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Thermal Properties of Matter

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Presentation on theme: "Thermal Properties of Matter"— Presentation transcript:

1 Thermal Properties of Matter
Chapter 16 Thermal Properties of Matter

2 Macroscopic Description of Matter

3 State Variables State variable = macroscopic property of thermodynamic system Examples: pressure p volume V temperature T mass m

4 State Variables State variables: p, V, T, m
I general, we cannot change one variable without affecting a change in the others Recall: For a gas, we defined temperature T (in kelvins) using the gas pressure p

5 Equation of State State variables: p, V, T, m
The relationship among these: ‘equation of state’ sometimes: an algebraic equation exists often: just numerical data

6 Equation of State Warm-up example:
Approximate equation of state for a solid Based on concepts we already developed Here: state variables are p, V, T Derive the equation of state

7 The ‘Ideal’ Gas The state variables of a gas are easy to study:
p, V, T, mgas often use: n = number of ‘moles’ instead of mgas

8 Moles and Avogadro’s Number NA
1 mole = 1 mol = 6.02×1023 molecules = NA molecules n = number of moles of gas M = mass of 1 mole of gas mgas = n M Do Exercise 16-53

9 The ‘Ideal’ Gas We measure:
the state variables (p, V, T, n) for many different gases We find: at low density, all gases obey the same equation of state!

10 Ideal Gas Equation of State
State variables: p, V, T, n pV = nRT p = absolute pressure (not gauge pressure!) T = absolute temperature (in kelvins!) n = number of moles of gas

11 Ideal Gas Equation of State
State variables: p, V, T, n pV = nRT R = J/(mol·K) same value of R for all (low density) gases same (simple, ‘ideal’) equation Do Exercises 16-9, 16-12

12 Ideal Gas Equation of State
State variables: p, V, T, and mgas= nM State variables: p, V, T, and r = mgas/V Derive ‘Law of Atmospheres’

13 Non-Ideal Gases? Ideal gas equation: Van der Waals equation: Notes

14 pV–Diagram for an Ideal Gas
Notes

15 pV–Diagram for a Non-Ideal Gas
Notes

16 Microscopic Description of Matter

17 Ideal Gas Equation pV = nRT n = number of moles of gas = N/NA
R = J/(mol·K) N = number of molecules of gas NA = 6.02×1023 molecules/mol

18 Ideal Gas Equation k = Boltzmann constant = R/NA = 1.381×10-23 J/(molecule·K)

19 Ideal Gas Equation pV = nRT pV = NkT k = R/NA
‘ RT per mol’ vs. ‘kT per molecule’

20 Kinetic-Molecular Theory of an Ideal Gas

21 Assumptions gas = large number N of identical molecules
molecule = point particle, mass m molecules collide with container walls = origin of macroscopic pressure of gas

22 Kinetic Model molecules collide with container walls
assume perfectly elastic collisions walls are infinitely massive (no recoil)

23 Elastic Collision wall: infinitely massive, doesn’t recoil molecule:
vy: unchanged vx : reverses direction speed v : unchanged

24 Kinetic Model For one molecule: v2 = vx2 + vy2 + vz2
Each molecule has a different speed Consider averaging over all molecules

25 Kinetic Model average over all molecules: (v2)av= (vx2 + vy2 + vz2)av
= (vx2)av+(vy2)av+(vz2)av = 3 (vx2)av

26 Kinetic Model (Ktr)av= total kinetic energy of gas due to translation
Derive result:

27 Kinetic Model Compare to ideal gas law: pV = nRT pV = NkT

28 Kinetic Energy average translational KE is directly proportional to gas temperature T

29 Kinetic Energy average translational KE per molecule:

30 Kinetic Energy average translational KE per molecule:
independent of p, V, and kind of molecule for same T, all molecules (any m) have the same average translational KE

31 Kinetic Model ‘root-mean-square’ speed vrms:

32 Molecular Speeds For a given T, lighter molecules move faster
Explains why Earth’s atmosphere contains alomost no hydrogen, only heavier gases

33 Molecular Speeds Each molecule has a different speed, v
We averaged over all molecules Can calculate the speed distribution, f(v) (but we’ll just quote the result)

34 Molecular Speeds f(v) = distribution function
f(v) dv = probability a molecule has speed between v and v+dv dN = number of molecules with speed between v and v+dv = N f(v) dv

35 Molecular Speeds Maxwell-Boltzmann distribution function

36 Molecular Speeds At higher T: more molecules have higher speeds
Area under f(v) = fraction of molecules with speeds in range: v1 < v < v1 or v > vA

37 Molecular Speeds average speed rms speed

38 Molecular Collisions? We assumed:
molecules = point particles, no collisions Real gas molecules: have finite size and collide Find ‘mean free path’ between collisions

39 Molecular Collisions

40 Molecular Collisions Mean free path between collisions:

41 Announcements Midterms: Returned at end of class
Scores will be entered on classweb soon Solutions available online at E-Res soon Homework 7 (Ch. 16): on webpage Homework 8 (Ch. 17): to appear soon

