[5] Halliday, David, Resnick, Robert, and Walker,. Jeart, Fundamentals of Physics
, 6th edition, John. Wiley, Singapore, 2001, ISBN 0-471-33236-4. [5], p.
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
