Theories applied to Chemistry
Quantum Mechanics (Planck, Einstein, Schrödinger

Ĥ
Y
=E
Y).
Applies to all chemical systems.
Al瑨t畧栠免=is=瑨攠灲pma特=瑨敯特=數灬ai湩湧=瑨t
=
扥桡i爠映
subatomic particles, atoms, and molecules it gives way to CM for
much of chemistry due to its more complex mathematical
foundations.
1
Wave

particle
duality allows mathematical wave function solutions to
predict chemical properties of systems.
Incorporates Heisenberg Uncertainty Principle.
The
behavior of individual particles are not certain, but represented as
outcome probabilities.
Statistical treatments make measurements on large systems essentially
certain.
Theories applied to Chemistry
Classical Mechanics (Newton’s Laws of Motion?)
Applies only to larger (atoms /molecules), slower (relative to c) particles.
Although CM leads to a number of inconsistencies in chemical behavior
it is still used for chemical topics where it provides accurate
predictions with more conceptually accessible explanations.
CM topics in chemistry include much of kinetics and
thermodynamics.
Theories applied to Chemistry
Classical Mechanics (Newton

e.g. F = ma)
There are two distinct approaches to studying classical
systems in chemistry ….
Macroscopic
Look at behavior of system as whole (P vs. V at
cst
T)
Summarize behavior as
a set of laws
(Boyle’s Law etc.)
Develop equations predicting the behavior of systems
. (P
1
V
1
= P
2
V
2
)
Microscopic (Classical)
Develop a physical model explaining molecular behavior.
Apply
Laws of motion to individual particles
.
Extrapolate
to collection of particles (statistical mechanics).
Derive
equations predicting the behavior of
systems
(CM).
Kinetic Molecular Theory (KMT
)
see handout
Assume
:
2. particles in constant, random motion
3. no attractive/repulsive forces
4. conservation of energy at every collision
x
y
z
1. gas particles have mass but no volume
x
y
z
Pressure
of Ideal
Gas
v
x
v
y
v
z
dp
x
=
?
dt
= ?
<v
2
> = 3nRT/
nM
= 3RT/M
v
2
=
v
x
2
+ v
y
2
+
v
z
2
& <v
2
> = 3<v
x
2
> & ….
<v
x
2
> = <v
2
>/3
PA
=
2mv
x
/(2x/
v
x
) = mv
x
2
/x
PAx
= PV = mv
x
2
PV =
nM
<v
2
>/3
For N molecules
― PV = Nm<v
x
2
> =
nM
<v
x
2
>
=
nRT
<v
2
>
1/2
=
v
rms
= (3RT/M)
1/2
P
=
F/A & PA = F =
ma = m(
dv
x
/
dt
) =
dp
x
/
dt
dp
x
= 2mv
x
dt
=
2x/
v
x
x
y
z
Pressure
of Ideal
Gas
v
x
v
y
v
z
PV = Nm<v
2
>/
3 =
nM
<v
2
>/3
=
nRT
Derive expression for <v
2
>
1/2
=
v
rms
P
=
F/A & F =
ma =
dp
x
/
dt
Derive expression for KE (
E
tr
) =
½nM<v
2
>
in terms of T.
use KE (
e
tr
) = ½mv
2
from Newton’s 2
nd
Law
E
tr
= ½Nm<v
2
> = ½nM<v
2
>
for N particles
E
tr
=
½
nM<v
2
> =
3nRT/2
e
tr
= 3kT/2
(
per
molecule)
since R =
N
A
k
or
nR
=
Nk
rms

