Chap10_Part2

Chap10_Part2 - GAS
LAWS‐2


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Unformatted text preview: GAS
LAWS‐2
 Measuring
Gas
Pressure
with
a
Manometer
 A
gas
(purple)
is
in
equilibrium

 with
atmospheric
pressure
(blue)

 through
a
glass
tube.


 In
the
tube
is
a
column
of

 mercury
(gray).


 The
height
difference
reports

 the
gas
pressure.
 For
a
given
gas
in
the
lab,
 the
height
difference
is
 h

=
8.9
cm.

What
is
the

 gas
pressure
in
units
of
atm?
 For
a
given
gas
in
the
lab,
 the
height
difference
is
 h

=
8.9
cm.

What
is
the

 gas
pressure
in
units
of
atm?
 10 mm Pgas = 1 atm + 8.9 cm( ) cm atm = 1 atm + 89 mm( ) 760 mm = 1 atm + 0.12 atm = 1.12 atm Blood
Pressure
 The
heart
pumps
blood
to
parts
of
the
 
body
through
the
arteries.
 The
blood
is
returned
to
the
heart
through

 the
veins.
 Blood
pressure
is
reported
as
two

 numbers
(e.g.,
120/80)
in
torr.
 The
first
number
is
the
systolic
pressure,

 the
maximum
pressure
when
the
heart
 
is
pumping.
 The
second
number
is
the
diastolic
pressure,

 the
pressure
when
the
heart
is
in
its
resMng
phase.
 Courtesy:
Web
search

 For
a
person
with
blood
pressure
of
130/90,
what
are
the
equivalents
 in
terms
of
atm?
 For
a
person
with
blood
pressure
of
130/90,
what
are
the
equivalents
 in
terms
of
atm?
 atm 130 torr ( ) = 0.17 atm 760 torr atm 90 torr ( ) = 0.12 atm 760 torr These
values
are
about
equal
to
the
pressure
exerted
by
the
earth
 on
the
human
body,
but
less
than
the
pressure
exerted
by
the

 atmosphere.
 The
Gas
Laws
 In
the
simplest
case,
four
variables
are
needed
to
describe
 the
condiMon,
or
state,
of
a
gas:
 Temperature
(T),
pressure
(P),
volume
(V)
&
amount
of
material
 The
amount
of
material
is
usually
expressed
in
terms
of
mols
(n).
 The
equaMons
that
relate
these
variables
are
called
the
gas
laws.
 Rela=onship
Between
Gas
Pressure
and
Volume
 Boyle’s
Law:

 “The
volume
of
a
fixed
quanMty
of
gas
maintained
at
constant
 temperature
is
inversely
proporMonal
to
the
pressure”.
 PV = C1 volume
 C1 P= V pressure
 C1 V= P The
value
of
the
constant
C1
depends
on
the
temperature
(T)

 and
the
amount
of
material
in
the
gas
(n).
We
will
find
the

 value
of
C1
later.


 Note
that,
for
two
different
condiMons
with
constant
T
and
n,
 P1V1 = C1 & P2V2 = C1 ⇒ P1V1 = P2V2 Higher
P
means
lower
V
and
higher
V
means
lower
P.
 At
constant
T
and
n,
a
plot
of
V
vs.
P
has
a
curved
shaped
where
 V
decreases
with
increasing
P.


A
plot
of
V
vs.
1/P
is
linear.
 Boyle’s
Law:
Experimental
Demo

 (a)
The
difference
in
height
of
the
 mercury
is
zero,
implying
that
the
 gas
pressure
is
1
atm.
 (b)
Enough
mercury
is
added
to
the
 tube
so
that
the
height
difference
is
 760
mm
=
1
atm.

Thus,
the
pressure
 on
the
gas
is
2
atm.

