Announcements 9/27/10
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5 days left to get your clicker registered
Exam review session: Thurs, 8-9:30 pm, room C460
Reading assignment for Wednesday:
a. New: Read the Colton “What is entropy?” handout (also
available on website). It’s complicated, but spend at least 10-15
minutes glancing over it. Don’t worry about the examples for
now. We’ll go over it in class on Wednesday.
– If it seems super complicated, don’t stress too much. You don’t
have to know details for the exam.
b. On syllabus: Also spend 5-10 minutes on the book section 22.8.
Read the marble example (Ex. 22.7, in my edition), but don’t
worry about the other example (Ex. 22.8, my edition).
Your reading quizzes on Wed. will be simply:
a. Did you spend 10-15 minutes looking over the “What is
Entropy” handout?
b. Did you spend 5-10 minutes reading through section 22.8?
Reading quiz
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Which of the following is a version of the
Second Law of Thermodynamics?
a. The entropy of any system decreases in all
real processes
b. The entropy of any system increases in all
real processes
c. The entropy of the Universe decreases in all
real processes
d. The entropy of the Universe increases in all
real processes
Time for some physics humor
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Xkcd comic:
Thermodynamics song:
a. http://www.uky.edu/~holler/CHE107/media/first_second_law.mp3
Second Law
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Clausius: Heat spontaneously flows from hot to
cold, not the other way around
Why? Order.
Which hand is more likely?
p.413a
Microstates vs Macrostates
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Hand on left
a. microstate = A spades, K spd, Q spd, J spd, 10 spd
b. macrostate = ?
c. How many microstates make up that macrostate?
Hand on right
a. microstate = 2 spades, 3 diam, 7 heart, 8 clubs, Q diam
b. macrostate = ?
c. How many microstates make up that macrostate?
The most common macrostates are those that…
p.413a
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Probability  Heat flow
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You separate a deck into two halves: one is
70% red, 30% black; the other is 30% red,
70% black. What will happen if you randomly
exchange cards between the two?
Thermodynamics
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For the air in this room, right now:
a. Microstate = ?
b. Macrostate = ?
Hold this thought until Wednesday
A New State Variable
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State variables we know: P, V, T, Eint
P
B
A
V
B
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Observation:
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A
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dQ
doesn’t depend on path
T
 Something is a state variable!
Assumption: path is well defined, T exists whole time
 “Internally reversible”
P
2P1
P1
“Proof” by example, monatomic gas
C
B
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A
V1 2V1
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D
V
4V1
Path 1: AC + CB
C
Path 2: AD + DB
C
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nCV dT
dQ

 nCV ln TC TA   nCV ln 2
T
T
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nCP dT
dQ

 nCP ln TB TC   nCP ln 2
T
T
A
B
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Path 1: ACB
Path 2: ADB
(DB = isothermal)
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C
D

A
B

C
D

dQ

T
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workon  nRT ln VB VD 
dQ 1
Q

dQ   

  nR ln 2
T
T
T
T
T
A
B
D
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A
nCP dT
 nCP ln TD TA   nCP ln 4
T
B

D
Equal?
Entropy: S
B
S AB
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dQ

T
Advertisement: On Wednesday I’ll
explain how/why this quantity is
related to microstates & macrostates
A
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Assume S = 0 is defined somewhere.
(That’s actually the Third Law, not mentioned in your
textbook.)
Integral only defined for internally reversible paths,
but…
S is a state variable!
…so it doesn’t matter what path you use to calculate
it!
S for “free expansion”
before
after
What is V2? T2? P2?
How to find S?
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S for adiabatic?
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Adiabats as constant entropy contours
(“isentropic” changes)
Wait… isn’t “free expansion” an adiabatic
process?
S for isothermal?
S for const. volume?
S for const. pressure?
S of Universe
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S of gas doesn’t depend on path (state variable):
B
P
S AB
B
A

dQ

T
A
Spath1  Spath 2
V
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What about S of surroundings?
What about Stotal = Sgas + Ssurroundings?
Thought Question
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What is required in order for the entropy of a
system to decrease?
a. It must undergo an irreversible process
b. The system must be isolated
c. The system must not be isolated
d. The initial temperature must be higher than
the final temp.
e. It is never possible for a system’s entropy
to decrease!