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Prentice-Hall © 2002 General Chemistry: Chapte r 6 Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8 th Edition

Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

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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

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

Page 2: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

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

Page 3: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

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

Page 4: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 4 of 41

6-1 Properties of Gases: Gas Pressure

• Gas Pressure

• Liquid Pressure

P (Pa) =

Area (m2)

Force (N)

P = g ·h ·d

Page 5: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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

Page 6: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 6 of 41

Manometers

Page 7: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 7 of 41

6-2 Simple Gas Laws

• Boyle 1662 P 1V

PV = constant

Page 8: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 8 of 41

Example 5-6

Relating Gas Volume and Pressure – Boyle’s Law.

P1V1 = P2V2 V2 = P1V1

P2

= 694 L Vtank = 644 L

Page 9: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 9 of 41

Charles’s Law

Charles 1787

Gay-Lussac 1802

V T V = b T

Page 10: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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

Page 11: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 12: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 12 of 41

Formation of Water

Page 13: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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:

Page 14: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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 nTP

Page 15: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 15 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

Page 16: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 16 of 41

The General Gas Equation

R = = P2V2

n2T2

P1V1

n1T1

= P2

T2

P1

T1

If we hold the amount and volume constant:

Page 17: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 17 of 41

6-4 Applications of the Ideal Gas Equation

Page 18: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 18 of 41

Molar Mass Determination

PV = nRT and n = mM

PV = mM

RT

M = m

PVRT

Page 19: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 20: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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

Page 21: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 21 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

Page 22: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 22 of 41

Gas Densities

PV = nRT and d = mV

PV = mM

RT

MPRTV

m= d =

, n = mM

Page 23: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 24: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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)

Page 25: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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

nRTV = =

(735 mm Hg)

(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)

760 mm Hg1.00 atm

Page 26: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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....

Page 27: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 27 of 41

Dalton’s Law of Partial Pressure

Page 28: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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

= aRecall

Page 29: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 29 of 41

Pneumatic Trough

Ptot = Pbar = Pgas + PH2O

Page 30: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 31: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 31 of 41

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

Page 32: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 32 of 41

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

Page 33: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 33 of 41

Pressure

M

3RTu

uM3RT

umRT3

um3

1PV

rms

2

2A

2A

N

NAssume one mole:

PV=RT so:

NAm = M:

Rearrange:

Page 34: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 34 of 41

Distribution of Molecular Speeds

M

3RTu rms

Page 35: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 35 of 41

Determining Molecular Speed

Page 36: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 36 of 41

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!

Page 37: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 38: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 39: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 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.

Page 40: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 40 of 41

Real Gases

Page 41: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 41 of 41

van der Waals Equation

P + n2a V2

V – nb = nRT

Page 42: Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General

Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 42 of 41

Chapter 6 Questions

9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.