Boyle
found
that
 the
volume
of
the
gas
was
reduced
 by
a
factor
of
two.
 (c)
The
result
that
the
volume
of
the
 gas
was
inversely
proporMonal
to
the
 pressure
on
the
gas
was
confirmed
for
 numerous
mercury
heights.
 (d)
Note
that
this
experiment
depends
 on
mercury
being
more
dense
than
gas.
 Boyle’s
Law:

 Doubling
the
pressure
reduces
the
volume
by
half.

 Conversely,
when
thevolume
doubles,
the
pressure
 decreases
by
half.
 h]p://exploraMon.grc.nasa.gov/educaMon/rocket/AnimaMon/frglab2.html
 Robert
Boyle
(1627‐1691)
 (images
from
Google)
 13
 (a)  A
gas
has
a
volume
of
600
mL
at
a
pressure
of
2
atm.

The
pressure
is

 increased
to
6
atm
with
constant
temperature.

What
is
the
final
volume?
 (b)
A
gas
at
5
atm
pressure
has
a
volume
of
50
mL.
The
volume
is

 increased
to
150
mL
with
constant
temperature.

What
is
the
final
P?
 14
 (a)  A
gas
has
a
volume
of
600
mL
at
a
pressure
of
2
atm.

The
pressure
is

 increased
to
6
atm
with
constant
temperature.

What
is
the
final
volume?
 P2V2 = P1V1 P1 2 atm ⇒ V2 = V1 = ( )(600 mL) = 200 mL < 600 mL P2 6 atm (b)
A
gas
at
5
atm
pressure
has
a
volume
of
50
mL.
The
volume
is

 increased
to
150
mL
with
constant
temperature.

What
is
the
final
P?
 15
 (a)  A
gas
has
a
volume
of
600
mL
at
a
pressure
of
2
atm.

The
pressure
is

 increased
to
6
atm
with
constant
temperature.

What
is
the
final
volume?
 P2V2 = P1V1 P1 2 atm ⇒ V2 = V1 = ( )(600 mL) = 200 mL < 600 mL P2 6 atm (b)
A
gas
at
5
atm
pressure
has
a
volume
of
50
mL.
The
volume
is

 increased
to
150
mL
with
constant
temperature.

What
is
the
final
P?
 P2V2 = P1V1 V1 50 mL ⇒ P2 = P1 = ( )(5 atm) = 1.7 atm < 5 atm V2 150 mL 16
 Rela=onship
Between
Gas
Temperature
and
Volume
 Charles
Law:

 “The
volume
of
a
fixed
quanMty
of
gas
maintained
at
constant
 pressure
is
proporMonal
to
its
absolute
temperature”.
 V = C 2T V T= C2 V = C2 T The
value
of
the
constant
C2
depends
on
the
pressure
(P)

 and
the
amount
of
material
in
the
gas
(n).
We
will
find
the

 value
of
C2
later.


 Note
that,
for
two
different
condiMons
with
constant
P
and
n,
 V1 V2 ⇒ = T1 T2 temperature
 V2 & = C2 T2 volume
 V1 = C2 T1 Higher
T
means
higher
V
and
higher
V
means
higher
T.
 17
 Charles
Law:

 As
the
temperature
increases,
the
volume
increases.

 Conversely,
when
the
temperature
decreases,
volume
 decreases.
 18
 Effect
of
Temperature
on
Volume
 As
liquid
nitrogen
(‐196oC
=
77
K)
is
poured
over
a
 balloon,
the
volume
decreases.
 19
 Important:

Note
that
T
must
be
expressed
in
units
of
absolute
temperature
so
that

 the
intercept
is
(0,0).

