Theories applied to Chemistry

Mechanics

Oct 29, 2013 (4 years and 6 months ago)

107 views

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
=

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

=

P

dP/P =
-
Mg/RT

0
z

dz

ln (P/P

=

-
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

=

k/k
o

&
D

(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
/

=
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

Pikes Peak

4302

C

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

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

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===

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
/

=
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

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===

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

3
/3

y

Scalar
v

represents

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

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
) =

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
)

=

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
.=========

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
=

=
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=ml散畬攠will=瑲t敬=扥景牥
=
======
units (m) colliding with another ‘like’ molecule,

l
=

Y

rel
>

=
z
b
(b)

= ms
-
1
/s
-
1

= 1/(A

N
b
/V) =
RT/(
p
d
2
P
b
N
A
)

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

dy
/
dx
)

Diffusion

J =
-
D dc/
dx

or
dn
i
/
dt

=
-
D
ik

A (
dc
i
/
dx
)

name

Property

Y

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

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
, 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

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