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Lecture # 5
Cassandra Paul
Physics 7A
Summer Session II 2008
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Quickly discuss ‘the race’
Ideal gases
What is Lennard-Jones/Pair-wise potential?
Particle Model of Bond Energy
The ‘Race’ Explained….
Case 1
M1
KEtrans
Speed
M2
PEgravity
Height
m1
KEtrans
Speed
m2
PEgravity
Height
m1
+d
-d
M1
½ M1 (vf2-0) +M1g(hf-0)+ ½ m1 (vf2-0) +
m1g(hf-0)=0
½ M1 vf2 +M1g(-d)+ ½ m1vf2 + m1g(d)=0
Combining PE and KE terms
(M1+m1)½vf2 + (m1-M1)gd =0
m2
M2
PE’s are the same for both systems (mass difference is the same)
So KE’s must be the same for both systems
But… M+m is bigger for case 1, therefore: vf must be smaller to make up for it!
Case 2
Ideal Gas
• In Intro Chemistry we always dealt with ‘Ideal’
gasses. What does that actually mean?
• Ideal gases:
– Have no intermolecular forces
– Have perfectly elastic collisions with each other
(and the sides of containers)
Like Billiards or Jezzball
What was the point of the N2
Activity?
• What did we calculate?
• Spacing of atoms is about 10σ.
• At what point of the pair-wise potential do
atoms/molecules have zero PE and Zero force?
• 3σ!
• What do we take away from this?
• The ideal gas approximation is useful for
gases!
Intro Particle Model of Matter
A graphical representation of the
energies associated with particles
Lennard-Jones (pair-wise) potential
We know the shape… but what
exactly is this a graph of?
A. The potential energy of one atom with
respect to a system of particles.
B. The potential energy of a system (many
particles)
C. The potential energy of one particle with
respect to another particle
D. The total energy of one particle with respect
to a system of atoms
E. The total energy of one atom with respect to
another
Remember the Anchor
But Cassandra when is one
particle ever ‘anchored’ in
space?
Good question! It’s
not, but our graph
is always drawn
with respect to one
particle at the origin,
even if the origin is
moving
Potential Energy between two atoms
“pair-wise potential” a.k.a. Lennard-Jones Potential
Energy
r
ro
 ~ 10-10m = 1Å


r (atomic diameters)


Do not need to memorize
 ~ 10-21 J
pair-wise
 is the atomic diameter
 is the well depth
ro is the equilibrium separation
Forces and the Potential
Force = -d(PE)/dx
Repulsive
Attractive
Or, negative
change in y over
change in x
Force has a
magnitude of slope,
and the direction of
decreasing PE!
If the curve only tells us about PE,
how do we find KE and Etot?
Let’s do a closed system…
Etot = KE + PE
-3ε = KE + -3ε
KE = 0
-3ε = KE + -4ε
KE = 1ε
-3ε = KE + -7ε
KE = 4ε
-3ε = KE + -8ε
KE = 5ε
-3ε = KE + -7ε
KE = 4ε
-3ε = KE + -4ε
KE = 1ε
Etot
-3ε = KE + -1ε
KE = -1ε
KE can’t be negative!!!!!
Turning Points
Where the Etot intersects the
PE curve, there are ‘turning
Points.’
Etot
The particle oscilates between
These two points.
How much work does it take to move one
particle from rest at equilibrium (1.12σ), to 3σ
with a minute (negligible but non zero) velocity?
i
f
A.
B.
C.
D.
E.
1ε
3ε
-1ε
2.88σ
Impossible
to tell
Same idea as before:
Initial: at 1.12σ, v=0
PE + KE = Etot
-1ε + 0 = -1ε
• Now what?
Is this a closed system?
NO! Adding energy:
Final: at 3σ, v~0
• So new Etot = 0
Must add 1ε to get there.
OK let’s draw an Energy System
Diagram:
System: Two Particles, one bond
Initial: v=0, r=1.12σ
Final: v~0 r=3σ
Work
f
PEpairwise
ΔPE = Work
Energy
Added
Wait! We don’t have an equation for PE pair-wise!
It’s ok, we have something better… a graph!
i
PEf – PEi = Work
0ε – (-1ε) = Work
Work = 1ε
DL sections
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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