Petrucci • Harwood • Herring • Madura
GENERAL
CHEMISTRY
Ninth
Edition
Principles and Modern Applications
Chapter 6: Gases
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2007
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General Chemistry: Chapter 6
Prentice-Hall © 2007
Contents
6-1
6-2
6-3
6-4
6-5
6-6
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Properties of Gases: Gas Pressure
The Simple Gas Laws
Combining the Gas Laws:
The Ideal Gas Equation and
The General Gas Equation
Applications of the Ideal Gas Equation
Gases in Chemical Reactions
Mixtures of Gases
General Chemistry: Chapter 6
Prentice-Hall © 2007
Contents
6-6
6-7
6-8
6-9
Mixtures of Gases
Kinetic—Molecular Theory of Gases
Gas Properties Relating to the
Kinetic—Molecular Theory
Nonideal (Real) Gases
 Focus On Earth’s Atmosphere
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General Chemistry: Chapter 6
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6-1 Properties of Gases: Gas Pressure
 The gaseous states of three halogens.
 Most common gases are colorless
 H2, O2, N2, CO and CO2
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General Chemistry: Chapter 6
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The Concept of Pressure
 The pressure exerted
by a solid.
 Both cylinders have the
same mass
 They have different
areas of contact
Force (N)
P (Pa) =
Area (m2)
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General Chemistry: Chapter 6
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Liquid Pressure
The pressure exerted by
a liquid depends on:
 The height of the
column of liquid.
 The density of the
column of liquid.
P = g ·h ·d
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General Chemistry: Chapter 6
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Barometric Pressure
Standard Atmospheric Pressure
1.00 atm, 760 mm Hg, 760 torr, 101.325 kPa, 1.01325 bar
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General Chemistry: Chapter 6
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Manometers
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General Chemistry: Chapter 6
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6-2 Simple Gas Laws
 Boyle 1662
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1
P
V
General Chemistry: Chapter 6
PV = constant
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EXAMPLE 6-4
Relating Gas Volume and Pressure – Boyle’s Law. The
volume of a large irregularly shaped, closed tank can be
determined. The tank is first evacuated and then connected to a
50.0 L cylinder of compressed nitrogen gas. The gas pressure in
the cylinder, originally at 21.5 atm, falls to 1.55 atm after it is
connected to the evacuated tank. What is the volume of the
tank?
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General Chemistry: Chapter 6
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EXAMPLE 6-4
P1V1 = P2V2
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P1V1
= 694 L
V2 =
P2
General Chemistry: Chapter 6
Vtank = 644 L
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Charles’s Law
Charles 1787
Gay-Lussac 1802
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VT
General Chemistry: Chapter 6
V=bT
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Standard Temperature and Pressure
 Gas properties depend on conditions.
 Define standard conditions of temperature
and pressure (STP).
P = 1 atm = 760 mm Hg
T = 0°C
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= 273.15 K
General Chemistry: Chapter 6
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Avogadro’s Law
Gay-Lussac 1808
 Small volumes of gases react in the ratio of
small whole numbers.
Avogadro 1811
 Equal volumes of gases have equal numbers of
molecules and
 Gas molecules may break up when they react.
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General Chemistry: Chapter 6
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Formation of Water
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General Chemistry: Chapter 6
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Avogadro’s Law
At an a fixed temperature and pressure:
Vn
or
V=cn
At STP
1 mol gas = 22.4 L gas
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General Chemistry: Chapter 6
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6-3 Combining the Gas Laws: The Ideal
Gas Equation and the General Gas
Equation
 Boyle’s law
V  1/P
 Charles’s law
VT
nT
V
P
 Avogadro’s law V  n
PV = nRT
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General Chemistry: Chapter 6
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The Gas Constant
PV = nRT
PV
R=
nT
= 0.082057 L atm mol-1 K-1
= 8.3145 m3 Pa mol-1 K-1
= 8.3145 J mol-1 K-1
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General Chemistry: Chapter 6
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Using the Gas Law
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General Chemistry: Chapter 6
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The General Gas Equation
P1V1
P2V2
R=
=
n1T1
n2T2
If we hold the amount and volume constant:
P1
T1
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=
P2
T2
General Chemistry: Chapter 6
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Using the Gas Laws
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General Chemistry: Chapter 6
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6-4 Applications of the Ideal Gas Equation
Molar Mass Determination
PV = nRT
and
n=
m
M
m
RT
PV =
M
m RT
M=
PV
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General Chemistry: Chapter 6
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EXAMPLE 6-10
Determining a Molar Mass with the Ideal Gas Equation.
Polypropylene is an important commercial chemical. It is used
in the synthesis of other organic chemicals and in plastics
production. A glass vessel weighs 40.1305 g when clean, dry
and evacuated; it weighs 138.2410 when filled with water at
25°C (δwater = 0.9970 g cm-3) and 40.2959 g when filled with
propylene gas at 740.3 mm Hg and 24.0°C. What is the molar
mass of polypropylene?
Strategy:
Determine Vflask. Determine mgas. Use the Gas Equation.
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General Chemistry: Chapter 6
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EXAMPLE 6-10
Determine Vflask:
Vflask = mH2O  dH2O = (138.2410 g – 40.1305 g)  (0.9970 g cm-3)
= 98.41 cm3 = 0.09841 L
Determine mgas:
mgas = mfilled - mempty = (40.2959 g – 40.1305 g)
= 0.1654 g
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General Chemistry: Chapter 6
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EXAMPLE 5-6
Use the Gas Equation:
PV = nRT
M=
m
RT
PV =
M
m RT
M=
PV
(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)
(0.9741 atm)(0.09841 L)
M = 42.08 g/mol
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General Chemistry: Chapter 6
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Gas Densities
PV = nRT
and
m
m
, n=
d=
M
V
m
RT
PV =
M
m
MP
=d=
V
RT
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General Chemistry: Chapter 6
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6-5 Gases in Chemical Reactions
Stoichiometric factors relate gas quantities to
quantities of other reactants or products.
Ideal gas equation relates the amount of a gas
to volume, temperature and pressure.
Law of combining volumes can be developed
using the gas law.
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General Chemistry: Chapter 6
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EXAMPLE 6-12
Using the Ideal gas Equation in Reaction Stoichiometry
Calculations. The decomposition of sodium azide, NaN3, at
high temperatures produces N2(g). Together with the necessary
devices to initiate the reaction and trap the sodium metal formed,
this reaction is used in air-bag safety systems. What volume of
N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0
g NaN3 is decomposed?
2 NaN3(s) → 2 Na(l) + 3 N2(g)
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General Chemistry: Chapter 6
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EXAMPLE 6-12
Determine moles of N2:
1 mol NaN3
nN2 = 70 g NaN3 
65.01 g NaN3
3 mol N2

