Transcript T - Himastron

AS3105
Proses Astrofisika 1
Bab1 Pendahuluan
Dhani Herdiwijaya
v1.50
Aturan
• Absensi >= 75%
− < 75% tidak boleh ikut UAS
• Prosentase nilai
− NA = 40% UAS + 30% UTS + 20% tugas+kuis +
10% absensi
• Tidak ada ujian susulan
• Jadwal
− Selasa jam 10.00-11.00, Ruang Seminar 2
− Kamis jam 09.00-11.00, Ruang Seminar 2
Referensi
• Reif, F., Fundamentals of Statistical and Thermal
Physics, McGraw Hill, 1987
• Reif, F., Statistical Physics, Berkeley Physics Course
Vol. 5, McGraw Hill, 1987
• Landau, Lifshitz, Statistical Physics 3 rd ed. Pergamon
Press (bacaan lanjut)
• Padmanabhan, T., Theoretical Astrophysics Vol. 1.
Astrophysical Processes, Cambridge University Press,
2000
• Mihalas, D. Stellar Atmosphere, W.H. Freeman, 1978
• Rybicky, G.B. and Ligthman, A.P., Radiative Processes
in Astrophysics, John Wiley & Sons, 1979 (bacaan
lanjut)
Materi
• Review termodinamika
• Pendahuluan
• Probabilitas
• Distribusi Maxwell-Boltzmann
• Distribusi Bose-Einstein
• Distribusi Fermi-Dirac
• Aplikasi
−
−
−
−
Radiasi Benda Hitam
Kondensat Bose
Gas Fermi terdegenerasi
Transfer Radiasi
Bab 1
• Review Hukum Termodinamika
• Termodinamika dan Fisika Statistik
• Konsep makroskopik dan mikroskopik
• Gas ideal
Contoh Pertanyaan Termodinamika
• Bagaimana kerja lemari es? Berapa
efisiensi maksimumnya?
• Berapa banyak energi diperlukan untuk
memanaskan air dalam panci sehingga
menjadi uap?
Hukum Termodinamika
• Hukum ke-0
• Hukum ke-1
• Hukum ke-2
• Hukum ke-3
Sistem
• Sebuah sistem terpisah dari lingkungannya oleh
dinding. Sistem tertutup, jika tidak ada perpindahan
materi antara sistem dan lingkungan.
• Dinding dibagi dua, yaitu
− Adiabatic: tidak ada pertukaran panas
− Diathermal: panas dapat mengalir lewat dinding
Sistem yang umum ditinjau adalah
• •sistem terisolasi dari lingkungan dengan dinding
adiabatik (sistem adiabatik)
• •sistem bersentuhan dengan sumber panas melalui
dinding diatermal (sistem isotermal)
Sebuah sistem terdiri dari variabel termodinamika, yang
terbagi menjadi variabel extensive dan intensive.
Variabel
• Variabel Intensive: tidak bergantung
ukuran sistem dan dapat berubah-ubah.
Misalkan tekanan, temperatur, densitas
• Variabel Extensive: bergantung kuran
sistem. Misalkan, volume, jumlah partikel,
massa, energi internal
• Variabel Extensive and intensive cenderung
berpasangan (conjugate variables).
Variabel Conjugate:
•
•
•
•
system
generalised
fluid
Wire
film
intensive variable
generalised force X
pressure p
tension F
surface tension g
extensive variable
generalised displacement x
volume V
length l
area A
Sistem Termodinamika
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).
Hukum ke-0
Jika sistem A setimbang dengan sistem B
dan sistem C, maka sistem B juga
setimbang dengan sistem C. Berlaku
kebalikannya
eksistensi keadaan
setimbang
A very important macro-parameter: Temperature
Temperature is a property associated with chaotic motion of many
particles (it would be absurd to refer to the temperature of a single
molecule, or to the temperature of many molecules moving with the
same speed in the same direction).
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.
The zeroth law implies the existence of some universal property of systems in
thermal equilibrium. This law, in conjunction with the concept of entropy, will
help us to mathematically define temperature in terms of statistical ideas.
Also, if system C is in thermal equilibrium with systems A and B, the latter two
systems, when brought in contact, also will be in thermal equilibrium.
