Transcript Slide 1

Lecture 10
Cassandra Paul
Physics 7A
Summer Session II 2008
Announcments
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Final is WEDNESDAY!
This room, 9:30-10:45
BRING YOUR STUDENT ID!
Be here at 9:15 so we can check you in.
The second half of today’s class is optional
To be posted:
– Quiz 5 grades, rubric, solutions for quiz 5
Today
• Finish where Yen left off
– Entropy
– Reversible vs. Irreversible
– Gibbs Free Energy
– Spontaneous vs. Non spontaneous
• Final Review
Recap…
What was entropy again?
Physical systems have an equilibrium state
(the one with the most microstates,
and therefore the highest entropy)
If we wait “long enough” the system will most likely
evolve to that equilibrium state. (by the laws of chance)
For large (i.e. moles of atoms) systems, the system is
(essentially) always evolving toward equilibrium.
Therefore the total entropy never decreases:
Second law of
Thermodynamics
This is only for total entropy!
heat flows
this way
air (200C)
Heat
ice
(00C=273K)
Ebond
ml 
Heat entering ice: dQice = Tice dSice > 0 =>Sice
increasing
Heat leaving air:
dQair = Tair dSair < 0 =>Sair
decreasing
Are the changes in entropy the same?
Heat entering ice: dQice = Tice dSice > 0 =>Sice
increasing
Heat leaving air:
dQair = Tair dSair < 0 =>Sair
decreasing
The magnitudes of the Q’s must be the same,
but the signs are opposite.
The T’s are different, but always positive.
The change in entropy for the lower
temperature object is always positive, and
always has a greater magnitude, than the
entropy of the lower temperature object.
dStot = dSice + dSair > 0
Let’s See Why….
Evolving Towards Thermal
equilibrium
Low temp
Tfinal
Energy leaves hot objects in the form of heat
Energy enters cold objects in the form of heat
High temp
Evolving AWAY FROM Thermal
equilibrium
Does this happen?
Low temp
Do
Tfinal
High temp
From everyday experience, we know that the process of heat leaving the
higher temperature object and entering the lower temperature object is
irreversible and spontaneous…. But how can we figure it out for less obvious
situations?
First Definitions:
• Reversible Process:
– No heat is transferred across a temperature
difference
– No friction occurs
ΔS=0
• Spontaneous Process:
– The process occurs in one specific direction.
ΔG<0
note: (ΔG=0 means both the forward and reverse processes are in equilibrium)
Let’s Get Quantitative!
A
B
The two blocks A and B are both made of copper, and have equal
masses: ma = mb = 100 grams. The blocks exchange heat with each
other, but a negligible amount with the environment.
a) What is the final state of the system?
b) What is the change in entropy of block A?
c) What is the change in entropy of block B?
d) What is the total change in entropy?
(Heat of melting)
A
B
The two blocks A and B are both made of copper, and have equal
masses: ma = mb = 100 grams. The blocks exchange heat with
each other, but a negligible amount with the environment.
a) What is the final state of the system?
What is the final equilibrium temperature of the system?


ΔEthA+ ΔEthB = 0

A
Ti=100 Etherm 
Tf=x
al
B
Etherm
al
Ti=200
Tf=x (.1)(385)(x-100)+ (.1)(385)(x-200)= 0
Tf=x=150
(Heat of melting)
A
B
Okay, now we have found the equilibrium state.
a) What is the final state of the system?
b) What is the change in entropy of block A?
c) What is the change in entropy of block B?
d) What is the total change in entropy?
How do we calculate the change in entropy?
What about the entropy change for this process?
Does the temperature remain constant for block A or block B?
dS=dQ/T
S=Q/T
S=Cln(Tf/Ti)
(IF constant temperature)
(IF temp is changing)
In this case Tf = 150 K = Tf,A = Tf,B
Ti,A = 100 K, Ti,B = 200 K
m c In (Tf,A/Ti,A) =
m c In (Tf,B/Ti,B) =
What about the entropy change for this process?
In this case Tf = 150 K = Tf,A = Tf,B
Ti,A = 100 K, Ti,B = 200 K
∆Stotal > 0
Irreversible and spontaneous process
Wait, how do we know it’s spontaneous? We haven’t talked about ΔG yet.
Gibbs Free Energy
G = H - TS
is a state function:
- H depends only on state of system
- T depends only on state of system
- S depends only on state of system
=> G depends only on state of system
Why is this useful?….
Gibbs Free Energy
G = H - TS
dG = dH - TdS + SdT
At Const T, dG = dH - TdS
or
∆G = ∆ H - T ∆S
Remember ∆H = Q for Const P process. (tells you whether
a reaction is endothermic or exothermic)
So then, Gibbs free energy reflects the change in the enthalpy
of the system as well as the change in the entropy of the system
Still …Why is this useful?
To find out if something happens spontaneously
we must know about the environment as well as
the system….
If something happens spontaneously, the total
change in entropy must be greater than 0:
Δ(entropy of system) +Δ(entopy of environment) > 0
Same as saying:
ΔS-ΔH/T> 0
Or, multiplying by –T:
-TΔS + ΔH < 0
The left side is equal to ΔG!!!!!
So for a process to happen spontaneously, ΔG must be
negative.
So How did I know our example
was spontaneous?
∆G = ∆ H - T ∆S
A
ΔH=0
T is positive
ΔS is positive
So ΔG must be negative.
B
Laws of Thermodynamics
Remember
conservation of energy?
(if there’s no change in ∆Emechanical),
∆U
First law of Thermodynamics
Total entropy
never decreases.
the system always
evolves toward
equilibrium.
spontaneous
changes occur
with an increase
in entropy.
∆Stotal  or  0
Second law of
Thermodynamics
Conserve energy?
Why, energy is always conserved!
Found in the PHYSICS department!
The rest of the story…
An inescapable truth defined by the 2nd law of
Thermodynamics
2nd Law of Thermodynamics tells you the direction of the energy flow.
Each time energy gets transferred, some of it gets less useful… until finally,
it becomes simply useless….
All of energy we use, sooner or later, ends up being low-grade thermal
energy…
Brake!
The rest of the story…
An inescapable truth defined by the 2nd law of
Thermodynamics
2nd Law of Thermodynamics tells you the direction of the energy flow.
Each time energy gets transferred, some of it gets less useful… until finally,
it becomes simply useless….
All of energy we use, sooner or later, ends up being low-grade thermal
energy…
Falling rock