speed is a function of both T and M ……
Kinetic energy is only a function of T
In the KMT all gases have an ‘equal’ impact on P at a fixed T, because
the greater force of larger particles is offset by their slower speed
Or ….
E
tr,
m
= 3RT/2
(per mole)
…to derive an expression for KE (
E
tr
) =
in terms of T.
PV = Nm<v
2
>/
3 =
nM
<v
2
>/3
=
nRT
x 3/2
<v
2
>
1/2
=
v
rms
= (3RT/M)
1/2
The Barometric Formula
This derivation is not in your text (see handout). However, it illustrates one
practical application of statistical mechanics …
The approximate atmospheric content as a function of altitude …
2
This model also serves as an example of the
Boltzman
Distribution Law,
a law derived using statistical mechanics, that illustrates how particles
distribute themselves over an energy gradient.
The
Boltzman
Distribution law is fundamental both to CM and QM.
In the latter it is required to explain the characteristics of spectroscopy
arising from the population distribution of quantum states.
We will apply the
Boltzman
Distribution law as a shortcut to avoid a
more exact but tedious derivation of the speed distribution function.
dz
= thickness of atmosphere layer
A = Surface area of layer
dm = mass of gas between z &
dz
dP
= Pressure difference between z & z + dz.
Barometric Formula
&
Boltzmann Distribution
z = altitude
F
up
= PA
F
down
= g • dm + (P +
dP
)A
What is
F
up
and
F
down
?
Barometric Formula
Let F
up
= F
down
PA = (dm)g + (P + dP)A
PV =
nRT
& n =
?
PA = MPVg/RT + PA + dP • A & dP =

MPVg/ART
integrate
=
Pº
P
dP/P =

Mg/RT
0
z
dz
ln (P/P
=
㴠

Mgz/RT==&...
=
Since V/A = dz ….
dP/P =

Mgdz/RT
P = P
=
exp(

Mgz/RT
=
dm = MPV/RT
dm
/M
Boltzmann Distribution Law
by analogy to Barometric Formula
P/P
=
==exp(

Mgz
/RT
=
P/P
=
縠
k/k
o
&
D
䔠
(J/mol)
=

Mgz
(kg•m
2
•s

2
/mol)
J = kg•m
2
•s

2
N/N
o
= exp(

D
E/RT==r....
=
N/N
o
= exp(

De
/
歔
=
The Boltzmann Distribution Law gives the relative
occupation for
any two energy levels.
e.g
height in gravitational field, MO’s
,
vibrational
energy states, translational energy of
gases with varying speeds.
N/N
o
=
g/g
o
•
exp
(

D
䔯R吩=
=
Mountains
ht (m)
l
ocation
Everest
8848
Himalayas
K2
8611
Kilamanjaro
5895
Africa
Eiger
3970
Alps
Matterhorn
4477
Alps
Denali
6194
Alaska
Pikes Peak
4302
C
olorado
Mauna Kea
4205
Hawaii
Harney Peak
2208
South Dakota
(Black Hills)
Structures
Burj
Dubai
818
United Arab Emirates
Eiffel Tower
324
Paris, France
Sears Tower
527
Chicago
Washington Monument
169
Washington, D.C
CN Tower
553
Toronto, Canada
P/P
=
㴠exp(

Mgz
/RT
=
What is the atmospheric pressure at the top of ……..
CHM 3460
Assignment #1
Due Monday, 9/12/11
Ball
–
chapter 19: #2 and #6
2.
KE of Hg atom and 1 mole of Hg atoms that has speed = 200 m/s?
From CM
e
tr
=
½mv
2
= 0.5 * 0.20059/6.022e23 * 200
2
=
6.664 x 10

21
J
For 1 mole
x by N
A
E
tr
= 4013 J
6.
P of interstellar space containing 10 molecules of H
2
per cm
3
at 2.7K.
What is
v
rms
?
(The book is in error in calling this <v> and the # they give as an
answer is
V
rm
s
.)
Use PV =
nRT
(n = N/N
A
& R =
N
A
k
)
P =
NkT
/V
P =
10/6.022 x 10
23
* 1.38 x 10

23
• 2.7
0.01
3
m
3
=
3.73 x 10

16
Pa
v
rms
=
(3RT/M)
½
= (3 • 8.314 • 2.7/0.002016)
0.5
=
183 m/s
if
on earth
v
rms
= 1920 m/s
alternately you could calculate
v
rms
first and then use P =
nM
<v
2
>/3V to get P = 3.71 x 10

16
Pa
SPEED DISTRIBUTION
What is the composition of air?
Do all of the oxygen molecules in air have the same speed?
Are all of the oxygen molecules in air identical?
Do we want to include the difference between O
2
molecules in our model?
<v
2
> = 3RT/M
16
O
99.76%
15.99491
17
O
0.04%
18
O
0.20%
15.999
Are all of the identical oxygen molecules (e.g.
16
O

