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
• 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
V
1
 1   (T  T0 )  ( p  p0 )
V0
B
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
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
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
pV 
mgas
M
RT
• State variables: p, V, T, and r = mgas/V
M p
r
R T
Derive ‘Law of Atmospheres’
Non-Ideal Gases?
• Ideal gas equation:
pV  nRT
• Van der Waals equation:

n2 
 p  a 2 (V  bn)  nRT
V 

Notes
pV–Diagram for an Ideal Gas
Notes
pV–Diagram for a Non-Ideal Gas
Notes
Microscopic Description
of 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
pV  nRT
N

RT
NA
 NkT
• 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:
2
pV  ( K tr ) av
3
Kinetic Model
2
pV  ( K tr ) av
3
• Compare to ideal gas law:
pV = nRT
pV = NkT
Kinetic Energy
3
( K tr ) av  nRT
2
3
 NkT
2
• average translational KE is directly
proportional to gas temperature T
Kinetic Energy
• average translational KE per molecule:
1
3
2
m(v ) av  kT
2
2
• average translational KE per mole:
1
3
2
M (v ) av  RT
2
2
Kinetic Energy
• average translational KE per molecule:
1
3
2
m(v ) av  kT
2
2
• independent of p, V, and kind of molecule
• for same T, all molecules (any m) have the
same average translational KE
Kinetic Model
3kT 3RT
(v ) av 

m
M
2
• ‘root-mean-square’ speed vrms:
vrms
3kT
3RT
 (v ) av 

m
M
2
Molecular Speeds
vrms
3kT
3RT
 (v ) av 

m
M
2
• 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
 m 
f (v)  4 

 2kT 
3/ 2
2
v 2e  mv / 2kT
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
vav  

0
8kT
v f (v) dv 
m
• rms speed
vrms  (v 2 ) av
(v ) av  
2

0
3kT
v f (v) dv 
m
2
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:
1

4 2 r 2 ( N / V )
kT

4 2 r 2 p
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  mc DT
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
DQ  mc DT
• Now introduce ‘molar heat capacity’ C
C = energy per (mol × unit DT) required
to change temperature of n moles by DT
DQ  nC DT
Heat Capacity Revisited
DQ  nC DT
• important case:
the volume V of material is held constant
• CV = molar heat capacity at constant volume
DQ  nCV DT
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
• note: your text just writes
Ktr instead of (Ktr)av
• 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
Then
But recall:
So we identify:
(Etot)av = (f/2) nRT
d(Etot)av = n (f/2)R dT
dQ = n CV dT
CV = (f/2)R
A Look Ahead
(Etot)av = (f/2) nRT
CV = (f/2)R
Monatomic gas:
Diatomic gas:
f=3
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 KE
in 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
A Look Ahead
(Etot)av = (f/2) nRT
CV = (f/2)R
Monatomic gas:
Diatomic gas:
f=3
f = 3, 5, 7
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
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:
Diatomic:
f = 3 (only)
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
Do Exercise 16-39
pT Phase Diagram
pV–Diagram for a Non-Ideal Gas
Notes
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