Physics 351
Thermal Physics
Spring 2011
Prerequisites: 01:750:227 or 272 or
permission of the instructor; 01:640:CALC3
Professor:
Office:
e-mail:
phone:
office hour:
lectures:
Grader:
Weida Wu
Serin Physics Building Room 117W
wdwu@physics.rutgers.edu
732-445-5500 X 8299 (office)
Friday 2:00-3:00 PM and by appointment
SEC-117, Wed. 10:20-11:40 AM
Fri. 3:20-4:40 PM
Bin Gao (bingao@physics.rutgers.edu)
Physics 351
Thermal Physics
Spring 2011
Textbook:
An Introduction to Thermal Physics,
D.V. Schroeder,
Addison-Wesley-Longman, 2000
Web Site Access:
http://www.physics.rutgers.edu/ugrad/351
Homework assignments, announcements and
notes will be posted here. The link must be
visited regularly.
Homework (30%)
~ 10 problems in each HW, 3 points each problem.
Due every Wednesday 4 PM in Grader’s mailbox.
No late homework will be accepted.
One lowest grade will be dropped.
3 points if the problem was done almost entirely correctly,
with at most a trivial error.
2 points if a generally correct approach was used, but some
mistakes were made along the way.
1 point if a good faith effort was made, but the method of
solution was significantly flawed.
0 points if little or no effort was made on the problem.
Exams (70%)
Midterm I (ch. 1-3)
Midterm II (ch. 4,5)
Final exam (ch. 1-7)
02/18
04/01
05/10
20%
20%
30%
~ 4 problems each midterm. ~6-8 problems for Final.
Only textbook is allowed. Neither homework nor
PRINT lecture note is allowed.
Tentative letter grade range:
A
B+
B
C+
C
85%
75%
70%
60%
55%
D
F
40% <40%
Thermal Physics = Thermodynamics + Statistical Mechanics
- conceptually, the most difficult subject of the undergraduate
physics program.
Thermodynamics provides a framework of relating the
macroscopic properties of a system to one another. It is
concerned only with macroscopic quantities and ignores the
microscopic variables that characterize individual molecules
(both strength and weakness).
Statistical Mechanics is the bridge between the microscopic
and macroscopic worlds: it links the laws of thermodynamics to
the statistical behavior of molecules.
* Lecture notes: Originally written by Prof. Gershenson *
Macroscopic Description is Qualitatively Different!
Why do we need to consider macroscopic bodies as a special
class of physical objects?
For a single particle: all equations of classical mechanics,
electromagnetism, and quantum mechanics are time-reversal
invariant (Newton’s second law, F = dp/dt, looks the same if
the time t is replaced by –t and the momentum p by –p).
For macroscopic objects: the processes are often irreversible (a
time-reversed version of such a process never seems to occur).
Examples: (a) living things grow old and die, but never get
younger, (b) if we drop a basketball onto a floor, it will bounce
several times and eventually come to rest - the arrow of
time does exist.
“More is different”, Phil Anderson, Science, 177, 393 (1972)
The Main Idea of the Course
Statistical description
of a large system
of identical (mostly,
non-interacting) particles
Equation of state
for macrosystems
(how macroparameters of the
system and the temperature
are interrelated)
all microstates of an isolated
system occur with the same
probability, the concepts of
multiplicity (configuration
space), Entropy
Irreversibility of
macro processes,
the 2nd Law of Thermodynamics
Thermodynamic Systems, Macroscopic Parameters
Open systems can exchange both
matter and energy with the environment.
Closed systems exchange energy but
not matter with the environment.
Isolated systems
can exchange
neither energy nor matter with the
environment.
Internal
and
external
macroscopic
parameters:
temperature, volume, pressure, energy, electromagnetic fields,
etc. (average values, fluctuations are ignored).
No matter what is the initial state of an isolated system,
eventually it will reach the state of thermodynamic
equilibrium (no macroscopic processes, only microscopic
motion of molecules).
A very important macro-parameter:
Temperature
Temperature is a property associated with random motion of
many particles.
Introduction of the concept of temperature in thermodynamics is
based on the the zeroth law of thermodynamics:
A well-defined quantity called temperature exists such
that two systems will be in thermal equilibrium if and
only if both have the same temperature.
Temperature Measurement
Properties of a thermoscope (any device that quantifies temperature):
1. It should be based on an easily measured macroscopic quantity a
(volume, resistance, etc.) of a common macroscopic system.
2. The function that relates the chosen parameter with temperature,
T = f(a), should be monotonic.
3. The quantity should be measurable over as wide a range of T as
possible.
The simplest case – linear dependence T = Aa (e.g., for the ideal gas
thermometer, T = PV/NkB).
the ideal gas thermometer, T = PV/NkB
T
the resistance thermometer with

