This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Physics 7A
The life and times of atoms, molecules and the energies associated with them (Fall 2006) Approach to problems
Try and ﬁt the problem into a particular category. Some problems require thinking of more than one concept. I am going to try and review the course, but to prepare for the exam I will be taking concepts from other parts of the course and mixing them. This is going to be a lecture; I suggest going to review sessions to ask questions. Best way to learn physics is by doing! http://student.physics.ucdavis.edu/~dmartin/ Other review sessions
There are many other review sessions this week. Don’t expect the TAs to plug in the numbers for you  that is not what the review sessions are for. It is completely appropriate to ask them what physics and assumptions go into a question. It is completely appropriate to ask the TA to write down a complete answer (i.e. what would you be expected to write down on the ﬁnal). This does not include the ﬁnal step of putting numbers in, however Threephase model
Temperature
(gas) Boiling/sublimation temperature Melting/Freezing temperature (solid) (liquid) (mixed phase) (mixed phase) Heat added •Putting heat in will change the temperature in a particular phase. •Putting heat in at a phase change will convert from one phase to another,
but will not change temperature. Question: Which phase has the smallest heat capacity? How can you tell? Which phase transition requires more energy, and why? Three phase model
Concepts deﬁned: ✦Phases: solid, liquid and gas ✦Ethermal ✦Heat capacity C = Q/T ✦Speciﬁc heat (per kg and per mole) ✦Ebond ✦Heat of melting, heat of vaporisation ✦Phase change occur at constant temperature Threephase model
Temperature
(gas) Boiling/sublimation temperature Melting/Freezing temperature (solid) (liquid) ∆Hmelt ∆m ∆Hvap ∆m Heat added Answering these questions should start by going off the graph. Explaining the results should use the particle model of matter (modes & bonds). Threephase model
Temperature
(gas) Boiling/sublimation temperature Melting/Freezing temperature (solid) (liquid) ∆Hmelt ∆m ∆Hvap ∆m Heat added Answering these questions should start by going off the graph. Explaining the results should use the particle model of matter (modes & bonds). Question: What would a threephase model diagram look like if vibrational mode(s) where becoming activated? Threephase model
Temperature
(high heat capacity) Boiling/sublimation temperature Melting/Freezing temperature (low heat capacity) # modes in solid & liquid are generally constant, so we expect lines. Heat added Note: This is a rather unconventional question for the threephase model. Normally the discussion of modes is in the particle model. Threephase model: Summary • Break processes up along straight line
segments. • Draw energy diagrams for each subprocess. • Heat capacity can be found from the graph
(the reciprocal of the slope) Mechanical energy
Type translational KE rotational massspring PE gravity Height Rotational speed Distance from “equilibrium” Indicator Velocity v ω y y These are the main types of (macroscopic) KE and PE we encounter in this course. All these energies increase as their indicator increases. Mechanical energy • Convention: Generally we ignore friction. The exceptions to this are when (i) The problem states otherwise (ii) The problem would be inconsistent without it. M 2 kg 10 kg Both of these systems start at rest. Which one goes fastest after the red block drops a distance d? 2 kg 10 kg (Assume nothing hits the ground or pulleys) 15 kg 40 cm
10 kg 2 kg 30 cm 30 cm
All these blocks are initially at rest and held in this position. Then they are let go. Assume energy losses to friction are negligible, and that the pulleys are essentially massless. (a) Will they move anywhere once let go? (b) Draw a complete energyinteraction diagram for the system. Let your ﬁnal point be after the red block has moved 0.1 m. (c) What is the speed of the red block at this ﬁnal point? (d) Draw an energy interaction diagram for the red block only between the same ﬁnal and initial points. (e) What would change if the pulleys were not massless? Would your answer to (c) be greater or smaller? 1 kg v v
20 kg
The ramps shown above are identical. Which goes higher up the ramp: the ball or the block? (Assume minimal friction, so that heat exchange is negligible) (a) The block goes higher (c) They reach the same height (b) The ball goes higher Spring’s equilibrium position 1 kg 1 kg Two 1 kg masses are being held at rest, compressing two identical springs the same distance from their equilibrium length. The two ramps shown are identical. When the springs are released, which mass has the greater maximum height? (Assume heat loss to friction negligible) (a) The block goes higher (c) They reach the same height (b) The ball goes higher A parachuter is falling with a speed of 55 m/s before the parachute opens. Shortly after the parachute opens the parachuter falls at a constant speed of 1.5 m/s.
