6) BTW: Your TA has Exam3. It should have been returned to you on Nov 16 (Mon) at Recitation if you

Transcription

1 Chap. 15: pv = nrt Mole and Avogadro s number. Equations of state. Kinetic theory of an ideal gas. Heat capacities. First Law of Thermodynamics. Thermodynamic processes. Properties of an ideal gas. 1 3 Exams and 2 More 1) DEC 1 (Tue): Lecture 2) DEC 3 (Thu): Exam 4 Chap. 12, 14 and 15 1) P.1 Five quick quizzes from 12, 14 and 15 2) P ) P ) P ) DEC 8 (Tue): Last class 4) DEC 8 (Tue): Common Makeup Test at 7 pm 5) DEC 11 (Fri): Final Exam Comprehensive (all chapters) Please start reviewing Eaxm1~Exam3 materials now! 6) BTW: Your TA has Exam3. It should have been returned to you on Nov 16 (Mon) at Recitation if you didn t pick up on Nov. 10 and 12. 2

2 Calendar You don t need to work on Week 13 & 14 Class Exam4 LAST Class Makeup4 Final Exam 3 Topics in Chap. 15 at First Glance Limited subjects KEEP checking my PHYS201 website. Special arrangement for MP you are not required to do all assigned HWs. 4

3 1 5 Equation of State: pv = nrt You are now familiar with Equation of motion : F = m a Now, imagine a device that we could vary temperature (T), volume (V), pressure (p), and the amount of sample. It would show that V is proportional to the moles of sample (n), that V varies inversely with p, and that p and/or V vary in proportion to T. Results lead to form one equation to describe the overall behavior of an ideal gas: pv = nrt Visualize the equation with the figure. 6

4 Key Numbers and Equations Equation of State in Per mole basis P V = n R T ; R = ideal gas constant in Per molecule basis Two types of equations. 7 H 2 O Molecules M = N A x m H2O m H2O = 29.9x10-24 g/molecule from Example 15.1 Be familiar with Molar mass. 8

5 Molecular Speed Distributions Molecules move at a distribution of speeds around a mean velocity for any given temperature (T). See Examples Check visually the locations of mp, avg and rms. 9 Kinetic Molecular Theory Gases may be treated as point particles undergoing rapid elastic collisions with each other and the container Pressure Check two equations Chap. 8 10

6 Especially useful to plot p and V at constant T for a range of temperatures. In this way we can generate a 3-D surface of isothermal lines and make predictions of an ideal gas s behavior. P, V, and T Plots Adiabatic: have no heat transfer in or out of the system Isochoric: have no volume change. Isobaric: no pressure change. Isothermal: no temperature change. Read p-v graphs 11 pv = nrt (Ideal Gas) M = N A x m H2O m H2O =29.9x10-24 g/molecule 12

7 Visual Examples of Work F x x = p (A x) = p V Check visually the work corresponds to the area in p-v graph. Note: they are signed numbers 13 [extra Q] total work from d to a? 14

8 [Q1] total work from a b d? [Q2] total work from d b a? 15 Example Questions Note that there is no dependence on m 16

9 Example p = gauge pressure + 1 atm 3 2 Molar mass 4 17 Example 15.3 (II) Absolute pressure is zero-referenced against a perfect vacuum, so it is equal to gauge pressure plus atmospheric pressure. Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure. 18

10 P P15-13 (Work Book) Fine print: Eq

11 P P15-23 (Work Book) 22

12 More Examples x10 23 molecules/mol: A number to describe a set count of atoms, like dozen is a standard set for eggs. Avogadro s Number Approximately 100 billion (1.0x10 11 ) stars in in Milky Way galaxy. It would take a trillion (1.0x10 12 ) Milky Way galaxies to contain as many stars as there are particles in a mole. Two dozens of eggs 24 Two moles of gas 12 x Molar mass: M= N A x m molecule Water (a liquid at 18g/mol), Nitrogen (a gas at 28g/mol), Table salt (sodium chloride, a solid at 58g/mol). m H2O =29.9x10-24 g/molecule from Example

13 Ref. 1: Model of Kinetic Property Chap. 7 Eq. 7.3 K av = (1/2) m v 2 T 2 Fig T 3 Chap. 8 Eq Ref. 2: Details of Kinetic Property Container V, N, T o o N = Number of particles in volume V Number of particles per unit volume is N/V v i,x = -30 m/s v i,x = 30 m/s Chap. 8 p = p f,i p i,x = (040 kg)(+30 m/s) (0.40 kg)(-30 m/s) =2 (Mass) x v x 26

