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CHAPTER 12
GASES
2
The Gas Laws
Describe
HOW gases behave.
Can be predicted by the theory.
Amount of change can be calculated
with mathematical equations.
The effect of adding gas.
When
we blow up a balloon we are
adding gas molecules.
Doubling the the number of gas
particles doubles the pressure.
(of the same volume at the same
temperature).
Pressure and the number of
molecules are directly related
More
molecules means more
collisions.
Fewer molecules means fewer
collisions.
Gases naturally move from areas of
high pressure to low pressure
because there is empty space to
move in.
If
you double the number of
molecules
1 atm
If
you double the number of
molecules
You double the pressure.
2 atm
4 atm
As
you remove
molecules from a
container
2 atm
As
you remove
molecules from a
container the pressure
decreases
1 atm
As
you remove
molecules from a
container the pressure
decreases
Until the pressure inside
equals the pressure
outside
Molecules naturally
move from high to low
pressure
Changing the size of the
container
In
a smaller container molecules
have less room to move.
Hit the sides of the container more
often.
As volume decreases pressure
increases.
1 atm
As
the
pressure on
a gas
increases
4 Liters
As
2 atm
2 Liters
the
pressure on
a gas
increases
the volume
decreases
Pressure and
volume are
inversely
related
Temperature
Raising
the temperature of a gas
increases the pressure if the
volume is held constant.
The molecules hit the walls
harder.
The only way to increase the
temperature at constant pressure
is to increase the volume.
300 K
If
you start with 1 liter of gas at 1
atm pressure and 300 K
and heat it to 600 K one of 2 things
happens
600 K
300 K
Either
the volume will
increase to 2 liters at 1
atm
300 K
•Or the pressure will increase
to 2 atm.
•Or someplace in between
600 K
Kinetic Molecular Theory
Particles
•
•
•
•
•
in an ideal gas…
have no volume.
have elastic collisions.
are in constant, random, straight-line
motion.
don’t attract or repel each other.
have an avg. KE directly related to
Kelvin temperature.
18
Real Gases
Particles
in a REAL gas…
• have their own volume
• attract each other
Gas
behavior is most ideal…
• at low pressures
• at high temperatures
• in nonpolar atoms/molecules
19
Characteristics of Gases
Gases
expand to fill any container.
• random motion, no attraction
Gases
are fluids (like liquids).
• no attraction
Gases
have very low densities.
• no volume = lots of empty space
20
Characteristics of Gases
Gases
can be compressed.
• no volume = lots of empty space
Gases
undergo diffusion & effusion.
• random motion
21
Temperature
Always
use absolute temperature
(Kelvin) when working with gases.
ºF
-459
ºC
-273
K
0
C 59 F 32
32
212
0
100
273
373
K = ºC + 273
22
Pressure
force
pressure
area
Which shoes create the most pressure?
23
Barometer
Pressure
• measures atmospheric pressure
Aneroid Barometer
Mercury Barometer
24
Pressure
Manometer
• measures contained gas pressure
U-tube Manometer
Bourdon-tube gauge 25
Pressure
KEY
UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
N
kPa 2
m
14.7 psi
26
STP
STP
Standard Temperature & Pressure
0°C
1 atm
273 K
-OR-
101.325 kPa
27
Boyle’s Law
P
Volume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0
20.0
30.0
40.0
760.0
379.6
253.2
191.0
7.60 x 103
7.59 x 103
7.60 x 103
7.64 x 103
PV = k
V
28
Boyle’s Law
The
pressure and volume of
a gas are inversely related
• at constant mass & temp
P
PV = k
V
29
Charles’ Law
V
T
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0
44.0
47.7
51.3
273.2
298.2
323.2
348.2
0.146
0.148
0.148
0.147
V
k
T
30
Charles’ Law
The
volume and absolute
temperature (K) of a gas are
directly related
• at constant mass & pressure
V
T
V
k
T
31
Gay-Lussac’s Law
Temperature
(K)
Pressure
(torr)
P/T
(torr/K)
248
273
298
373
691.6
760.0
828.4
1,041.2
2.79
2.78
2.78
2.79
P
k
T
P
T
32
Gay-Lussac’s Law
The
pressure and absolute
temperature (K) of a gas are
directly related
• at constant mass & volume
P
k
T
P
T
33
Combined Gas Law
P
V
PV
PV = k
T
P1V1
P2V2
=
T1
T2
P1V1T2 = P2V2T1
34
Gas Law Problems
gas occupies 473 cm3 at 36°C. Find
its volume at 94°C.
