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Prentice-Hall General Chemistry: Chapter 6Slide 1 of 41
Chapter 6: Gases
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th
Edition
Prentice-Hall General Chemistry: Chapter 6Slide 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
Prentice-Hall General Chemistry: Chapter 6Slide 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
Prentice-Hall General Chemistry: Chapter 6Slide 4 of 41
6-1 Properties of Gases: Gas Pressure
• Gas Pressure
• Liquid Pressure
P (Pa) =
Area (m2
)
Force (N)
P = g ·h ·d
Prentice-Hall General Chemistry: Chapter 6Slide 5 of 41
Barometric Pressure
Standard Atmospheric
Pressure
1.00 atm
760 mm Hg, 760 torr
101.325 kPa
1.01325 bar
1013.25 mbar
Prentice-Hall General Chemistry: Chapter 6Slide 6 of 41
Manometers
Prentice-Hall General Chemistry: Chapter 6Slide 7 of 41
6-2 Simple Gas Laws
• Boyle 1662 P 
1
V
PV = constant
Prentice-Hall General Chemistry: Chapter 6Slide 8 of 41
Example 5-6
Relating Gas Volume and Pressure – Boyle’s Law.
P1V1 = P2V2 V2 =
P1V1
P2
= 694 L Vtank = 644 L
Prentice-Hall General Chemistry: Chapter 6Slide 9 of 41
Charles’s Law
Charles 1787
Gay-Lussac 1802
V  T V = b T
Prentice-Hall General Chemistry: Chapter 6Slide 10 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
Prentice-Hall General Chemistry: Chapter 6Slide 11 of 41
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.
Prentice-Hall General Chemistry: Chapter 6Slide 12 of 41
Formation of Water
Prentice-Hall General Chemistry: Chapter 6Slide 13 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:
Prentice-Hall General Chemistry: Chapter 6Slide 14 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 
nT
P
Prentice-Hall General Chemistry: Chapter 6Slide 15 of 41
The Gas Constant
R =
PV
nT
= 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
Prentice-Hall General Chemistry: Chapter 6Slide 16 of 41
The General Gas Equation
R = =
P2V2
n2T2
P1V1
n1T1
=
P2
T2
P1
T1
If we hold the amount and volume constant:
Prentice-Hall General Chemistry: Chapter 6Slide 17 of 41
6-4 Applications of the Ideal Gas Equation
Prentice-Hall General Chemistry: Chapter 6Slide 18 of 41
Molar Mass Determination
PV = nRT and n =
m
M
PV =
m
M
RT
M =
m
PV
RT
Prentice-Hall General Chemistry: Chapter 6Slide 19 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.
Prentice-Hall General Chemistry: Chapter 6Slide 20 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
Prentice-Hall General Chemistry: Chapter 6Slide 21 of 41
Example 5-6Example 5-6
Use the Gas Equation:
PV = nRT PV =
m
M
RT M =
m
PV
RT
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
Prentice-Hall General Chemistry: Chapter 6Slide 22 of 41
Gas Densities
PV = nRT and d =
m
V
PV =
m
M
RT
MP
RTV
m
= d =
, n =
m
M
Prentice-Hall General Chemistry: Chapter 6Slide 23 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.
Prentice-Hall General Chemistry: Chapter 6Slide 24 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)
Prentice-Hall General Chemistry: Chapter 6Slide 25 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
nRT
V = =
(735 mm Hg)
(1.62 mol)(0.08206 L atm mol-1
K-1
)(299 K)
760 mm Hg
1.00 atm
Prentice-Hall General Chemistry: Chapter 6Slide 26 of 41
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....
Prentice-Hall General Chemistry: Chapter 6Slide 27 of 41
Dalton’s Law of Partial Pressure
Prentice-Hall General Chemistry: Chapter 6Slide 28 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
= χaRecall
Prentice-Hall General Chemistry: Chapter 6Slide 29 of 41
Pneumatic Trough
Ptot = Pbar = Pgas + PH2O
Prentice-Hall General Chemistry: Chapter 6Slide 30 of 41
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.
Prentice-Hall General Chemistry: Chapter 6Slide 31 of 41
Pressure – Assessing Collision Forces
• Translational kinetic energy,
• Frequency of collisions,
• Impulse or momentum transfer,
• Pressure proportional to
impulse times frequency
2
k mu
2
1
e =
V
N
u=v
mu=I
2
mu
V
N
P ∝
Prentice-Hall General Chemistry: Chapter 6Slide 32 of 41
Pressure and Molecular Speed
• Three dimensional systems lead to: 2
um
V
N
3
1
P =
2
u=
um is the modal speed
uav is the simple average
urms
Prentice-Hall General Chemistry: Chapter 6Slide 33 of 41
Pressure
M
3RT
u
uM3RT
umRT3
um
3
1
PV
rms
2
2
A
2
A
=
=
=
=
N
NAssume one mole:
PV=RT so:
NAm = M:
Rearrange:
Prentice-Hall General Chemistry: Chapter 6Slide 34 of 41
Distribution of Molecular Speeds
M
3RT
urms =
Prentice-Hall General Chemistry: Chapter 6Slide 35 of 41
Determining Molecular Speed
Prentice-Hall General Chemistry: Chapter 6Slide 36 of 41
Temperature
(T)
R
2
3
e
e
3
2
RT
)um
2
1
(
3
2
um
3
1
PV
A
k
k
22
A
N
N
NN
A
A
=
=
==Modify:
PV=RT so:
Solve for ek:
Average kinetic energy is directly proportional to temperature!
