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Entropy, probability and disorder. Thermal equilibrium. Experience tells us: two objects in thermal contact will attain the same temperature and keep that temperature Why? More than just energy conservation! Involves concept of entropy. Entropy and disorder. - PowerPoint PPT Presentation
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Entropy, probability and disorder
Thermal equilibrium
Experience tells us: two objects in thermal contact will attain the same temperature and keep that temperature
Why? More than just energy conservation!
Involves concept of entropy
Entropy and disorder
It is often said that entropy is a measure of disorder, and hence every system in isolation evolves to the state with “most disorder”
Consider a box sliding on a floor: internal energy due to disorderly motion of the
molecules kinetic energy (of the box) due to the collective,
orderly, motion of all the molecules
Entropy and disorder II
Now the box comes to rest due to friction
Temperature rise in both floor and box so the internal energy increases
No more collective motion: all K.E. has been transferred into internal energy
More disorder, so entropy has increased
A vessel of two halves
Large number of identical molecules – distribution?
About 50% in left half, 50% in right halfWhy?
Definitions
Microstate: position and momentum of each molecule accurately specified
Macrostate: only overall features specified
Multiplicity: the number of microstates corresponding to the same macrostate
Fundamental assumption
Statistical Mechanics is built around this one central assumption:
Every microstate is equally likely to occur
This is just like throwing dice:
A throw of the diceRoll one die: 1/2/3/4/5/6 all equally likely
Roll a pair of dice: for each 1/2/3/4/5/6 equally likely the sum 7 is most likely, then 6 and 8, etc.
Why? 6 combinations (microstates) give 7 (the macrostate): 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. There are 5 combinations that give 6 or 8, etc.
Four identical molecules
4 molecules ABCD5 macrostates:
Four identical molecules (2)
left: right: A&B&C&D -
A&B&C D A&B&D C A&C&D B B&C&D A
multiplicity: 1
multiplicity: 4
Four identical molecules (3)
left: right: A&B C&D A&C B&D A&D B&C B&C A&D B&D A&C C&D A&B
multiplicity: 6
Four identical molecules (4)
#left #right multiplicity probability 4 0 1 1/16 3 1 4 4/16 2 2 6 6/16 1 3 4 4/16 0 4 1 1/16
16
Ten identical molecules
Multiplicity to find 10 –...– 0 molecules on left 1–10–45–120–210–252–210–120–45–10–1
Probability of finding #left = 4, 5 or 6:
For large N: extremely likely that #left is very close to N/2
66.01024
210252210
Generalisation
Look at a gas of N molecules in a vessel with two “halves”.
The total number of microstates is 2N: two possible locations for each molecule we’ve just seen the N=4 example
Binomial distribution I
A gas contains N molecules, N1 in the left half (“state 1”) and N2 = N – N1 in state 2 (the right half). How many microstates correspond to this situation?
N1 N2
Binomial distribution II
Pick the molecules one by one and place them in the left hand side:
choose from N molecules for the first molecule choose from N – 1 for the second choose from N – 2 for the third, … choose from N – N1 + 1 for the N1-th molecule
Binomial distribution III
Number of ways of getting N1 molecules into left half:
The macrostate doesn’t depend on the order of picking these molecules; there are N1! ways of picking them. Multiplicity is mathematical “combination”:
)!(!
)1(...)2()1(1
1 NNN
NNNNN
!)!(!
111 NNNN
N
NC
Verification
Look at a gas with molecules A,B,C,D,E.
Look at the number of ways of putting 2 molecules into the left half of the vessel.
So: N = 5, N1 = 2, N – N1 = 3
Verification II
The first molecule is A, B, C, D, or E.
Pick the second molecule. If I first picked A then I can now pick B, C, D or E, etc:
AB BA CA DA EA
AC BC CB DB EB
AD BD CD DC EC
AE BE CE DE ED
That is possibilities
!3!5
12312345
45
Verification III
In the end I don’t care which molecule went in first. So all pairs AB and BA, AC and CA, etc, really correspond to the same situation. We must divide by 2!=2.
AB A
B=
Binomial distribution plotted
Look at N=4, 10, 1000:
0 1 2 3 4
P(N1)
N1
0 2 4 6 8 10
N1
0 200 400 600 8001000
N1
Probability and equilibrium
As time elapses, the molecules will wander all over the vessel
After a certain length of time any molecule could be in either half with equal probability
Given this situation it is overwhelmingly probable that very nearly half of them are in the left half of the vessel
Second Law of Thermodynamics
Microscopic version:
If a system with many molecules is permitted to change in isolation, the system will evolve to the
macrostate with largest multiplicity and will then remain in that macrostate
Spot the “arrow of time”!