a semi- conductor sensor, R  exp T
 
a
Thermometer  a thermoscope
calibrated to a standard temp. scale
The Absolute (Kelvin) Temperature Scale
The
absolute
(Kelvin)
temperature scale is based
on fixing T of the triple point
for water (a specific T =
273.16 K and P = 611.73 Pa
where water can coexist in
the solid, liquid, and gas
phases in equilibrium).
T,K
273.16
0
absolute zero
 P
T  273 .16 K 
 PTP
PTP
P



- for an ideal gas
constant-volume
thermoscope
PTP – the pressure of the gas in a
constant-volume gas thermoscope
at T = 273.16 K
Our first model of a many-particle system: the Ideal Gas
Models of matter:
gas models
(random motion of particles)
lattice models (positions of particles are fixed)
Air at normal conditions:
~ 2.71019 molecules in 1 cm3 of air (Pr. 1.10)
Size of the molecules ~ (2-3)10-10 m, distance
between the molecules ~ 310-9 m
The average speed - 500 m/s
The mean free path - 10-7 m (0.1 micron)
The number of collisions in 1 second - 5 109
The ideal gas model - works well at low densities (diluted gases)
• all the molecules are identical, N is huge;
• the molecules are tiny compared to their average separation (point masses);
• the molecules do not interact with each other;
• the molecules obey Newton’s laws of motion, their motion is random;
• collisions between the molecules and the container walls are elastic.
The Equation of State of Ideal Gases
An equation of state - an equation that relates macroscopic
variables (e.g., P, V, and T) for a given substance in
thermodynamic equilibrium.
In equilibrium ( no macroscopic motion), just a few
macroscopic parameters are required to describe the state of
a system.
The ideal gas
equation of state:
P
V
n
T
–
–
–
–
pressure
volume
number of moles of gas
the temperature in Kelvins
R – a universal constant
PV  nRT
[Newtons/m2]
[m3]
[mol]
[K]
R  8.315
J
mol  K
The Ideal Gas Law
In terms of the total number of molecules, N = nNA
Avogadro’s number
NA  6.022045×1023
PV  Nk BT
the Boltzmann constant kB = R/NA  1.3810-23 J/K
(introduced by Planck in 1899)
The equations of state cannot be derived within the frame of
thermodynamics: they can be either considered as experimental
observations, or “borrowed” from statistical mechanics.
Avogadro’s Law: equal volumes of different
gases at the same P and T contain the same
amount of molecules.
The P-V diagram – the projection of the surface
of the equation of state onto the P-V plane.
isotherms
Connection between Ktr and T for Ideal Gases
?
T of an ideal gas  the kinetic energy of molecules
Pressure – the result of collisions between the molecules and walls of the
container.
Momentum
Strategy: Pressure = Force/Area = [p / t ]/Area
px = 2 m vx
Intervals between collisions: t = 2 L/vx
For each (elastic) collision:
Piston
area A
vx
Volume = LA
L
N
For N molecules -
2m vx 1
1
2 1
Pi 
 px vx  m vx
2 L / vx A
V
V
PV   m vx2  N m vx2
i
no-relativistic
motion
Connection between Ktr and T for Ideal Gases (cont.)
N
PV   m vx2  N m vx2
i
PV  Nk BT
Average kinetic energy of
the translational motion of
molecules:
K tr
3
 k BT
2
3
U  K tr  Nk BT
2
2
PV  U
3
K tr
m v x2  k BT
1
1
3
2
2
2
2
 m v  m vx  v y  vz  m vx2
2
2
2
- the temperature of a gas is a direct measure of the
average translational kinetic energy of its molecules!
The internal energy U of a monatomic ideal gas
is independent of its volume, and depends only on
T (U =0 for an isothermal process, T=const).
- for an ideal gas of non-relativistic particles,
kin. energy  (velocity)2 .
Comparison with Experiment
dU/dT(300K)
(J/K·mole)
3
U  Nk BT
2
Monatomic
Helium
12.5
Argon
12.5
Neon
12.7
Krypton
12.3
Diatomic
H2
20.4
N2
20.8
O2
21.1
CO
21
Polyatomic
H20
27.0
CO2
28.5
- for a point mass
with three degrees
of freedom
Testable prediction: if we put a known
dU into a sample of gas, and measure
the resulting change dT, we expect to
get
dU 3
 Nk B
dT 2
3
 6 1023 mole-1 1.3810 23 J/K
2
 12.5 J/K  mole



