Chapter 11 Lecture

```Chapter 11
Gases
2006, Prentice Hall
CHAPTER OUTLINE
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Properties of Gases
Pressure & Its Measurement
The Gas Laws
Vapor Pressure & Boiling Point
Combined Gas Law
STP & Molar Volume
Ideal Gas Law
Partial Pressures
2
PROPERTIES
OF GASES
 Gases occupy
are the much
least dense
greater
and
space
mostthan
mobile
the of
same
the three
amount
phases
of liquid
of matter.
or solid.
 This
Particles
is because
of matter
the gas
in the
particles
gas phase
are are
spaced
spaced
apart
far apart
fromfrom
one one
another
another
and and
are therefore
move rapidly
compressible.
and collide with each other often.
 Solid or liquid particles are spaced much
closer and cannot be compressed.
3
PROPERTIES
OF GASES
 Gases are characterized by four properties.
 These are:
1. Pressure (P)
2. Volume
(V)
3. Temperature (T)
4. Amount
(n)
4
KINETIC-MOLECULAR
THEORY
 The
Scientists
KMT use
consists
the kinetic-molecular
of several postulates:
theory
(KMT) to describe the behavior of gases.
1. Gases consist of small particles (atoms or molecules)
that move randomly with rapid velocities.
2. Gas particles have little attraction for one another.
Therefore attractive forces between gas molecules
can be ignored.
3. The distance between the particles is large compared
to their size. Therefore the volume occupied by gas
molecules is small compared to the volume of the gas.
5
KINETIC-MOLECULAR
THEORY
4. Gas particles move in straight lines and collide with
each other and the container frequently. The force of
collision of the gas particles with the walls of the
container causes pressure.
5. The average kinetic energy of gas molecules is
directly proportional to the absolute temperature
(Kelvin).
Animation
6
Kinetic Molecular Theory
Tro's Introductory Chemistry,
Chapter
7
PRESSURE &
ITS MEASUREMENT
 Pressure is the result of collision of gas particles
with the sides of the container. Pressure is defined
as the force per unit area.
 Pressure is measured in units of atmosphere (atm)
or mmHg or torr. The SI unit of pressure is
pascal (Pa) or kilopascal (kPa).
1 atm = 760 mmHg = 101.325 kPa
1 mmHg = 1 torr
8
Common Units of Pressure
Unit
Average Air Pressure at
Sea Level
pascal (Pa)
101,325
kilopascal (kPa)
101.325
atmosphere (atm)
1 (exactly)
millimeters of mercury (mmHg)
inches of mercury (inHg)
torr (torr)
760 (exactly)
29.92
760 (exactly)
pounds per square inch (psi, lbs./in2)
Tro's Introductory Chemistry,
Chapter
14.7
9
PRESSURE &
ITS MEASUREMENT
 Atmospheric pressure can
be measured with the use
of a barometer.
 Mercury is used in a
barometer due to its high
density.
 At sea level, the mercury
stands at 760 mm above its
base.
10
Atmospheric Pressure & Altitude
• the higher up in the atmosphere you go,
the lower the atmospheric pressure is
around you
at the surface the atmospheric pressure is
14.7 psi, but at 10,000 ft is is only 10.0 psi
• rapid changes in atmospheric pressure
may cause your ears to “pop” due to an
imbalance in pressure on either side of
Tro's Introductory Chemistry,
Chapter
11
Pressure Imbalance in Ear
If there is a difference
in pressure across
the eardrum membrane,
the membrane will be
pushed out – what we
commonly call a
“popped eardrum.”
Tro's Introductory Chemistry,
Chapter
12
Example 1:
The atmospheric pressure at Walnut, CA is 740. mmHg.
Calculate this pressure in torr and atm.
1 mmHg = 1 torr
740. mmHg = 740. torr
740. mmHg x
1
atm
= 0.974 atm
760 mmHg
13
Example 2:
The barometer at a location reads 1.12 atm. Calculate
the pressure in mmHg and torr.
1.12 atm x
760 mmHg
= 851 mmHg
atm
1
1 mmHg = 1 torr
851 mmHg = 851 torr
14
PRESSURE &
MOLES OF A GAS
 The pressure of a gas is directly proportional to
the number of particles (moles) present.
The greater the moles of the gas, the greater the pressure
15
BOYLE’S LAW
 At
Theconstant
relationship
temperature,
of pressure
the and
volume
volume
of a in
fixed
amount
gases is called
of gas Boyle’s
is inversely
Law.proportional to its
pressure.
As pressure
Pressure
and
increases,
volume
of athe
volume
gas areof the
gas
decreases
inversely
proportional
16
BOYLE’S LAW
 Boyle’s Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V
under
initial
condition
P and V
under
final
condition
17
Example 1:
A sample of gas has a volume of 12-L and a pressure of
4500 mmHg. What is the volume of the gas when the
Pressure
pressure is reduced to 750 mmHg?