16
O = 99.52%) in air
traveling at the same speed?
3
dN
v
/N = fraction of molecules with speed = v
SPEED DISTRIBUTION
G(v) = Speed Distribution Function =
dN
v
/N
Fractional probability = ∫
v1
v2
G(v)
dv
Distribution functions must be normalized such that…
∫
0
∞
G(v)
dv
= 1
Plot G(v) vs. v
G(v) or
dN
v
/N
v
What should the graph look like?
Molecules with speed = 0?
Molecules with speed =
∞?
Expand to 3D velocity distribution function
(瘩=
=
†††††††††††=
(瘩===
木
x
) g(
v
y
) g(
v
z
)
Find 1D velocity distribution = g(
v
x
)
DERIVATION STRATEGY
Change to speed distribution function G(v)
removing dependence on specific direction of motion
What is the average
v
x
for a gas molecule?
What should the function g(
v
x
) look like?
→
→
→
→
→
→
→
→
Find
the value of
A
(normalization constant
)
see page 793 …..
∫
0
∞
exp(

bx
2
)
dx
=
½(
p
/b
1/2
Find 1D velocity distribution = g(
v
x
)
Find g(
v
x
) =
dN
(
v
x
)/N
from Boltzmann
Distribution….
N
vx
/N
= exp(

De
/
歔
=
g(
v
x
) =
dN
(
v
x
)/N
= A exp(

mv
x
2
/2kT)
A = normalization constant
∫

∞
∞
g(
v
x
)
dx
= A
∫

∞
∞
exp(

mv
x
2
/2kT)
= 2A ∫
0
∞
exp(

mv
x
2
/2kT) = 1
(b = m/2kT)
A
= 1/(2∫
0
∞
exp(

mv
x
2
/2kT
))
A = 1/
(
p
/b
1/2
g(
v
x
) = (m/2
p
kT
½
exp(

mv
x
2
/2kt)
KE = ½mv
x
2
→
A = 1/
(2
p
歔km
1/2
CO2
He
CH4
O2
g(
v
x
) = (m/2
p
kT
1/2
exp(

mv
x
2
/2kT)
Graph g(
v
x
) vs.
v
x
….?
g(
v
x
) =
(M/2
p
RT
1/2
exp
(

Mv
x
2
/2RT
)
Find
(瘩===
x
) g(
v
y
) g(
v
z
):
extrapolate to 3D ‘ray’
v
2
= v
x
2
+ v
y
2
+ v
z
2
= 3v
x
2
g(
v
x
) = (M/2
p
RT
½
exp(

Mv
x
2
/2RT)
[(M/2
p
RT
½
]
3
=
[exp(

Mv
x
2
/2RT)]
3
(v
) =
(M/2
p
RT
3/2
exp(

Mv
2
/2RT
)
dv
= [exp(

3Mv
x
2
/2RT)]
<v
x
2
> = <v
2
>/3
= [exp(

Mv
2
/2RT)]
For a large collection of particles ……
v
x
2
= v
y
2
= v
z
2
and m/k = M/R
4
→
→
Expand to 3D velocity distribution function
(瘩=
=
(瘩===
木
x
) g(
v
y
) g(
v
z
)
Find 1D velocity distribution = g(
v
x
)
DERIVATION STRATEGY
Change to speed distribution function G(v)
removing dependence on specific direction of motion
→
→
→
→
→
→
g(
v
x
) = (M/2
p
RT
½
exp(

Mv
x
2
/2RT)
⡶
===
(M⼲
p
RT
3/2
exp(

Mv
2
/2RT
)
dv
→
→
→
z
x
dV
dV
(cube)
=
dv
x
dv
y
dv
z
= ?
dv
G(v) =
(M/2
p
RT
3/2
exp(

Mv
2
/2RT
)
4
p
2
dv
(v)
expanded to
G(v)
(v
) =
(M/2
p
RT
3/2
exp(

Mv
2
/2RT
)
dv
dV
=
4
p
(
+摶
3
/3
–
4
p
3
/3 =
4
p
2
dv
4
p
3
/3 +
4
p
2
dv
+ 4
p
摶
2
+ 4
p
摶
3
/3
y
→
→
→
Scalar
v
represents
radius of sphere
but also represents speed
v = r
dv
=
dr
He
CH
4
O
2
CO
2
G(v)
v
ms