Conclusion: diatomic and polyatomic
gases can store thermal energy in forms
other than the translational kinetic
energy of the molecules.
Degrees of Freedom
The degrees of freedom of a system are a
collection of independent variables required to
characterize the system.
Polyatomic molecules: 6
(transl.+rotat.) degrees of freedom
Diatomic molecules: 3 + 2 = 5
transl.+rotat. degrees of freedom
Degrees of Freedom (cont.)
Plus all vibrational degrees of freedom. The one-dimensional vibrational
motion counts as two degrees of freedom (kin. + pot. energies):
K  U x  
1
1
m x 2  k x 2
2
2
For a diatomic molecule (e.g., H2), 5 transl.+rotat. degrees of freedom
plus 2 vibrational degrees of freedom = total 7 degrees of freedom
Among 7 degrees of freedom, only 3 (translational) degrees correspond to a
continuous energy spectrum (classical) , the other 4 – to a discrete energy
spectrum (Quantum).
4
U(x)
U(x) 3
E4
Energy
2
1
0
Approx.
-1
E3
E2
-2
E1
-3
1.5
2.0
2.5
3.0
x
distance
3.5
4.0
x
“Frozen”
degrees of
freedom
U /kBT
one mole of H2
7/2N
Vibration
5/2N
Rotation
For an ideal gas
3/2N
PV = NkBT
U = f/2 NkBT
Example of H2:
Translation
10
100
1000
T, K
An energy available to a H2 molecule colliding U(x)
E4
with a wall at T=300 K: 3/2kBT ~ 40meV. If the
E3
difference between energy levels is >> kBT,
E2
then a typical collision cannot cause transitions kBT
E1 x
to the higher (excited) states and thus cannot
transfer energy to this degree of freedom: it is
“frozen out”.
The rotational energy levels are ~15 meV apart, the difference between
vibrational energy levels ~270 meV. Thus, the rotational degrees start
contributing to U at T > 200 K, the vibrational degrees of freedom - at T >
3000 K.
Equipartition of Energy
“Quadratic” degree of freedom – the corresponding energy = f(x2, vx2)
[ translational motion, (classical) rotational and vibrational motion, etc. ]
Equipartition Theorem:
At temperature T, the average energy of any
“quadratic” degree of freedom is 1/2kBT.
- holds only for a system of particles whose kinetic energy is a quadratic
form of x2, vx2 (e.g., the equipartition theorem does not work for photons, E =
cp)
Piston – a mechanical system with one degree of
freedom. Thus,
vx
m v 2x
2

M u2
2
1
 k BT
2
M – the mass of a piston, u2 the average u2,
where u is the piston’s speed along the x-axis.
Thus, the energy that corresponds to the one-dimensional translational
motion of a macroscopic system is the same as for a molecule (in this
respect, a macrosystem behaves as a giant “molecule”).
For the “next” lecture:
Recall the ideal gas laws, look at the ideal gas problems (some problems
are posted on the course Web page).
reference books
The root-mean-square speed
vrms 
v
2
3k BT

m
- not quite the average speed, but close...
For H2 molecules (m ~21.710-27 kg ) at 300K: vrms~ 1.84 103 m/s
For N2 – vrms (Pr. 1.18), for O2 – vrms= 461 m/s
This speed is close to the speed of sound in the gas – the sound wave
propagates due to the thermal motion of molecules.
D(v)
vrms
v
Problem 1.16
The “exponential” atmosphere
Consider a horizontal slab of air whose thickness is dz. If this slab is at rest,
the pressure holding it up from below must balance both the pressure from
above and the weight of the slab. Use this fact to find an expression for the
variation of pressure with altitude, in terms of the density of air, . Assume
that the temperature of atmosphere is independent of height, the average
molecular mass m.
P(z+dz)A
P( z  dz)  A  Mg  P( z )  A
area A
z+dz
z
P(z)A
Mg
Mg
P ( z  dz )  P ( z )  
A
dP
M  Adz 
  g
dz
M Nm 
PV  Pm

 N 
the density of air:  

V
V
k BT  k BT

Assuming T is independent of z
dP
mg


P

dz
kT
 m gz
P( z )  P(0) exp
 kT 