The document discusses gases and gas laws. It provides an overview of the composition and importance of Earth's atmosphere. Key topics covered include gas pressure, how it is measured using devices like the barometer, and how pressure varies with weather and altitude. The document then explains Boyle's, Charles', and Avogadro's laws regarding the behavior of gases. It introduces the ideal gas law and provides examples of using it to solve gas problems involving changes in pressure, volume, temperature, and moles of gas. It also covers gas stoichiometry, partial pressures of gases in mixtures, and the kinetic molecular theory of gases.

1. Gases
Advanced Chemistry Chapter 5

2. Intro
Earth’s atmosphere is a gaseous
solution that consists mainly of
nitrogen (N2) and oxygen (O2). This
atmosphere supports life and acts as
a waste receptacle for many
industrial processes. The chemical
reactions that follow often lead to
various types of pollution, including
smog and acid rain.

3. Intro
The gases in the atmosphere also
shield us from harmful radiation
from the sun and keep the earth
warm by reflecting heat radiation
back toward the earth. In fact, there
is now great concern that an increase
in atmospheric carbon dioxide, a
product of the combustion of fossil
fuels , is causing a dangerous
warming of the earth.

4. Pressure

5. Pressure
Gas uniformly fills a container, is
easily compressed, and mixes
completely with any other gas.
One of the most important
properties is that it exerts pressure
on its surroundings equally.

6. Pressure
Incredible Tank Implosion

7. Pressure

8. Pressure

9. Pressure

10. Pressure
Barometer - A device to
measure atmospheric pressure,
was invented in 1643 by Torricelli
(a student of Galileo).
Torricelli’s barometer is
constructed by filling a glass
tube with liquid mercury and
inverting it in a dish of mercury.

11. Pressure
Barometer - A device to
measure atmospheric pressure,
was invented in 1643 by Torricelli
(a student of Galileo).
Notice that a large quantity of
mercury stays in the tube. In
fact, at sea level the height of
this column of mercury averages
760 mm.

12. Pressure
Barometer - A device to
measure atmospheric pressure,
was invented in 1643 by Torricelli
(a student of Galileo).
Atmospheric pressure results
from the mass of the air being
pulled toward the center of the
earth by gravity.

13. Pressure
Barometer - A device to
measure atmospheric pressure,
was invented in 1643 by Torricelli
(a student of Galileo).
Atmospheric pressure varies
with weather changes and
altitude.

14. Pressure
Manometer – an instrument for measuring pressure
often below that of atmospheric pressure.

15. Units of Pressure
Pressure = force/area
torr – in honor of Torricelli is equal
to a mm Hg.
760 mm Hg = 760 torr
1 atm = 760 torr
Pascal = N/m2
1atm = 101,325 Pa

16. Practice Problems
Page 225 #35, 37, 39

17. The Gas
Laws of
Boyle,
Charles and
Avogadro

18. Boyles Law
Boyle (1627-1691) –
performed the first quantitative
experiments on gases. Used a
J tube to measure pressures.
PV = k; P1V1=P2V2
k is a constant for a given
sample of air at a specific
temperature.

19. Boyles Law
Pressure and volume are often plotted.
P vs V – gives a hyperbola and an
inverse relationship.
Boyles law rearranged is
V=k/P=k1/P; when plotted as V vs
1/P – gives a straight line with the
intercept of zero

20. Boyles Law

21. Boyles Law
Boyles’s law holds precisely at
very low temperatures, but
varies at higher pressures. PV
will vary as pressure is varied.
An ideal gas is a gas that
strictly obeys Boyles’ law.

22. Boyles Law

23. Charles Law
Charles (1746-1823) – the first person to
fill a balloon with hydrogen gas and who
made the first solo balloon flight.
Charles found in 1787 that the volume of
a gas at constant pressure increases
linearly with the temperature of the gas.
V = bT ; V1/T1 = V2/T2
T is in Kelvin, b is a proportionality
constant

24. Charles Law
Temperature vs Volume plots a
straight line.
Slope will vary will type of gas.
All gas plots of T vs V will
extrapolate to zero at the same
temperature.
-273° or 0 K

25. Charles Law

26. Avogadro’s Law
Avogadro (1811) – postulated
that equal volumes of gases at
the same temperature and
pressure contain the same
number of particles (moles).
V = an; V1/n1=V2/n2
n is number of moles; a is a
proportionality constant.

