Download THERMODYNAMICS [5] Halliday, David, Resnick, Robert, and

THERMODYNAMICS
[5] Halliday, David, Resnick, Robert, and Walker,
Jeart, Fundamentals of Physics, 6th edition, John
Wiley, Singapore, 2001, ISBN 0-471-33236-4
[5], p. 426-427 thermodynamics is the study of the
thermal energy of systems. The central concept of
thermodynamics is temperature
Discuss intuitive notions of temperature. Why are
they unreliable ? What is a thermometer ?
Zeroth Law of Thermodynamics: if bodies A and B
are each in thermal equilibrium with a third body T,
then they are in thermal equilibrium with each other
MEASURING TEMPERATURE
[5], p. 427-429 We define the triple point of water to
have a temperature value of T3 = 273.16 K
We define the temperature of any body by
T = T3 lim ( p / p 3 )
gas → 0
here p is the pressure of the gas in thermal equilibrium
with the body and p 3 is the pressure of the gas at T3
T
Gasfilled
bulb
h
p = p atm − ρgh
Reservoir
that can be
raised and
lowered
TEMPERATURE AND HEAT
[5] 9. 433 Heat is energy transferred to a system from
its environment because of a temperature difference
that exists between them
Q = C∆T = C(Tf − Ti )
here C equals the heat capacity of an object
One calorie (cal) is the amount of heat required that
would raise the temperature of 1 g of water from 14.5
degrees Celsius to 15.5 degrees Celsius (degrees
Celsius = degrees Kelvin – 273.15) (1 cal = 4.1860J)
i
n
s
u
l
a
t
i
o
n
lead shot
W
WORK
∫
W = dW =
Vf
∫V
pdV
i
Pressure
State
diagram
Q
thermal reservoir
Volume
Q and W are path dependent, not functions of state
FIRST LAW OF THERMODYNAMICS
There exists a function of state, called the internal
energy and denoted by E int , such that
∆E int = Q - W
or
dE int = dQ - dW
Special cases of thermodynamic processes include
Adiabatic, Constant-volume, Cyclical, and Free
expansions.
SECOND LAW REVISITED
Lord Kelvin : A transformation whose only final
result is to transform into work heat extracted from a
source which is at the same temperature throughout is
impossible
Rudolph Clausius : A transformation whose only
final result is to transfer heat from a body at a given
temperature to a body at a higher temperature is
impossible (this principle implies the previous one)
Martian Skeptic : What temperature ?
SECOND LAW REVISITED
Definition : Body A has higher temperature than body
B (A f B ) if, when we bring them into thermal
contact, heat flows from A to B. Body A has the same
temperature as B ( A ≈ B ) if, when we bring them
into thermal contact, no heat flows from A to B and no
heat flows from B to A. ( [A] := {U | U ≈ A} )
Enrico Fermi : (Clausius Reformulated) If heat
flows by conduction from a body A to another
body B, then a transformation whose only final
result is to transfer heat from B to A is impossible.
SECOND LAW REVISITED
Sadi Carnot : If A f B then we can use
a reversible engine to absorb a quantity
Q A > 0 of heat from A, surrender a
quantity of heat Q A > Q B > 0 to B,
and perform a quantity of work W > 0.
Carnot & Kelvin implies Clausius, and
Q A Q B = f ([A], [B]), and for B f C,
f ([A], [B]) f ([B], [C]) = f ([A], [C])
SECOND LAW REVISITED
Definition : Absolute Thermodynamic Temperature
Choose a body D
If A f D
A≈D
DfA
then T( A ) = f ([ A ], [ D ])
then T( A ) = 1
then T( A ) = 1 f ([ D ], [ A ])
In a reversible process
QA
T( A ) − Q B T( B) = 0
SECOND LAW REVISITED
System : Cylinder that has a movable piston and
contains a fixed amount of homogeneous fluid
States (Macroscopic) : Region in positive quadrant
of the (V = volume, T = temperature) plane.
Functions (on region) : V, T, p = pressure
Paths (in region) : Oriented curves Γ
Differential Forms : can be integrated over paths
WORK
W( Γ) =
HEAT
∫
Γ
pdV
Q( Γ) =
∫
Γ
?
SECOND LAW REVISITED
Definition : Entropy Function S
Choose a state ( V1 , T1 )
Define S(V, T) = Q( Γ1 ) T1
where Γ1 ∪ Γ2 joins ( V1 , T1 ) to (V, T),
Γ1 is isothermal , and Γ2 is adiabatic
(by thermal equilibrium and by thermal isolation)
Carnot' s cyle
Γ1 ∪ Γ2 is
and Q( Γ) =
2
∫
Γ
TdS
FIRST LAW REVISITED
There exists an (internal energy) function U
such that
TdS - pdV = dU
Therefore
 ∂U