Volume
decreases
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 = ???
increases
P1 x V1 = P2 x V2
P1V1
(4500 mmHg)(12 L)
V2 =
=
P2
(750 mmHg)
V2 = 72 L
18
Example 2:
A sample of hydrogen gas occupies 4.0 L at 650 mmHg.
What volume would it occupy
at 2.0 atm?
Pressure
P1= 650 mmHg
P2= 2.0 atm
V1 = 4.0 L
V2 = ???
Volume
Pincreases
=
2.0
atm
=
1520
mmHg
2
decreases
P1 x V1 = P2 x V2
P1V1
(650 mmHg)(4.0 L)
V2 =
=
P2
(1520 mmHg)
Pressures
must be in
the sameV2 = 1.7 L
unit
19
CHARLES’S LAW
 At
Theconstant
relationship
pressure,
of temperature
the volume and
of a volume
fixed in
amount
gases is called
of gas Charles’
is directlyLaw.
proportional to its
absolute temperature.
Temperature
and
As temperature
volume
of a gas
increases,
the
are
directly
volume
of the
proportional
gas increases
20
CHARLES’S LAW
 Charles’ Law can be mathematically expressed as
V1 V2
=
T1 T2
V and T
under
T must be in
initial
units of K
condition
V and T
under final
condition
21
Example 1:
A 2.0-L sample of a gas is cooled from 298K to 278 K, at
constant pressure. What is the new volume of the gas?
Temperature
Volume
decreasesV1 V2
decreases
=
V1 = 2.0 L
V2 = ???
T1 = 298 K
T2 = 278 K
T1
T2
T2
V2 = V1 x = 2.0 L x 278 K
T1
298 K
V2 = 1.9 L
22
Example 2:
If 20.0 L of oxygen is cooled from 100ºC to 0ºC, what is
the new volume?
Temperature
s must
be in + 273 = 373
Volume
Temperature
T
=
100ºC
K
V1 = 20.0 L
1
K
decreases
decreases
V2 = ???
T1 = 100ºC
T2 = 0ºC
T2= 0ºC + 273 = 273 K
T2
273 K
V2 = V1 x
= 20.0 L x
T1
373 K
V2 = 14.6 L
23
GAY-LUSSAC’S
LAW

 The
At constant
relationship
volume,
of temperature
the pressureand
of apressure
fixed
in
amount
gases is
ofcalled
gas is Gay-Lussac’s
directly proportional
Law. to its
absolute temperature.
As temperature
Temperature
and
increases,
the
pressure
of a gas
pressure
of the
are
directly
gas increases
proportional
24
GAY-LUSSAC’S
LAW
 Gay-Lussac’s Law can be mathematically
expressed as
P1 P2
=
T1 T2
P and T
under
initial
T must be in
condition
units of K
P and T
under final
condition
25
Example 1:
An aerosol spray can has a pressure of 4.0 atm at 25C.
What pressure will the can have if it is placed in a fire
Temperatures
and reaches temperature of 400C ?
must be in K
P1 = 4.0 atm
P2 = ???
T1= 25ºC + 273 = 298 K
T2= 400ºC + 273 = 673 K
T1 = 25ºC
T2 = 400ºC
26
Example 1:
An aerosol spray can has a pressure of 4.0 atm at 25C.
What pressure will the can have if it is placed in a fire
Temperature
and reaches temperature
of 400C ?
Pressure
increases
increases
P1 P2
=
P1 = 4.0 atm
T1 T2
P2 = ???
T1 = 298 K
T2
673 K
P2 = P1 x
= 4.0 L x
T1
298 K
T2 = 673 K
P2 = 9.0 atm
27
Example 2:
A cylinder of gas with a volume of 15.0-L and a pressure
of 965 mmHg is stored at a temperature of 55C. To
what temperature
must the cylinder be cooled to reach a
Pressure
Temperature
pressure of 850 decreases
mmHg?
decreases
P1 = 965 mmHg
T1= 55ºC + 273 = 328 K
P2 = 850 mmHg
850 mmHg
P2
T2 = T1 x = 328 K x
965 mmHg
P1
T1 = 55ºC
T2 = ???
TT22==289
16ºC
K
28
VAPOR PRESSURE
& BOILING POINT
 In an open container, liquid molecules at the surface
that possess sufficient energy, can break away from
the surface and become gas particles or vapor.
 In a closed container, these gas particles can
accumulate and create pressure called vapor
pressure.
 Vapor pressure is defined as the pressure above a
liquid at a given temperature. Vapor pressure
varies with each liquid and increases with
temperature.
