Chapter 2
The Behaviour of Gases
2.1 Pressure
2.2 Describing Gases
2.3 Gas Mixtures
2.4 Gas Stoichiometry
2.5 Molecular View of Gases
2.6 Additional Gas Properties
2.7 Non-Ideal (Real) Gases
2.8 Chemistry of the Earth’s Atmosphere
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
2.1 Pressure
Learning Objective:
Understand gas pressure and expressing pressure in various units
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2.1 Pressure
 Pressure (p) – in a gas it is
caused by molecular
collisions with a container.
 The air around us is a huge
reservoir of gas that exerts
pressure on the Earth’s
Surface
 The atmospheric pressure
can be measured with a
mercury barometer.
 Why does pressure have
units of mm Hg?
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How Do We Measure the Pressure of a
Gas in a Closed Container?
 We use a manometer.
 In the U-shaped manometer,
the difference between the
two mercury levels gives the
pressure (in mm Hg) of the
gas.
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Units of Pressure
 Because the pressure at sea level is 760 mm Hg we
define another unit called the atmosphere.
1 atm = 760 mm Hg
 Another common pressure unit is Torr
1 atm = 760 Torr
 The official SI unit is the pascal (Pa)
1 atm = 101325 Pa
 Another SI unit is the bar (1 bar = 100,000 Pa)
1 atm = 1.01325 bar
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Example 2 – 1 Measuring Pressure
A scientist collected an
atmospheric gas sample. A
manometer attached to the
gas sample gave the reading
shown in the figure, and the
barometric pressure in the
laboratory was 752 mmHg.
Calculate the pressure of the
sample in atmospheres and in
bars.
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2.2 Describing Gases
Learning Objective:
Relate pressure, volume, temperature and amount of gas
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2.2 Describing Gases
 Molecules in a gas move
freely throughout the
entire volume of a
container, changing
direction whenever they
collide with other
molecules or with a
wall.
 The line traces a
possible path of a single
molecule.
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Variations in Gas Volume
Boyle
investigated
gases in a Jtube and
determined
that volume
was inversely
proportional
to pressure.
1
Vgas 
pgas
(fixed temperature and fixed amount)
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Charles reported studies of gas
volume as a function of
temperature. He determined
that the volume of gas is
directly proportional to its
temperature.
Vgas  Tgas
(fixed pressure and fixed amount)
The volume of gas also changes when the amount changes.
Gas volume is directly proportional to the amount of gas.
Vgas  ngas
(fixed pressure and fixed temperature)
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The Ideal Gas Equation
All four variables (p, T, V, and n) can be related through
the Ideal Gas Equation:
pV = nRT
Units: n (mol), p (bar or kPa), V (L) and T (K)
Units of the ideal gas constant, R:
R = 0.08314 L bar mol-1 K-1
R = 8.314 L kPa mol-1 K-1
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Example 2 – 2 Calculation of Gas Pressure
A 10 m3 steel storage tank contains 8.85 kg of methane
(CH4). If the temperature is 25 °C, what is the pressure
inside the tank?
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Example 2 – 3 Calculation of Gas Amount
A pressure gauge on a tank of molecular oxygen reads
5.67 bar. If the tank holds 75.0 L and the temperature
in the lab is 27.6 °C, how many grams of molecular
oxygen are in the tank?
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Example 2 – 4 Using the Molar Volume of Gas
A cement factory produces large volumes of CO2 gas by
the reaction CaCO3 (s) → CaO (s) + CO2 (g). If the
factory processes 10.0 tonnes of CaCO3 (s) per hour,
what volume of CO2 (g) per hour is produced at 25.0 oC
and 100.0 kPa?
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Example 2 – 5 Pressure-Volume Variations
A sample of helium gas is held at constant temperature inside a
cylinder whose volume is 0.80 L when a piston exerts a pressure
of 1.5 bar. If the external pressure on the piston is increased to
2.1 bar, what will be the new volume?
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Variations on the Gas Equation
During transformations, any of the four variables in the
ideal gas equation may change, and any of them
remain constant.
p1 V1 p2 V2
=
n1 T1 n2 T2
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Example 2 – 6 Changing Gas Conditions
A sample of carbon dioxide in a 10.0-L gas cylinder at
25°C and 1.00 bar pressure is compressed and
heated. The final temperature and volume are 55 °C
and 5.00 L. Compute the final pressure
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Example 2 – 7 Gas Calculations
Two natural gas storage tanks, with volumes of 1.5 x 104
and 2.2 x 104 L, are at the same temperature. The tanks
are connected by pipes that equalize their pressures.
What fraction of the stored natural gas is in the larger
tank?
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2.3 Gas Mixtures
Learning Objective:
Use the concept of partial pressures in gas mixtures
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2.3 Gas Mixtures
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Dalton’s Law of Partial Pressures