Boltzmann’s Epitaph: S = k logW
Boltzmann linked heat, temperature, and multiplicity (!)
Entropy defined by S = k ln
W: multiplicity; k: Boltzmann’s constant
= “dimensionless entropy” = ln
Second Law of Thermodynamics
Macroscopic version:
A system evolves to attain the state with maximum entropy
Spot the “arrow of time”!
Question 1
Is entropy a state variable?
a) Yes
b) No
c) Depends on the system
Question 2
The total entropy of two systems, with respective entropies S1 and S2, is given by
a) S = S1 + S2
b) S = S1 · S2
c) S = S1 – S2
d) S = S1 / S2
Entropy and multiplicity
Motion of each molecule of a gas in a vessel can be specified by location and velocity multiplicity due to location and velocity
Ignore the velocity part for the time being and look at the multiplicity due to location only
Multiplicity due to location I
Divide the available space up into c small cells. Put N particles inside the space: =cN.
For c=3, N=2: =32=9
AB
A B
A B
AB
AB
A B AB
AB
AB
Multiplicity due to location II
Increasing the available space is equivalent to increasing the number of cells c.
The volume is proportional to the number of cells c
Hence VN
“Slow” and “fast” processes
Slow processes are reversible: we’re always very close to equilibrium so we can run things backwards
Fast processes are irreversible: we really upset the system, get it out of equilibrium so we cannot run things backwards (without expending extra energy)
Slow isothermal expansion
Slow isothermal expansion of ideal gas; small volume change
“velocity part” of multiplicity doesn’t change since T is constant
N
N
N
VV
V
VV
1initial
final
V
V
Slow isothermal expansion (2)
Use the First Law:
Big numbers take logarithm
V
V
VkTVN
VpQ
NN
NkTQ
VV
11initial
final
Slow isothermal expansion (3)
Manipulation:
or
1ln
1lnlninitial
final
kTQ
NkTQ
NNkTQ
N
NkTQ N
V
V
TQ
kk initialfinal lnln
Slow isothermal expansion (4)
Use definition of entropy:
valid for slow isothermal expansion
V
V
TQ
SSS
TQ
kk
initialfinal
initialfinal lnln
Example
To melt an ice cube of 20 g at 0 °C we slowly add 6700 J of heat. What is the change in entropy? In multiplicity?
24.5 J K-1; 000,000,000,000,000,000,000,770
initial
final 10
Very fast adiabatic expansion
Expand very rapidly into same volume V+V which is now empty
Isothermal: same #collisions, #molecules, etc.
Entropy change:
NO! Entropy is a state variable and therefore
S = same as for slow isothermal expansion
???0TQ
S
TQ
S
Slow adiabatic expansion
Same volume change, but need to push air out of the way so temperature drops
Again we ask:YES!
The “location part” of multiplicity increases as with slow isothermal expansion
The “velocity part” decreases as temperature drops The two exactly cancel
???0TQ
S
Constant volume process
Heat is added to any (ideal or non-ideal) gas whose volume is kept constant. What is the change in entropy?
Integrate (assuming CV is constant)
TTnC
TdQ
STnCQ VV
dd ;dd
1
2lnd
d2
1
2
1TT
nCT
TnCSS V
T
T
VT
T
Constant pressure processes
Heat is added to an ideal gas under constant pressure. What is the change in entropy?
a) b)
c) d) 0
1
2lnTT
nCS V
1
2lnVV
nCS p
1
2lnpp
nCS p
Entropy and isothermal processes
An ideal gas expands isothermally. What is the change in entropy?
Constant temperature so
First Law: (done previously)
Therefore
TQ
S
1
2lnVV
nRTWQ
2
1
1
2 lnlnpp
nRVV
nRS
Entropy and equilibrium
We have established a link between multiplicity and thermodynamic properties such as heat and temperature
Now we see how maximum entropy corresponds to maximum probability and hence to equilibrium
Equilibrium volume
In general the number of microstates depends on both the volume available and the momentum (velocity) of the molecules
Let’s ignore the momentum part and look at the spatial microstates only.
Equilibrium volume II
Say we have 3 molecules in a vessel which we split up into 6 equal parts. A partition can be placed anywhere between the cells. One molecule is on the left-hand side, the other two on the right-hand side. What is the equilibrium volume?
Look for maximum entropy!
Equilibrium volume III
Number of cells on the left c1, on the right c2.
We’ll look at c1=4, c2=2:
A
A
A
A
BC
B C
C B
BC
Equilibrium volume IV
Left: 1= c1=4.
Right: 2= (c2)2=4.
