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THERMODYNAMICS
A GENERALLY GLOOMY SUBJECTTHAT TELLS US THAT THE UNIVERSEIS RUNNING DOWN, EVERYTHING ISGETTING MORE DISORDERED ANDGENERALLY GOING TO HELL IN AHANDBASKET
James Trefil - NY Times
THE LAWS OF THERMODYNAMICSTHE THIRD OF THEM,THE SECOND LAW,WAS RECOGNIZED FIRST
THE FIRST, THE ZEROTH LAW,WAS FORMULATED LAST
THE FIRST LAW WAS SECOND
THE THIRD LAW IS NOT REALLY A LAW
P. Atkins ~ The Second Law
Zero’th and third laws TemperatureFirst law Conservation of energy
THE SECOND LAW ENTROPY
QUESTIONS
Define T, S.
What is thermodynamics?
What are the laws of thermodynamics?
∆∆Gm – Wha t i s i t?
THERMODYNAMICS
Expresses relationships between the macroscopic properties of a system without regard to the underlying physical (i.e., molecular) structure.
E.g., EQUATION OF STATE OF AN IDEAL GAS
PV = nRT
•Where does this come from?
•What is the molecular machinery? ie. can we obtain this equation from a molecular theory ?
•Can we also obtain equations of state for liquids and solutions?
•How do we describe interactions?
DEFINITIONS
1. VOLUME AND PRESSURE
Cross section area = A
BOYLE’S LAW
Volume V
Force F P = FA
P V = constant
P1
~ V
(at constant T,n)
E X P E R I M E N T
T H E O R Y
1/V
P
DEFINITIONS2. TEMPERATURE - -What is it?
Thermodynamic definition: - something that determines the direction heat will flow
Normal scales — °C, °F arbitrary
If there is no heat flow, the two bodies are at the sametemperature (zero’th law of thermodynamics)
HEAT
CHARLES LAWV ~ T (at constant n, P)
Note: third law. Can’t get to absolute zero in a finite number of steps
Is there anabsolutescale oftemperature ?
-273 CO
V
T
0 CO
DEFINITIONS3. The amount of stuff; n.
• Don’t confuse mass and weight
• Numbers of molecules - - large ! • Use moles (mol) as a unit (Latin - - massive heap)
1 mol of particles =# of atoms in 12 gms of 12C1 mol = 6.022 x 1023
AVOGADRO’S PRINCIPLE V ~ n (constant T,P)
FINAL DEFINITIONS
ENERGY AND ENTROPY
Energy
• Started with the concepts of HEAT and WORK
• ~200 years ago these were considered to be different things and had different units
HEAT Caloric A weightless form of matter that flowed in and out of materials
WORK NEWTON Force x distance moved
FINAL DEFINITIONS
UNITED BY CONCEPT OF ENERGY, defined as the capacity to do work
1Joule = 1 kg m2 s-2
Joule - a wierd Manchester Brewer interested in the mechanical equivalent of heat
Thermodynamics grew up around the question of the transformation of energy, particularly
HEAT MECHANICAL WORK
NOTE: It’s easy to turn mechanical work into heat
Turning heat into mechanical work is much harder. Cannot turn 100% of the heat into work
THERE IS A MISSING QUANTITY
THE FIRST LAW
Conservation of EnergydE = dQ - PdV
ENTHALPY: HEAT SUPPLIED AT CONSTANT PRESSURE
Heat Q F dl = P dV(F dl = P Adl)
dl
p dQ = d H
H = E + P V
GETTING WORK OUT OF HEAT WORK HEAT - easy
HEAT WORK - harder
CARNOT - Impossible to take heat at a certain temperature and convert it to work with no other changes in the system or the surroundings
HOT
COLD
NO USEABLE WORK
USEABLE WORK
THE CONCEPT OF ENTROPY AROSE FROM THIS ANALYSIS
∆∆S =∆∆Q
T__ rev
ENTROPY - THEMODYNAMICDEFINITION
∆S = ∆Qrev
T and
∆S > ∆Q
irrev
T
From Carnot’s analysis
AnalogyW = – PdVQ = – T∆S
(note sign change - direction of Q)
i.e., S is the quantity that defines the relationship between heat and temperature
ENTROPY
. . ..
. ..
.. .. ..
16TONS
16TONS
16TONS
THUD !!
PE = mgh
heat
doesn'tlevitate !
.
.
.
. .
. .
...