In a mixture of gases, each gas contributes to the total
pressure the amount that it would exert if the gas were
present in the container by itself.
ptotal = p1 + p2 + p3 + p4 + …
and
ntotal = n1 + n2 + n3 + n4 + …
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Mole Fraction and Partial Pressures
Mole fraction, X – the number of moles of a particular
substance in a mixture divided by the total moles
nA
nA
XA =
=
nA +nB +nC +... ntotal
and consequently
pA = XA ptotal
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Example 2 – 8 Gas Mixtures
The amount of gas introduced into a diving tank can be
determined by weighing the tank before and after
charging the tank with gas. A diving shop placed 80.0 g
of O2 and 20.0 g of He in a 5.00 L tank at 298 K.
Determine the total pressure of the mixture and find
the partial pressures and mole fractions of the two
gases.
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ppm and ppb
 When we are talking about mixtures (gas or liquid) that have a
very small amount of solute, we may use parts per million (ppm)
as a unit for concentration.
 Parts per million measures how many molecules of a substance
are present in one million molecules of a sample.
1 ppm = 1 molecule out of every 106 molecules
 Another common unit for concentration is parts per billion (ppb).
1 ppb = 1 molecule out of every 109 molecules
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Example 2 – 9 Working with Concentrations in ppm
The exhaust gas from an average automobile contains
206 ppm of the pollutant nitrogen oxide, NO. If an
automobile emits 125 L of exhaust gas at 100 kPa and
350 K, what mass of NO has been added to the
atmosphere? Assume the atmospheric pressure is 100
kPa.
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2.4 Gas Stoichiometry
Learning Objective:
Use stoichiometry to solve problems involving gas-phase
chemical reactions
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2.4 Gas Stoichiometry
 Stoichiometry applies to solids, liquids and gases.
 Notice that the ideal gas law contains moles.
 Remember that in order to use stoichiometry, you must
be using units of moles.
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Example 2 – 10 Gas Stoichiometry
This example describes the synthesis of acetylene (C2H2) from calcium carbide
(CaC2). Modern industrial production of acetylene is based on a reaction of
methane (CH4) under carefully controlled conditions. At temperatures
greater than 1600 K, two methane molecules rearrange to give three
molecules of hydrogen and one molecule of acetylene:
A 50.0 L steel vessel, filled with CH4 to a pressure of 10.0 bar at 298 K, is heated
to 1600 K to convert CH4 into C2H2. What mass of C2H2 can be produced?
What pressure does the reactor reach at 1600 K?
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Example 2 – 11 Limiting Reagents in a Gas Mixture
Margarine can be made from natural oils, such as coconut oil, by
hydrogenation:
200oC, 7 atm, Ni catalyst
C57H104O6 (l) + 3 H2 (g) 
 C57H110O6 (s)
oil
margarine
An industrial hydrogenator with a volume of 2.50 x 102 L is charged
with 12.0 kg of oil and 7.09 bar (= 709 kPa) of hydrogen (H2) at
473 K (200 °C), and the reaction goes to completion. What is the
final pressure of H2 and how many kilograms of margarine will be
produced?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 2 – 12 General Stoichiometry
The reaction of a metal with an acid generates hydrogen gas and
an aqueous solution of ions. Suppose that 3.50 g of magnesium
metal is dropped into 0.150 L of 6.00 M HCl in a 5.00-L cylinder
with an initial gas pressure of 1.00 bar at 25.0 °C, and the
cylinder is immediately sealed. Find the final partial pressure of
hydrogen, the total pressure in the container, and the
concentrations of all ions in solution.
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2.5 Molecular View of Gases
Learning Objective:
Explain the basic concepts of kinetic molecular theory: molecular
speed, energy, and the effects of temperature and volume on
gas pressure
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2.5 Molecular View of Gases
How do gases behave at a molecular level?
Let’s examine Molecular Speed.
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Example 2 – 13 A Molecular Beam Experiment
The figures below represent mixtures of neon atom and hydrogen
molecules. One of the gas mixtures was used in a pulsed
molecular beam experiment. The result of the experiment is
shown below. Which of the two gas samples, A or B, was used
for this experiment?
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Speed and Energy
 How would we calculate the kinetic energy of a molecule of
hydrogen if a molecule has a most probable speed of 1.57 x
103 m/s at 300 K? (You will need to calculate the mass of a
molecule of H2 )
 How about for methane, CH4 at 300 K? (v = 5.57 x 102 m/s)
1 2
Ekinetic = mv
2
 What conclusion can be drawn from these calculations?
At a given temperature, all gases have the same
molecular kinetic energy distribution.
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Average Kinetic Energy
The average kinetic energy of gas molecules depends on
the temperature of the gas.
Ekinetic
3RT
=
2NA
R = 8.314 J mol-1 K-1
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What is an Ideal Gas?
 An ideal gas has the following two characteristics:

