prentice-hall © 2008general chemistry: chapter 6slide 1 of 41 chapter 6: gases philip dutton...

Prentice-Hall © 2008 General Chemistry: Chapter 6 Slide 1 of 41

Chapter 6: Gases

Philip DuttonUniversity of Windsor, Canada

Prentice-Hall © 2008

General ChemistryPrinciples and Modern Applications

Petrucci • Harwood • Herring

8th Edition


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


Contents

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


6-1 Properties of Gases: Gas Pressure

• Gas Pressure

• Liquid Pressure

P (Pa) =

Area (m2)

Force (N)

P (Pa) = g ·h ·d

d(Hg) = 13.5951 g/cm3 (0°C)

g = 9.807 m/s2 - acceleration due to gravity

h - height of the liquid column

d - density of the liquid


Barometric Pressure

Standard Atmospheric Pressure

1.00 atm

760 mm Hg, 760 torr

101.325 kPa

1.01325 bar

1013.25 mbar

Atmospheric Pressure: isobars in hPa

General Chemistry: Chapter 6 Slide 6 of 41


Manometers


6-2 Simple Gas Laws, Isotherm: T=const

• Boyle 1662 P ~ 1V PV = constant

T=const


Example 5-6

Relating Gas Volume and Pressure – Boyle’s Law.

P1V1 = P2V2 V2 = P1V1

P2

= 694 L Vtank = 644 L


Charles’s Law, Isobar, P=const

Charles 1787

Gay-Lussac 1802

V ~ T V = b T


STP

• Gas properties depend on conditions.• Define standard temperature and pressure

(STP).• International Union of Pure and Applied Chemistry

(IUPAC):

Temperature Absolute pressure Publishing or establishing entity

°C kPa 0 100 IUPAC (present definition)

0 101.325 IUPAC (former definition), NIST, ISO 10780

20 101.325 EPA, NIST25 101.325 EPA


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.


Formation of Water


Avogadro’s Law

V ~ n or V = c n

At STP

1 mol gas = 22.41 L gas

At an a fixed temperature and pressure:


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


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

P-V-T Surface



The General Gas Equation

R = = P2V1

n1T2

P1V1

n1T1

= P2

T2

P1

T1

If we hold the amount and volume constant:


6-4 Applications of the Ideal Gas Equation


Molar Mass Determination

PV = nRT and n = mM

PV = mM

RT

M = m

PVRT


Example 6-4

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 g 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.


Example 6-4

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 dm3


Example 5-6Example 6-4, Molar Mass with the Ideal Gas Equation

Use the Ideal Gas Equation:

PV = nRT PV = mM

RT M = m

PVRT

M = (P kPa)(0.09841 dm3)

(0.1654 g)(8.3145 J mol-1 K-1)(297.2 K)

M = 42.08 g · mol-1

P = 740.3/760 * 101.325 kPa

[Pa · m3] = [kPa · dm3] = [J]


Gas Densities

PV = nRT and d = mV

PV = mM

RT

MPRTV

m= d =

, n = mM


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.


Example 6-5

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)


Example 6-5

Determine moles of N2:

Determine volume of N2:

nN2 = 70 g ·

1 mol NaN3

65.01 g·

3 mol N2

2 mol NaN3

= 1.62 mol N2

= 41.1 l

P

nRTV = =

(735 mm Hg)

(1.62 mol)(8.3145 J mol-1 K-1)(299 K)

760 mm Hg101.325 kPa

Airbag design



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....


Dalton’s Law of Partial Pressure


Partial pressure and volume

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

= xaRecall

Average molar mass of ideal gas mixtures

The average molar mass is a sum of the products of the components molar fractions (xi) and masses (Mi).

n

iiinn MxMxMxMxM

12211 ...

The athmosphere

Average

molar mass?


Pneumatic Trough

Ptot = Pbar = Pgas + PH2O

Equilibrium vapor pressure of H2O vs temperature

Csonka,G. I. © 2008 General Chemistry: Chapter 6 Slide 36 of 41

0 10 20 30 40 50 60 70 80 90 1000

20

40

60

80

100

t (C)

P*

(kP

a)


Water Vapor

• nH2O PH2O in air.

Relative Humidity =PH2O (actual)

PH2O (max)· 100%

Relative humidity map

General Chemistry: Chapter 6 Slide 38 of 41 Csonka,G. I. © 2013


Chemicals from the Atmosphere


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.


Pressure – Assessing Collision Forces

• Translational kinetic energy,

• Frequency of collisions,

• Impulse or momentum transfer,

• Pressure proportional to impulse times frequency

2

2

1muek

V

Nuv

muI

2muV

NP~


Pressure and Molecular Speed

• Three dimensional systems lead to: 2

3

1um

V

NP

2u

um is the modal speed

uav is the simple average

urms


Pressure

M

RTu

uMRT

umNRT

umNPV

rms

A

A

3

3

3

3

1

2

2

2

Assume one mole:

PV=RT so:

NAm = M:

Rearrange:


Distribution of Molecular Speeds

M

RTurms

3


Determining Molecular Speed


Temperature

TN

Re

eNRT

)um(NumNPV

Ak

kA

AA

2

33

22

1

3

2

3

1 22

Modify:

PV=RT so:

Solve for ek:

Average kinetic energy is directly proportional to temperature!


6-8 Gas Properties Relating to the Kinetic-Molecular Theory

• Diffusion– Net rate is proportional to

molecular speed.

• Effusion– A related phenomenon.


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

RT/MB

RT/M

)(u

)(u

Bofeffusionofrate

Aofeffusionofrate

3

3

• Ratio used can be:– Rate of effusion (as above)– Molecular speeds– Effusion times

– Distances traveled by molecules– Amounts of gas effused.


6-9 Real Gases

• Compressibility factor PV/nRT = 1 – Ideal gas• Deviations occur for real gases.

– PV/nRT > 1 - molecular volume is significant.– PV/nRT < 1 – intermolecular forces of attraction.

The green molecule is attracted by the red molecules.

Compression factor, Z

• Product of pressure (P) and molar volume (Vm) divided by the gas constant (R) and thermodynamic temperature (T).

• For an ideal gas it is equal to 1.

• As n = 1


TR

VPZ m

Gas Vm (dm3)

He 22.436

Ne 22.421

H2 22.429

CO2 22.295

H2O 22.199

Real gases at STP

P0 and T = 0 C°

Csonka, G. I. © 2008


van der Waals Equation

TRnbnVVn

aP

2

2

Vn

abnV

TRnP

van der Waals Coefficients

Gas a (Pa m6/mol2) b(m3/mol)

Helium 0.0036 24.1 · 10-6

Neon 0.0212 17.1 · 10-6

Hydrogen 0.0245 26.6 · 10-6

Carbon dioxide 0.366 42.8 · 10-6

Water vapor 0.552 30.4 · 10-6

CH4 0.228 42.8· 10-6

2mm V

a

bV

TRP

van der Waals vs ideal curves



Chapter 6 Questions

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

prentice-hall © 2008general chemistry: chapter 6slide 1 of 41 chapter 6: gases philip dutton...

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