1
G(v) = (M/2
p
RT
3/2
exp(

Mv
2
/2RT) 4
p
2
300K
600K
1000K
G(v) for CO2
As T
↑ G(v) becomes lower
and broader.
Open Excel file
–
“speed graph”
0
0.0005
0.001
0.0015
0.002
0
250
500
750
1000
Neon: Speed
Distribution (300 K)
<v> = ∫
0
∞
vG
(v)
dv
= (8RT/
p
M
½
=
561
m s

1
V
mp
= v where
dG
(v)/
dv
= 0
= (2RT/M)
½
=
497
ms

1
v
rms
= ∫
0
∞
(v
2
G(v)
dv
)
½
= (3RT/M)
½
=
609
m s

1
Finding the average speed for any gas from ……
G(v) = (M/2
p
RT
3/2
exp(

Mv
2
/2RT) 4
p
2
<v> = ∫
0
∞
v • G(v)
dv
General formula
∫
0
∞
x
2n+1
exp(

cx
2
)
dx
= n!/2c
n+1
integral tables
Let n = 1 and c = M/2RT
substitutions
v • G(v)
dv
=
4
p
⡍/2
p
RT
3/2
•
v
3
•
exp(

Mv
2
/2RT)
4p
{M/(
2p
RT⥽
3/2
• ∫
0
∞
v
3
exp(

cv
2
)
dv
=
4p
{M/(
2p
RT}
3/2
• 1/(2c
2
)
=
4p
{M/(
2p
RT⥽
3/2
• 4R
2
T
2
/(2M
2
)
=
8
p
=
R
2
T
2
M
3/2
= (8RT/
p
M
1/2
2
3/2
p
㌯3
R
3/2
T
3/2
M
2
Neon:
frac
. with 400 < v <
440
T = 300 & M = 0.02018 kg/mol
fraction of molecules with speed
between 400 ms

1
and 440 ms

1
=
㐰4
㐴4
G(v)
dv
G(v) = (M/2
p
RT
3/2
exp(

Mv
2
/2RT) 4
p
2
~
G(v) *
dv
(sub in v = 420 &
dv
= 40)
better estimate if
dv
is small)
Speed Distribution
19.17
Will they always have the same relative values? ….
… or will variations in either T or M change their
relative magnitudes?
v
mp
(2RT/M)
½
<v>
(8RT/
p
M
½
v
rms
(3RT/M)
½
v
mp
(2RT/M)
½
(2)
½
1
<v>
(8RT/
p
M
½
(8/
p
½
1.13
v
rms
(3RT/M)
½
(3)
½
1.22
Assignment
–
19.14
plus
make graph of G(v) vs. v from 0 to 1000 m/s
using increments of 100 m/s.
Show <
v
x
> = 0 from
g(
v
x
) =
(M/2
p
RT
1/2
exp
(

Mv
x
2
/2RT
)
Collisions with Wall:
dN
w
/
dt
=
…
what factors?
1. Area of wall (A
) =
yz
2.
average speed in direction of wall
<
v
x
> =
0
v
x
g(
v
x
)
dv
x
= <v>/4
3.
The gas density (N/V)
= PN
A
/RT
dN
w
/
dt
= A • <v>/4 • PN
A
/RT
= A • (RT/2
p
M)
½
• PN
A
/RT
Effusion Rate:
dN
/
dt
same as collisions with wall
except replace area of wall with
area of hole
x
z
y
v
x
v
y
v
z
5
2.
average speed in direction of wall
<
v
x
> =
0
v
x
g(
v
x
)
dv
x
= <v>/4
0
x
exp
(

cx
2
) dx =
1/(2c)
c = M/(2RT)
<
v
x
> =
(M/2
p
RT
1/2
• RT/M
= {RT/(2
p
M}
½
x 8
½
/8
½
=
= {8RT/(16
p
M
⥽
½
= ¼
{
8RT
/(
p
M
⥽
½
= <v>/4
g(
v
x
) =
(M/2
p
RT
1/2
exp
(