Prentice-Hall General Chemistry: Chapter 6Slide 37 of 41
6-8 Gas Properties Relating to the
Kinetic-Molecular Theory
• Diffusion
– Net rate is proportional to
molecular speed.
• Effusion
– A related phenomenon.
Prentice-Hall General Chemistry: Chapter 6Slide 38 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.
Prentice-Hall General Chemistry: Chapter 6Slide 39 of 41
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.
Prentice-Hall General Chemistry: Chapter 6Slide 40 of 41
Real Gases
Prentice-Hall General Chemistry: Chapter 6Slide 41 of 41
van der Waals Equation
P +
n2
a
V2
V – nb = nRT
Prentice-Hall General Chemistry: Chapter 6Slide 42 of 41
Chapter 6 Questions
9, 13, 18, 31, 45,
49, 61, 63, 71, 82,
85, 97, 104.

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Ch06 121227103404-phpapp01

  • 1. Prentice-Hall General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8th Edition
  • 2. Prentice-Hall General Chemistry: Chapter 6Slide 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
  • 3. Prentice-Hall General Chemistry: Chapter 6Slide 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
  • 4. Prentice-Hall General Chemistry: Chapter 6Slide 4 of 41 6-1 Properties of Gases: Gas Pressure • Gas Pressure • Liquid Pressure P (Pa) = Area (m2 ) Force (N) P = g ·h ·d
  • 5. Prentice-Hall General Chemistry: Chapter 6Slide 5 of 41 Barometric Pressure Standard Atmospheric Pressure 1.00 atm 760 mm Hg, 760 torr 101.325 kPa 1.01325 bar 1013.25 mbar
  • 6. Prentice-Hall General Chemistry: Chapter 6Slide 6 of 41 Manometers
  • 7. Prentice-Hall General Chemistry: Chapter 6Slide 7 of 41 6-2 Simple Gas Laws • Boyle 1662 P  1 V PV = constant
  • 8. Prentice-Hall General Chemistry: Chapter 6Slide 8 of 41 Example 5-6 Relating Gas Volume and Pressure – Boyle’s Law. P1V1 = P2V2 V2 = P1V1 P2 = 694 L Vtank = 644 L
  • 9. Prentice-Hall General Chemistry: Chapter 6Slide 9 of 41 Charles’s Law Charles 1787 Gay-Lussac 1802 V  T V = b T
  • 10. Prentice-Hall General Chemistry: Chapter 6Slide 10 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
  • 11. Prentice-Hall General Chemistry: Chapter 6Slide 11 of 41 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.
  • 12. Prentice-Hall General Chemistry: Chapter 6Slide 12 of 41 Formation of Water
  • 13. Prentice-Hall General Chemistry: Chapter 6Slide 13 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:
  • 14. Prentice-Hall General Chemistry: Chapter 6Slide 14 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  nT P
  • 15. Prentice-Hall General Chemistry: Chapter 6Slide 15 of 41 The Gas Constant R = PV nT = 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
  • 16. Prentice-Hall General Chemistry: Chapter 6Slide 16 of 41 The General Gas Equation R = = P2V2 n2T2 P1V1 n1T1 = P2 T2 P1 T1 If we hold the amount and volume constant:
  • 17. Prentice-Hall General Chemistry: Chapter 6Slide 17 of 41 6-4 Applications of the Ideal Gas Equation
  • 18. Prentice-Hall General Chemistry: Chapter 6Slide 18 of 41 Molar Mass Determination PV = nRT and n = m M PV = m M RT M = m PV RT
  • 19. Prentice-Hall General Chemistry: Chapter 6Slide 19 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.
  • 20. Prentice-Hall General Chemistry: Chapter 6Slide 20 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
  • 21. Prentice-Hall General Chemistry: Chapter 6Slide 21 of 41 Example 5-6Example 5-6 Use the Gas Equation: PV = nRT PV = m M RT M = m PV RT 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
  • 22. Prentice-Hall General Chemistry: Chapter 6Slide 22 of 41 Gas Densities PV = nRT and d = m V PV = m M RT MP RTV m = d = , n = m M
  • 23. Prentice-Hall General Chemistry: Chapter 6Slide 23 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.