The volume occupied by the molecules of an ideal gas is
negligible compared with the volume of its container.

The energies generated by forces among ideal gas molecules
are negligible compared with molecular kinetic energies.
(Small gas molecules are not attracted to one another)
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2.6 Additional Gas Properties
Learning Objective:
Calculate gas densities and molar masses from pressure-volumetemperature data
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2.6 Additional Gas Properties
 Determination of molar mass using pV=nRT
 Gas density
 Rates of gas movement
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Molar Mass and the Ideal Gas Law
 Easy to manipulate the ideal gas law to include the
molar mass:
pV = nRT and
m
pV = RT
M
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m
n=
M
Example 2 - 15
Calcium Carbide (CaC2) is a hard, gray-black solid that has
a melting point of 2000 °C. This compound reacts
strongly with water to produce a gas and a solution
containing OH- ions. A 12.8 g sample of CaC2 was
treated with excess water. The resulting gas was
collected in an evacuated 5.00 L glass bulb with a mass
of 1254.49 g. The filled bulb had a mass of 1259.70 g
and a pressure of 98.8 kPa atm when its temperature
was 26.8 °C. Calculate the molar mass and determine
the formula of the gas.
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Density and the Ideal Gas Law
 Easy to manipulate ideal gas law to determine the
density.
m
pV = RT
M
m
pM = RT
V
pM = ρgasRT or ρgas
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pM
=
RT
Example 2 - 16
A hot-air balloon will rise when the density of its air is
15% lower than that of the atmospheric air. Calculate
the density of air at 295 K, 100 kPa (assume that dry air
is 78% N2 and 22% O2) and determine the minimum
temperature of air that will cause a balloon to rise.
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Rates of Gas Movement
Effusion – movement of
gas through a tiny
opening into a vacuum
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Rates of gas
movement
Effusion – movement of
gas through a tiny
opening into a vacuum
Diffusion – mixing of two
gases due to their
molecular motion
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Rates of Gas Movement
 When you are driving, do you follow a semitrailer up a
mountain or do you pass it?
 Similarly, the heavier the gas molecule, the slower it
will diffuse or effuse. It just moves slower.
 We can relate the mass of a molecule to speed using
the root-mean-squared speed,v .
 3RT 
v=

 M 
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1
2
Graham’s Law
Heavier molecules diffuse (and effuse) more slowly
rate1
M2
=
rate2
M1
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Example 2 – 17 Graham’s Law:
Relative Rates of Diffusion
Refer to Figure 2-17, and calculate how far from the HCl
end of the tube the NH4Cl (s) forms. Assume the
distance between the HCl and NH3 is initially 60 cm.
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Example 2 – 18 Graham’s Law:
Measuring Molecular Weights
A sample of krypton gas (Kr) escapes through a small hole
in 64.4 s. The same amount of a gas whose identity is
unknown escapes through the hole in 31.6 s, under
identical conditions. Calculate the molecular weight of
the unknown gas.
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2.7 Non-Ideal (Real) Gases
Learning Objective:
Calculate the pressure of a gas under non-ideal conditions, and
explain the deviations from ideality
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2.7 Non-Ideal (Real) Gases
 Intermolecular forces: forces of attraction between
molecules which result in liquids and solids.
 At STP, only 11 elements are gases.
 Intermolecular forces are important in all elements,
except for those which are gases.
 Intermolecular forces become important in gases at
high pressure and/or low temperature
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Real Gases
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Real Gases
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van der Waals Equation
During conditions of high pressure and/or low
temperature and when forces exist between
molecules, the ideal gas equation does not work
At high pressure, molecular volume is important
 At low temperature, the attractive potential energies may be
larger than the kinetic energies
 The van der Waals Equation corrects for these problems