. .
. . ..
. .
... .. ..
doesn't happenspontaneously
doesn't happenspontaneously
OTHER MANIFESTATIONS OF ENTROPY
Also happens spontaneously
ENERGY AND MATTER TEND TO DISPERSE CHAOTICALLY - THE SECOND LAW
HOT
Happens spontaneously
COLD
THE SECOND LAW OF THERMODYNAMICSAND FREE ENERGY
For a process to occur spontaneously,S must increase; eg mixing
∆∆Stot = ∆∆S sys + ∆∆Ssurr
But,if heat is released by the system
∆∆ G = -T∆∆S tot = ∆∆ H -T∆∆ S
Define free energy
∆∆Ssurr = - ∆∆Q sysT
= - ∆∆H sysT
FREE ENERGY
∆G → advantage of using the free energy is that it contrives a way to relate overall changes to changes in the system alone
WHY IS IT CALLED THE FREE ENERGY?
– Because it is the maximum amount ofnon-expansion work that can be obtainedfrom a system (T,P const)
∆G – ∆H → heat tax!
SUMMARY
PV = nRT∆Gm = ∆Hm - T∆Sm
NOW WHAT?
WE HAVE DISCUSSED THE ORIGIN OF THETHERMODYNAMIC EQUATIONS
NEXT STEP INTRODUCE THE MOLECULES
1. From F = ma derive the gas laws and provide a molecular machinery
2. Show that by looking at averages and distributionsof “mechanical” properties we can obtain S, Getc.
FEYNMAN - LECTURES ON PHYSICSFEYNMAN - “If in a cataclysm all human knowledge was destroyedexcept one sentence that could be passed on to future generations – – – what would contain the most information in the fewest words ?”
“ALL THINGS ARE MADE OF ATOMS – LITTLEPARTICLES THAT MOVE AROUND INPERPETUAL MOTION, ATTRACTING EACHOTHER WHEN THEY ARE A LITTLE DISTANCEAPART, BUT REPELLING UPON BEINGSQUEEZED INTO ONE ANOTHER .”
BASIC LAWS OF PARTICLES
Mechanics (various forms) Electricity + magnetism
BASIC LAWS OF STUFF
Thermodynamics
THE PROBLEM IS TO LINK THEM
PRESSURE
If there is no opposing force onthe piston, each collision movesit a little bit.
How much force, F, do we need to put on the piston to prevent movement?
First: Remember P = F/A
Second: How much force is imparted to the piston by the collisions ?
F = mx. .
= ddt (mx
. )i.e., WE NEED TO CALCULATE HOW MUCH MOMENTUM PER SEC IS DELIVERED BY THE COLLISIONS
PRESSURETHIS IS EASY!
a) Calculate momentum delivered by one collisionb) Count # of collisions/sec
Assume perfectly elastic conditions
Then the particles have + mvx momentum before - mvx momentum after
Change = mvx – (-mvx) = 2mvx
v x
PRESSURE
Now calculate the # of collisions in time t
FIRST: Define the number density
(# of particles/unit vol)
In a time t, only those particles that are close Enough and have sufficient velocity will hit the piston
Vol occupied by the molecules that will make it is
The number hitting the piston = N vx t A# per sec = N vx A
N =
nV
vx = xt
v x t A
this doesn’t get there in t
v x
v x
PRESSUREHENCE
BUT a) All molecules don’t have the same vxb) Some are moving away from the piston
F = NvxA.2mvx
P = FA
= 2 Nm v x2
S o re p l a c e v x2 w i th
12 v 2
x
P = N m
N o w
= v 2
3
v 2x
v 2yv 2
x v 2z= =
= 13 v x
2 + v y2 + v z
2 v 2x
THE IDEAL GAS LAW- KINETIC THEORY
HENCE
Recall the thermodynamic definition
TWO BODIES ARE AT THE SAME T IF THERE IS NO HEAT FLOW BETWEEN THEM.
2
3 2P =
2 N m v
2 = 23
nV
m v2
OR P V = 23 n KE
P V = n'R T COMPARED TO
IS n’ THE # OF MOLES ?IS KE ~ T ?
TEMPERATURE - KINETIC THEORY
WHAT HAPPENS AT EQUILIBRIUM?
•Forces balance•Atoms gain or lose energy depending on wether the piston is moving towards them or away from them during the collisions
FOR COLLISIONS OF PAIRS OF MOLECULES
THEN – IF TWO GASES ARE AT THE SAME T, THE MEAN KE OF THE CENTER OF MASS MOTIONS ARE EQUAL.
