chapter 2 the second law. why does q (heat energy) go from high temperature to low temperature? cold...
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Chapter 2
The Second Law
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Why does Q (heat energy) go from high temperature to low temperature?
coldcold hothot
Q flow
Thermodynamics explains the direction of time.The Big Bang
32 NkBT 1
2 mv 2
Hot objects are faster so they are more quick to move to the cold side.
BUT in a solid objects aren’t actually moving from one side to the other. ? . ? . ?
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Why does Q (heat energy) go from high temperature to low temperature?
coldcold hothot
Q flow
Let’s look at how probability tells us which way energy should flow.
How many ways can energy be arranged?
Which arrangements are most likely?EE
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EE
EE
EE
EE
EE EE EE
EE
EEEEEE
EE EE
ENTROPY
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Simple ProbabilityIn the mid-1960s, the Adams gum company acquired American Chicle and introduced a new slogan for Trident: "4 out of 5 Dentists surveyed would recommend sugarless gum to their patients who chew gum." The phrase became strongly associated with the Trident brand.
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Some New Terms• Mutually Exclusive
– Outcomes of events have a single possibility, others are excluded• A coin flip is heads or tails, not both
• Collectively Exhaustive– The full set of propositions or outcomes
• Heads and tails are all possible outcomes
• Independent– An event or outcome does not depend of previous or future events or
outcomes• A previous heads does not determine the next coin flip
• Multiplicity– The number of ways to get a particular outcome
• W or W is typically used as the variable
• Conditional Probability– An outcome depends on a previous event
• Drawing colored balls from a bag
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Distributions Continued• Discrete
– Coin flips• List the macrostates, probability, microstates, & multiplicity
One coin
Two coins
Three coins
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Distributions Continued• Discrete
– Four coin flip
How can we predict what will happen?
How can we talk about outcomes in percentages?
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Distributions Continued• Discrete
– Four coin flip
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Pascal’s Triangle
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Paramagnets
B = 0
B ≠ 0
(N ) N!
N !N !
N!
N !(N N )!
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Cards
)!(!
!)(
nNn
NN
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Harmonic Oscillators – Einstein Solid
)( 21 nEn
U 12 kx 2
Schrodinger Eqn. solutions for energy are
n = 0
n = 1
n = 2
n = 3
n = 4
n = 5
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Einstein Solid
3,2,1,0
3
A
A
n
NTotal energy
0
1
Oscillator #1 #2 #3 W
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Einstein Solid
3,2,1,0
3
A
A
n
NTotal energy
0
1
2
0 0 0
Oscillator #1 #2 #3 W
100
010
001
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Einstein Solid
Total energy
3
Oscillator #1 #2 #3 W
3,2,1,0
3
A
A
n
N
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Einstein Solid
!)!1(
)!1()(
nN
nNn
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Einstein Solids in Thermal Equilibrium
1
3
3
3
B
B
A
A
n
N
n
N
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Einstein Solids in Thermal Equilibrium
1
3
3
3
B
B
A
A
n
N
n
N
n
iBA
BA
innP
0
)()(
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Einstein Solids in Thermal Equilibrium
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Einstein Solids in Thermal Equilibrium
3059
P70/3059 = 0.023350/3059 = 0.114825/3059 = 0.2701100/3059 = 0.360714/3059 = 0.233
1.000sum
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Einstein Solids in Thermal Equilibrium
Normalized probability for an energy arrangement
n
iBA
BA
innP
0
)()(
For NA = NB = 100 oscillatorsP(nA=30) = P(nA=70) = 0.004 = 0.4%
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Einstein Solids in Thermal Equilibrium
Normalized probability for an energy arrangement
n
iBA
BA
innP
0
)()(
For NA = NB = 100 oscillatorsP(nA=30) = P(nA=70) = 0.004 = 0.4%
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Numbers• Small
– 1, 5, 10, 235, etc.• Large
– 1010, 1023, 10114, etc.• Very Large
The Universe: 1018 s old (Big Bang)~1080 atoms
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Computer Numbers and the 2nd Law• 64 bit floating point numbers
– 52 bit mantissa– 11 bit exponent– 1 bit sign
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The Second Law
The second law of thermodynamics: "You can't even break even, except on a very cold day."