14 Ref. 3: Details of Kinetic Property V, N, T, p o Container N = Number of particles in volume V K av = (1/2) m v 2 (3/2) nr T = N K av = K trans o o Number of particles per unit volume is N/V pv = n R T pv = N k T K av = (3/2) k T v i,x = -30 m/s v i,x = 30 m/s Chap. 8 p = p f,i p i,x = (040 kg)(+30 m/s) (0.40 kg)(-30 m/s) =2 (Mass) x v x 27 Chap. 15 for Exam4 Chap Chap Chap Chap (only Eq.15 & 15.16) Chap

15 Where We Are as of Nov 24? Exam 4 29 Q&A after Nov.24 Class [Q] How do I study? Repeat MP problems? [A] If you have completed MPs with your own notes on problems, then review my slides with your notes. My slides are reflection of what I was/am thinking. See reflection for each exam at URL. [URL] /class/2015c/2015c_until_exam3.html 30

16 2 31 Part 2: (Recap) Heat Capacity Substances have an ability to hold heat that goes to the atomic level. Q = m c T Per kg basis [J] = [kg] [?] [K] c = specific heat capacity [J / (kg * K)] c water = 4.19 x 10 3 J/(kg*K) vs. c copper = 0.39 x 10 3 J/(kg*K) [Q] What is c? How effectively the substance can hold heat. [Q] Why Q is proportional to m? The vibration of each atom is a reason for holding heat. Many atoms means holding more heat. [Q] Why Q is proportional to T? It also depends on how much the heat transfer is made. c m 32

17 Molar Heat Capacities Per mole basis Q = m c T C = Heat capacity for 6 x molecules 33 Finding C V Q = K 34

18 C V = (3/2) R Q = K C V = (3/2) R Monatomic molecules Why? Diatomic molecules 35 C V = (3/2) R + R = (5/2) R Per molecule basis Per mole basis Molecules can store heat energy in translation, rotation and vibration. Table 15.3 to guide you through a calculation. 36

19 3 37 Part 3: Thermodynamic Processes A process can be isochoric and have no volume change. A process can be isobaric and have no pressure change. A process can be isothermal and have no temperature change. A process can be adiabatic and have no heat transfer in or out of the system. 38

20 The First Law of Thermodynamics In simple terms, the heat added to a system will be distributed between internal energy ( U) and work (W). Q = U+ W Work is defined differently than we did in earlier chapters, here it refers to a p V (a pressure increasing a volume). Q = n C V T + p (V 2 V 1 ) 39 The First Law of Thermodynamics In simple terms, the heat added to a system will be distributed between internal energy ( U) and work (W). Q = U+ W Work is defined differently than we did in earlier chapters, here it refers to a p V (a pressure increasing a volume). Note on the sign convention. Q = n C V T + p (V 2 V 1 ) 40

21 The First Law of Paycheck = + Paycheck = Saving + Spending In simple terms, the income to a household will be distributed between Saving and Spending. Spending is defined differently than we did in earlier chapters, here it refers to a p V (unit price times volume). Q = n C V T + p (V 2 V 1 ) 41 Blank Page 42

22 Work = Area in p-v graph Example 15.8(a) Incremental, reproducible changes may be summed. Example More Visual Examples 44

23 P15-78 A process can be adiabatic and have no heat transfer in or out of the system Q = 0 Q = U+ W W = U U = n C V T 45 P15-78 A process can be adiabatic and have no heat transfer in or out of the system Q = 0 Q = U+ W W = U U = n C V T 46

24 Adiabatic Process 47 P15-78 Etrxa T (b) = K Can you find T (c)? 48

25 Adiabatic Compression 49 Blank Page 50

26 Work in Isothermal Process W = n R T ln(v 2 / V 1 ) 51 Isothermal Expansion Examples 15.8, 15.9, and

27 You heat a sample of air to twice its original temperature in a constant-volume container. The average translational kinetic energy of the molecules is A. half the original value. B. unchanged. C. twice the original value. D. four times the original value. K av = (3/2) k T Q = n C V T + p (V 2 V 1 ) 53 You heat a sample of air to twice its original temperature in a constant-volume container. The average translational kinetic energy of the molecules is C. twice the original value. 54

28 You compress a sample of air slowly to half its original volume, keeping its temperature constant. The internal energy of the gas A. decreases to half its original value. B. remains unchanged. C. increases to twice its original value. K av = (3/2) k T Q = n C V T + p (V 2 V 1 ) 55 You compress a sample of air slowly to half its original volume, keeping its temperature constant. The internal energy of the gas B. remains unchanged. 56

29 Example 57 MC Example 58

30 Where We Are? Final 59 60