∂U
TdS =
dT + 
+ p dV
∂T
 ∂V

FIRST & SECOND LAWS COMBINED
∂S 1 ∂U
=
;
∂T T ∂T

∂S 1  ∂U
= 
+ p
∂V T  ∂V

Therefore, the basic (but powerful) calculus identity
∂ ∂S ∂ ∂ S
=
∂V ∂T ∂T ∂V
Yields (after some tedious but straightforward algebra)
∂p
∂U
=T
-p
∂V
∂T
IDEAL GAS LAW
(Chemists) Boyl, Gay-Lussac, Avogardo
p(V, T) = n R g / V
n = amount of gas in moles
R = ideal gas constant
(8.314 joules / degree Kelvin)
g = ideal gas temperature in Kelvin
(water freezes at 373.16 degrees)
JOULE’S GAS EXPANSION EXPERIMENT
We substitute the expression for p
(given by the ideal gas law) to obtain

∂U nR  dg
=
− g
T
∂V V  dT

and observe that the outcome of
Joule’s gas expansion experiment
∂U
=0 ⇔ T∝g
∂V
IDEAL GAS LAW
(Physicists)
p(V, T) = N k T V
N = number of molecules of gas
23
( 6.0225 × 10 molecules / mole)
k = Boltzmann’s constant
(1.38 × 10
−23
joules / deg _Kelvin)
GAS THERMODYNAMICS
Experimental Result : (dilute gases)
dU
= Nk ( γ − 1)
−1
dT
Therefore
TV
γ −1
γ
& p V constant on adiabatic paths
S = Nk ( γ − 1)
−1
ln(T / T1 ) + Nk ln(V / V1 )
GAS KINETICS
Monatomic dilute gas, m = molecular mass
U
=
N
1
2
mν
2
1
2
=3
mν x
2
average kinetic
energy / molecule
 N   Aν x ∆t 
F × ∆t =   
 (2mν x )
 V  2 
F
1
p = = N ( γ − 1) mν 2 V
2
A
5
γ=
1
2
3
kT = ( γ − 1) mν
2
GAS KINETICS
Photon gases E = c × momentum
4
γ=
3
Maxwell Equipartition of Energy
 m1ν1 + m 2 ν 2 
 = 0
(ν1 − ν 2 ) ⋅ 
 m1 + m 2 
kinetic energy
1
= kT
in each direction
2
2
γ = 1+
degrees of freedom
EQUIPARTITION
Number of ways of partitioning N objects into
m bins with relative frequencies (probabilities)
p1 , p 2 ,..., p m is C = N! ( p1 N )!... (p m N )!
Stirling’s formula ( ln N! ≈ N ln N - N ) yields
ln C ≈ N H ( p1 ,..., p m ) where H ( p1 ,..., p m )
denotes Shannon’s information-theoretic entropy
H ( p1 ,..., p m ) = − [p1 ln p1 + L + p m ln p m ]
EQUIPARTITION
If the bins correspond to energies, then
H ( p1 ,..., p m ), and therefore (nearly) C,
is maximized, subject to an energy constraint
E = N ( p1E1 + L + p m E m ), by the Gibbs
distribution p i = exp( − E i kT ) Z(T )
1 T = dS dE
S = k ln C
and free energy E − TS = − NkT ln Z(T )
2
Maxwell dist. prob(x, ν ) ∝ exp (-mν / 2 kT )
Therefore
THIRD LAW
Nernst : The entropy of every system at absolute
zero can always be taken equal to zero
inherently quantum mechanical
discrete microstates, a quart bottle of air has about
10
24
10 22
molecules & 10
microstate s
1 bit of information = kT ln 2 energy
Maxwell’s demon : may he rest in peace
Time’s arrow : probably forward ???
REFERENCES
V. Ambegaokar, Reasoning about Luck
H. Baeyer, Warmth Disperses and Time Passes
F. Faurote, The How and Why of the Automobile
E. Fermi, Thermodynamics
R. Feynman, Lectures on Physics, Volume 1
REFERENCES
H. S. Green and T. Triffet, Sources of Consciousness,
The Biophysical and Computational Basis of Thought
K. Huang, Statistical Mechanics
N. Hurt and R. Hermann, Quantum Statistical
Mechanics and Lie Group Harmonic Analysis
C. Shannon and W. Weaver, The Mathematical
Theory of Communication