- this allows us to obtain the temperature of a system without a direct
comparison to some standard.
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) Temp. 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  - for an ideal gas
 constant-volume
T  273.16 K 
 PTP 
thermoscope
PTP
P
PTP – the pressure of the gas in a
constant-volume gas thermoscope
at T = 273.16 K
Hukum ke-1
Energi internal sistem terisolasi, E
dapat berubah jika terjadi aliran
panas, Q atau melakukan usaha, W
(tanda bisa pos. atau neg. bergantung konvensi)
Dirumuskan sebagai
ΔE=ΔQ+ΔW
Banyak panas masuk ke sistem
Usaha yg dilakukan dalam sistem
Dq =CP dT atau dq = CV dT
Dw = - pdV
Temperatur dan Energi Kinetik
The temperature T of a system of N particles is a
quantity related to the averaged kinetic energy of
the disordered motion of the particle in the frame
of reference.
T  Ek ,ave
1
1 2
  mi vi
N i 2
Mass of particle i
Velocity of
particle i
Thermal equilibrium:
Average kinetic energy is the same in all regions of the system
Hukum ke-2
• Clausius: Panas mengalir dari panas ke dingin
• Kelvin: Tidaklah mungkin mengubah semua
energi panas menjadi usaha
• Carathéodory: lingkungan suatu keadaan stabil
dari sistem yang terisolasi secara termal,
terdapat suatu keadaan yang tidak teramati
• Callen: Entropi tidak pernah turun dalam
sistem terisolasi dan proses spontan
Hukum ke-2
The entropy S of an isolated system increases
during any spontaneous change.
Spontaneous
Low
entropy
gas
S  0
Not observed
High
entropy
Hukum ke-3
Nernst, Simon, dkk.
• Entropi dalam suatu sistem akan
mendekati nol, jika temperatur
mendekati nol absolut
Nernst’s Theorem (1906)
The Third Law of Thermodynamics
CV (T ) dT 
S (T )  S (0)  
T
0
T
A non-trivial issue –
what’s the value of S(0)?
The answer is provided by Q.M. (discreteness of quantum states), it cannot be
deduced from the other laws of thermodynamics – thus, the third law:
Nernst’s Theorem: The entropy of a system at T = 0 is a well-defined constant.
For any processes that bring a system at T = 0 from one equilibrium state to
another,  S = 0 .
This is because a system at T = 0 exists in its ground state, so that its entropy is
determined only by the degeneracy of the ground state.
In other words, S(T) approaches a finite limit at T = 0 , which does not depend
on specifics of processes that brought the system to the T = 0 state.
A special case – systems with unique ground state, such as elementary
crystals : S(0) = 0 ( = 1). However, the claim that S(0) is "usually" zero or
negligible is wrong.
Residual Entropy
If S(0)  0, it’s called residual entropy.
1. For compounds, for example, there is frequently a significant amount of
entropy that comes from a multiplicity of possible molecular orientations in the
crystal (degeneracy of the ground state), even at absolute zero.
Structure of the hexagonal ice:  - H
atoms, --- - covalent O-H bonds, - - - weak H bonds. Six H atoms can “tunnel”
simultaneously through a potential barrier.
2. Nernst’s Theorem applies to
the equilibrium states only!
Glasses
aren’t
really
in
equilibrium, the relaxation time –
huge. They do not have a welldefined T or CV. Glasses have a
particularly large entropy at T = 0.
S
supercooled
liquid
liquid
glass
residual
entropy
crystal
T
Example (Pr. 3.14)
For a mole of aluminum, CV = aT + bT3 at T < 50 K (a = 1.35·10-3 J/K2, b =
2.48·10–5 J/K4). The linear term – due to mobile electrons, the cubic term –
due to the crystal lattice vibrations. Find S(T) and evaluate the entropy at
T = 1K,10 K.


