3.5 Thermodynamic Identity
General Approach
(N,U,V) S(N,U,V) U(N,S,V)
When all macroscopic quantities S,V,N,U are allowed to
vary:
dS S d U S d V S d N
U
N ,V
N
U ,V
V
N ,U
2nd law: at equilibrium, S reaches max.
We have abbreviated one of
S
Blackbody Radiation
The Ultraviolet Catastrophe
Consider the electromagnetic radiation inside a box at a
given temperature. In classical physics, we can think of
radiation as a superposition of various standing wave patterns
as shown in the figure.
3 =
2L
Energy in Thermal
Physics
Equipartition of Energy
PHYS 4050: Thermodynamics and Statistical Physics
Prof. B. A. Foreman
Translational kinetic energy
We have just shown that the average translational kinetic energy in
the x direction is given by
By symme
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PHYS 4050 (Spring 2012) Tutorial 2: Quiz 1, 27 February 2009
Estimated time: 10:30 am 10:55 am (10 points)
Consider an Einstein solid with N oscillators and q energy quanta. The energy of each
energy quantum is . Each oscillator can take any
PHYS4050 (Spring 2012) Quiz 4 Solution
1. dU = TdS pdV. For constant U, dU = 0= TdS pdV
2. Given , by differentiating the expression:
Substitute
3.
By using
We get
By using
We get
Remark: The general way to determine which thermodynamics identity to use
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PHYS 4050 (Spring 2012) Tutorial 6: Quiz 4 Estimated time: 10:30 am 10: 50 am
1. (4 points) Show that:
S
>0
V U
.
2. (2 points)
Based on dU = T dS PdV + dN , derive the thermodynamic identity about the Gibbs free energy G =
U TS + PV.
3.
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PHYS 4050 Tutorial 4: Quiz 3 Estimated time: 10:25 am 10: 50 am, 20 March 2012
An ideal gas is expanded adiabatically from (p1, V1) to 3V1, then compressed isobarically
(constant pressure) to V1. Finally the pressure increases to p1 at const
PHYS 4050 Tutorial 4: Quiz 3 Solution
P
A(P1,V1
)
B(P2,3V
C(P2,V1)
1
)
V
The cycle is shown in the P-V diagram.
A to B: adiabatic process,
,
B to C: isobaric compression process, gas release heat to the surrounding, the values of
heat released is
C to A:
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PHYS 4050 (Spring 2012) Tutorial 7: Quiz 5
24 April 2012 Estimated time: 11:30 am 11:50 am
A particle in a heat bath can be at three energy levels: , 2, 3. The degeneracies of these three
levels are 1, 3, 5 respectively.
(a) Calculate the pa
PHYS321 Spring 2010, Quiz 5 Solution.
(a)
(b) Probability to getis:
(c) The average energy is:
(d)
Method 1: The energy levels are shifted by 10, which should not affect the probability.
Hence is the same as the result in (b). The average energy is the re
1. Solution:
(a)Zeroth law : If system A and system B are each in thermal equilibrium with system C,
system A is also in thermal equilibrium with system B.
(b) Version1, Any large (isolated) system in equilibrium will be found in the macrostate
with the g
Due time: 10:30am on Monday, May 14, 2012 (in the instructors office room
4448).
7.13(a,d) (10 points)
For a system of bosons at room temperature, compute the average occupancy of a
single-particle state and the probability of the state containing 0, 1, 2
PHYS 4050 (Spring 2011) Homework 8
Due date: Thursday 10:35am, 8 May. 2012
Problem 1(P6.41 in text book)(10 points)
Imagine a world in which space is two-dimensional, but the laws of
physics are otherwise the same. Derive the speed distribution formula fo
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PHYS 4050 Quiz 2, Mar.13 2012 Estimated time: 11:30 11:55 am (10 points)
1. Sketch a graph of the entropy of water as a function of temperature , at fixed
pressure. Indicate the solid, liquid and gas phases. (5 points)
2. A mass M of a liqui
1. Lecture note chapter 3, slide 60:
At T 0, the graph goes to 0 with zero slope (3rd law: S=0, C=0).