Draw an energy diagram for the parachuter. Take the initial point as just before the chute opens, and the ﬁnal point as the parachuter is travelling at 1.5 m/s. Note: you may neglect the change in height from where the ‘chute opens and constant speed is obtained. If the mass of the parachuter is 65 kg, how much energy is released in the above process? Plotting energies
Kinetic energy is always nonnegative Potential energy found from graph Speciﬁc examples: LennardJones Nuclear bonds Etot (generically) does not depend on r (ignore friction, etc) Force can be found from the slope of PE vs distance. B A C Babe Ruth hits a 150g ball in the “world” series. The ball leaves the bat with an initial velocity of 10 m/s and goes to a maximum height of 3 m above the point it was hit with the bat. The bat hit the ball 1 metre above the ground.
Draw a graph of energy versus vertical distance. On this graph put PE, KE and Etot of the ball. Indicate where the points A, B and C are on the graph. Deﬁne y=0 to be the height of point A. (Point A is just after the ball leaves the bat, C is just before the ball hits the ground). 10 J Energy 5J 0J y (m) 5 J 10 J 4 3 2 1 0 1 2 3 4 5 To the right is a graph of PE vs. centretocentre particle separation. The bottom of the potential is at  ε , and the top of the energy bump (point (e)) has an energy of 0.2ε . The energy asymptotes to zero as the separation becomes inﬁnite. E
(e) (a) (d) (f) r
(b) (c) a) Which point(s) (a)(f) is the force between the particles repulsive? b) Which point(s) (a)(f) is the force between the particles attractive? c) Which point(s) (a)(g) is the force between the particles zero? d) If all we know about a particle is that its Etot is 0.1ε , do we know if it is bounded or unbounded? If it is bounded, what points is it bounded between? e) If all we know is that the separation corresponds to point (d) and we know that the particles are moving away from each other, what can we say about i.The potential energy? ii.The kinetic energy? iii.Etot? f) Rank the magnitude of forces. (If two forces seem roughly the same size, treat them as equal) LennardJones: Potential energy vs centrecentre separation 1 0.5 (Recall r0=1.12 σ ) Energy (") 0 0.5 1 0.5! ! r0 Find the maximum kinetic energy when Etot = 1.0x1021J. At what centretocentre separation does this occur. Give your answer in physical units! For this potential the welldepth is 2.0x1021J, and the atom size is 1.1x1010 m. Also draw KE vs. r and Etot vs. r on the graph above. 1.5! Distance 2! 3! LennardJones: Potential energy vs centrecentre separation 1 0.5 Energy (") (ε)
0 0.5 1 0.5! ! r0 1.5! Distance 2! 3! What is the force (i.e. attractive or repulsive) between two atoms separated by 1.5 sigma? Is it possible for two atoms this far apart to be moving closer together? Is it possible for two atoms this far apart to be moving away from one another? LennardJones: Potential energy vs centrecentre separation 1 0.5 Energy (") (ε)
0 0.5 1 0.5! ! r0 1.5! Distance 2! 3! What is the force (i.e. attractive or repulsive) between two atoms separated by 1.5 sigma? Is it possible for two atoms this far apart to be moving closer together? Is it possible for two atoms this far apart to be moving away from one another? LennardJones: Potential energy vs centrecentre separation 1 0.5 Energy (") (ε)
0 0.5 0.8
1 0.5! ! r0 1.5! Distance 2! 3! What is the force (i.e. attractive or repulsive) between two atoms separated by 1.5 sigma? Is it possible for two atoms this far apart to be moving closer together? Is it possible for two atoms this far apart to be moving away from one another? LennardJones: Potential energy vs centrecentre separation 1 0.5 Energy (") (ε)
0 0.5 0.8
1 0.5! ! r0 1.5! Distance 2! 3! 2.333 What is the force (i.e. attractive or repulsive) between two atoms separated by 1.5 sigma? Is it possible for two atoms this far apart to be moving closer together? Is it possible for two atoms this far apart to be moving away from one another? Substance A Substance B The two substances shown above have the same molar heat of vaporisation, and the same molecular size. These are two dimensional packings (no atoms are stacked into or out of the page). (i) If the energy required to break a single bond of substance A is 1x1021 J, what is the energy required to break a single bond of substance B? (ii) Draw a LennardJones potential for the energy between a pair of atoms for each substance on the same graph. (iii) Given one mole of substance A and one mole of substance B, which requires more energy to break all the bonds? (iv) Which substance will have more modes per atom in its solid phase? (v) Which substance will have more modes per atom in its gaseous phase? • The twin ramp problem from DL
(Mechanical) functions. • Something about work/heat vs state • Nuclear? • LennardJones? • Equipartition
In DL, we argued that for a massspring system that the average KE was equal to the average PE. PE = KE PE + KE = Etot PE + KE = Etot PE = KE (Only have KE and PE) (True for masssprings) Equipartition
•
Because solids and liquids effectively have 3 springs, the kinetic and potential energy in any given spring are equal.