Mv
x
2
/2RT
)
z
b
(b)
(s

1
) = how many collisions will one ‘b’ molecule make with all other
‘b’ molecules per unit of time?
Molecular collisions (like molecules)
z
b
(b)
(s

1
) =
function of ......
dN
/
dt
= A • <v>/4 • N/V
dN
b
(b)
= the # of collisions of 1 ‘b’ molecule with other ‘b’ molecules.
How do these 3 factors change for molecular collisions?
<v>/4 was speed in direction of wall/hole (<
v
x
>).
How should we represent speed in direction of another molecule that
is also moving?
N/V = density of molecules near hole.
How should we represent N/V for collisions?
A = How do we represent the area of a collision?
Molecular collisions (like molecules)
z
b
(b)
(s

1
) =
function of ......
density of
molecules
N/V
=
P
b
N
A
/RT
Relative speed
<
v
rel
>
<
v
rel
>
(assume 90º∟) =
(<
v>
2
+ <
v>
2
)
1/2
<
v
rel
>
<
v>
<
v>
<
v
rel
> = 2
½
•
<v>
Collision area
p(
r
cir
)
2
=
p
d
2
<
v
rel
>
dt
2 • d
Molecular collisions (like molecules)
z
b
(b)
= A
• <
v
rel
> •
N
b
/V
z
b
(b)
=
p
d
2
• 2
½
<v> •
P
b
N
A
/RT
<v>
= (
8RT/
p
M
½
<
v
rel
> = 4(RT/
p
M
½
Z
b
(b)
=
the total # of collisions per unit volume
=
Z
b
(b)
=
z
b
(b)
• ?
N
b
/V
• ½
l
(mean free path)
= the average distance a molecule will travel before
units (m) colliding with another ‘like’ molecule,
l
=
<
v
rel
>
=
z
b
(b)
= ms

1
/s

1
= 1/(A
•
N
b
/V) =
RT/(
p
d
2
P
b
N
A
)
or …
z
b
(b)
= ms

1
/m
=
<
v
rel
>
/l
Collisions between like molecules
–
Density

# molecules are available for collision (m

3
):
N/V = PN
A
/(RT)
relative speed
–
effective collision speed (ms

1
):
<
v
rel
> = 4{RT/(
p
M
⥽
1/2
Area of molecular displacement (m
2
):
A =
p
d
2
Collision frequency (s

1
):
z = <
v
rel
>
•
A•
(
N/V
)
= 4
p
1/2
PN
A
d
2
/{MRT}
1/2
Total collisions (s

1
m

3
):
Z = ½•z•
(
N/V
) =
2
p
1/2
(
PN
A
d
)
2
/{M
1/2
(RT)
3/2
}
Mean free path (m):
l
==Y
rel
>/z = RT/{
p
d
2
PN
A
} =
kT
/(
p
d
2
P)
Collisions between like unlike molecules
–
Collisions between like unlike molecules

<
v
rel
> (ms

1
) = (8RT/
p
1/2
•
(1/M
b
2
+ 1/M
c
2
)
1/2
A (m
2
) =
p
(
r
b
+
r
c
)
2
z
bc
(s

1
) = <
v
rel
>•
A•
(
N
c
/V
)
= {8
p
•(1/M
b
2
+ 1/M
c
2
)}
1/2
•(
r
b
+
r
c
)
2
•P
c
N
A
/{RT}
1/2
Z
bc
(s

1
m

3
) = ½•z
bc
•
(
N
b
/V
) =
笲
p
•(1/M
b
2
+ 1/M
c
2
)}
1/2
•(
r
b
+
r
c
)
2
•P
c
•P
b
•N
A
2
/{RT}
3/2
l
bc
(m) = <
v
rel
>/
z
bc
=
p
(
r
b
+
r
c
)
2
•
P
c
N
A
/RT
Collisions between like unlike molecules

<
v
rel
> (ms

1
) = (8RT/
p
1/2
•
(1/M
b
2
+
1/M
c
2
)
1/2
A
(m
2
) =
p
(
r
b
+
r
c
)
2
z
bc
(s

