prentice-hall © 2002general chemistry: chapter 6slide 1 of 41 chapter 6: gases philip dutton...
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![Page 1: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General](https://reader036.vdocuments.net/reader036/viewer/2022081512/56649e755503460f94b762ed/html5/thumbnails/1.jpg)
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
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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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 3 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 4 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 5 of 41
Manometers
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 6 of 41
6-2 Simple Gas Laws
• Boyle 1662 P 1V
PV = constant
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 7 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 8 of 41
Charles’s Law
Charles 1787
Gay-Lussac 1802
V T V = b T
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 9 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 10 of 41
Avogadro’s Law
• Avogadro 1811– Equal volumes of gases have equal numbers of
molecules and– Gas molecules may break up when they react.
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 11 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 12 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 13 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 14 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 15 of 41
6-4 Applications of the Ideal Gas Equation
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 16 of 41
Molar Mass Determination
PV = nRT and n = mM
PV = mM
RT
M = m
PVRT
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 17 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 18 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 19 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 20 of 41
Gas Densities
PV = nRT and d = mV
PV = mM
RT
MPRTV
m= d =
, n = mM
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 21 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 22 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 23 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 24 of 41
Dalton’s Law of Partial Pressure
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 25 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 26 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 27 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 28 of 41
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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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 29 of 41
Chapter 6 Questions
9, 13, 18, 31, 45, 49, 61, 63,