A
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
35
Gas Law Problems
A
gas occupies 100. mL at 150. kPa.
Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
36
Gas
Law
Problems
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
T2 = 273 K
37
Gas Law Problems
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the pressure
be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P T
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1V1T2 = P2V2T1
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C
38
Your Turn
A
balloon is filled with 25 L of air at
1.0 atm pressure. If the pressure is
change to 1.5 atm what is the new
volume?
A balloon is filled with 73 L of air at
1.3 atm pressure. What pressure is
needed to change to volume to 43
L?
Your Turn
What
is the temperature of a gas that
is expanded from 2.5 L at 25ºC to
4.1L at constant pressure.
What is the final volume of a gas that
starts at 8.3 L and 17ºC and is heated
to 96ºC?
Your Turn
What
is the pressure inside a 0.250 L
can of deodorant that starts at 25ºC
and 1.2 atm if the temperature is
raised to 100ºC?
At what temperature will the can
above have a pressure of 2.2 atm?
Your Turn
A
15 L cylinder of gas at 4.8 atm
pressure at 25ºC is heated to 75ºC
and compressed to 17 atm. What is
the new volume?
If 6.2 L of gas at 723 mm Hg at 21ºC
is compressed to 2.2 L at 4117 mm
Hg, what is the temperature of the
gas?
Avogadro’s Principle
Equal
volumes of gases
contain equal numbers of
moles
• at constant temp & pressure
• true for any gas
V
k
n
V
n
43
Ideal Gas Law
Merge the Combined Gas Law with Avogadro’s Principle:
PV
V
k
=R
nT
T
n
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!
44
Ideal Gas Law
PV=nRT
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!
45
The Ideal Gas Law
Pressure
times Volume equals the
number of moles times the Ideal Gas
Constant (R) times the temperature
in Kelvin.
R does not depend on anything, it is
really constant
R = 0.0821 (L atm)/(mol K)
R = 62.4 (L mm Hg)/(K mol)
Ideal Gas Law Problems
Calculate
the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25
L.
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
47
Ideal Gas Law Problems
Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
WORK:
V=?
85 g 1 mol = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
48
3
V = 64 dm
Examples
How
many moles of air are there in a
2.0 L bottle at 19ºC and 747 mm Hg?
What is the pressure exerted by 1.8 g
of H2 gas exert in a 4.3 L balloon at
27ºC?
Dalton’s Law
The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + ...
When a H2 gas is
collected by water
displacement, the gas in
the collection bottle is
actually a mixture of H2
and water vapor.
50
Dalton’s Law
Hydrogen
gas is collected over water at
22.5°C. Find the pressure of the dry gas if
the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
for 22.5°C.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Sig Figs: Round to least number
of decimal places. 51
Dalton’s Law
A gas is collected over water at a temp of 35.0°C
when the barometric pressure is 742.0 torr. What is
the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Look up water-vapor pressure
for 35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
of decimal places. 52
Density
The
Molar mass of a gas can be
determined by the density of the gas.
D=
mass = m
Volume
V
Molar mass = mass =
m
Moles
n
n = PV
RT
Molar
Mass =
m
(PV/RT)
Molar mass = m RT
V P
Molar mass = DRT
P
You might need to review
Chapter 8
At STP
At
STP determining the amount of
gas required or produced is easy.
22.4 L = 1 mole
For example
How many liters of O2 at STP are
required to produce 20.3 g of
H2O?
Not At STP
Chemical
reactions happen in
MOLES.
If you know how much gas - change
it to moles
Use the Ideal Gas Law
n = PV/RT
If you want to find how much gas use moles to figure out volume
V = nRT/P
Example #1
HCl(g)
can be formed by the
following reaction
2NaCl(aq) + H2SO4 (aq)
2HCl(g) + Na2SO4(aq)
What mass of NaCl is needed to
produce 340 mL of HCl at 1.51 atm at
20ºC?