For
example,
37.0oC
=
(273.1
+
37.0)
K
=
310.1
K.
 The
dashed
line
is
an
extrapolaMon
at
which
the
material
is
no
longer
a
gas.
 20
 Jacques
Charles
(1746‐1823)
 (images
from
Google)
 21
 (a)  A
gas
has
a
volume
of
600
mL
at
a
temperature
of
300
K.
The
temperature
 is
decreased
to
250
K
with
constant

pressure.
What
is
the
final
volume?
 (b)
A
gas
has
a
volume
of
20
mL
at
a
temperature
of
10oC.
The
temperature
 is
increased
to
20oC
with
constant

pressure.
What
is
the
final
volume?
 22
 (a)  A
gas
has
a
volume
of
600
mL
at
a
temperature
of
300
K.
The
temperature
 is
decreased
to
250
K
with
constant

pressure.
What
is
the
final
volume?
 V2 V1 = T2 T1 T2 250 K ⇒ V2 = V1 = ( )(600 mL) = 500 mL < 600 mL T1 300 K (b)
A
gas
has
a
volume
of
20
mL
at
a
temperature
of
10oC.
The
temperature
 is
increased
to
20oC
with
constant

pressure.
What
is
the
final
volume?
 23
 (a)  A
gas
has
a
volume
of
600
mL
at
a
temperature
of
300
K.
The
temperature
 is
decreased
to
250
K
with
constant

pressure.
What
is
the
final
volume?
 V2 V1 = T2 T1 T2 250 K ⇒ V2 = V1 = ( )(600 mL) = 500 mL < 600 mL T1 300 K (b)
A
gas
has
a
volume
of
20
mL
at
a
temperature
of
10oC.
The
temperature
 is
increased
to
20oC
with
constant

pressure.
What
is
the
final
volume?
 T2 273.1 + 20 K V2 = V1 = ( )(20 mL) = 20.7 mL > 20 mL T1 273.1 + 10 K 24
 Rela=onship
Between
Amount
of
Gas
and
Volume
 Avogadro’s
Law:

 “The
volume
of
a
gas
maintained
at
constant
temperature
and
 pressure
is
proporMonal
to
the
number
of
moles
of
the
gas”.
 V = C3 n V n= C3 V = C3 n The
value
of
the
constant
C3
depends
on
the
pressure
(P)

 and
the
temperature
(T).
We
will
find
the
value
of
C3
later.


 Note
that,
for
two
different
condiMons
with
constant
P
and
T,
 V1 = C3 n1 V2 & = C2 n2 V1 V2 ⇒ = n1 n2 Higher
n
means
higher
V
and
higher
V
means
higher
n.
 25
 Avagadro’s
Law:

 As
the
number
of
moles
of
a
gas
increases,
the
volume
 increases,
at
constant
P
and
T
 26
 The
Law
of
Combining
Volumes
 At
a
given
temperature
and
pressure,
the
volumes
of
gases
 that
react
with
one
another
are
in
the
raMos
of
small
whole
 numbers
[Joseph
Louis
Gay‐Lussac].
 27
 Joseph
Louis
Gay‐Lussac
(1778‐1850)
 (images
from
Google)
 28
 Avogadro’s
Hypothesis
 “Equal
volumes
of
gases
at
the
same
temperature
and
pressure
 contain
equal
numbers
of
molecules.”
(or
atoms)
 22.4
L
of
any
gas
at
0oC
and
1
atm
contain
6.02
x
1023
gas
 molecules
or
atoms
(i.e.,
1
mol).

Note
that
the
masses
differ.
 29
 Amedeo
Avogadro
(1776‐1856)
 (images
from
Google)
 30
 Exercise:

Nitrogen
and
hydrogen
gases
react
to
form
ammonia
gas.
 
 
N2(g)
+
3
H2(g)
→
2
NH3(g)
 At
a
certain
temperature
and
pressure,
1.2
L
of
N2
reacts
with
3.6
L
of
H2.
 What
volume
of
NH3
is
produced?
 31
 Exercise:

Nitrogen
and
hydrogen
gases
react
to
form
ammonia
gas.
 
 
N2(g)
+
3
H2(g)
→
2
NH3(g)
 At
a
certain
temperature
and
pressure,
1.2
L
of
N2
reacts
with
3.6
L
of
H2.
 What
volume
of
NH3
is
produced?
 nN
=
number
of
mols
of
N2. nH
=
number
of
mols
of
H2. nf
=
number
of
mols
of
NH3. Vf VN ⇒ ⇒ = nf nN Vf VN Vf VH & Vf VH = nf nH 
VN
=
volume
of
N2
=
1.2
L.