KE = Eth 2 PEth = Eth 2 (Liquids and solids only) Need to use Eth because we no longer have springs at equilibrium and at rest having Etot = 0. In a substance, all atoms at equilibrium and all atoms at rest mean Etot = Ebond.
KE = Eth 2 Eth 2 (Liquids and solids only) PE = Ebond + • This applies to microscopic random motion, not to an overall motion of the object. Equipartition
Equipartition theorem:
1 Energy per mode = kBT 2 This told us not only <PE> = <KE> for any given spring but also: •The <KE> of different springs was the same •The <PE> of different springs was the same •The amount of energy in a particular KE or PE more
depends only on temperature. The result of equipartition: we can look at gases as well! Solids & Liquids: New langauge • We have 6 active modes altogether: • We have 3 PE modes & 3 KE modes: • Therefore:
1 KE = 3 × kBT 2 1 PEthermal = 3 × kBT 2
Eth KE = 2 1 Ethermal = 6 × kBT 2 Eth PE = Ebond + 2 • Monatomic:
(leftright) Gases 3 KE modes, 0 potential modes (updown) (inout) • Diatomic: 5 KE modes: 3 translational: 2 rotational: 2 vibrational modes (possibly frozen out) (1 KE + 1 PE) Relating old and new
1 Ethermal = (# modes) × (Eth per mode) = (# modes per atom)N × kBT 2 • • • • New concept: modes! Relationship change in Eth, assuming the number of modes is constant: 1 ∆Ethermal = (# modes per atom)N kB ∆T 2 ∆U = ∆Eth = Q Q If no work is being done, 1st law tell us Eth
T Therefore: Q 1 C= ⇒ C = (# modes per atom)N × kB ∆T 2 We see that the particle model can give us heat capacities! Thermodynamics
State functions: U, T, P V, H , (Entropy as well; but S is discussed separately) Process dependent functions: Q, W State diagrams: PV diagrams UT diagrams HT diagrams TS diagrams (discussed separately) Thermodynamics
State functions (or variables): Depend only on the state of the system at a given time, not on how it got there. e.g. P V PV (if I know P and know V, I know PV) Changes in state functions between two states do not depend on the path taken. e.g. ∆(PV ) is the same for all paths connecting the initial and ﬁnal points. It is o.k. to talk about pressure, temp, U, (state variable) of a system “increasing” or “decreasing” because it is a property of the system. Thermodynamics
Path dependent functions/variables: Are not properties of the system, but rather properties of a process. Can take different values for different processes. The typical examples are work and heat. For small processes, W and Q are found from a state function and a change in a state function:
W = −P dV ⇒ Work = area on PV diagram Q = +T dS ⇒ Heat = +area on TS diagram It is not o.k. to talk about Q or W “increasing” or “decreasing” because they are not a property of the system.They are a property of a process. Instead, refer to them as “positive” or “negative” P (Pa)
1.2x105 1x105 0.02 0.06 V (m3) One mole of an ideal monatomic gas is taken through the process shown above. (a)Calculate UA, UB and UC. (b)Calculate HA, HB and HC. (c) Calculate ∆H for the process A>B and B>C. (d)Calculate the heat for the two processes: A>B and B>C (e)Does ∆H = Q for each process? Why or why not? Entropy (S)
Deﬁnition of entropy:
S = kB ln Ω This deﬁnition is pretty good for explaining entropy, and if it goes up or down. This deﬁnition is awful for calculating entropy, and for calculating the change in entropy. Microstates of a box