At the phase transition point, S = Latent heat / T.
Cp is almost constant in each phase, so the slope ~1/T because
dS =
dQ CP dT
CP
S
=
=
T
T
T P T
S
Thermodynamics
Thermal Physics Statistical Mechanics
Deals with a collection of a large number of particles
It is effectively impossible to follow the motion and
trajectory of each particle
In thermal physics, we combine two approaches
thermodynamics
The Second Law
Multiplicity of a Single Particle
PHYS 4050: Thermodynamics and Statistical Physics
Prof. B. A. Foreman
Multiplicity of an ideal gas
Lets now use what we have learned about the geometry of a
hypersphere to calculate the multiplicity of an
Microstates,
Macrostates, and
Multiplicity
Two-State Systems
PHYS 4050: Thermodynamics and Statistical Physics
Prof. B. A. Foreman
Motivation
Why does heat always flow from hot objects to cold objects?
The ultimate reason is that this is the most probab
Degenerate Fermi Gas
Consider a gas of fermions (electrons, 3He atoms,
protons, etc) at low temperature. To begin with, lets discuss
what is meant by the condition of low temperature.
By low temperature, we mean that the validity condition
for Boltzmann S
Quantum Statistics
The Gibbs Factor
In chapter 6, the Boltzmann factor was derived for a
system in equilibrium with a reservoir at constant temperature.
The system and reservoir were allowed to exchange energy,
but not particles. Now, consider a system an
Equipartition Theorem
Consider a system whose energy depends upon a particular
degree of freedom q in quadratic form
E (q) = cq2
where c = constant
1
For example: Harmonic Oscillator potential energy: E ( x) = kx 2
2
1
&
&
Any form of kinetic energy: E (
Boltzmann Statistics
In Chapter 2 and 3, we learned how to calculate some
thermodynamic quantities using a few statistical mechanical
principles:
All microstates have equal probability
The macrostate with the largest number of
microstates has the highes
Free Energy and Chemical Thermodynamics
Chemical reactions and other transformations are
constrained by the laws of thermodynamics. Reactions and
transformations most often occur with some interactions with
the surroundings. Usually, the energy and volume
Engines and Refrigerators
Heat engines
A heat engine is any device that absorbs heat and converts
part of that energy into work.
When an engine absorbs heat from a hot reservoir ( Th ), it
also absorbs entropy. The engine must expel that entropy. It does
Interactions and Implication
Temperature (a thermodynamic definition)
A The Second Law of thermodynamics says that when two
objects are in thermodynamic equilibrium, then their total
entropy has reached a maximum possible value.
B Previously (in Chapter 1
Multiplicity of a Large Einstein Solid
N > 1, q > 1
In the high temperature (classical) limit, q > N , the multiplicity is
( N , q) =
( q + N 1)!
q !( N 1)!
( q + N )!
~
q! N !
Take the natural log and use Stirlings approximation
ln ( N , q ) = ln ( q +
The Second Law
Two State Systems: Microstate, Macrostate
Consider flipping 3 coins:
$1
H = heads
T = tails
The possible outcomes are:
$1
$2
$10
H
H
H
H
H
T
T
H
H
T
T
H
T
T
T
T
H
T
3H (0T)
H
H
Macrostates
T
T
23 = 8
microstates
H
$2 $10
T
2H (1T)
1H (2T)
0
PHYS4050 HW7 Solution
Problem 6.20
e)
For ,
For
Problem 6.31
The
energy
levels
are
(with
To count the states, assume the states are separated by small intervals.
Then the partition function is
.
ranging
from
to).
When is very small, we can approximate the
HW7 Marker: Lam Yat Hong
Mean 90.09091
SD: 14.85618
Comment
Q1. Easy. Some students forget to calculate the excited state
Q2. Some students make mistake about the temperature.
Q3. Good
Q4. Many people fail to explain low temperature case in part e
Q5. Man