2 mol NaN3
= 1.62 mol N2
Determine volume of N2:
V=
nRT
P
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(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
= 41.1 L
=
(735 mm Hg) 
1.00 atm
760 mm Hg
General Chemistry: Chapter 6
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6-6 Mixtures of Gases
Gas laws apply to mixtures of gases.
Simplest approach is to use ntotal, but....
Partial pressure
 Each component of a gas mixture exerts a
pressure that it would exert if it were in the
container alone.
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General Chemistry: Chapter 6
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Dalton’s Law of Partial Pressure
The total pressure of a mixture of gases is the sum of
the partial pressures of the components of the mixture.
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General Chemistry: Chapter 6
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Partial Pressure
Ptot = Pa + Pb +…
Va = naRT/Ptot
and
Vtot = Va + Vb+…
na
naRT/Ptot
Va
=
=
ntotRT/Ptot
ntot
Vtot
Recall
na
= a
ntot
na
naRT/Vtot
Pa
=
=
ntotRT/Vtot
ntot
Ptot
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General Chemistry: Chapter 6
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Pneumatic Trough
Ptot = Pbar = Pgas + PH2O
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General Chemistry: Chapter 6
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6-7 Kinetic Molecular Theory
 Particles are point masses in constant,
random, straight line motion.
 Particles are separated by great
distances.
 Collisions are rapid and elastic.
 No force between particles.
 Total energy remains constant.
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General Chemistry: Chapter 6
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Pressure – Assessing Collision Forces
 Translational kinetic energy,
1
e k  mu 2
2
 Frequency of collisions,
N
vu
V
 Impulse or momentum transfer,
I  mu
 Pressure proportional to
impulse times frequency
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N
P  mu 2
V
General Chemistry: Chapter 6
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Pressure and Molecular Speed
 Three dimensional systems lead to:
1N
P
m u2
3V
um is the modal speed
uav is the simple average
urms  u 2
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General Chemistry: Chapter 6
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Pressure
Assume one mole:
1
2
PV  N A m u
3
PV=RT so:
3RT  N A m u
NAm = M:
3RT  M u
Rearrange:
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u rms 
2
2
3RT
M
General Chemistry: Chapter 6
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Distribution of Molecular Speeds
u rms 
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General Chemistry: Chapter 6
3RT
M
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Determining Molecular Speed
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General Chemistry: Chapter 6
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Temperature
Modify:
1
2
1
2
2
PV  N A m u  N A ( m u )
2
3
3
PV=RT so:
2
RT  N A ek
3
Solve for ek:
3 R
(T)
ek 
2 NA
Average kinetic energy is directly proportional to temperature!
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General Chemistry: Chapter 6
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6-8 Gas Properties Relating to the
Kinetic-Molecular Theory
 Diffusion
 Net rate is proportional to
molecular speed.
 Effusion
 A related phenomenon.
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General Chemistry: Chapter 6
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Graham’s Law
rate of effusion of A (u rms ) A


rate of effusion of B (u rms ) B
3RT/M A

3RT/MB
MB
MA
 Only for gases at low pressure (natural escape, not a jet).
 Tiny orifice (no collisions)
 Does not apply to diffusion.
 Ratio used can be:
 Rate of effusion (as above)
 Molecular speeds
 Effusion times
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 Distances traveled by molecules
 Amounts of gas effused.
General Chemistry: Chapter 6
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6-9 Nonideal (Real) Gases
 Compressibility factor PV/nRT = 1
 Deviations occur for real gases.
 PV/nRT > 1 - molecular volume is significant.
 PV/nRT < 1 – intermolecular forces of attraction.
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General Chemistry: Chapter 6
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Real Gases
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General Chemistry: Chapter 6
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van der Waals Equation
n2a
P+
V2
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V – nb
= nRT
General Chemistry: Chapter 6
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End of Chapter Questions
A problem is like a knot in a ball of wool:
 If you pull hard on any loop:
◦ The knot will only tighten.
◦ The solution (undoing the knot) will not be achieved.
 If you pull lightly on one loop and then another:
◦ You gradually loosen the knot.
◦ As more loops are loosened it becomes easier to
undo the subsequent ones.
Don’t pull too hard on any one piece of
information in your problem, it tightens.
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General Chemistry: Chapter 6
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