42 Heat Capacity Revisited

43 Heat Capacity Revisited
DQ = energy required to change temperature of mass m by DT c = ‘specific heat capacity’ = energy required per (unit mass × unit DT)

44 Heat Capacity Revisited
Now introduce ‘molar heat capacity’ C C = energy per (mol × unit DT) required to change temperature of n moles by DT

45 Heat Capacity Revisited
important case: the volume V of material is held constant CV = molar heat capacity at constant volume

46 CV for the Ideal Gas Monatomic gas:
molecules = pointlike (studied last lecture) recall: translational KE of gas averaged over all molecules (Ktr)av = (3/2) nRT

47 CV for the Ideal Gas Monatomic gas: (Ktr)av = (3/2) nRT
note: your text just writes Ktr instead of (Ktr)av Consider changing T by dT

48 CV for the Ideal Gas Monatomic gas: (Ktr)av = (3/2) nRT
d(Ktr)av = n (3/2)R dT recall: dQ = n CV dT so identify: CV = (3/2)R

49 In General: If (Etot)av = (f/2) nRT Then d(Etot)av = n (f/2)R dT
But recall: dQ = n CV dT So we identify: CV = (f/2)R

50 A Look Ahead (Etot)av = (f/2) nRT CV = (f/2)R Monatomic gas: f = 3
Diatomic gas: f = 3, 5, 7

51 CV for the Ideal Gas What about gases with other kinds of molecules?
diatomic, triatomic, etc. These molecules are not pointlike

52 CV for the Ideal Gas Diatomic gas: molecules = ‘dumbell’ shape
its energy takes several forms: (a) translational KE (3 directions) (b) rotational KE (2 rotation axes) (c) vibrational KE and PE Demonstration

53 (Etot)av = (Ktr)av + (Krot)av + (Evib)av
CV for the Ideal Gas Diatomic gas: Etot = Ktr + Krot + Evib (Etot)av = (Ktr)av + (Krot)av + (Evib)av we know: (Ktr)av = (3/2) nRT what about the other terms?

54 Equipartition of Energy
Can be proved, but we’ll just use the result Define: f = number of degrees of freedom = number of independent ways that a molecule can store energy

55 Equipartition of Energy
It can be shown: The average amount of energy in each degree of freedom is: (1/2) kT per molecule i.e. (1/2) RT per mole

56 Check a known case Monatomic gas:
only has translational KE in 3 directions: vx, vy, vz f = 3 degrees of freedom (Ktr)av = (f/2) nRT = (3/2) nRT

57 CV for the Ideal Gas Diatomic gas:
more forms of energy are available to the gas as you increase its T: (a) translational KE (3 directions) (b) rotational KE (2 rotation axes) (c) vibrational KE and PE

58 A Look Ahead (Etot)av = (f/2) nRT CV = (f/2)R Monatomic gas: f = 3
Diatomic gas: f = 3, 5, 7

59 CV for the Ideal Gas Diatomic gas: low temperature
only translational KE in 3 directions: vx, vy, vz f = 3 degrees of freedom (Etot)av = (f/2) nRT = (3/2) nRT

60 CV for the Ideal Gas Diatomic gas: higher temperature
translational KE (in 3 directions) rotational KE (about 2 axes) f = 3+2 = 5 degrees of freedom (Etot)av = (f/2) nRT = (5/2) nRT

61 CV for the Ideal Gas Diatomic gas: even higher temperature
translational KE (in 3 directions) rotational KE (about 2 axes) vibrational KE and PE f = =7 degrees of freedom (Etot)av = (f/2) nRT = (7/2) nRT

62 Summary of CV for Ideal Gases
(Etot)av = (f/2) nRT CV = (f/2)R Monatomic: f = 3 (only) Diatomic: f = 3, 5, 7 (with increasing T)

63

64 CV for Solids Each atom in a solid can vibrate about its equilibrium position Atoms undergo simple harmonic motion in all 3 directions

65 CV for Solids Kinetic energy : 3 degrees of freedom K = Kx+ Ky + Kz
Kx = (1/2) mvx2 Ky = (1/2) mvy2 Kz = (1/2) mvz2

66 CV for Solids Potential energy: 3 degrees of freedom U = Ux+ Uy + Uz
Ux = (1/2) kx x2 Uy = (1/2) ky y2 Uz = (1/2) kz z2

67 CV for Solids f = 3 + 3 = 6 degrees of freedom (Etot)av = (f/2) nRT
CV = (f/2)R = 3 R

68

69 Phase Changes Revisited

70 Phase Changes ‘phase’ = state of matter = solid, liquid, vapor
during a phase transition : 2 phases coexist at the triple point : all 3 phases coexist

71 Do Exercise 16-39 pT Phase Diagram

72 pV–Diagram for a Non-Ideal Gas
Notes

73 Announcements Midterms: Returned at end of class
Scores will be entered on classweb soon Solutions available online at E-Res soon Homework 7 (Ch. 16): on webpage Homework 8 (Ch. 17): to appear soon


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