Transcript Physics 7A -- Lecture 3 Winter 2008

Physics 7A – Lecture 8
Winter 2009
Prof. Robin D. Erbacher
343 Phy/Geo Bldg
[email protected]
• Quiz 5 is today. Next week is review.
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• Final Exam review schedule to be posted soon.
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• Turn off cell phones during lecture.
Depends only on properties of the system
at a particular time: “Snapshot”
Example: Ethermal of a gas
For an ideal gas, Ethermal depends only on:
• Temperature
• Number of modes
Not a property of a particular object.
Instead a property of a particular process, or
“way of getting from the initial state
to the final state”
LHS: depends only
on i and f
Q,W depend on
process between i and f
∆Etotal must include all changes of energy associated
with the system…
∆Etotal = ∆Ethermal+ ∆Ebond+ ∆Eatomic+ ∆Enuclear+ ∆Emechanical
Energy associated with
the motion of a body
as a whole
∆U : Internal energy
Energy associated with the
atoms/molecules inside
the body of material
∆U
First law of Thermodynamics
State functions
•
•
•
Depend only on what
the object is doing at
the time.
Change in state
function depends only
on start and end
points.
Examples:
T, P, V, modes, bonds,
mass, position, KE,
PE...
Process-dependent
•
Depend on the
process.
•
Not a property of
an object
•
Examples:
Q, W, learning ,
......
P
P
initial
initial
final
final
V
Along any given segment:
V
OR
= - Area under curve
(sign positive when V decreases,
sign negative when V increases)
W = -PDV
(if P constant)
final
Is the work done in the
process to the right
positive or negative?
A) Negative
B) Positive
C) Zero
D) Impossible to tell.
P
initial
V
P
fina
l
initial
V
First section:
W1 < 0 (volume
expands)
Second section: W2 = 0 (volume
constant)
Third section:
W3 > 0 (volume
contracts)
final
Here are two separate
processes acting on two
different ideal gases.
P
initia
l
Which one has a greater
magnitude of work? The initial
and final points are the same.
A) Magnitude of work in top
process greater
B) Magnitude of work in lower
process greater
C) Both the same
D) Need more info about the
gases.
V
P
final
initia
l
V
P
initial
final
V
We can read work directly off this graph
(i.e. don’t need to know anything about modes, U, T, etc.)
If we know something about the gas, we can figure out
Ui, Uf and Uf - Ui
P
initial
final
V
5 moles of a monatomic gas has its pressure increased from
105 Pa to 1.5x105 Pa. This process occurs at a constant
volume of 0.1 m3. Determine:
* work
* change in internal energy DU
* heat involved in this process.
P
final
HW:
There are two
ways to solve
this using the
ideal gas law…
initial
V
Work depends on the process
Heat depends on the process
DU only depends on initial and final
W = 0 for all constant volume processes
H = U + PV
Is it a state function?
- U depends only on state of system
- P depends only on state of system
- V depends only on state of system
=> H depends only on state of system
(Hess’s law)
Why is this useful?!?
Constant volume
P
Constant pressure
P
final
initial
final
initial
V
V
W=0
*Derivation in book
Note: nothing about gases used –
works for solids and liquids too!
In unit #1 we taught you:
But the gas is expanding, so work is being done!
Now we know better….
But in a large room, pressure is roughly constant…
Microstates versus
States, and Entropy
So far we have described systems using P,V,T,…
Incomplete information about the system
A microstate is a particular configuration of everything
(such as atoms/molecules) in the system.
For example,
For a box of gas, you need to specify
where all the atoms are: ie- position
(x,y,z), and how fast they are moving,
ie- velocity (vx, vy, vz).
Things we worry about:
Microstates
Constraints
States
Constraints: Tell us which microstates are allowed.
States:
Examples
•The volume of a box constrains the possible positions
of gas atoms.
•The energy of the box constrains the possible speeds
of gas atoms.
Groups of microstates that share some average
properties,
i.e. A collection of states that “look” the same macroscopically.
Examples
gases: P ~ average density, V~volume filled, T~average
Example- Flipping a coin 2
times:
Constraints: some combinations not possible
Microstates: all possible combinations of coin flips
States: total number of heads
Every microstate is equally likely.
The state that is most likely is the one with the most microstates
Example: Flipping a coin 3
times:
Constraints: some combinations not possible (e.g. HHTHHH)
Microstates: all possible combinations of coin flips
States: total number of heads
States Microstates
Prob.
Every microstate is equally likely.
