Chapter 18 Lecture

1. Chapter 18 Lecture

2. Chapter 18 The Micro/Macro Connection Chapter Goal: To understand a macroscopic system in terms of the microscopic behavior of its molecules. Slide 18-2

3. Chapter 18 Preview Slide 18-3

4. Chapter 18 Preview Slide 18-4

5. Chapter 18 Preview Slide 18-5

6. Chapter 18 Preview Slide 18-6

7. Chapter 18 Preview Slide 18-7

8. Chapter 18 Preview Slide 18-8

9. Chapter 18 Reading Quiz Slide 18-9

10. Reading Question 18.1 What is the name of the quantity represented as vrms? • Random-measured-step viscosity. • Root-mean-squared speed. • Relative-mean-system velocity. • Radial-maser-system volume. Slide 18-10

11. Reading Question 18.1 What is the name of the quantity represented as vrms? • Random-measured-step viscosity. • Root-mean-squared speed. • Relative-mean-system velocity. • Radial-maser-system volume. Slide 18-11

12. Reading Question 18.2 What additional kind of energy makes CV larger for a diatomic than for a monatomic gas? • Charismatic energy. • Translational energy. • Heat energy. • Rotational energy. • Solar energy. Slide 18-12

13. Reading Question 18.2 What additional kind of energy makes CV larger for a diatomic than for a monatomic gas? • Charismatic energy. • Translational energy. • Heat energy. • Rotational energy. • Solar energy. Slide 18-13

14. Reading Question 18.3 The second law of thermodynamics says that • The entropy of an isolated system never decreases. • Heat never flows spontaneously from cold to hot. • The total thermal energy of an isolated system is constant. • Both A and B. • Both A and C. Slide 18-14

15. Reading Question 18.3 The second law of thermodynamics says that • The entropy of an isolated system never decreases. • Heat never flows spontaneously from cold to hot. • The total thermal energy of an isolated system is constant. • Both A and B. • Both A and C. Slide 18-15

16. Reading Question 18.4 In general, • Both microscopic and macroscopic processes are reversible. • Both microscopic and macroscopic processes are irreversible. • Microscopic processes are reversible and macroscopic processes are irreversible. • Microscopic processes are irreversible and macroscopic processes are reversible. Slide 18-16

17. Reading Question 18.4 In general, • Both microscopic and macroscopic processes are reversible. • Both microscopic and macroscopic processes are irreversible. • Microscopic processes are reversible and macroscopic processes are irreversible. • Microscopic processes are irreversible and macroscopic processes are reversible. Slide 18-17

18. Chapter 18 Content, Examples, and QuickCheck Questions Slide 18-18

19. Molecular Speeds and Collisions • A gas consists of a vast number of molecules, each moving randomly and undergoing millions of collisions every second. • Shown is the distribution of molecular speeds in a sample of nitrogen gas at 20C. • The micro/macro connection is built on the idea that the macroscopic properties of a system, such as temperature or pressure, are related to the averagebehavior of the atoms and molecules. Slide 18-19

20. Mean Free Path • A single molecule follows a zig-zag path through a gas as it collides with other molecules. • The average distance between the collisions is called the mean free path: • (N/V) is the number density of the gas in m−3. • r is the the radius of the molecules when modeled as hard spheres; for many common gases r ≈ 10−10 m. Slide 18-20

21. QuickCheck 18.1 The temperature of a rigid container of oxygen gas (O2) is lowered from 300C to 0C. As a result, the mean free path of oxygen molecules Increases. Is unchanged. Decreases. Slide 18-21

22. QuickCheck 18.1 The temperature of a rigid container of oxygen gas (O2) is lowered from 300C to 0C. As a result, the mean free path of oxygen molecules Increases. Is unchanged. Decreases. λ depends only on N/V, not T. Slide 18-22

23. Example 18.1 The Mean Free Path at Room Temperature Slide 18-23

24. Pressure in a Gas • Why does a gas have pressure? • In Chapter 15 we suggested that the pressure in a gas is due to collisions of the molecules with the walls of its container. • The steady rain of a vast number of molecules striking a wall each second exerts a measurable macroscopic force. • The gas pressure is the force per unit area (p = F/A) resulting from these molecular collisions. Slide 18-24

25. Pressure in a Gas • The figure shows a molecule which collides with a wall, exerting an impulse on it. • The x-component of the impulse from a single collision is: • If there are Ncoll such collisions during a small time interval t, the net impulse is: Slide 18-25

26. Pressure in a Gas • The figure reminds you that impulse is the area under the force-versus-time curve and thus Jwall = Favgt. • The average force on the wall is: • where the rate of collisions is: Slide 18-26

27. Pressure in a Gas • The pressure is the average force on the walls of the container per unit area: • (N/V) is the number density of the gas in m−3. • Note that the average velocity of many molecules traveling in random directions is zero. • vrms is the root-mean-square speed of the molecules, which is the square root of the average value of the squares of the speeds of the molecules: Slide 18-27

