = Ebond (PEbond )+ (1/2)Ethermal (PEoscillation) Etotal = KEall atoms + PEall atoms = Ethermal + Ebond Gases (monoatomic): The gas phase has no springs: no PEoscillation or PEbond

Summary of Ebond and 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.

Equipartition of Energy

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.

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”.

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

kB = 1.38  10-23 Joule per degree Kelvin 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  10-23 Joule per degree Kelvin

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

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

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

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

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 PE modes Total number of modes is 3PE + 3KE = 6 Ethermal = 6(1/2)kBT

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

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

Modes of a Molecule in a Diatomic Gas 3 KEtranslational modes 2 KErotational 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)

Equipartition of Energy: Ethermal per Mode = ½ kBT KE mode PE mode Total Solids 3 6 Liquids Monatomic gases Diatomic gases 3+2+1 1 7

Equipartition of Energy: Ethermal per Mode = ½ kBT KE mode PE mode Total Solids 3 6 Liquids Monatomic gases Diatomic gases 3+2+1 1 7 When energy is added to a system, what does it mean to have more places (modes) to store it in?

Clicker Question: Energy for Temperature Does it take more or less energy to raise the temperature of a diatomic gas than a monatomic gas? KE mode PE mode Total Solids 3 6 Liquids Monatomic gases Diatomic gases 3+2+1 1 7 More Less Equal I don’t even know how to start guessing

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

Answer: Energy for Temperature Does it take more or less energy to raise the temperature of a diatomic gas than a monatomic gas? KE mode PE mode Total Solids 3 6 Liquids Monatomic gases Diatomic gases 3+2+1 1 7 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

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

Specific Heat of Solids All measurements at 25 C unless listed otherwise (-100 C)

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

Molar Specific Heat of Solids 6 modes All measurements at 25 C unless 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)

Specific Heat of Liquids All measurements at 25 C unless listed otherwise

Molar Specific Heat of Liquids All measurements at 25 C unless listed otherwise 6 modes

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 C unless listed otherwise 6 modes

Specific Heat for Gases All measurements at 25 C unless listed otherwise

Molar Specific Heat for Gases Oops! What’s going on??! All measurements at 25 C unless listed otherwise diatomic (no vibrations) monatomic The dotted lines are the values you’d get if you calculated using 1/2kT or ½ nRT. Doesn’t match the measured values as it does with solids. Why? (500 C) (100 C)

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

Discrepancy Explained… Closed box of gas Open box of gas Question So then does it take More energy to raise the Temperature of Closed box of gas Or Open box of gas? CV measurement CP measurement Takes more energy to heat an open box of gas, which explains why the Cp values are higher than you would calculate using ½ kT Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way

Molar Specific Heat for Gases Whew! Now it Makes more sense. diatomic (no vibrations) All measurements at 25 C unless listed otherwise monatomic (500 C) (100 C)

In this example of thermal phenomena (i.e.,measuring heat capacity) , “Constant volume” “Constant pressure” (definition of heat capacity) “Process” seems to matter… => Chapter 4 Models of Thermodynamics

Summary of Equipartition of Energy Equipartition tells us that the energy per mode Is ½ kbT. For solids and liquids, moving the atom in any direction is like moving a mass on a spring. Three directions means 3 KE modes/atom and 3 PE modes/atom. Nothing to do with # of bonds! For gases things are more complicated: need to count carefully- some KE modes don’t have PE modes.

Summary of the Particle Model Atom-atom interactions determine how much energy we need to
 pull our system apart. Sum over all pairs to find Ebond Ebond depends on the strength of the interactions (well-depth) and
 the number of interactions (determined by packing) Ethermal depends on the number of modes. 
 Equipartition tells us that each (active) mode has the same 
 amount of energy, and that the energy per mode depends 
 only on temperature.
 Our mode counting works well for solids and gases. For gases 
 we need to distinguish between work and heat carefully (next)
 Our mode counting does not work for liquids.