29
VAPOR PRESSURE
& BOILING POINT
 Listed below is the vapor
pressure of water at
various temperatures.
30
VAPOR PRESSURE
& BOILING POINT
 A liquid reaches its boiling point
when its vapor pressure becomes
equal to the external pressure
(atmospheric pressure).
 For example, at sea level, water
reaches its boiling point at 100C
since its vapor pressure is 760
mmHg at this temperature.
31
VAPOR PRESSURE
& BOILING POINT
 At higher altitudes, where
atmospheric pressure is lower,
water reaches boiling point at
temperatures lower than 100C.
 For example, in Denver, where
atmospheric pressure is 630
mmHg, water boils at 95C,
since its vapor pressure is 630
mmHg at this temperature.
32
COMBINED
GAS LAW
 This
All pressure-volume-temperature
law is useful for studying the effect
relationships
of
changes
can be combined
in two variables.
into a single relationship called
the combined gas law.
P1V1 P2 V2
=
T1
T2
Initial
condition
Final
condition
33
COMBINED
GAS LAW
 The individual gas laws studied previously are
embodied in the combined gas law.
P1 V1 P2 V2
=
T1
T2
Charles’s
Law
Boyle’s
Gay-Lussac’s
Law
Law
34
Example 1:
A 25.0-mL sample of gas has a pressure of 4.00 atm at a
temperature of 10C. What is the volume of the gas at a
pressure of 1.00 atm and a temperature of 18C?
P1= 4.00 atm
V1= 25.0 mL
T1= 283 K
P2= 1.00 atm
V2= ???
T2 = 291 K
P1V1 P2 VBoth
pressureP1 T2
2
V2 = V1 x x
=
T1
T2 and temp. P2 T1
change
4.00 atm 291 K
V2 = 25.0 mL x
x
1.00 atm 283 K
Must use
V2 =combined
103 mL
gas law
35
LAW
 At
Theconstant
relationship
temperature
of molesand
andpressure,
volume inthe
gases
volume
a fixed amount
Law.of gas is directly
proportional to the number of moles.
As number of
moles increases,
Number
of moles
thevolume
volumeof
ofa
and
the directly
gas
gas are
increases
proportional
36
Tro's Introductory Chemistry,
Chapter
37
LAW
 As a result of Avogadro’s Law, equal volumes of
different gases at the same temp. and pressure
contain equal number of moles (molecules).
2 Tanks of gas of equal
volume at the same T
& P contain the same
number of molecules
38
LAW
 Avogadro’s Law also allows chemists to relate
volumes and moles of a gas in a chemical reaction.
This relationship is
For example:
only valid for
gaseous
substances
(g) 
2H O
2 H2 (g) + 1 O2
2
(g)
2 molecules
1 molecule
2 molecules
2 moles
1 mole
2 moles
2 Liters
1 Liter
2 Liters
39
Example 1:
A sample of helium gas with a mass of 18.0 g occupies
1.6 Liters at a particular temperature and pressure.
Same
Temp. L at the same
What mass of oxygen would
occupy1.6
temperature and pressure?& Pressure
1 mol
mol of He = 18.0 g x
= 4.50 mol
4.00 g
mol of O2 = mol of He = 4.50 mol
32.0 g
mass of O2 = 4.50 mol x
= 144 g
1 mol
40
Example 2:
How many Liters of NH3 can be produced from reaction
of 1.8 L of H2 with excess N2, as shown below?
N2 (g) + 3 H2 (g)  2 NH3 (g)
1.8 L H2
x
2 L NH 3
= 1.2 L NH3
3 L H2
41
STP &
MOLAR VOLUME
 To better understand the factors that affect
gas behavior, a set of standard conditions have
been chosen for use, and are referred to as
Standard Temperature and Pressure (STP).
STP =
0ºC (273 K)
1 atm (760 mmHg)
42
STP &
MOLAR VOLUME
 At STP conditions, one mole of any gas is
observed to occupy a volume of 22.4 L.
V = 22.4 L
Molar
Volume at
STP
43
Example 1:
If 2.00 L of a gas at STP has a mass of 3.23 g, what is
the molar mass of the gas?
1 mol
mol of gas = 2.00 L x
22.4 L = 0.0893 mol
g
3.23 g
=
Molar
Molar mass
=
= 36.2 g/mol
mol 0.0893 mol
volume at
STP
44
Example 2:
A sample of gas has a volume of 2.50 L at 730 mmHg
and 20C. What is the volume of this gas at STP?
P1= 730 mmHg
V1= 2.50 L
T1= 293 K
P2= 760 mmHg
V2= ???