28. QuickCheck 18.2 A rigid container holds oxygen gas (O2) at 100C. The average velocity of the molecules is Greater than zero. Zero. Less than zero. Slide 18-28

29. QuickCheck 18.2 A rigid container holds oxygen gas (O2) at 100C. The average velocity of the molecules is Greater than zero. Zero. Less than zero. Slide 18-29

30. Example 18.2 Calculating the Root-Mean-Square Speed Units of velocity are m/s. Slide 18-30

31. Example 18.2 Calculating the Root-Mean-Square Speed Slide 18-31

32. Example 18.2 Calculating the Root-Mean-Square Speed Slide 18-32

33. Example 18.3 The RMS Speed of Helium Atoms Example 18.3 The rms Speed of Helium Atoms Slide 18-33

34. QuickCheck 18.3 A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100C. Which statement describes their rms speeds? vrmsofH2 < vrmsof N2. vrmsofH2 = vrmsof N2. vrmsofH2 > vrmsof N2. Slide 18-34

35. QuickCheck 18.3 A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100C. Which statement describes their rms speeds? vrms of H2 < vrms of N2. vrms of H2 = vrms of N2. vrms of H2 > vrms of N2. Slide 18-35

36. Temperature in a Gas • The thing we call temperature measures the average translational kinetic energy єavg of molecules in a gas. • A higher temperature corresponds to a larger value of єavgand thus to higher molecular speeds. • Absolute zero is the temperature at which єavg= 0 and all molecular motion ceases. • By definition, єavg = ½mvrms2, where vrms is the root mean squared molecular speed; using the ideal-gas law, we found єavg = 3/2 kBT. • By equating these expressions we find that the rms speed of molecules in a gas is: Slide 18-36

37. QuickCheck 18.4 A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100C. Which statement describes the average translational kinetic energies of the molecules? єavg of H2 < єavg of N2. єavg of H2 = єavg of N2. єavg of H2 > єavg of N2. Slide 18-37

38. QuickCheck 18.4 A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100C. Which statement describes the average translational kinetic energies of the molecules? єavg of H2 < єavg of N2. єavg of H2 = єavg of N2. єavg of H2 > єavg of N2. Slide 18-38

39. The Micro/Macro Connection for Pressure and Temperature Slide 18-39

40. Example 18.4 Total Microscopic Kinetic Energy Slide 18-40

41. Example 18.5 Calculating an RMS Speed Slide 18-41

42. Example 18.6 Mean Time Between Collisions Slide 18-42

43. Thermal Energy and Specific Heat • The thermal energy of a system is Eth = Kmicro + Umicro. • The figure shows a monatomic gas such as helium or neon. • The atoms in a monatomic gas have no molecular bonds with their neighbors, hence Umicro = 0. • Since the average kinetic energy of a single atom in an ideal gas is єavg = 3/2 kBT, the total thermal energy is: Slide 18-43

44. Thermal Energy and Specific Heat • If the temperature of a monatomic gas changes by T, its thermal energy changes by: • In Chapter 17 we found that the change in thermal energy for any ideal-gas process is related to the molar specific heat at constant volume by: • Combining these equations gives us a prediction for the molar specific heat for a monatomic gas: • This prediction is confirmed by experiments. Slide 18-44

45. The Equipartition Theorem • Atoms in a monatomic gas carry energy exclusively as translational kinetic energy (3 degrees of freedom). • Molecules in a gas may have additional modes of energy storage, for example, the kinetic and potential energy associated with vibration, or rotational kinetic energy. • We define the number of degrees of freedom as the number of distinct and independent modes of energy storage. Slide 18-45

46. QuickCheck 18.5 A mass on a spring oscillates back and forth on a frictionless surface. How many degrees of freedom does this system have? 1. 2. 3. 4. 6. Slide 18-46

47. QuickCheck 18.5 A mass on a spring oscillates back and forth on a frictionless surface. How many degrees of freedom does this system have? 1. 2. 3. 4. 6. It can hold energy as kinetic energy or potential energy. Slide 18-47

48. Thermal Energy of a Solid • The figure reminds you of the “bedspring model” of a solid with particle-like atoms connected by spring-like molecular bonds. • There are 3 degrees of freedom associated with kinetic energy + 3 more associated with the potential energy in the molecular bonds = 6 degrees of freedom total. • The energy stored in each degree of freedom is ½ NkBT, so: Slide 18-48

49. Diatomic Molecules • In addition to the 3 degrees of freedom from translational kinetic energy, a diatomic gas at commonly used temperatures has 2 additional degrees of freedom from end-over-end rotations. • This gives 5 degrees of freedom total: Slide 18-49

50. Thermal Energy and Specific Heat Slide 18-50