COLD H O T
12 m 1 v 1
2
= 12 m 2 v 2
2
High densityLow v
Low densityHigh v
PISTON
KINETIC THEORY- THE CONSTANTS k and R
HENCE
At same T, P, V, n is a constant!
Absolute scale of temp
K E ~ T
12 m v 2 = 3
2 k T
P V = n'R T
zero is w hen
= 0!
V
T
PV = 23 n = n k T1
2 m v 2
12 m v 2
IMPORTANT POINTS
•A simple consideration of the motion of particles gives a fundamental understanding of P, T, the ideal gas law, absolute T and absolute zero, etc.
•Can we stretch this approach and ultimately get a molecular interpretation of S , ∆∆G , etc.?
THE DISTRIBUTION OFENERGY AND MATTER
a) The distribution of velocities?
b) The distribution of molecules in space?
i.e., THIS WILL LEAD TO A DESCRIPTION OF ENTROPY, ∆∆G etc. IN TERMS OF STATISTICAL ARGUMENTS
All of the thermodynamic quantities we have dealtwith so far deal with how much material is presentand,on average,how fast the molecules are moving
WHAT ABOUT
EXAMPLEThe distribution of molecules in the atmosphere
ASSUMPTIONS Constant T !No wind !
i.e., if we know P, we know n, if P is a constantBUT, in the atmosphere it varies
P V = n k T o r P = Nk T ( N = nV )
hh + dh
Area A
P at (h + dh) must be less than P at h by an amount that is proportional to the weight of gas in Adh
# of moles in Adh = NAdh
ForceArea
Nadh A
=
Ph+dh - Ph = dP = - mgNdh
P = NkT, or dP = kTdN
EXAMPLE (cont.)
hh + dh
Area A
THIS IS A GENERAL RESULTBOLTZMANN’S LAW
SIMILARLY
HENCE
N = Noe -m gh/kT = Noe -P E /kT
N ~ e -P E /kT
n >u ~ e -K E /kT
ForceArea
Nadh A
=
dNdh = -
mgN kT
STATISTICAL MECHANICS
PROPERTIES OF MOLECULES
Classical mechanicsQuantum mechanics
PROPERTIES OFSTUFF (Bulk Materials)
Thermodynamics(P, T, V, G, S etc+ relationships between them)
Can bridge the “gap”by considering the average properties of all theparticles of the system and the distribution of matter and energy.
ENTROPY A measure of the distribution of energy and matter in a system
How do weget from the
molecules tobulk properties ?
o b ta i n e d P V = 23 n
m v 2
2
BOLTZMANN’S TOMBWe have seen that Entropy is associated with the distribution of energy and matter in a system.Thiscan be expressed formally in terms of the equationcarved on Boltzmann’s tomb
S = k lnW
Which,today,is normally written
S = k ln Ω
Where Ω is the number of arrangementsAvailable to the system[At a given V, E, N]
ENTROPY
S = kln ΩΩ Why does the relationship have this form ?
Consider an ordered pack of cards- 1 arrangement
Now shuffle; what is the number of Possible arrangements ?
ΩΩ = 52! (~4.45 X 10 )66
Now consider two packs shuff led separately
ΩΩ = ΩΩ1ΩΩ2
BUT S IS A THERMODYNAMIC PROPERTY THAT MUST BE ADDITIVE = S 1 + S 2
S = kln ΩΩ = kln ΩΩ1ΩΩ2= klnΩΩ1 + klnΩΩ2
FREE ENERGY REVISITED
BUT,IF HEAT IS RELEASED BY THE SYSTEM
For this to occur S must increase ie. ∆∆S > 0
∆∆Stot = ∆∆S sys + ∆∆Ssurr
∆∆ G = -T∆∆ S tot = ∆∆ H -T∆∆ S
∆∆Ssurr = - ∆∆Q sysT
= - ∆∆H sysT
WHAT YOU SHOULD KNOW
THERMODYNAMICS•Describes relationships between macroscopic properties
•2nd law: In a spontaneous process S always increases
•Free energy: What is it?
STATISTICAL MECHANICSRelates the “mechanical” properties ofatoms and molecules to macroscopic orThermodynamic quantities
EXAMPLES:PressureTemperature(Heat is motion!)
WHAT IS ENTROPY?