Chapter 16

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

)(1

)(1 000

ppB

TTV

V

Derive the equation of stateDerive 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

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 16-53Do 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


• 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, 16-12Do Exercises 16-9, 16-12


• State variables: p, V, T, and mgas= nM

• State variables: p, V, T, and = mgas/V

RTM

mpV gas

T

p

R

M

Derive ‘Law of Atmospheres’Derive ‘Law of Atmospheres’

Non-Ideal Gases?

• Ideal gas equation:

• Van der Waals equation:

nRTbnVV

nap

)(

2

2

nRTpV

NotesNotes

pV–Diagram for an Ideal Gas

NotesNotes

pV–Diagram for a Non-Ideal Gas

NotesNotes

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)

NkT

RTN

N

nRTpV

A

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 + vy

2 + vz2

• Each molecule has a different speed

• Consider averaging over all molecules

Kinetic Model

• average over all molecules:

(v2)av= (vx2 + vy

2 + vz2)av

= (vx2)av+(vy

2)av+(vz2)av

= 3 (vx2)av

Kinetic Model

• (Ktr)av= total kinetic energy of gas due to translation

• Derive result:

avtr )(3

2KpV

Kinetic Model

• Compare to ideal gas law:

pV = nRT

pV = NkT

avtr )(3

2KpV

Kinetic Energy

• average translational KE is directly proportional to gas temperature T

NkT

nRTK

2

3

2

3)( avtr

Kinetic Energy

• average translational KE per molecule:

• average translational KE per mole:

kTvm2

3)(

2

1av

2

RTvM2

3)(

2

1av

2

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

kTvm2

3)(

2

1av

2

Kinetic Model

• ‘root-mean-square’ speed vrms:

M

RT

m

kTv

33)( av

2

M

RT

m

kTvv

33)( av

2rms

Molecular Speeds

• For a given T, lighter molecules move faster

• Explains why Earth’s atmosphere contains alomost no hydrogen, only heavier gases

M

RT

m

kTvv

33)( av

2rms

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

kTmvevkT

mvf 2/2

22/3

24)(

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 speedm

kTdvvfvv

8

)( 0av

m

kTdvvfvv

vv

3 )( )(

)(

0

2av

2

av2

rms

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

Molecular Collisions

• Mean free path between collisions:

pr

kT

VNr

2

2

24

)/( 24

1

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


Q = energy required to change temperature of mass m by T

c = ‘specific heat capacity’

= energy required per (unit mass × unit T)

TmcQ


• Now introduce ‘molar heat capacity’ C

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

TmcQ

TnCQ


• important case:the volume V of material is held constant

• CV = molar heat capacity at constant volume

TnCQ

TnCQ V

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


• Monatomic gas:

(Ktr)av = (3/2) nRT

• note: your text just writesKtr instead of (Ktr)av

• Consider changing T by dT


• 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

A Look Ahead

(Etot)av = (f/2) nRT

CV = (f/2)R

Monatomic gas: f = 3

Diatomic gas: f = 3, 5, 7


• What about gases with other kinds of molecules?

• diatomic, triatomic, etc.

• These molecules are not pointlike


• 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

DemonstrationDemonstration


• 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 KE in 3 directions: vx, vy, vz

• f = 3 degrees of freedom

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


• 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

A Look Ahead


CV = (f/2)R

Monatomic gas: f = 3

Diatomic gas: f = 3, 5, 7


• 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


• Diatomic gas:

higher temperature

• translational KE (in 3 directions) • rotational KE (about 2 axes)

• f = 3+2 = 5 degrees of freedom



• 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


Summary of CV for Ideal Gases


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


= 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

pT Phase DiagramDo Exercise 16-39Do Exercise 16-39

pV–Diagram for a Non-Ideal Gas

NotesNotes

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

Documents

Chapter 16