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)

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

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