Chapter 16

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# Chapter 16 - PowerPoint PPT Presentation

Chapter 16. Thermal Properties of Matter. Macroscopic Description of Matter. State Variables. State variable = macroscopic property of thermodynamic system Examples: pressure p volume V temperature T mass m. State Variables.

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### Chapter 16

Thermal Propertiesof Matter

### Macroscopic Descriptionof Matter

State Variables
• State variable = macroscopic property of thermodynamic system
• Examples: pressure p

volume V

temperature T

mass m

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
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
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

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
• 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

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!

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
Ideal Gas Equation of State
• State variables: p, V, T, n

pV = nRT

• R = 8.3145 J/(mol·K)
• same value of R for all (low density) gases
• same (simple, ‘ideal’) equation

Do Exercises 16-9, 16-12

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’

Non-Ideal Gases?
• Ideal gas equation:
• Van der Waals equation:

Notes

### Microscopic Descriptionof Matter

Ideal Gas Equation

pV = nRT

• n = number of moles of gas = N/NA
• R = 8.3145 J/(mol·K)
• N = number of molecules of gas
• NA = 6.02×1023 molecules/mol
Ideal Gas Equation
• k = Boltzmann constant = R/NA = 1.381×10-23 J/(molecule·K)
Ideal Gas Equation

pV = nRT

pV = NkT

• k = R/NA
• ‘ RT per mol’ vs. ‘kT per molecule’

### Kinetic-Molecular Theory of an Ideal Gas

Assumptions
• gas = large number N of identical molecules
• molecule = point particle, mass m
• molecules collide with container walls= origin of macroscopic pressure of gas
Kinetic Model
• molecules collide with container walls
• assume perfectly elastic collisions
• walls are infinitely massive (no recoil)
Elastic Collision
• wall:

infinitely massive, doesn’t recoil

• molecule:

vy: unchanged

vx : reverses direction

speed v : unchanged

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

(v2)av= (vx2 + vy2 + vz2)av

= (vx2)av+(vy2)av+(vz2)av

= 3 (vx2)av

Kinetic Model
• (Ktr)av= total kinetic energy of gas due to translation
• Derive result:
Kinetic Model
• Compare to ideal gas law:

pV = nRT

pV = NkT

Kinetic Energy
• average translational KE is directly proportional to gas temperature T
Kinetic Energy
• average translational KE per molecule:
• average translational KE per mole:
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
Kinetic Model
• ‘root-mean-square’ speed vrms:
Molecular Speeds
• For a given T, lighter molecules move faster
• Explains why Earth’s atmosphere contains alomost no hydrogen, only heavier gases
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)
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

Molecular Speeds
• Maxwell-Boltzmann distribution function
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
Molecular Speeds
• average speed
• rms speed
Molecular Collisions?
• We assumed:
• molecules = point particles, no collisions
• Real gas molecules:
• have finite size and collide
• Find ‘mean free path’ between collisions
Molecular Collisions
• Mean free path between collisions:
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

### Heat Capacity Revisited

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)

Heat Capacity Revisited
• Now introduce ‘molar heat capacity’ C

C = energy per (mol × unit DT) required to change temperature of n moles by DT

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

CV for the Ideal Gas
• Monatomic gas:

(Ktr)av = (3/2) nRT

• Consider changing T by dT
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
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

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7

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

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?
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

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

Check a known case
• Monatomic gas:
• only has translational KEin 3 directions: vx, vy, vz
• f = 3 degrees of freedom

(Ktr)av = (f/2) nRT = (3/2) nRT

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

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7

CV for the Ideal Gas
• Diatomic gas:

low temperature

• only translational KEin 3 directions: vx, vy, vz
• f = 3 degrees of freedom

(Etot)av = (f/2) nRT = (3/2) nRT

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

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 = 3+2+2 =7 degrees of freedom

(Etot)av = (f/2) nRT = (7/2) nRT

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)

CV for Solids
• Each atom in a solid can vibrate about its equilibrium position
• Atoms undergo simple harmonic motion in all 3 directions
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
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
CV for Solids
• f = 3 + 3 = 6 degrees of freedom

(Etot)av = (f/2) nRT

= 3 nRT

CV= (f/2)R = 3 R

### Phase Changes Revisited

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