27. Ideal Gas
Law

28. Ideal Gas Law
The relationships that Boyle, Charles
and Avogadro presented can be
combined to show how the volume of a
gas depends on pressure, temperature,
and number of moles of gas present.
V = R(Tn/P)
R is the universal gas constant.
A combination of proportionality
constants

29. Ideal Gas Law
The equation is often rearranged to
form the more common:
PV=nRT
R=.08206 Latm/Kmol

30. Ideal Gas Law
Limitations
A gas that obeys this equation is said
to behave ideally. The ideal gas
equation is best regarded as a limiting
law, it expresses behavior that real
gases approach at low pressures and
high temperatures.
Most gases behave ideally at pressures
below 1 atm.

31. Example Problem
A sample of hydrogen gas (H2) has a
volume of 8.56 L at a temperature of
0°C and a pressure of 1.5 atm.
Calculate the moles of H2 molecules
present in this gas sample.

32. Example Problem
V = 8.56 L at a temperature of 0°C
P =1.5 atm
Calculate the moles
PV=nRT; R = .08206 L atm/K mol
.57 mol

33. Example Problem 2
You have a sample of ammonia gas with
a a volume of 7.0ml at a pressure of
1.68 atm. The gas is compressed to a
volume of 2.7 ml at a constant
temperature. Use the ideal gas law to
calculate the final pressure.

34. Example Problem 2
V1 = 7.0 ml at a pressure of 1.68 atm
V2 =2.7 ml at a constant temperature.
Calculate the final pressure.
PV=nRT; but nRT are constant: PV=PV
4.4 atm

35. Example Problem 3
A sample of methane gas that has a
volume of 3.8 L at 5°C is heated to 86°C
at constant pressure. Calculate its new
volume.

36. Example Problem 3
V1 = 3.8 L and T15°C
T2=86°C at constant pressure.
Calculate its new volume.
PV=nRT but n, R and P are constant: V1/T1 = V2/T2
4.9 L

37. Example Problem 4
A sample of diborane gas (B2H6), a
substance that burst into flame when
exposed to air, has a pressure of 345 torr
at a temperature of -15°C and a volume
of 3.48 L. If conditions are changed so
that the temperature is 36°C and the
pressure is 468 torr, what will be the
volume of the sample.

38. Example Problem 4
P1=345 torr, T1=-15°C and V1 = 3.48L
T2=36°C and P2=468 torr,
What is the volume?
PV = nRT; nR are constant: P1V1/T1 =
P2V2/T2
3.1 L

39. Example Problem 5
A sample containing 0.35 mol argon gas at
a temperature of 13°C and a pressure of
568 torr is heated to 56°C and a pressure
of 897 torr. Calculate the change in
volume that occurs.

40. Example Problem 5
n=0.35, T1=13°C, P1=568 torr
T2=56°C, P2=897 torr
Calculate the change in volume that occurs.
-3 L
PV=nRT; R = .08206 L atm/K mol

41. Practice Problems
Page 226 #41, 43, 45,
47, 49, 51, 53, 57, 59,
61

42. Gas
Stoichiometry

43. Molar Volume
One mole of an ideal gas at:
 0°C (273K)
1atm
V=nRT/P = 22.42L

44. Gas Stoichiometry
We use STP or standard
temperature and pressure of an ideal
gas to make calculations with a gas.
1 atm
0°C (273K)
1 mole = 22.42 L becomes a
conversion factor for dimensional
analysis.

45. Gas Stoichiometry
Example
A sample of nitrogen gas has a volume
of 1.75 L at STP. How many moles of
N2 are present?
1.75L N2 x
1mole N2
22.42L N2
=
7.81 x 10-2
mol N2

46. Gas Stoichiometry
Example 2
Quicklime (CaO) is produced by the
thermal decomposition of calcium
carbonate (CaCO3). Calculate the
volume of CO2 at STP produced from
the decomposition of 152g CaCO3 by
the reaction
CaCO3(s) CaO(s) + CO2(g)

47. Gas Stoichiometry
Example 2
152g CaCO3
22.42 L = 1 mol of gas at STP
Calculate the volume of CO2
CaCO3(s) CaO(s) + CO2(g)
34.1 L CO2 at STP

48. Gas Stoichiometry
Example 3
A sample of methane gas having a
volume of 2.80 L at 25°C and 1.65
atm was mixed with a sample of
oxygen gas having a volume of 35.0 L
at 31°C and 1.25 atm. The mixture
was then ignited to form carbon
dioxide and water. Calculate the
volume of CO2 formed at a pressure of
2.50 atm and a temperature of 125°.