Thermodynamics: Microscopic to Macroscopic

What is Thermodynamics? In a nutshell, we study the transfer of energy between systems and how the energy instills movement, i.e., how the system responds. Ex. If we heat something, it expands. Sadi Carnot (1796-1832) Father of Thermodynamics Lord Kelvin (1824-1907) James Joule (1818-1889)

Thermodynamics: Goals Carefully distinguish between heat and work. Learn what a state function is. Think about things that depend on the process, and ask about processes. (e.g. ice melts/water freezes, eggs fry)  Why doesn’t the fried egg turn into raw egg again?  Why can carbon exist as a diamond as well as graphite but not ice and vapor?

State Function Depends only on properties of the system at a particular time Example: Ethermal of a gas For an ideal gas, Ethermal depends only on: Temperature Number of modes

Work and Heat Not a property of a particular object. Instead a property of a particular process, or “way of getting from the initial state to the final state” LHS: depends only on i and f Q,W depend on process between i and f

Conservation of Energy (Review) ∆Etotal must include all changes of energy associated with the system… ∆Etotal = ∆Ethermal+ ∆Ebond+ ∆Eatomic+ ∆Enuclear+ ∆Emechanical ∆U : Internal energy Energy associated with the atoms/molecules inside the body of material Energy associated with the motion of a body as a whole So then, if there’s no change in ∆Emechanical ∆ U …If no change in DEmechanical First law of Thermodynamics

State v. Process-dependent Functions State functions Process-dependent Depend only on what the object is doing at the time. Change in state function depends only on start and end points. Examples: T, P, V, modes, bonds, mass, position, KE, PE... Depend on the process. Not a property of an object Examples: Q, W, learning , ......

Along any given segment: Work initial final P V Along any given segment: W = -PDV (if P constant) OR = - Area under curve (sign positive when V decreases, sign negative when V increases)

Clicker: Calculation of Work final P Is the work done in the process to the right positive or negative? initial A) Negative B) Positive C) Zero D) Impossible to tell. V

Work P V First section: W1 < 0 (volume expands) final initial V First section: W1 < 0 (volume expands) Second section: W2 = 0 (volume constant) Third section: W3 > 0 (volume contracts)

Clicker: Process Comparisons final Here are two separate processes acting on two different ideal gases. Which one has a greater magnitude of work? The initial and final points are the same. P initial V Magnitude of work in top process greater B) Magnitude of work in lower process greater C) Both the same D) Need more info about the gases. P final initial V

Heat and the First Law of Thermo P initial final V We can read work directly off this graph (i.e. don’t need to know anything about modes, U, T, etc.) PV = nRT U = 5/2 nRT We have PiVi from the graph… If we know something about the gas, we can figure out Ui, Uf and Uf - Ui

Heat and the First Law of Thermo P initial final V

Example 5 moles of a monatomic gas has its pressure increased from 105 Pa to 1.5x105 Pa. This process occurs at a constant volume of 0.1 m3. Determine: * work * change in internal energy DU * heat involved in this process. initial final P V W=0 dU=3/2nRdT = 3/2dPV = 3/2 0.5x10^5 (0.1) = 7500 J = dQ = 3/2 (5)RdT = 3/2(5)R(120) =3/2 (5)(8.314)(120) = 7482.6 J = dQ HW: There are two ways to solve this using the ideal gas law…

Comments Work depends on the process Heat depends on the process DU only depends on initial and final W = 0 for all constant volume processes

Enthalpy Why is this useful?!? H = U + PV Is it a state function? - U depends only on state of system - P depends only on state of system - V depends only on state of system => H depends only on state of system (Hess’s law) Why is this useful?!?

Enthalpy Constant volume Constant pressure P V P V W = 0 initial final P V initial final P V Enthalpy useful for constant pressure processes: dH = dU + PdV dU = dQ – dW = dQ – PdV dH = dQ – PdV + PdV OR dH = dQ (constant pressure process and only PV work) W = 0 *Derivation in P.84 Note: nothing about gases used – works for solids and liquids too!

Unit 1 Revisited In unit #1 we taught you: But the gas is expanding, so work is being done! Now we know better…. But in a large room, pressure is roughly constant…

Unit 1 Revisited

Next Time: Microstates versus States, and Entropy