Example #2
2NaCl(aq)
+ H2SO4 (aq)
2HCl(g) + Na2SO4 (aq)
What volume of HCl gas at 25ºC and
715 mm Hg will be generated if 10.2 g
of NaCl react?
Ideal Gases don’t exist
Molecules
do take up space
There are attractive forces
otherwise there would be no liquids
Real Gases behave like Ideal
Gases
When
the molecules are
far apart
The molecules do not
take up as big a
percentage of the space
We can ignore their
volume.
This is at low pressure
Real Gases behave like Ideal
gases when
When
molecules are moving fast.
Collisions are harder and faster.
Molecules are not next to each other
very long.
Attractive forces can’t play a role.
Diffusion
Molecules
moving from areas of high
concentration to low concentration.
Perfume molecules spreading across
the room.
Effusion Gas escaping through a tiny
hole in a container.
Depends on the speed of the molecule.
Graham’s Law
The
rate of effusion and diffusion is
inversely proportional to the square
root of the molar mass of the
molecules.
Kinetic energy = 1/2 mv2
m is the mass v is the velocity.
Chem Express
Graham’s Law
bigger
molecules move slower at the
same temp. (by Square root)
Bigger molecules effuse and diffuse
slower
Helium effuses and diffuses faster
than air - escapes from balloon.
Graham’s Law
Diffusion
• Spreading of gas molecules
throughout a container until evenly
distributed.
Effusion
• Passing of gas molecules through a
tiny opening in a container
67
Graham’s Law
Speed
of diffusion/effusion
• Kinetic energy is determined by the
temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
– Larger m smaller v
KE =
2
½mv
68
Graham’s Law
Graham’s
Law
• Rate of diffusion of a gas is inversely
related to the square root of its molar
mass.
• The equation shows the ratio of Gas A’s
speed to Gas B’s speed.
vA
vB
mB
mA
69
Graham’s Law
Determine
the relative rate of diffusion for
krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
vA
vB
v Kr
v Br2
m Br2
m Kr
mB
mA
159.80g/mol
1.381
83.80g/mol
Kr diffuses 1.381 times faster than Br702.
Graham’s Law
A molecule of oxygen gas has an average speed of
12.3 m/s at a given temp and pressure. What is the
average speed of hydrogen molecules at the same
conditions?
vA
vB
mB
mA
vH 2
12.3 m/s
32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2
mO2
mH 2
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
3.980
vH2 49.0m/s
71
Graham’s Law
An unknown gas diffuses 4.0 times faster than O2.
Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA
vB
vA
v O2
mB
mA
mO2
mA
Square both
sides to get rid
of the square
root sign.
32.00 g/mol
g/mol
32.00
4.0
m
A
A
32.00 g/mol
16
mA
32.00 g/mol
mA
2.0 g/mol
72
16
2
Gas Stoichiometry
Liters of a Gas:
• STP - use 22.4 L/mol
• Non-STP - use ideal gas law
Moles
Non-STP
• Given liters of gas?
– start with ideal gas law
• Looking for liters of gas?
– start with stoichiometry conv.
73
Gas Stoichiometry Problem
What
volume of CO2 forms from 5.25 g of
CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g
CaO
+
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g
CaCO3
1 mol
CaCO3
1 mol
CO2
100.09g
CaCO3
1 mol
CaCO3
CO2
?L
non-STP
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.
74
Gas Stoichiometry Problem
What
volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 1.26 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
75
Gas Stoichiometry Problem
How many grams of Al2O3 are formed from 15.0
L of O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2
15.0 L
non-STP
2 Al2O3
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT
n = 0.597 mol O2
76
Gas Stoichiometry Problem
How
many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al
+
Use stoich to convert moles
of O2 to grams Al2O3.
0.597
mol O2
3 O2
15.0L
non-STP
2 mol
Al2O3
101.96 g
Al2O3
3 mol O2
1 mol
Al2O3
2 Al2O3
?g
= 40.6 g Al2O3
77