Energy will "flow" until the state of maximum multiplicity is obtained.
S = kB ln (W)
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Very Large NumbersStirling’s Approximation
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Very Large NumbersStirling’s Approximation
n
nxn
e
nndxexn 2!
0
See appendix B
A more rigorous derivation comes from
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Multiplicity of a Large Einstein Solid
n
nxn
e
nndxexn 2!
0
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Multiplicity of a Large Einstein Solid
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Multiplicity of a Large Einstein SolidHigh temperature limit: Assume n >> N
TNknE Bf
n 221 )(
N
N
en
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Multiplicity of a Large Einstein SolidHigh temperature limit: Assume n >> N
N
N
en
This can’t be plotted for very large systems 25
23
23
10
102
10
total
B
A
n
N
N
N
N
en
ln)ln( Neither can this.
N
enN ln)ln( But this can.
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Multiplicity of a Large Einstein SolidHigh temperature limit: Assume n >> N
A
AAA N
enN ln)ln(
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Multiplicity of a Large Einstein SolidHigh temperature limit: Assume n >> N
B
ABB N
nneN
)(ln)ln(
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Multiplicity of a Large Einstein SolidHigh temperature limit: Assume n >> N
B
AB
A
AABABA N
nneN
N
enN
)(lnln)ln()ln()ln(
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W is a Gaussian Function
NA=NB=100n=500
NN
B
B
N
A
Atot N
en
N
en
N
enBA 2
2
2s Nnw 2
222222
maxln2
nx
nxn NNNN
Ne eee
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W is a Gaussian Function
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Multiplicity of a Monatomic Ideal Gas
z
yx
A container of monatomic gas. How can we describe the ways to arrange the atoms and the energy they contain?
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Multiplicity of a Monatomic Ideal Gas
z
yx
In 2D momentum space, constant energy is defined by a circle.
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Multiplicity of a Monatomic Ideal Gas
z
yx
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Multiplicity of a Monatomic Ideal Gas
z
yx
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Multiplicity of a Monatomic Ideal Gas
z
yx
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Multiplicity of a Monatomic Ideal Gas
z
yx
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Multiplicity of an Ideal GasWhat is the relative probability of a gas taking the full volume of its container to the probability of taking half the volume of its container?
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Multiplicity of an Ideal Gas
A 1
NA
VANA
h3NA
23NA
2
3NA2 ! 2mUA
3NA2
B 1
NB
VBNB
h3NB
23NB
2
3NB2 ! 2mUB
3NB2
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Multiplicity of an Ideal GasExchanges possibleNA NB : Diffusive EquilibriumVA VB : Pressure EquilibriumUA UB : Thermal Equilibrium
Utotal
2 3N2
Vtotal
2 N
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Entropy
S kB ln()
S NkB lnV
N
4mU
3Nh2
32
5
2
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Entropy – Ideal Gas
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Creating Entropy
Free Expansion
W = 0 because there is nothing to push against in a vacuum.
Q = 0 because this is an adiabatic process, and insulated from the surroundings.
DU = Q + W = 0.
BUT there is a volume change!
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Entropy of Mixing
S NkB lnV
N
4mU
3Nh2
32
5
2
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Entropy of Mixing
S NkB lnV
N
4mU
3Nh2
32
5
2
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Entropy – Einstein Solid
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Entropy
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Entropy
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W vs. ln(W)
NA=NB=1000n=1000
NA=NB=1023
n=1024
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Creating Entropy
Free Expansion
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Entropy
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• Experiment• Flip n=10 coins• Count and record number of heads, nH• Repeat N=1000 times• Create a histogram from 0 to 10 of nH
• Computer Simulation• Generate n=10 random 1 or 0 (heads or tails)• 1 1 1 0 0 1 1 1 0 1
• Count and record number of heads, nH• Repeat N=1000 times• Create a histogram from 0 to 10 of nH
• Fit a gaussian to the data• My results
• x0 = 4.93 ± 0.04• = 2.27 ± 0.09