The state that is most likely is the one with the most microstates
“Ordered microstate”
equally likely as
random microstate
Are we likely to find a system in an “ordered state”
or a “random state”?
…ummm…. Almost all the microstates look pretty random!
Define states by “total number of heads”
Different states contain different # of
microstates
Therefore even though each microstate is
equally likely, some states are more likely than
others.
1024 microstates, each microstate equal width
We can also consider our physical system to be two “subsystems”:
* Sub-system A: the first two coin flips
* Sub-system B: the final eight coin flips
HH
HT
TH
TT
Any of the 256 micro.
Any of the 256 micro.
Any of the 256 micro.
Any of the 256 micro.
where kB is Boltzman’s constant
f our system is composed of two sub-systems A and B:
We can add the entropy of the subsystems to get the total
entropy.
Omega  is
always increasing.
As the number of
microstates
available increases,
so does entropy!
Splitting 10 coin flips into the first 2 then 8 is arbitrary.
But in calculating the heat needed to to raise the temperature of
this system to 10oC we would split into subsystems: ice and water.
ie- Q = DEice + DEwater
Similarly, we would calculate DStotal = DSice + DSwater
We should divide our box up into
“atom-sized” chunks. But how big
should our velocity microstates be?
How can we get a definite answer for
the number of microstates in this
Water (00 C)
system?
ice (00
C)
Relating entropy to microstates is useful for conceptually
understanding what entropy is.
At this level, it is not useful for calculating the change in
entropy
For slow, reversible processes:
final
T
P
initial
initial
final
S
V
Q > 0 when Delta S > 0
Q < 0 when Delta S < 0
W > 0 when Delta V < 0
W < 0 when Delta V > 0
For slow, reversible processes:
To get to entropy we can “turn this expression around”
If temperature is constant, then we can easily integrate:
Q
(isothermal only!)
This last equation is not generally true;
as heat enters or leaves a system the temperature often changes.
If the process is not slow or reversible, or it is very
ifficult, you can use the fact that entropy is a state functio
DS depends only on initial and final points!
If you can find any process from the
initial to final states, you can use this path
to calculate DS for the process in question!
(As in the calculation of enthalpy in DLM14)
• In dealing with physical systems, often convenient to deal with
subsystems.
• The total number of microstates with subsystems A, B, …. Is
Ωtotal = ΩA + ΩB
• We introduce entropy S=kBlnΩ. It is a state function, and the
logarithm makes it additive:
Stotal = SA + SB
• It is difficult to calculate entropy directly from the microstates.
Instead we have two alternatives:
- For small amounts of heat (T is approximately constant)
calculate dS = dQ/T
- Find a different process which is easier. Because S is a
state function, DS only depends on the endpoints.
Why do we care about entropy?
That is our next topic..... the quest for equilibrium
• Equilibrium is the most likely state
• Each microstate is equally likely, so the
•
•
equilibrium state has the most microstates,
or the greatest entropy.
Therefore the equilibrium state has the highest
entropy.
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)
ice
(00C=273K)
Heat entering ice: dQice = Tice dSice > 0 =>Sice
increasing
Heat leaving air:
dQair = Tair dSair < 0 =>Sair
decreasing
This is okay, because dStot = dSice + dSair > 0
Low temp
Tfinal
High temp
Energy leaves hot objects in the form of heat
Energy enters cold objects in the form of heat
We need to determine the process… and the
final state of the system.
TA=150K
TB=150K
TA = TB = 150 K
Ok, now we have found the equilibrium state.
From our daily experience, and the activity in DL, we
expect this process to be
irreversible and spontaneous….
How do we calculate the change in entropy?
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? Temperature!
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
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?
In this case Tf = 150 K = Tf,A = Tf,B
Ti,A = 100 K, Ti,B = 200 K
∆Stotal > 0
Irreversible and spontaneous process
∆Stotal = 0
Reversible process
• State functions (S,H,U,P,V,T, …) depend only on
the initial and final states of the system
• Therefore we can join our
initial and final states by any
process. Most convenient
to use reversible processes.
Environment
system
Note: reversible means that
DStotal=0
NOT DSsystem=0
(DSsystem can be +, -, or zero for a reversible process)
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
Next Time:
Physics 7A
Course Review!
For slow, reversible processes:
final
T
P
initial
initial
S
Q > 0 when Delta S > 0
Q < 0 when Delta S < 0
final
V
W > 0 when Delta V < 0
W < 0 when Delta V > 0
Rules of thumb, similar to Work case….