Chap10-Part4

Chap10-Part4 - LAST
CLASS…….
 (a) 

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Unformatted text preview: LAST
CLASS…….
 (a)  Calculate
the
average
kine2c
energy
of
a
gas
molecule
 at
25oC
in
units
of
Joules.
 3RT (3)(8.314 J / K − mol )(298.1 K ) J − 21 ε= = = 6.2 x10 23 2N molecule 2(6.02 x10 molecules / mol ) (b)
Calculate
the
rms
speed
for
H2
and
for
O2
at
25oC
in
m/s.
 u= 3RT M (3)(8.314 J / K − mol )(298.1 K ) kg − m 2 / s 2 10 3 gm m u(H 2 ) = [[[ = 1,900 2 gm / mol J kg s (3)(8.314 J / K − mol )(298.1 K ) kg − m 2 / s 2 10 3 gm m u (O2 ) = [[[ = 500 32 gm / mol J kg s 2 Graham’s
Law
of
Effusion
 Effusion
is
the
process
in
which
gas
molecules
 escape
through
a
small
hole
into
an
evacuated
 space.
 Graham’s
Law
states
that,
for
two
gases
at
equal
 temperatures
and
pressures,
the
relaBve
rates
of

 effusion
(r)
are
inversely
proporBonal
to
the
raBo
 of
the
square
root
of
their
molecular
masses.
 The
faster
the
molecules
are
moving,
the
more
 likely
that
they
will
encounter
the
hole
and
escape.

 r1 u1 3RT / M 1 = = r2 u 2 3RT / M 2 ⇒ r1 = r2 M2 M1 The
2me
required
for
25‐mL
samples
 of
different
gasses
to
diffuse
through
a

 pinhole
into
a
vacuum.
 Diffusion
‐
The
rate
at
which
two
gases
mix.

 Effusion
‐
The
rate
at
which
a
gas
escapes
through
a
pinhole
into
a
 vacuum.

 The
KineBc
Molecular
Theory
&Graham's
Law
 ‐ 
Since
KEavg
is
dependent
only
upon
T,
two
different
gases
at
the



 

same
temperature
must
have
the
same
KEavg.
 ½
mHVH2
=
½
mOVO2
 mHVH2
=
mOVO2
 mO
 mH =
 
 mO
 mH 
 =
 VH2
 VO2
 V H
 V O
 If
equal
amounts
of
helium
and
argon
are
placed
in
a
porous
container

 and
allowed
to
escape,
which
gas
will
escape
faster
and
how
much
faster?
 If
equal
amounts
of
helium
and
argon
are
placed
in
a
porous
container

 and
allowed
to
escape,
which
gas
will
escape
faster
and
how
much
faster?
 Which
gas
escapes
fast:



Helium

(Low
Molecular
Weight)
 If
equal
amounts
of
helium
and
argon
are
placed
in
a
porous
container

 and
allowed
to
escape,
which
gas
will
escape
faster
and
how
much
faster?
 Which
gas
escapes
fast:



Helium

(Low
Molecular
Weight)
 How
much
Faster?
 Let
rate
of
effusion
of
Helium
=

rHe


and
Argon
=
rAr

 Molecular
weight
of
Helium
=
MHe=
4.0,

Argon=
MAr=
39.95
g /mol

 If
equal
amounts
of
helium
and
argon
are
placed
in
a
porous
container

 and
allowed
to
escape,
which
gas
will
escape
faster
and
how
much
faster?
 Which
gas
escapes
fast:



Helium

(Low
Molecular
Weight)
 How
much
Faster?
 Let
rate
of
effusion
of
Helium
=

rHe


and
Argon
=
rAr

 Molecular
weight
of
Helium
=
MHe=
4.0,

Argon=
MAr=
39.95
g /mol
 We
know,

 mAr
 mHe
 mAr
 mHe
 =
 rHe
 rAr
 =
 VHe
 VAr
 Velocity
and
rate
are
referring
the
same
 39.95
 rHe
 =
 
 4 rAr
 rHe
=
3.16
rAr
 Thomas
Graham
(1805‐1869)
 
‐
Law
of
Diffusion
 ‐

Study
on
‘Dialysis’
 ‐
Founder
of
Colloid
Chemistry
 (images
from
Google)
 Diffusion
and
Mean
Free
Path
 The
average
speeds
of
molecules
 in
gases
are
very
high
(500
m/s).
 However,
the
molecules
do
not
travel
 very
far
on
the
average
because
they
 are
con2nually
colliding
with
other
 molecules
(about
1010
per
sec
for
 each
molecule).
 The
average
distance
traveled
between
 collisions
is
called
the
mean
free
path,
 and
depends
on
the
density
of
molecules.
 The
path
of
the
molecule
begins
at

 At
sea
level,
the
mean
free
path
is
about
 the
dot.

Each
short
segment
shows
 600
A
=
60
nm

=
6
x
10‐6
cm
=
6
x
10‐8
m.
 travel
between
collisions.