CV (T ) dT 
aT   bT 3 dT 
b
S (T )  

 aT  T 3
T
T
3
0
0
T
T
1
S (1K )  1.35 10 3 J/K 2 1K  2.48 10 5 J/K 4 1K 3  1.36 10 3 J/K
3
- at low T, nearly all the entropy comes from the mobile electrons
T = 1K
T = 10K
1
S (10 K )  1.35 10 3 J/K 2 10K  2.48 10 5 J/K 4 103 K 3  2.18 10  2 J/K
3
- most of the entropy comes from lattice vibrations
S (1K ) 1.35 103 J/K
20


10
kB
1.38 1023 J/K
S (10 K ) 2.18 10 2 J/K
21


1
.
6

10
kB
1.38 1023 J/K
- much less than the # of particles, most degrees of freedom are still frozen out.
C  0 as T  0
CV (T ) dT 
S (T )  S (0)  
T
0
T
C (T )  0 as T  0
For example, let’s fix P :
 Q  CP (T )dT
T

0
CP (T ) dT 

T 0
T
Q
T
-
finite,

CP(0) = 0
Similar, considering V = const - CV(0) = 0 .
Thus, the specific heat must be a function of T. We know that at high T, CV
approaches a universal temperature-independent limit that depends on N
and # of degrees of freedom. This high - T behavior cannot persist down to
T = 0 – quantum effects will come into play.
Entropi
dS surroundings
dSuniverse  dS  dS surroundings
dSuniverse  0
dS
System
Heat:
dqsurroundings
T
Surroundings
dSuniverse  0
For an equilibrium
process
System + Surroundings = Universe
The universe seeks higher entropy
Overall entropy production
Entropi
dq
dS 
0
T
dq : heat supplied to the system
dS : Change of entropy of system due to internal processes.
Termodinamika
Termodinamika merupakan kerangka
umum dengan hanya melihat sifat-sifat
makroskopik atau variabel termodinamika
dan mengabaikan sifat mikroskopik (setiap
molekul).
Makroskopik
• Kondisi makroskopik memperhatikan sistem
secara keseluruhan. Sistem yang setimbang
dinyatakan dalam beberapa variabel,
misalkan: tekanan, volume, temperatur, energi
internal dan entropi.
• Kondisi makro dapat berubah dengan 2 cara,
yaitu usaha pada sistem dan/atau panas yang
mengalir dalam sistem.
• Variabel makroskopik  termodinamika
• Misalkan, tabung gas He, maka dapat diketahui
volume, jumlah molekul, energi internalnya
Macroscopic Description is Qualitatively Different!
Why do we need to consider macroscopic bodies as a special class of
physical objects? Because the behavior of macroscopic bodies differs from
that of a single particle in a very fundamental respect:
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 timereversed 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.
Conservation of energy does not explain why the time-reversed process
does not occur - such a process also would conserve the total energy.
Though the total energy is conserved, it has been transferred from one
degree of freedom (motion of ball’s center of mass) to many degrees of
freedom (associated with the individual molecules of the ball and the floor) –
the distribution of energy changes in an irreversible manner.
Contoh Pertanyaan lain
• Mengapa sifat air berbeda dengan uap,
meskipun molekul keduanya sama?
• Mengapa besi kehilangan sifat
kemagnetannya pada temperatur
tertentu?
 Mempertanyakan variabel mikroskopik
 Sifat dari molekul dengan jumlah
sangat besar  diperlukan statistika
atau fisika statistik
Mikroskopik
• Kondisi mikroskopik dinyatakan oleh banyak sekali
variabel, karena skala molekul. Misalkan kondisi gas
dinyatakan dari posisi dan momentum setiap saat.
Untuk N partikel mempunyai N variabel. Untuk 1 cm3
gas at STP, N ~ 1019.
• Posisi dan momentum setiap partikel memenuhi
hukum mekanika. Praktis, tidaklah mungkin
melakukan komputasi, sehingga diperlukan distribusi
statistik  mekanika statistik
• Dalam tabung Helium, perlu diketahui posisi dan
kecepatan setiap (dan fungsi gelombang) molekul.
Energi total merupakan jumlah energi semua
molekul
• Gaya setiap molekul, F=ma
- terdapat 6N koordinat saat t0: ri and vi
- dengan N ≡ moles (1023) !
Ruang fase
{p}
momenta
N particles