= ln 4 + ln 4 = ln 16 = 2.77
A
A
A
A
BC
B C
C B
BC
Question
The dimensionless entropy of this system of 6 cells and one partition dividing it into c1 and c2 cells is
a) = ln (c1+ c2)
b) = ln (c1+ c22)
c) = ln c1+ln 2c2
d) = ln c1+2·ln c2
Equilibrium volume V
c1 P()
1 25 3.22 0.25
2 32 3.47 0.31
3 24 3.18 0.24
4 16 2.77 0.16
5 5 1.61 0.05
total 102 1
Maximum entropy and probability
1 2 3 4 50
1
2
3
4
dimensionless entropy 1
2
c1
1 2 3 4 50.0
0.1
0.2
0.3
0.4
Probability
c1
Maximum probability
Probability maximum coincides with entropy maximum
Volume V1 = c1·V where V is the cell size
Most likely situation when
Same density on both sides:21
42
;21
1
1
1
1 VN
VN
0dd
1
V
Question
Which relationship holds for the probabilities of finding a the system in a microstate corresponding to c1=2,3,4 ?
a) P(2) < P(3) < P(4)
b) P(2) = P(3) = P(4)
c) P(2) > P(3) > P(4)
Entropy and mixing
Suppose we remove the partition. What is the entropy of this system?
Answer: ln 63 = ln 216 = 5.38
The additional entropy of 1.91 is called “the entropy of mixing”
Generalisation I
Look at N1 particles occupying c1 cells on the left, N2 particles occupying c2 cells on right. Volume of each cell = V.
Multiplicity:
N
NN
NN
NN
V
VVV
VV
VV
cc
)(
)( 21
21
21
11
21
2121
V1 V2
N1 N2
Generalisation II
Entropy:
Maximum entropy for equal densities:
V1 V2
N1 N2
VNVVNVNV
VVVN
NN
ln)ln(ln)(
)(lnln
1211
1121
2
2
1
1
1
2
1
1
1 0
dd
VN
VN
VVN
VN
V
Multiplicity and energy
According to quantum mechanics, atoms in a crystal have energies 0, , 2… (This is called the Einstein model of solids)
Say we have three atoms with total energy 3
Microstates are distinguished by the different energies E1, E2, E3.
The microstates
Energy
23
0
23
0
23
0
23
0
E1=3
E1=2
E1=
E1=0
E1,E2,E3 E1,E2,E3 E1,E2,E3 E1,E2,E3
Question
What is the probability for any of these three atoms to have energy 0? ? 2? 3?
a) 4/10,3/10,2/10,1/10
b) 1/4,1/4,1/4,1/4
c) 1/10,2/10,3/10,4/10
d) not sure
Generalisation
If there are n atoms and the total energy is q, then the number of microstates is given by
Works for previous example (n=3,q=3):
)!1(!)!1(
),(
nqnq
nq
10!2!3
!5
Partition I
Look at 10 particles, with energy 20.
n1=3 particles on the left-hand side, n2=7 on right-hand-side
What is the most likely energy distribution? Plot as a function of q1.
Partition II
0 2 4 6 8 10 12 14 16 18 20
0
500000
1000000
against energy on left-hand side
1
2
q1
Partition III
0 2 4 6 8 10 12 14 16 18 200
5
10
15
Entropy against energy on left-hand side
1
2
q1
Partition IV
You expect the atoms on the left to have the same energy on average as the atoms on the right
Calculation/plot shows this: maximum for
2
2
1
121
21
11 )(
nq
nq
qqnn
nq
Entropy, energy, temperature I
The internal energy on the left U1 = q1.
Equilibrium/entropy maximum when
Use = 1 + 2:
Use U2 = U – U1:
0dd
1
U
0dd
dd
1
2
1
1 UU
1
2
2
1
1
2
2
2
dd
dd
dd
dd
UU
UUU
Entropy, energy, temperature II
It follows that the most likely distribution of energy corresponds to a situation where
We know that in this situation T1 = T2
Clearly the two are linked!
2
2
1
1
2
2
1
1
dd
dd
or dd
dd
US
US
UU
Entropy, energy, temperature III
Remember: we kept V, N constant so the only way in which energy could be exchanged was through heat transfer
Remember:
only heating todue
TU
TQ
S
parameters external fixedonly heating todue
1
US
US
T
Entropy, energy, temperature IV
In our example the only external parameter was the volume
In general, gravitational, electric or magnetic fields, elastic energy etc. could all change
The definition of temperature only holds if all of these are held fixed
PS225 – Thermal Physics topicsThe atomic hypothesisHeat and heat transferKinetic theoryThe Boltzmann factorThe First Law of ThermodynamicsSpecific HeatEntropyHeat enginesPhase transitions