Transcript Lecture # 7

Lecture # 7
Cassandra Paul
Physics 7A
Summer Session II 2008
Today
• Equation Confusion
• Heat capacities and their role in Particle
model
• Start Intro to Thermodynamics
• Quiz
Equation confusion…
The reason I want you to memorize a few
equations is not (contrary to popular belief) so
you can do physics somewhere you don’t have
access to your notes…
Equations contain Ideas
• What is gravitational potential energy?
– We can increase potential energy by increasing
the height or mass of an object.
– We can also increase potential energy if we have
greater gravity.
• Hence, PE=mgh
• The equation reminds you of the concepts.
What about Kinetic Energy?
• KE= ½ mv2
• Faster objects have more kinetic energy
• More massive objects have more kinetic
energy
• Kinetic energy can’t be negative.
Ok… so you have to remember the ½ . But that’s
not the important part
Bond Energy
• Why do we have three different equations for
bond energy?
• Because each are useful at different times.
• If we want to talk about the Macroscopic
properties of solids liquids and gasses, we want
to use ΔEbond= ±lΔmΔHl .
• Why?
• Because we aren’t talking about individual
particles here, only the energy associated with
changing the phase of the system.
Bond Energy
• When do we use the equation:
Ebond = Σall pairs (PEpair-wise)
• When we know something about the number
of particles in the system and we have a small
number of bonds where we can count.
• Why?
• That’s the exact definition of bond energy at
the microscopic level.
Bond Energy
• When do we use this equation?
Ebond = (tot # n-n bonds) (-ε)
• When there are so many bonds in the system
that we can’t count so we estimate.
Cassandra, the blue pages in my book say:
Ebond = (tot # n-n bonds) (-ε)
But in DL we were taught:
Ebond = (nn)/2 (-ε) (tot # of atoms)
Which is correct!?
Let’s ask your classmates…..
Which is the correct approximation?
A.
B.
C.
D.
E.
Ebond = (tot # n-n bonds) (-ε)
Ebond = (# of n-n)/2 (-ε) (tot # of atoms)
Ebond = (# of n-n)/2 (-ε) (Avogadro's #)
A B and C are all correct
A and B are correct
Which is the correct approximation?
A.
B.
C.
D.
E.
Ebond = (tot # n-n bonds) (-ε)
Ebond = (# of n-n)/2 (-ε) (tot # of atoms)
Ebond = (# of n-n)/2 (-ε) (Avogadro's #)
A B and C are all correct
A and B are correct
A and B are both correct, and say EXACTLY the same thing.
When is C correct?
When the total number of atoms is equal to 1 mole.
Why do you have us memorize any
equations?
• If you know the ideas you can work out the (derive) the
equations.
• If you know the equations, you can work out the
concepts.
• This way you aren’t stuck. Even if you forget
something, you can check it against what you know.
• Notice… many times I will give you the equations even
if I tell you that you must memorize them. This is not
because I’m a ‘push over’, it’s to reduce test anxiety.
• If you don’t know what I’m going to give you on the
exam, you’ll work a little harder to organize your
thoughts.
Let’s talk Ethermal now.
• From the beginning of the course, we know:
ΔEthermal = mcΔT
• This tells us that the amount of energy it takes
to change the temperature of a substance
depends on how massive it is, how much of a
temperature change we are talking about and
finally what the heat capacity of that
substance is.
In the Particle Model of Thermal
Energy…
• Ethermal = (amount of energy per mode) (Total
number of Ethermal modes)
• Ethermal =( ½ kb T)(tot # particles)(tot # Eth
modes per particle)
What if we want to know the change in Eth?
• ΔEthermal =( ½ kb ΔT)(tot # particles)(tot # Eth
modes per particle)
We place a delta in front of the Temperature because this
is our indicator for when Eth is changing!
Cassandra, in the beginning of the
course we talked about heat
capacity being a part of Eth. In the
particle model of thermal energy, it
doesn’t seem like heat capacity is
represented, why is this?
Ahhh but heat capacity is represented… it is
just hidden in the concept of MODES.
Let’s see how…
Equipartition tells us that the energy per mode
is 1/2 kBT. Also it tells us that each mode gets
the same amount of energy.
KE mode
PE mode
Total
Solids
3
3
6
Liquids
3
3
6
Monatomic gasses
3
0
3
Diatomic gasses
3+2+1
1
7
Equipartition tells us that the energy per mode
is 1/2 kBT.
Question:
Does it take More or
Less energy to raise
the temperature of
diatomic gas
compared to
monatomic gas?
(a) More
(b) Less
(c) Equal
(d) Can’t Determine
KE
PE
Total
mode mode
Solids
3
3
6
Liquids
Monatomic
gasses
Diatomic gasses
3
3
6
3
0
3
3+2+1
1
7
Let’s think of modes as ‘buckets’ and
Energy and Temperature as Water.
If I have two systems, one system has two buckets
and another has three buckets, which system is it
harder to raise the water level of?
(Remember, equipartition says that all modes get
the same amount of energy!)
System 1
E
n
e
r
g
y
a
d
d
e
d
System 2
System 1
System 2
If the same amount of energy is added
to each system…
Less modes, lower heat
capacity, easier to change
T
More modes, higher heat
capacity, harder to
change T
ΔT1
ΔT2
In other words…
Lots of Modes = smaller temp
change…
ΔEth = ½ kb ΔT
(tot # of modes)
Less Modes = larger temp
change…
ΔEth = ½ kb
ΔT (tot # of modes)
It’s like the red and blue car!
Let’s Get a more quantitative Idea
about how heat capacity and modes
are related….
In DL on Friday…
• Think back to the beginning of the course…
• Heat Capacity Definition: C=Q/ΔT
• We want to make a definition of heat capacity at
the microscopic level, where should we start?
• We know that Eth is present in both the
microscopic and macroscopic models
• And we know that Q sometimes = ΔEth
• What assumptions must we make to assume that
Q=ΔEth?
ΔEth + ΔEbond = Q + W
•
•
•
•
•
•
No Change in Ebond
No work Done
ONLY then is ΔEth = Q
So operating under those assumptions:
C = Q/ΔT 
Cv = ΔEth /ΔT = ( ½ kb ΔT)(tot # particles) (tot # Eth modes per
particle)*/ΔT
• Cv = (½ kb)(tot # of particles)(# Eth modes per particle)
This is Heat Capacity, what if we want specific heat in MOLAR
units? (per mole)
• Cvm = (½ kb)(Avagadro’s #)(# Eth modes per particle)
*From slide 13
Deriving Molar Specific Heat Definition
Cvm = (½ kb)( Avagadro’s #)(# Eth modes per particle
We could be done here but Na (Avagadro’s #) multiplied
by kb = R (gas constant)
(1.38x10-23)*(6.02x1023) = 8.3076 (or 8.31 J/K!)
So…
Cvm = (½ R)(# Eth modes per particle)
And now we have a definition of molar
specific heat for the particle model of
matter.
Cvm = (½ R)(# Eth modes per particle)
DL sections
•
•
•
•
Swapno:
11:00AM Everson Section 1
Amandeep: 11:00AM Roesller Section 2
Yi:
1:40PM Everson Section 3
Chun-Yen: 1:40PM Roesller Section 4