  • 24. Prentice-Hall General Chemistry: Chapter 6Slide 24 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)
  • 25. Prentice-Hall General Chemistry: Chapter 6Slide 25 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 nRT V = = (735 mm Hg) (1.62 mol)(0.08206 L atm mol-1 K-1 )(299 K) 760 mm Hg 1.00 atm
  • 26. Prentice-Hall General Chemistry: Chapter 6Slide 26 of 41 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....
  • 27. Prentice-Hall General Chemistry: Chapter 6Slide 27 of 41 Dalton’s Law of Partial Pressure
  • 28. Prentice-Hall General Chemistry: Chapter 6Slide 28 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 = χaRecall
  • 29. Prentice-Hall General Chemistry: Chapter 6Slide 29 of 41 Pneumatic Trough Ptot = Pbar = Pgas + PH2O
  • 30. Prentice-Hall General Chemistry: Chapter 6Slide 30 of 41 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.
  • 31. Prentice-Hall General Chemistry: Chapter 6Slide 31 of 41 Pressure – Assessing Collision Forces • Translational kinetic energy, • Frequency of collisions, • Impulse or momentum transfer, • Pressure proportional to impulse times frequency 2 k mu 2 1 e = V N u=v mu=I 2 mu V N P ∝
  • 32. Prentice-Hall General Chemistry: Chapter 6Slide 32 of 41 Pressure and Molecular Speed • Three dimensional systems lead to: 2 um V N 3 1 P = 2 u= um is the modal speed uav is the simple average urms
  • 33. Prentice-Hall General Chemistry: Chapter 6Slide 33 of 41 Pressure M 3RT u uM3RT umRT3 um 3 1 PV rms 2 2 A 2 A = = = = N NAssume one mole: PV=RT so: NAm = M: Rearrange:
  • 34. Prentice-Hall General Chemistry: Chapter 6Slide 34 of 41 Distribution of Molecular Speeds M 3RT urms =
  • 35. Prentice-Hall General Chemistry: Chapter 6Slide 35 of 41 Determining Molecular Speed
  • 36. Prentice-Hall General Chemistry: Chapter 6Slide 36 of 41 Temperature (T) R 2 3 e e 3 2 RT )um 2 1 ( 3 2 um 3 1 PV A k k 22 A N N NN A A = = ==Modify: PV=RT so: Solve for ek: Average kinetic energy is directly proportional to temperature!
  • 37. Prentice-Hall General Chemistry: Chapter 6Slide 37 of 41 6-8 Gas Properties Relating to the Kinetic-Molecular Theory • Diffusion – Net rate is proportional to molecular speed. • Effusion – A related phenomenon.
  • 38. Prentice-Hall General Chemistry: Chapter 6Slide 38 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.
  • 39. Prentice-Hall General Chemistry: Chapter 6Slide 39 of 41 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.
  • 40. Prentice-Hall General Chemistry: Chapter 6Slide 40 of 41 Real Gases
  • 41. Prentice-Hall General Chemistry: Chapter 6Slide 41 of 41 van der Waals Equation P + n2 a V2 V – nb = nRT
  • 42. Prentice-Hall General Chemistry: Chapter 6Slide 42 of 41 Chapter 6 Questions 9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.

Editor's Notes

  1. Iron rusts Natural gas burns Represent chemical reactions by chemical equations. Relationship between reactants and products is STOICHIOMETRY Many reactions occur in solution-concentration uses the mole concept to describe solutions
  2. Difficult to measure force exerted by gas molecules Measure gas pressure indirectly by comparing it with a liquid pressure. Liquid pressure depends only on the height of the liquid column and the density of the liquid.
  3. 1 atm = 760 mm Hg when δHg = 13.5951 g/cm3 (0°C) g = 9.80665 m/s2 Mention here that Pbar refers to MEASURED ATMOSPHERIC PRESSURE in the text.
  4. Difficult to place a barometer inside a gas to be measured. Manometers compare gas pressure and barometric pressure.
  5. Three different gases show this behavior with temperature. Temperature at which the volume of a hypothetical gas becomes 0 is the absolute zero of temperature. The hypothetical gas has mass, but no volume, and does not condense into a liquid or solid.
  6. Dalton thought H + O → HO so ratio should have been 1:1:1. So the second part of Avogadro’s hypothesis is critical. This identifies the relationship between stoichiometry and gas volume.
  7. Any gas whose behavior conforms to the ideal gas equation is called an ideal or perfect gas. R is the gas constant. Substitute and calculate.
  8. This gas must be ketene. There really are no other possibilities.
  9. Partial pressure is the pressure of a component of gas that contributes to the overall pressure. Partial volume is the volume that a gas would occupy at the total pressure in the chamber. Ratio of partial volume to total volume, or of partial pressure to total pressure is the MOLE FRACTION.
  10. Total pressure of wet gas is equal to atmospheric pressure (Pbar) if the water level is the same inside and outside. At specific temperatures you know the partial pressure of water. Can calculate Pgas and use it stoichiometric calculations.
  11. Natural laws are explained by theories. Gas law led to development of kinetic-molecular theory of gases in the mid-nineteenth century.
  12. Lighter gases have faster speeds.