 
VH
=
volume
of
H2
=
3.6
L.

 
Vf
=
volume
of
NH3.
 & n f = 2n N & 3n f = 2n H = 2 ⇒ V f = 2V N = 2(1.2 L) = 2.4 L = 2 2 2 ⇒ V f = VH = (3.6 L) = 2.4 L 3 3 3 32
 The
Ideal
Gas
Law
 Boyle’s
Law: 
 
V
is
proporMonal
to
1/P

(constant
n,T)
 Charles’s
Law: 







V
is
proporMonal
to
T





(constant
n,P)
 Avogadro’s
Law: 
V
is
proporMonal
to
n





(constant
P,T)
 Thus
 nT V P Define
the
proporMonality
constant
as
R.

Then
 V = R( nT ) P ⇒ PV = nRT ideal
gas
law
 The
temperature
is
always
expressed
as
the
absolute
temperature
(K).
 33
 The
Gas
Constant
R
 Which
constant
is
used
depends
on
the
units
of
P,
V,
n
&
T.
 34
 Sample
Calcula=ons
for
the
Ideal
Gas
Law
and
R
 (a)  What
is
the
volume
of
2.5
mol
of
gas
at
3
atm
pressureand
 10oC?
 (b)
How
many
mols
of
gas
are
contained
in
1.4
L
when
the
 pressure
is
3.1
bars
and
the
temperature
is
250
K?
 35
 Sample
Calcula=ons
for
the
Ideal
Gas
Law
and
R
 (a)  What
is
the
volume
of
2.5
mol
of
gas
at
3
atm
pressure
and
 10oC?
 L − atm (2.5 mol )(0.08206 )(283.1 K ) nRT mol − K V= = = 19.4 L P 3 atm (b)
How
many
mols
of
gas
are
contained
in
1.4
L
when
the
 pressure
is
3.1
bars
and
the
temperature
is
250
K?
 36
 Sample
Calcula=ons
for
the
Ideal
Gas
Law
and
R
 (a)  What
is
the
volume
of
2.5
mol
of
gas
at
3
atm
pressure
and
 10oC?
 L − atm (2.5 mol )(0.08206 )(283.1 K ) nRT mol − K V= = = 19.4 L P 3 atm (b)
How
many
mols
of
gas
are
contained
in
1.4
L
when
the
 pressure
is
3.1
bars
and
the
temperature
is
250
K?
 PV (3.1)(1.4) bar − L 10 5 Pa kg / m − sec 2 10 3 cm 3 m n= = ( )( )( )( )( )3 RT (250)( R) K bar Pa L 100 cm 1.736 J / K 1.736 kg − m 2 / sec 2 J J /K n= ( )( ) = 1.736 = = 0.21 mol 2 2 R K R 8.314 J / K − mol kg − m / sec 37
 The
Ideal
Gas
Law
is
Not
Exact
 According
to
the
ideal
gas
law,
one
mol
of
gas
at
standard

 temperature
and
pressure
(0oC
and
1
atm)
occupies
22.41
L.
 38
 The
Constants
C1,
C2
&
C3
 Boyle’s
Law: 
 
 
 
V
=
C1/P
 
(constant
n,T)
 
then



C1
=
nRT
 Charles’s
Law: 
V
=
C2T
(constant
n,P)
 
 
 
then



C2
=
nR/P
 Avogadro’s
Law: 
V
=
C3n
(constant
P,T)
 
 
 
then



C3
=
RT/P
 Gas
Densi=es
 mass density = volume m d= V mass = ( molar mass)(# mol) ⇒ n P d = M( ) = M( ) V RT gm (e.g., ) liter m = Mn ⇒ mass ( recall M = ) mol MP d= RT 40
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