6N dimensional
space
{q}
Coordinates
Trajectory
Up
One-dimensional
harmonic oscillator 
2N dimensional
phase space
Alam Mikroskopik dan
Makroskopik
Mekanika
klasik
Mekanika
Kuantum
Interaksi
Microscopic
Ruang Phase
Fungsi
gelombang
Statistical Averaging
Fisika
Statistik
Makroskopik
Observables:
Dynamic and
Thermodynamic
quantities
Fisika Statistika
• Fisika Statistik merupakan jembatan
antara alam mikroskopik dan
makroskopik  ada hubungan hukum
termodinamika dan sifat statistik
molekul.
• Fisika statistik menjelaskan sifat
makroskopik dalam perhitungan detil
mikroskopik.
• Fisika statistik mempunyai kemampuan
prediksi, karena adanya model
mikroskopik yang dapat menghitung
sesuatu yang terukur.
Fisika Statistika sbg Jembatan
• Dalam mikroskopik, energi setiap atom bernilai dikrit
ata kontinu
• Dalam kesetimbangan, atom terdistribusi dalam
tingkatan energi. Dalam distribusi tersebut dapat
diketahui berapakah atom yang mempunyai energi
dalam rentang E dan E + dE. Probabilitas distribusi
atom ini cukup untuk menentukan variabel
makroskopik termodinamika
• Jadi Fisika Statistik menghubungkan probabilitas
mikroskopik menjadi variabel makroskopik dalam
suatu sistem.
• Orang yang banyak berjasa menyatukan fisika termal
(termodinamika) dan mekanika adalah Boltzmann dan
Gibbs
Fisika Statistik dan Termodinamika
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
Concepts of Statistical Physics
1. The macrostate is specified by a sufficient number of
macroscopically measurable parameters.
2. The microstate is specified by the quantum state of each
particle in a system.
3. The multiplicity is the number of microstates that are
consistent with a given macrostate. For each macrostate,
there is an extremely large number of possible microstates
that are macroscopically indistinguishable.
4. The Fundamental Assumption: for an isolated system,
all accessible microstate are equally likely.
5. The probability of a macrostate is proportional to its
multiplicity. This will sufficient to explain irreversibility.
Hukum ke-1 dalam bentuk
Probabilitas  Fisika Statistik
Boltzmann hypothesis: the entropy of a system is related to
the probability of its state; the basis of entropy is statistical.
Fisika Kuantum
• Dalam fisika kuantum, keadaan
mikroskopik dinyatakan dalam fungsi
gelombang
• Mikroskopik dinyatakan dalam bilanganbilangan kuantum
Quantum numbers
• The state of the system is fully described by
the wave function.
• The state of the system is fully described by a
set of quantum numbers (n, m, l, …..).
• The total energy of the system is given from
Schrödinger equation. An observable such as
the energy E corresponds to an operator such as
H.
r1 , r2 ,..., t 
H  E
Model Gas Ideal
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 quantum version of the ideal gas model helps to understand the
blackbody radiation, electrons in metals, the low-temperature behavior of
crystalline solids, etc.
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.
Geometrical
representation of the
equation of state:
P
an equilibrium
state
the equationof-state surface
f (P,V,T) = 0
V
T
The ideal gas
equation of state:
PV  nRT
P
V
n
T
–
–
–
–
pressure
volume
number of moles of gas
the temperature in Kelvins
R – a universal constant
[Newtons/m2]
[m3]
[K]
R  8.31
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  NkBT
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
The van der Waals model of real gases
For real gases – both quantitative and qualitative deviations from the ideal
gas model
4
U(r)
3
Electric interactions between
electro-neutral molecules :
Energy
2
1
0
-1
-2
-3
Veff  V  Nb
repulsion
attraction
aN 2
Peff  P  2
V
1.5
2.0
2.5
r
3.0
3.5
4.0
distance
van der Waals
attraction (~ r -7)

aN 2 
 P  2 V  Nb  NkBT
V 

The van der Waals equation of state for real gases
(the only model with interactions that we’ll consider)
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 -
2mvx 1
1
2 1
Pi 
 p x v x  mvx
2 L / vx A
V
V
PV   mvx2  N m v x2
i
no-relativistic
motion
Connection between Ktr and T for Ideal Gases (cont.)
N
PV   mvx2  N m v x2
i
PV  NkBT
Average kinetic energy of
the translational motion of
molecules:
3
K tr  k BT
2
3
U  K tr  NkBT
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 .
Units for Energy, Temperature
3
K tr  k BT
2
- the kinetic energy is proportional to the temperature,
and the Boltzmann constant kB is the coefficient of
proportionality that provides one-to-one correspondence
between the units of energy and temperature.
Theorists usually assume that kB = 1 - the same units for energy and
temperature – which makes a lot of sense.
If the temperature is measured in Kelvins, and the energy – in Joules:
kB = 1.3810-23 J/K
In many sub-fields of physics that deal with microscopic particles (including
condensed matter physics, physics of high energies, astrophysics, etc.), a
convenient unit for energy is an electron-Volt (the kinetic energy acquired
by an electron accelerated by the electrostatic potential difference in 1 V).
K = eV  1 eV = 1.6 10-19 J
At T = 300K
1.38 1023 J / K  300 K
k BT 
 26 meV
19
1.6 10 J / eV
Pressure of a photon gas
1
2
2
PV  N m v  N K tr
- valid for non-relativistic particles
3
3
1
PV  N vp
- valid also for relativistic particles (m = f(v))
3
In particular, it applies to the gas of photons that move randomly within
some cavity with mirror walls. For photons,
p
E ph
c
E
ph
 cp  h 

