Announcements 9/29/10
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Three days to register your clicker.
Exam starts Saturday
Exam review session: Thurs, 8-9:30 pm,
room C460
Reduced Tutorial Lab hours on Saturday due to
General Conference: only open 12 – 2pm.
Increased Testing Center hours on Saturday due
to G.C.: it opens at 8 am instead of the usual 10
am.
Thought Question
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A gas in contact with a thermal reservoir undergoes an
isothermal expansion. The gas and the thermal reservoir are
isolated from the rest of the universe. Which of the following
is true?
a. The entropy of both the gas and the reservoir will increase.
b. The entropy of both the gas and the reservoir will decrease.
c. The entropy of both the gas and the reservoir will not
change.
d. The entropy of the gas will go up, the entropy of the
reservoir will go down, and the total entropy of the system
will not change.
e. The entropy of the gas will go up, the entropy of the
reservoir will go down, and the total entropy of the system
will increase.
Reading quiz
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Did you spend at least 10-15 mins looking over
Dr Colton’s “What is Entropy?” handout?
a. Yes
b. No
Did you spend at least 5-10 mins reading the
book section 22.8?
a. Yes
b. No
Microstates vs Macrostates
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Reminder
a. Left microstate: part of the “royal flush”
macrostate
b. Right microstate: part of the “garbage”
macrostate
Dice
You roll two dice. What are the microstates?
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),…
 How many microstates are there?
 How does that compare to the number of
microstates for rolling one die?
 How many microstates if we roll 3 dice?
 What are the macrostates for 2 dice? (sum of
numbers)
 What is the most likely macrostate?
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MANY Dice
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You roll 1023 dice with your left hand.
How many microstates are there?
You roll 1023 dice with your right hand.
How many microstates are there?
How many microstates are there in the
COMBINED system?
Isn’t this ridiculous?
Solution: Use logarithms
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S = some constant  ln(#microstates)
[units of J/K]
a. Much more manageable numbers.
b. Combining two systems:
Stot = C  ln(#microstates1  #microstates2)
= C  ln(#microstates1) + C  ln(#microstates2)
= S1 + S2
c. System in macrostate with most microstates
 system in macrostate with largest S
System and Reservoir
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System: E1
Reservoir: E2
Etot = E1 + E2 (const. volume so no work)
0 = dE1 + dE2
Want to maximize S: take dS/dE1, set = 0
dS
d

 S1  S2 
dE1 dE1
dS1 dS 2


dE1 dE1
dS1 dS 2


dE1 dE2
dS1 dS 2


dE1 dE2
Temperature
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dS/dE is the same for two systems in thermal contact!
Temperature is the same for two systems in thermal
contact!
a. dS/dE has units of 1/K…
Compare to
dS 1

dE T
dQ
dS 
T
We are assuming no
work, so dE=dQ (First
Law)
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This “works” if the constant is chosen properly:
S = kB ln(#microstates)
Small system with 2 possible energies:
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E1A vs E1B
(1 microstate each)
Probability of system 1 being in state A vs state B?
P1A ~ (#microstates of system 1 having energy E1A) 
(#microstates of system 2 having energy E2A = E – E1A)
Let #microstates of E1A = 1 for now.
Same thing for state 1B…
P1A # microstates of 2 for case A

P1B # microstates of 2 for case B
P1A # microstates of 2 for case A e S2 A

 S
P1B # microstates of 2 for case B e 2 B
kB
kB
Math…
P1A # microstates of 2 for case A e S2 A

 S
P1B # microstates of 2 for case B e 2 B
kB
kB
S 2 A  S 2 ( E  E1 A )
“of”, not
“times”
dS 2
 S2 ( E ) 
 E1 A
dE
E1 A
 S2 ( E ) 
T
(same with S 2 B )
 E1 A / k BT
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P1A e
Result:
 E /k T
P1B e 1B B
The Boltzmann Factor
P~e
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 E / kBT
The
Boltzmann
Factor
Prob ~ BF; Prob = BF/(sum of all BFs)
Worked Problem: Suppose an atom has only
two available energy levels, which are
separated by 210-23 J. If the temperature is
1.5 K, what is the probability the atom is in the
lower state?
Maxwell-Boltzmann Velocity Distribution
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E = ½mv2
What’s probability of having speed 5 vs speed
10?
P1A e E1 A / kBT
 E /k T
P1B e 1B B
v
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2
1
 mv 2 / k BT
e 2
1
 mv A2 / k BT
2 2
e
P1 A v A

1
P1B
 mvB 2 / k BT
2 2
vB e
Multiplicities
 (Number of states with speed v) ~ v2
Maxwell-Boltzmann Velocity Distribution
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The result:
P(speed v) 
v
2
1
 mv 2 / k BT
e 2
1 2
 mv / k BT
2 2
v e


1 2
 mv / k BT
2 2
v e
dv
0
 Exactly the equation given for the
velocity distribution in your textbook! (after
you do the integral in the denominator with,
e.g. Mathematica)