49. Gas Stoichiometry
Example 3
CH4 V=2.80 L at 25°C and 1.65 atm
Oxygen V=35.0 L at 31°C and 1.25 atm.
Calculate the volume of CO2 at 2.50 atm and
125°C.
The mixture was then ignited to form carbon
dioxide and water.

50. Gas Stoichiometry
Example 3
CH4 V=2.80 L at 25°C and 1.65 atm
Oxygen V=35.0 L at 31°C and 1.25 atm.
Calculate the volume of CO2 at 2.50 atm and
125°C.
CH4(g) + O2(g)  CO2(g) + H2O(g)
2.47 L

51. Practice Problems
Page 227 #65, 69

52. Molar Mass of a Gas
 One use of the ideal gas law is in the calculations of the
molar mass of a gas from its measured density.
 n =
 P
 D
 P =
grams of gas
molar mass
=
µ
µολαρµασσ
=
nRT
V
=
m / molar mass( )RT
V
=
mRT
V(molar mass)
=
m
V
dRT
molar mass
; Molar mass =
δΡΤ
Π

53. Gas Density/Molar
Mass Example
The density of a gas was measured
at 1.50 atm and 27°C and found to
be 1.95 g/L. Calculate the molar
mass of the gas.
32.0 g/mol

54. Dalton’s
Law of
Partial
Pressures

55. John Dalton
John Dalton formed his atomic theory from his
experiments and studies of the mixture of gases.
His observations car be summarized as follows:
For a mixture of gases in a container, the total
pressure exerted is the sum of the pressures
that each as would exert if it were alone.

56. John Dalton
Ptotal=P1+P2+P3+….
Subscripts refer to the individual gases
and Px refers to partial pressure that a
particular gas would exert if it were
alone in the container.
Each Partial pressure can be derived
from the ideal gas law and added
together to determine the total.

57. John Dalton
Ptotal=P1+P2+P3+….
Since each partial pressure can be
broken down into ; the Ptotal can be
represented by:
Ptotal=
Ptotal=
nx RT
V
(n1 + ν2 + ν3 + ...)
ΡΤ
ς




ntotal
RT
V





58. For a mixture of ideal gases, it is the total number of
moles of particles that is important, not the identity
or composition of the involved gas particle.

59. Dalton’s Law Example
Mixtures of helium and oxygen can be
used in scuba diving tanks to help
prevent “the bends.” For a particular
dive, 46 L He at 25° and 1.0 atm and
12 L O2 at 25° and 1.0 atm were
pumped into a tank with a volume of
5.0 L. Calculate the partial pressure of
each gas and the total pressure in the
tank at 25° C.

60. Mole Fraction
The ratio of the number of
moles of a given component
in a mixture to the total
number of moles in the
mixture.
 = nχ x/ntotal
 = Pχ 1/Ptotal

61. Dalton’s Law Example
The partial pressure of oxygen was
observed to be 156 torr in air with a
total atmospheric pressure of 743
torr. Calculate the mole fraction of O2
present.

62. Dalton’s Law Example
The mole fraction of nitrogen in the air
is 0.7808. Calculate the partial
pressure of N2 in air when the
atmospheric pressure is 760 torr.

63. Collecting Gas over
Water
A mixture of gases results whenever a
gas is collected by displacement of water.
In this situation, the gas in the bottle is a
mixture of water vapor and the oxygen
being collected.

64. Collecting Gas over
Water
Water vapor is present because
molecules of water escape from the
surface of the liquid and collect in the
space above the liquid.

65. Collecting Gas over
Water
Molecules of water also return to the
liquid. When the rate of escape equals
the rate of return, the number of water
molecules in the vapor state remain
constant.

66. Collecting Gas over
Water
When the number of water molecules in
the vapor state remain constant the
pressure of the water vapor remains
constant.