PV 
1
E ph
3
pre-factor 1/3 (instead 2/3) - due to a different relation between
E and p for relativistic particles
We will use this equation later in the course when we consider
the thermal radiation.
Degrees of Freedom
Indeed, the polyatomic molecules have more than just three degrees of
freedom – they rotate and vibrate.
The degrees of freedom of a system are a collection of independent
variables required to characterize the system.
Thus, for instance, the degrees of freedom of an ideal gas, in a
thermodynamic description, are P and V (or a pair of other variables).
The independent coordinates that describe the position of a mechanical
system in space - e.g., for a rigid body, it is sufficient to specify the
coordinates of its center of mass (x, y, z) and three angles (x, y, z).
K

 

Ix – the moment of inertia
for rotations around the
x-axis, etc.
Polyatomic molecules: 6 transl.+rotat. degrees of freedom
1
1
M x 2  y 2  z 2  I x x2  I y y2  I z z2
2
2
Diatomic molecules: 3 + 2 = 5 transl.+rotat. degrees of
freedom (QM: degrees of freedom corresponding to rotations
that leave the molecule completely unchanged do not count )
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, the other 5 – to a discrete energy spectrum.
U(x)
U(x)
4
E4
3
Energy
2
1
E3
0
E2
-1
-2
-3
1.5
2.0
2.5
3.0
x
distance
3.5
4.0
E1
x
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”).
“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.
Adiabatic Process in an Ideal Gas
adiabatic (thermally isolated system)
dU  W12
Q12  0
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
V2
W12    P(V , T )dV
P
V1
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
to calculate W1-2 , we need to know P (V,T)
for an adiabatic process
U
f
Nk BT
2
 2  dP
1   
0
f
P




f
Nk B dT   PdV
2
( f – the # of “unfrozen” degrees of freedom )
PV  NkBT  PdV  VdP  NkB dT
dV
V
 dU 
,
2
  1
f
PdV  VdP  
2
PdV
f
V
P
dV
dP


0
V P1 P
V1
V 
P 

ln    ln  1   PV   P1V1  const
P
 V1 
 PV
Adiabatic Process in an Ideal Gas (cont.)

PV   P1V1  const
P
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
An adiabata is “steeper” than an isotherma:
in an adiabatic process, the work flowing
out of the gas comes at the expense of its
thermal energy  its temperature will
decrease.
P1V1
   P(V , T )dV     dV
V
V1
V1
V2
W12
V2
1  1
1 

 P1V1
  1 
 1

  1  V2
V1 


 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
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= 493 m/s (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)
hydrogen molecules
in the “tail” escape the
Earth gravitation field
vrms
v
Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bike’s tire). What is its final temperature?
Rapid compression – approx. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
P1V1  Nk BT1
P2V2  Nk BT2
P2 
P1V1  P2V2

P1V1
V2
P1V1
P1V1

Nk
T

T2
B 2
V2 1
T1
For adiabatic processes:
 V1 
T2  T1  
 V2 
 1

 1
 V1 
  
 V2 
 1
T1 V1
 T2 V2
also
P 1 / T   const
 1

 const
 295 K  40.4  295 K 1.74  514 K
- poor approx. for a bike pump, works better for diesel engines
T2
T1
Homework 1: Thermodynamics
Let p be the pressure and V be the volume for
a real gas of n moles. If the equation of state
is given by pV = φ(T), where T is the
temperature of the system. The internal
energy of the system U is also a function of T
only, i.e. (∂U/∂V )T = 0. Show that the
function φ(T) must be linear function of the
temperature, i.e. φ(T) = cT, where c is a
constant.
Solution HW1
Homework 2: 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
area A
z+dz
z
P(z)A
Mg
M=rAdz
Solution HW 2: 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
area A
z+dz
z
P(z)A
Mg
P( z  dz )  A  Mg  P( z )  A
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

 mgz 
P( z )  P(0) exp 
 kT 
dP
mg


P

dz
kT