The
blue
 arrow
shows
the
net
distance

 At
100
km,
where
the
air
density
is
much

 traveled
by
the
molecule.
 lower,
the
mean
free
path
is
about
10
cm.
 DeviaBon
from
Ideal
Behavior
with
Pressure
 Real
Gases:

 PV =n RT Ideal
gas
 Real
gases
do
not
behave
according
to
the
ideal
gas
law
at
high
pressure
(above
 about
10
atm
at
room
temperature).

This
plot
shows
the
ra2o
PV/RT
for
1
mol
of
 gas
as
a
func2on
of
pressure.

The
ra2o
equals
1
only
at
low
pressures
(and
a
 given
single
intermediate
pressure).

Data
are
for
300
K
except
for
CO2
which
is
 shown
for
313
K
because
it
liquifies
under
high
pressure
at
300
K.
 DeviaBon
from
Ideal
Behavior
with
Temperature
 Real
Gases:

 PV =n RT Ideal
gas
 Real
gases
do
not
behave
according
to
the
ideal
gas
law
at
lower
temperatures
 where
the
gas
approaches
the
phase
transi2on
to
liquid.
This
plot
shows
the
ra2o
 PV/RT
for
1
mol
of
N2
gas
as
a
func2on
of
pressure
and
at
different
temperatures.


 Why
the
DeviaBons
from
Ideal
Behavior?
 1)
Gas
molecules
are
not
point
 par2cles.

They
have
finite
volumes.
 In
(a),
at
low
pressures,
the
combined
 volume
of
the
gas
molecules
is
much
 smaller
than
the
container
volume.
 The
empty
space
between
the
molecules
 is
about
equal
to
the
container
volume.
 In
(b),
at
higher
pressures,
the
combined
 volume
of
the
gas
molecules
is
not
 negligible
and
must
be
accounted
for.
 Why
the
DeviaBons
from
Ideal
Behavior?
 2)
Gas
molecules
do
not
act
 completely
independently
of
each

 other.

They
experience
short‐range
 afrac2ve
forces.
 The
molecule
that
is
about
to
hit
the
 wall
experiences
afrac2ve
forces

 from
nearby
gas
molecules,
its
 impact
on
the
wall
is
lessened,
and

 the
pressure
is
lowered.
 The
afrac2ve
forces
are
significant

 only
at
high
densi2es
where
the
 molecules
are
close
together.
 The
van
der
Waals
EquaBon
for
Real
Gases
 1)
 2)
 nRT n2a P= −2 V − nb V 1)
This
term
represents
the
correc2ons
for
volume
effects.

The
effec2ve
 volume
is
reduced
by
nb,
where
n
is
the
number
of
mols
and
‘b’
is
a

 Constant.
The
constant
b
has
units
of
volume/mol,
depends
on
the

 iden2ty
of
the
gas
molecules,
and
represents
their
finite
volume.
 2)
This
term
describes
intermolecular
interac2ons.

The
afrac2ve
 
interac2ons
increase
as
the
square
of
the
number
of
molecules
per

 volume,
(n/V)2.
Constant
“a”
is
a
propor2onality
constant
which
depends
 
on
the
iden2ty
of
the
gas
molecules
and
has
units
of
 pressure(volume/mol)2.
 Larger
molecules
tend
to
have
higher
values
of
the
constant
b
because
of
 their
larger
volumes.

Larger
molecules
also
tend
to
have
higher
values
of
 the
constant
a
because
they
are
more
likely
to
interact.
 When
constants
a
&
b
equal
zero,
 [ P ] a = 0 ,b = 0 nRT n2a nRT n 2 (0) nRT =[ − 2 ] a = 0 ,b = 0 = − = 2 V − nb V V − n(0) V V nRT ⇒ P= V The
van
der
Waals
equa2on
is
oken
rewrifen
as
 n2a ( P + 2 )(V − nb) = nRT V Why
does
PV/RT
first
decrease
with
pressure
and
then
increase?
 nRT n2a P= −2 V − nb V ⇒ PV nV n2a = − RT V − nb VRT At
low
pressures,
the
second
term,
describing
the
reduc2on
in
pressure
 from
afrac2ve
forces,
is
dominant.
 At
higher
pressures,
the
first
term,
describing
the
decreased
effec2ve
 volume,
is
dominant.
 Johannes
Van
der
Waals
(1837‐1923)
 ‐ Proposed
‘intermolecular
forces’
first
2me
 ‐ 
‘Equa2on
of
State’:

modifica2on
of
ideal
gas
 


law
and
approximate
behavior
of

real
fluids
 


…
NOBEL
PRIZE
in
1910
in
Physics
 ‐
Applica2ons:
‘Liquefac2on’
of
Hydrogen
by

 

Dewar
and

Helium
by
K’Onnes
 (images
from
Google)
 Problems
in
Gas
laws?
 We
will
discuss
them
in
Review
Session
on
Thursday
6‐7:30
pm
 I
will
post
the
review
session
lecture
notes
in
Bb
 if
you
can
not
come
to
it!


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This note was uploaded on 09/13/2011 for the course CHEM 101 taught by Professor Farahh during the Spring '02 term at UNC.

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