1
) = <
v
rel
>•
A•
(
N
c
/V
)
=
{8
p
•(1/M
b
2
+ 1/M
c
2
)}
1/2
•(
r
b
+
r
c
)
2
•P
c
N
A
/{
RT}
1/2
Z
bc
(s

1
m

3
) = ½•z
bc
•
(
N
b
/V
)
=
笲
p
•(1/M
b
2
+ 1/M
c
2
)}
1/2
•(
r
b
+
r
c
)
2
•P
c
•P
b
•N
A
2
/{
RT}
3/2
l
bc
(m) = <
v
rel
>/
z
bc
=
p
(
r
b
+
r
c
)
2
•
P
c
N
A
/RT
Z
bc
(s

1
m

3
) = ½•z
bc
•
(
N
b
/V
)
=
z
bc
(s

1
) = <
v
rel
> • A •
(
N
c
/V
)
z
bc
(s

1
) =
(<
v
b
>
2
+ <
v
c
>
2
)
1/2
•
p
(
r
b
+
r
c
)
2
•
(
N
c
/V
)
6.
P of interstellar space containing 10 molecules of H
2
per cm
3
at 2.7K.
What is
v
rms
?
Use PV =
nRT
(n = N/N
A
& R =
N
A
k
) P =
NkT
/V
P =
10/6.022 x 10
23
* 1.38 x 10

23
• 2.7
0.01
3
m
3
=
3.73 x 10

16
Pa
19.20
Diameter (H
2
) = 1.10
Å = 1.10 x 10

10
m
determine z, Z, and
l
.=========
If finished answer #21/22
Collisions between like molecules
–
Density

# molecules are available for collision (m

3
):
N/V = PN
A
/(RT)
relative
speed
–
effective collision speed (ms

1
):
<
v
rel
> = 4{RT/(
p
M
⥽
1/2
Area of molecular displacement (m
2
):
A
=
p
d
2
Collision frequency (s

1
):
z = <
v
rel
>
•
A•
(
N/V
)
= 4
p
1/2
PN
A
d
2
/{MRT}
1/2
Total collisions (s

1
m

3
):
Z = ½•z•
(
N/V
)
=
2
p
1/2
(
PN
A
d
)
2
/{M
1/2
(RT)
3/2
}
Mean free path (m):
l
=‼
re l
>/z
=
1
/{A
•
(
N/V
)}
= RT/{
p
d
2
PN
A
} =
kT
/(
p
d
2
P)
v
G(v)
v
G(v) • dv
0
0
100
0.00034
15
8.24E

05
200
1.13E

03
105
0.003766
300
1.85E

03
1005
0.000568
400
2.10E

03
5005
1.5E

69
500
1.84E

03
10005
4.9E

278
600
1.31E

03
700
0.00078
v
mp
395
800
0.00039
<v>
446
900
1.64E

04
v
rms
484
1000
6.00E

05
Ball
–
19.14 with speed distribution graph from 0 to 1000 m/s
0
0.0005
0.001
0.0015
0.002
0.0025
0
100
200
300
400
500
600
700
800
900
1000
G(v)
v (m/s)
Gas = Oxygen at 300K
MW = 0.032 kg/mol
G(v) =
4p
{M/(
2p
RT⥽
3/2
• v
2
• exp {

Mv
2
/(2RT)}
19
z
b
(b)
= A
• <
v
rel
> •
N
b
/V
3.8 x 10

20
238
1 x 10
7
<
v
rel
> = 4(RT/
p
M
½
=
4
• {8.314 • 2.7/(
p
• 0.002)}
½
= 238 m/s
Z
b
(b)
=
z
b
(b)
• ?
N
b
/V
• ½ = 4.53 x 10

4
m

3
l
=
<
v
rel
>
=
z
b
(b)
= ms

1
/s

1
= 1/(A
•
N
b
/V) =
RT/(
p
d
2
P
b
N
A
)
= 2.63 x 10
12
m
20
H2
MW
(kg/mol)
T (K)
d (m)
<v
rel
> (m/s)
A (m
2
)
z (s

