Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 1 of 41
Chapter 6: Gases
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 2 of 41
Contents
6-1 Properties of Gases: Gas Pressure
6-2 The Simple Gas Laws
6-3 Combining the Gas Laws:
The Ideal Gas Equation and
The General Gas Equation
6-4 Applications of the Ideal Gas Equation
6-5 Gases in Chemical Reactions
6-6 Mixtures of Gases
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Contents
6-6 Mixtures of Gases
6-7 Kinetic—Molecular Theory of Gases
6-8 Gas Properties Relating to the Kinetic—Molecular Theory
6-9 Non-ideal (real) Gases
Focus on The Chemistry of Air-Bag Systems
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6-1 Properties of Gases: Gas Pressure
• Gas Pressure
• Liquid Pressure
P (Pa) =
Area (m2)
Force (N)
P = g ·h ·d
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Barometric Pressure
Standard Atmospheric Pressure
1.00 atm
760 mm Hg, 760 torr
101.325 kPa
1.01325 bar
1013.25 mbar
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Manometers
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6-2 Simple Gas Laws
• Boyle 1662 P 1V
PV = constant
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Example 5-6
Relating Gas Volume and Pressure – Boyle’s Law.
P1V1 = P2V2 V2 = P1V1
P2
= 694 L Vtank = 644 L
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Charles’s Law
Charles 1787
Gay-Lussac 1802
V T V = b T
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STP
• Gas properties depend on conditions.
• Define standard conditions of temperature and pressure (STP).
P = 1 atm = 760 mm Hg
T = 0°C = 273.15 K
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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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Formation of Water
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Avogadro’s Law
V n or V = c n
At STP
1 mol gas = 22.4 L gas
At an a fixed temperature and pressure:
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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• Avogadro’s law V n
PV = nRT
V nTP
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The Gas Constant
R = PVnT
= 0.082057 L atm mol-1 K-1
= 8.3145 m3 Pa mol-1 K-1
PV = nRT
= 8.3145 J mol-1 K-1
= 8.3145 m3 Pa mol-1 K-1
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The General Gas Equation
R = = P2V2
n2T2
P1V1
n1T1
= P2
T2
P1
T1
If we hold the amount and volume constant:
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6-4 Applications of the Ideal Gas Equation
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Molar Mass Determination
PV = nRT and n = mM
PV = mM
RT
M = m
PVRT
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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 (δ=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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Example 5-6
Determine Vflask:
Vflask = mH2O dH2O = (138.2410 g – 40.1305 g) (0.9970 g cm-3)
Determine mgas:
= 0.1654 g
mgas = mfilled - mempty = (40.2959 g – 40.1305 g)
= 98.41 cm3 = 0.09841 L
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Example 5-6Example 5-6
Use the Gas Equation:
PV = nRT PV = mM
RT M = m
PVRT
M = (0.9741 atm)(0.09841 L)
(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)
M = 42.08 g/mol
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Gas Densities
PV = nRT and d = mV
PV = mM
RT
MPRTV
m= d =
, n = mM
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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 used to relate 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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Example 6-10
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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Example 6-10
Determine moles of N2:
Determine volume of N2:
nN2 = 70 g N3
1 mol NaN3
65.01 g N3/mol N3
3 mol N2
2 mol NaN3
= 1.62 mol N2
= 41.1 L
P
nRTV = =
(735 mm Hg)
(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
760 mm Hg1.00 atm
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6-6 Mixtures of Gases
• Partial pressure– Each component of a gas mixture exerts a
pressure that it would exert if it were in the container alone.
• Gas laws apply to mixtures of gases.
• Simplest approach is to use ntotal, but....
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Dalton’s Law of Partial Pressure
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Partial Pressure
Ptot = Pa + Pb +…
Va = naRT/Ptot and Vtot = Va + Vb+…
Va
Vtot
naRT/Ptot
ntotRT/Ptot= =
na
ntot
Pa
Ptot
naRT/Vtot
ntotRT/Vtot= =
na
ntot
na
ntot
= aRecall
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Pneumatic Trough
Ptot = Pbar = Pgas + PH2O
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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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Pressure – Assessing Collision Forces
• Translational kinetic energy,
• Frequency of collisions,
• Impulse or momentum transfer,
• Pressure proportional to impulse times frequency
2k mu
2
1e
V
Nuv
muI
2muV
NP
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Pressure and Molecular Speed
• Three dimensional systems lead to: 2umV
N
3
1P
2u
um is the modal speed
uav is the simple average
urms
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Pressure
M
3RTu
uM3RT
umRT3
um3
1PV
rms
2
2A
2A
N
NAssume one mole:
PV=RT so:
NAm = M:
Rearrange:
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Distribution of Molecular Speeds
M
3RTu rms
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Determining Molecular Speed
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Temperature
(T)R
2
3e
e3
2RT
)um2
1(
3
2um
3
1PV
Ak
k
22A
N
N
NN
A
A
Modify:
PV=RT so:
Solve for ek:
Average kinetic energy is directly proportional to temperature!
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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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Graham’s Law
• Only for gases at low pressure (natural escape, not a jet).
• Tiny orifice (no collisions)
• Does not apply to diffusion.
A
BA
Brms
Arms
M
M
3RT/MB
3RT/M
)(u
)(u
Bofeffusionofrate
Aofeffusionofrate
• Ratio used can be:– Rate of effusion (as above)
– Molecular speeds
– Effusion times
– Distances traveled by molecules
– Amounts of gas effused.
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6-9 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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Real Gases
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van der Waals Equation
P + n2a V2
V – nb = nRT
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Chapter 6 Questions
9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.