I want to put 2 particles in a box, then calculate the total number of microstates. Adam labels microstates by {1,1} {1,2} {2,1} {2,2} Beth labels microstates by {1,1} {1,2} {1,3} {1,4} {2,1} {2,2} {2,3} {2,4} {3,1} {3,2} {3,3} {3,4} {4,1} {4,2} {4,3} {4,4} Constantine labels microstates using 56 boxes! Total microstates = 4 Total microstates = 16 Total microstates = 3136 Microstates of a box
I want to put 56 particles in a box, then calculate S for the most probable state based off this information: Adam’s most probable state is 28 on each side Beth’s most probably state is 14 in each subbox Constantine’s most probable state is one in each box. Ωmost probable = 7.7 × 1015 Smost probable = 5.0 × 10−22 J/K Ωmost probable = 1.2 × 1031 Smost probable = 9.9 × 10−22 J/K Ωmost probable = 7.1 × 1074 Smost probable = 2.3 × 10−21 J/K There entropy’s do not agree  even on the same box! Entropy S: what is the point?
This is a good explanation: “Water vapour at 100 C has more entropy than liquid water at 100 C because the gases can have more available position microstates.” DO NOT DO A CALCULATION THIS WAY! Instead, turn to the deﬁnition of heat:
dQ dQ = T dS ⇒ dS = T (equilibrium processes) Two important special cases • Constant temperature: • Constant heat capacity: Q ∆S = T Tf ∆S = C ln Ti n or m, depending on if you use molar c or per kg c Tf = nc ln Ti Example: 1.0 kg of ice has been sitting for a long time in a large
vat of liquid nitrogen at 196 C. The block of ice is then removed and thrown into a very large body of water at 20 C. a) What is the heat required to bring the ice into thermal equilibrium b) c)
with the water? What is the change of entropy of the ice? What is the change of entropy of the water? (Ignore any heat exchange with the environment) Useful info:
N2 boiling point
H20 melting point Ice speciﬁc heat Water speciﬁc heat N2 (gas) speciﬁc heat H20 heat of melting
N2 heat of vaporisation 196 C 0C 2.05 kJ/kg K 4.18 kJ/kg K 1.04 kJ/kg K 333.5 kJ/kg 25.7 kJ/kg Pressure
Initial point Final points Volume
Two identical containers of monatomic gas are initially in identical states. Both of these gases are expanded to the same ﬁnal volume, but one is expanded isothermally (constant T) and the other is expanded adiabatically. a) Which ﬁnal point has a greater internal energy? b) Which process is isothermal, and which is adiabatic? c) Which ﬁnal point has more entropy? If you do not have enough information to answer any of these questions, state what else you would need to know to answer it. T S
Shown above is an unknown amount of ideal gas.You may assume that no gas enters or leaves the box. If you don’t have enough information, say what else you would need.
• A student argues that because the initial and ﬁnal entropy is the same,
the heat is zero. Are they correct? If not, is the heat absorbed or released? •Is the work done by the system or done on the system? T 473 K 373 K S
Shown above is an unknown amount of an ideal gas.You may assume that no gas enters or leaves the box. If you cannot tell, what other thing(s) would you need to know?
• A student argues that because the initial and ﬁnal entropy is the same,
the heat is zero. Are they correct? If not, is the heat absorbed or released? •Is the work done by the system or done on the system? ...
View
Full
Document
This note was uploaded on 10/22/2010 for the course PHY 7A 56192 taught by Professor ? during the Summer '09 term at UC Davis.
 Summer '09
 ?

Click to edit the document details