67. Collecting Gas over
Water
This pressure, which depends on
temperature, is called vapor pressure of
water.

68. Collecting Gas over
Water Example
A sample of solid potassium chlorate
(KClO3) was heated in a test tube and
decomposed by the reaction:
2KClO3(s) → 2ΚΧλ(σ) + 3Ο2(γ )

69. Collecting Gas over
Water Example
The oxygen produced was collected by
displacement of water at 22°C at a total
pressure of 754 torr. The volume of gas
collected was .650L, and the vapor
pressure of water at 22°C is 21 torr.
Calculate the partial pressure of O2 in the
gas collected and the mass of KClO3 in
the sample that was decomposed.
2KClO3(s) → 2ΚΧλ(σ) + 3Ο2(γ )

70. Collecting Gas over
Water Example
oxygen T=22°C and V=.650L
Total pressure = 754 torr.
vapor pressure at 22°C is 21 torr.
Calculate the partial pressure of O2 and
the mass of KClO3 in the sample
2KClO3(s) → 2ΚΧλ(σ) + 3Ο2(γ )
2.59 x 10-2
mol O2 2.12 g KClO3

71. The Kinetic
Molecular
Theory of
Gases

72. Kinetic Molecular Theory
KMT
A simple model that attempts to
explain the properties of an ideal
gas. This model is based on
speculations about the behavior of
the individual gas particles (atoms or
molecules).

73. Kinetic Molecular Theory
KMT
1. The particles are so small
compared with the distances
between them that the volume of
the individual particles can be
assumed to be negligible (zero).

74. Kinetic Molecular Theory
KMT
2. The particles are in constant
motion. The collisions of the
particles with the walls of the
container are the cause of the
pressure exerted by the gas.

75. Kinetic Molecular Theory
KMT
3. The particles are assumed to exert
no forces on each other; they are
assumed neither to attract nor to
repel each other.

76. Kinetic Molecular Theory
KMT
4. The average kinetic energy of a
collection of gas particles is
assumed to be directly
proportional to the Kelvin
temperature of the gas.

77. KMT and Boyle’s Law
 Because a decrease in volume, the
gas particles will hit the walls more
often, thus increasing the pressure

78. KMT and Charles Law
 When the gas is heated to a higher
temperature, the speeds of its molecules
increase and thus hit the walls more
often and with more force. Volume
and/or pressure will increase.

79. KMT and Advogadro’s Law
 An increase in the number of gas
particles at the same temperature would
cause the pressure to increase if the
volume were constant.

80. KMT and Advogadro’s Law
 The volume of a gas (at constant T and
P) depends only on the number of gas
particles present. The individual particles
are not a factor because the particle
volumes are so small compared with the
distances between the particles.

81. KMT and Dalton’s Law
 All gas particles are independent of each
other and that the volumes of the
individual particles are unimportant.
Identities of the gas particles do not
matter.

82. The Meaning of Temperature
Kelvin temperature indicates the
average kinetic energy of the gas
particles.
The exact relationship between
temperature and average kinetic
energy can be expressed:
(KE)avg=3/2 RT

83. The Meaning of Temperature
The Kelvin temperature is an index
of the random motions of the
particles of a gas, with higher
temperature meaning greater
motion.

84. Root Mean Square Velocity
u2
=the average of the squares of
the particle velocities.
The square root of u2
is called the
root mean square velocity and is
symbolized with urms
urms= =
M= mole of gas particles (kg)
R = ; J = kgm2
/s2
3RT
Mu2
8.3145
J
Kgmol

85. Root Mean Square Velocity
Example
Calculate the root mean square
velocity for the atoms in a sample
of helium gas at 25°C.
1.36 x 103
m/s

86. Mean Free Path
The average distance a particle travels
between collisions in a particular gas
sample.
1 x 10-7
m for O2 at STP
urms=500 m/s

87. Mean Free Path
A velocity distribution that show the effect
of temperature on the velocity distribution
in a gas.

88. Effusion
and
Diffusion

89. Diffusion
Diffusion describes the
mixing of gases.
The rate of mixing gases, is
the same as the rate of
diffusion
Dependent upon urms

90. Effusion
Effusion describes the
transfer of gas from one
chamber to another (usually
through a small hole or
porous opening).
The rate of transfer is said
to be the rate of effusion.

91. Effusion
The rate of effusion of a gas is
inversely proportional to the square
root of the mass of its particles.

92. Effusion
Rate of Effusion for gas 1
Rate of Effusion for gas 2
=
M2
M1
Temperature must be the same for
both gases.
M represents the molar masses of the
gases.
Units can be in g or kg since the
units will cancel out.
This is called Graham’s law of effusion:

93. Effusion Example
Calculate the effusion rates of
hydrogen gas (H2) and Uranium
hexafluoride (UF6), a gas used in the
enrichment process to produce fuel
for nuclear reactors.
13.2 : 1

94. Real
Gases

95. Real Gases
An ideal gas is a hypothetical
concept. No gas exactly
follows the ideal gas law,
although many gases come
very close at low pressures
and/or high temperatures.