1
)
N/V
l
=
(m
=
扯潫
=
Z
=
6
0.002016
2.7
1.10E

10
238
3.80E

20
9.05E

11
10000000
2.63E+12
2.63E+12
4.53E

04
19.20
A =
p
d
2
=
p
=
• (1.10 x 10

10
)
2
N/V = PN
A
/(RT
) =
3.73 x
10

16
• 6.022 x 10
23
/(8.314 • 2.7) =
Brownian
Motion
The
movement of a particle through a medium
l
=
⡭敡渠晲敥灡瑨⤠
==瑨攠a敲eg攠摩s瑡湣攠a=ml散畬攠will=瑲t敬=扥景牥
=
======
units (m) colliding with another ‘like’ molecule,
l
=
㴠
Y
rel
>
=
z
b
(b)
= ms

1
/s

1
= 1/(A
•
N
b
/V) =
RT/(
p
d
2
P
b
N
A
)
http://www.youtube.com/watch?v=ZAGloLXO9L0
Diffusion Demonstration
1D: (
D
x
===(2Dt
½
3D: (
D
x
===(6Dt
½
D
gas
= 3
p
/16 •
l
• <
v
rel
>
Transport
Properties
dY
/
dt
=

cst
• Area • gradient (
dy
/
dx
)
Diffusion
–
matter over [ ] gradient
J =

D dc/
dx
or
dn
i
/
dt
=

D
ik
A (
dc
i
/
dx
)
name
Property
Y
Gradient
dy
/
dx
equation
thermal conductivity
heat
q
temperature
T
dq
/
dt
=

k A (
dT
/
dx
)
diffusion
moles
n
i
concentration
n
i
dn
i
/
dt
=

D
ik
A (
dc
i
/
dx
)
viscosity
Momentum
p
x
Velocity
v
y
dp
x
/
dt
=

h
A (
dv
y
/
dx
)
Electrical
conductivity
Charge
Q
Potential
dQ/dt =

k
A (d
f
/dx)
A theoretical model is developed to predict the constant value
similar to barometric formula and effusion
Diffusion:
change of concentration with time
Dependent on concentration gradient and
diffusion coefficient.
[ ]
t
dn
i
/dt =

D
ik
A (dc
i
/dx)
Fick’s 1
st
Law
Fick’s
1st law:
dn
i
/
dt
=

D
ik
A (
dc
i
/
dx
)
Fick’s
1st law:
the rate of change in the
amount of solute
present at a position in a solution is
a function
of the area
observed, the
concentration gradient
, and the intrinsic ‘ease
of movement
’ of
the solute in the given solvent.
Fick’s
2
nd
law: (
dc
i
/
dt
)
x
=

D
ik
A (d
2
c
i
/dx
2
)
t
Fick’s
2nd
law:
the rate of change
in concentration with
change in time is a function of the diffusion coefficient of the
solvent, the area, and the 2
nd
derivative of
D
c
i
with respect to
position.
if r
i
> r
s
then D
is
= kT/(6
ph
s
r
i
)
if r
i
~ r
s
then D
is
= kT/(4
ph
s
r
i
)
Diffusion
–
matter over [ ] gradient
dn
i
/
dt
=

D
is
A (
dc
i
/
dx
)
In liquids ……
D
is
depends on the absolute and relative size of solute
(
i
) vs. solvent (s), and the viscosity of the solvent.
Gases : ~ 3 cm
2
/min
Liquids : ~ 0.03 cm
2
/min
solids : < 10

8
cm
2
/min
32
W
MW (kg/mol)
T (K)
P (torr)
P (Pa)
A (m
2
)
<v>/4
N/V
dN/dt (s

1
)
dN/dt (g/hr)
0.18385
4500
4.98
664
0.000001
180
1.07E+22
1.923E+18
2.113
dN
/
dt
= A • <v>/4 • N/V
Density

# molecules are available for collision (m

3
):
N/V = PN
A
/(RT)
relative speed
–
effective collision speed (ms

1
):
<v> = {8RT
/(
p
M
⥽
1/2
19.32
Tungsten effusion
–
MW = 0.18385 kg/
mol
Given: T = 4500 K

dN
/
dt
= 2.113 g/hour

A = 0.10 mm
2
.
find P
W
at 4500K?
Convert all units to SI and find <v>/4
Find N/V from effusion equation
Solve for P
–
which will represent Tungsten vapor pressure
For Friday do 19.28 and 19.31
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