n 
RT
p=
- a 2 
V-nb
V 
2
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van der Waals Equation Constants
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Example 2 – 19 Real Gas Pressures
Gases such as methane are supplied in steel cylinders. A
15.0 L cylinder contains 62.0 mol CH4 when full, and
0.620 mol CH4 after prolonged use. Compare the ideal
and real (van der Waals) pressures before and after use
if the cylinder is at 27 °C.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
2.8 Chemistry of the Earth’s Atmosphere
Learning Objective:
Do calculations involving water vapour pressure and relative
humidity and describe some of the basic chemistry of the
troposphere
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2.8 Chemistry of the Earth’s Atmosphere
 Our atmosphere behaves as an
ideal gas, where gravity
determines its volume
 Pressure varies with atmospheric
conditions:
nRT
p=
V
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Troposphere
 Layer of the atmosphere closest to the earth.
 99% (N2 and O2)
 H2O, Ar and CO2 are the only other gases present in
amounts greater than 0.01%
 H2O concentration can vary depending on the
atmospheric conditions.
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Vapour Pressure
 In a closed container, a gas will evaporate until it
reaches a dynamic equilibrium.
 At dynamic equilibrium, the pressure of the gas in the
closed container is called the vapour pressure, pvap.
 So what about for water in the earth’s atmosphere?
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Relative Humidity
 Most of the time, the atmosphere contains less water
vapour than the maximum amount it can hold.
pH2O < p vap,H2O
 The amount of water in the earth’s atmosphere is
called relative humidity:
 pH2O 
% Relative Humidity= 
× 100
 p vap,H O 

2

where pvap,H2O is the partial pressure of water present in
the atmosphere
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Dew Point
 The formation of dew and fog are the result of relative
humidity.
 Warm air with a high relative humidity may cool.
 When the air temperature falls below a certain
temperature, some water must condense from the
atmosphere.
 This temperature is called the dew point.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 2 - 20
Fog forms when humid warm air from above a body of
water moves inland and cools. What is the highest
temperature at which fog could form from air that is at
65% relative humidity when its temperature is 27.5 °C?
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Chemistry in the Troposphere
 The majority of the pollution in the earth’s atmosphere
is due to fossil fuels.
 Some important reactions involve:
Oxides of nitrogen
and
 Oxides of sulphur

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Oxides of Nitrogen
Normally, N2 is very stable. However, under extreme conditions (like
in a car engine), it reacts with oxygen to form NO:
NO can then react in the atmosphere with O2 to form nitrogen dioxide.
NO2 is a red-brown gas that can be seen over many large cities where
the concentration can reach 0.9 ppm. (intolerable > 5 ppm)
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Formation of Ozone
 Nitrogen dioxide absorbs energy from the sunlight and
decomposes to NO and O atoms
 O atoms are very reactive and will react with O2 to form ozone,
O3 (highly toxic and reactive)
 Both O2 and O3 react with hydrocarbons to produce harmful
pollutants sometimes referred to as photochemical smog.
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Oxides of Sulphur
SO2 is generated from burning sulphur-containing fuels:
S (organic) + O2 (g) → SO2 (g)
or from metal refining operations:
NiS (s) + 1.5 O2 (g) → NiO (s) + SO2 (g)
and in the presence of dust particles or UV light, SO2will react with
oxygen to form SO3:
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Oxides of Sulphur
 In humans, prolonged exposure to SO2 diminishes lung capacity




and aggravates respiratory problems.
At 0.15 ppm, people with existing problems will be incapacitated.
At 5 ppm, everyone will experience breathing difficulties.
Also, SO2 and SO3 can react with water to produce acid rain.
SO2 (g) + H2O (g) → H2SO3(aq)
SO3 (g) + H2O (g) → H2SO4 (aq)
SO2 can be scrubbed from flu gases before it reaches the
atmosphere:
SO2(g) + CaO (s) → CaSO3 (s)
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Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 2 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.