T2 = 273 K
P1 T2
V2 = V1 x x
P2 T1
730 mmHg 273 K
V2 = 2.50 L x
x
760 mmHg 293 K
V2 = 2.24 L
45
IDEAL GAS LAW
 Combining all the laws that describe the behavior
of gases, one can obtain a useful relationship that
relates the volume of a gas to the temperature,
pressure and number of moles.
n R T
V=
P
Universal
gas constant
R = 0.0821 L atm/mol K
46
IDEAL GAS LAW
 This relationship is called the Ideal Gas Law,
and commonly written as:
P V = n
R T
Temp.
in K
Pressure
in atm
Volume
in Liters
Number
of moles
47
Example 1:
A sample of H2 gas has a volume of 8.56 L at a temp. of
0C and pressure of 1.5 atm. Calculate the moles of gas
present.
PV = nRT
P = 1.5 atm
V = 8.56 L
T = 273 K
n = ???
R = 0.0821
(1.5 atm)(8.56 L)
PV
=
n=
RT (0.0821 L atm )(273 K)
mol K
= 0.57 mol
48
Example 2:
What volume does 40.0 g of N2 gas occupy at 10C and
750 mmHg?
Moles must
1 mol
be
calculated
mol of N2 = 40.0 g x
= 1.43 mol
P =750 mmHg
from mass 28.0 g
V = ???
T = 283 K
n = not given
m = 40.0 g
R = 0.0821
1 atm
x
P = 750 mmHgPressure
760bemmHg
must
converted to
P = 0.987 atm
atm
49
Example 2:
What volume does 40.0 g of N2 gas occupy at 10C and
750 mmHg?
nRT
V=
P
P =0.987 atm
V = ???
T = 283 K
n = 1.43
R = 0.0821
L atm
(1.43 mol)(0.0821
)(283 K)
mol K
V=
0.987 atm
V = 33.7 L
50
Ideal vs. Real Gases
•
•
Real gases often do not behave like ideal
gases at high pressure or low temperature
Ideal gas laws assume
1) no attractions between gas molecules
2) gas molecules do not take up space
 based on the Kinetic-Molecular Theory
•
at low temperatures and high pressures
these assumptions are not valid
Tro's Introductory Chemistry,
Chapter
51
Ideal vs. Real Gases
Tro's Introductory Chemistry,
Chapter
52
PARTIAL
PRESSURES
 Many gas samples are mixture of gases. For
example, the air we breathe is a mixture of mostly
oxygen and nitrogen gases.
 Since gas particles have no attractions towards one
another, each gas in a mixture behaves as if it is
present by itself, and is not affected by the other
gases present in the mixture.
 In a mixture, each gas exerts a pressure as if it was
the only gas present in the container. This
pressure is called partial pressure of the gas.
53
DALTON’S
LAW
 In a mixture, the sum of all the partial pressures of
gases in the mixture is equal to the total pressure of
the gas mixture.
Ptotal = P1 + P2 + P3 + ···
Total pressure of = Sum of the partial pressures
a gas mixture
of the gases in the mixture
 This is called Dalton’s law of partial pressures.
54
PARTIAL
PRESSURES
+
PHe = 2.0 atm
PAr = 4.0 atm
PTotal = ???
PHe + PAr = 6.0 atm
Total pressure of a gas mixture is the sum of
partial pressures of each gas in the mixture
55
PARTIAL
PRESSURES
 The partial pressure of each gas in a mixture is
proportional to the amount (mol) of gas present in
the mixture.
 For example, in a mixture of gases consisting of 1
mole of nitrogen and 1 mol of hydrogen gas, the
partial pressure of each gas is one-half of the total
pressure in the container.
56
PARTIAL
PRESSURES
Twice as many
moles of Ar
compared to He
+
PHe = 2.0 atm
PPAr
= 2 PHe
Ar = 4.0 atm
PTotal = 6.0 atm
The partial pressure of each gas is proportional to
the amount (mol) of the gas present in the mixture
57
Example 1:
A scuba tank contains a mixture of oxygen and helium
gases with total pressure of 7.00 atm. If the partial
pressure of oxygen in the tank is 1140 mmHg, what is
the partial pressure of helium in the tank?
Ptotal
Poxygen
= Poxygen + Phelium
= 1140 mmHg x
1 atm
= 1.50 atm
760 mmHg
PTotalatm
- P–oxygen
1.50 atm = 5.50 atm
Phelium ==7.00
58
Example 2:
A mixture of gases contains 2.0 mol of O2 gas and
4.0 mol of N2 gas with total pressure of 3.0 atm. What
Twice
as many
is the partial pressure of each gas in
the mixture?
moles of N2
compared to O2
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Pnitrogen = 2/3 (Ptotal ) = 2.0 atm
Poxygen = 1/3 (Ptotal ) = 1.0 atm
59
THE END
60
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