96. Real Gases
Thus ideal gas behavior can
best be thought of as the
behavior approached by real
gases under certain
conditions.

97. Real Gases
Plots of PV/nRT vs. P for several gases
(200K). Ideal behavior only at low
pressures.

98. Real Gases
Plots of PV/nRT vs. P for N2 at three
temperatures. Ideal behavior at higher
temperatures.

99. KMT Modifications
Johannes van der Walls (1837-
1923), a physics professor at the
University of Amsterdam started
work in the area of ideal vs real gas
behavior. He won the nobel prize in
1910 for his work.

100. KMT Modifications
van der Waals modifications to the ideal
gas law accounted for the volume of
particle space. Therefore adjusting for the
volume actually available to a give gas
molecule.
V-nb
n is number of moles
b is an empirical constant

101. KMT Modifications
van der Waals modifications to the ideal
gas law allowed for the attractions that
occur among particle in a real gas which
is dependent upon the concentration of
the particles.
 , pressure correction
a is proportionality constant.
Pobs = Π∋
− α
ν
ϖ




2

102. van der Waals Equation
Insert both corrections and the equation
can be written as:
Rearranged for van der Waals:
Pobs =
νΡΤ
ς − νβ
− α
ν
ς




2
Pobs + α
ν
ς




2








ξ ς − νβ( ) = νΡΤ

103. van der Waals
Equation
a and b values are
determined for a given gas
by fitting experimental
behavior. That is a and b
are varied until the best fit
of the observed pressure is
obtained under all
conditions.

104. Characteristics of Real
Gases
A low value for a reflects
weak intermolecular forces
among gas molecules.

105. van der Waals
Ideal behavior at low
pressure (large volume)
makes sense because the
small amount of volume that
the particles consume are not
a factor.

106. van der Waals
Ideal behavior at high
temperatures also makes
sense because particles are
moving at such a high rate
that their interparticle
interactions are not very
important.

107. Chemistry in the
Atmosphere

108. Chemistry in the
Atmosphere
The most important gases to us are
those in the atmosphere that surround
the earth’s surface.
The principal components are N2 and
O2, but many other important gases,
such as H2O and CO2, are also
present.

110. Chemistry in the
Atmosphere
Because of gravitational effects, the
composition of the earth’s atmosphere
is not constant; heavier molecules
tend to be near the earth’s surface,
and light molecules tend to migrate to
higher altitudes, with some eventually
escaping into space.

111. Chemistry in the
Atmosphere
The chemistry occurring in the higher
levels of the atmosphere is mostly
determined by the effects of high-
energy radiation and particles from
the sun and other sources in space.
The upper atmosphere serves as a
shield to prevent this radiation from
reaching earth.

112. Chemistry in the
Atmosphere
The troposphere (closest to earth) is
strongly influenced by human
activities. Millions of tons of gases
and particulates are released into the
troposphere by our highly industrial
civilization.

113. Chemistry in the
Atmosphere
Severe air pollution is found around
many large cities. The two main
sources of pollution are transportation
and the production of electricity. The
combustion of petroleum in vehicles
produces CO, CO2, NO, NO2.

114. Chemistry in the
Atmosphere
The complex chemistry of polluted air
appears to center around the nitrogen
oxides (NOx). At high temperatures
found in the gasoline and diesel
engines of cars and trucks, N2 and O2
react to form a small quantity of NO
that is emitted into the air with the
exhaust gases. NO is immediately
oxidized in air to NO2.

115. Reactions in the
Atomsphere
NO2(g)
radiant
energy
→ NO(g) + O(g)
O(g) + O2(g) → O3(g)
Ozone is very reactive and can react directly with
other pollutants, or the ozone can absorb light and
break up to form an energetically excited O2 molecule
(O2*) and excited O (O*).

116. Reactions in the
Atomsphere
O*
+ H2O → 2OH
OH + NO2 → HNO3
The end product of this whole process is often referred
to as photochemical smog, so called because light is
required to initiate some of the reactions.

117. Reactions in the
Atomsphere
S(coal) + O2(g) → SO2(g)
2SO2(g ) + O2 → 2SO3(g )
SO3(g) + H2O(l) → H2SO4(aq)
Sulfuric acid is very corrosive to both living things and
building materials. Another result of this type of
pollution is called acid rain.

118. THE
END