Notes Chapter 05

8/12/2019 Notes Chapter 05

1/21

Chapter 5: Gases and the

Kinetic - Molecular Theory

5.1 An Overview of the Physical States of Matter5.2 Gas Pressure and its Measurement

5.3 The Gas Laws and Their Experimental Foundations

5.4 Further Applications of the Ideal Gas Law

5.5 The Ideal Gas Law and Reaction Stoichiometry

5.6 The Kinetic - Molecular Theory: A Model for Gas

Behavior

5.7 Real Gases: Deviations from Ideal Behavior

Important Characteristics of Gases

1) Gases are highly compressibleAn external force compresses the gas sample and decreases its

volume, removing the external force allows the gas volume to

increase.

2) Gases are highly thermally expandableWhen a gas sample is heated, its volume increases, and when it is

cooled its volume decreases.

3) Gases have low viscosityGases flow much easier than liquids or solids.

4) Most gases have low densitiesGas densities are on the order of grams per liter whereas liquids

and solids are grams per cubic cm, 1000 times greater.

5) Gases are infinitely miscibleGases mix in any proportion such as in air, a mixture of many gases.

Helium He 4.0

Neon Ne 20.2

Argon Ar 39.9

Hydrogen H2 2.0

Nitrogen N2 28.0

Nitrogen Monoxide NO 30.0

Oxygen O2 32.0

Hydrogen Chloride HCL 36.5

Ozone O3 48.0

Ammonia NH3 17.0

Methane CH4 16.0

Substances that are Gases underNormal Conditions (300 K 1 atm)

Substance Formula MM(g/mol)

Chapter 5: Gases and the

Kinetic - Molecular Theory

5.2 The Gas Laws of Boyle, Charles, and Avogadro

5.3 The Ideal Gas Law

5.4 Gas Stoichiometry5.5 Daltons Law of Partial Pressures

5.6 The Kinetic - Molecular Theory of Gases

5.7 Effusion and Diffusion

5.8 Collisions of Gas Particles with Container Walls

5.9 Intermolecular Collisions

5.10 Real Gases

5.11 Chemistry in the Atmosphere

Gas pressure can

be measured

using a

manometer

760. mm Hg =1 atm

Boyles Law: At constant T, the product PV for a

given amount of a gas is constant


2/21

CHARLESS LAW: V/T = constant

if P, n held constant

AVOGADROSLAW:

V/n = constant

at constant T, P

Moles gas in B is

double that in A

so V is double

Ideal Gas Law An ideal gas is defined as one for which both the

volume of molecules and forces between themolecules are so small that they have no effect onthe behavior of the gas.

Independent of substance! In the limit that P0,

all gases behave ideally

The ideal gas law is:

PV=nRT T is in kelvin

R = ideal gas constant = 8.314 J / mol K = 8.314 Jmol-1 K-1

R = 0.08206 L atm mol-1 K-1

The Ideal Gas Law Subsumes the Other Gas Laws

-- for a fixed amount at constant temperature

PV = nRT = constant PV = nRT

Boyles Law

for a fixed amount at constant volume

P/T = nR/V = constant P = (nR/V)T

Amontons Law

for a fixed amount at constant pressure

V/T = nR/P = constant V = (nR/P)T

Charless Law

for a fixed volume and temperature

P/n = RT/V = constant P = (RT/V)n

Avogadros Law

Standard Temperature and Pressure (STP)

A set of Standard conditions have been chosen to make it easier to

understand the gas laws, and gas behavior.

Standard Temperature = 00 C = 273.15 K

Standard Pressure = 1 atmosphere = 760. mm Hg (Mercury)

or 760. torr

At these standard conditions, 1.0 mole of a gas will occupy a

standard molar volume of 22.4 L.


3/21

Many gas law problems involve a change of

conditions, with no change in the amount of gas.

= constant Therefore, for a changeof conditions :

T1 T2

P x V

T

P1 x V1=

P2 x V2

EP100: Change of Three Variables - I

A gas sample in the laboratory has a volume of45.9 L at 25.0 oC and a pressure of 743 mm Hg.If the temperature is increased to 155 oC bycompressing the gas to a new volume of 31.0 Lwhat is the pressure in atm?

P1= 743 mm Hg x1 atm/ 760. mm Hg=

P2 = ?

V1 = 45.9 L V2 = 31.0 L

T1 = 25.0oC + 273 =

T2

= 155 oC + 273 =

1 1 2 2

1 2

PV PV

T T=

EP101: Gas Law

Calculate the pressure (in atm) in a container filled with 5.038kg of

xenon at a temperature of 18.8 oC, whose volume is 87.5 L.

Strategy:

(1)Convert all information into the units required, and

(2)substitute into the ideal gas equation.

EP102: Sodium Azide Decomposition - ISodium azide (NaN3) is used in air bags in automobiles. Write

a balanced chemical equation for the decomposition of

NaN3 into N2(g) and Na(s). Calculate the volume in liters of

nitrogen gas generated at 21.0 oC and 823 mm Hg by the

decomposition of 60.0 g of NaN3.

Strategy:

(1) Write balanced equation.

(2) Use stoichiometric coefficients to calculate moles of N2.

(3)Convert given parameters into appropriate units.

(4) Calculate V using the ideal gas law.


4/21

mol NaN3 =

mol N2=

T = 273.15 +

P = 823 mm x

EP102a: Calculate the density of ammonia gas(NH3) in grams per liter at 752 mm Hg and

55.0 oC.

Strategy:(1) Assume one mole and calculate V for given

conditions.

(2) Use density = mass/volume to calculate density.

Calculation of Molar Mass

n = =P x V

R x T

Mass

Molar Mass

Molar Mass = MM =Mass x R x T

P x V


5/21

EP103: A volatile liquid is placed in 590.0 ml flask and allowed toboil until only vapor fills the flask at a temperature of 100.0 oC and

736 mm Hg pressure. If the mass of the flask before and after the

experiment was 148.375g and 149.457 g, what is the molar mass of

the liquid?

Strategy:Use the gas law to calculate the molar mass of the liquid.

Pressure = 736 mm Hg x 1atm/760. mm Hg = 0.968 atm

Gas Mixtures

Gas behavior depends on the number,

not the identity, of gas molecules.

The ideal gas equation applies to each

gas individually and to the mixture as

a whole.

All molecules in a sample of an ideal

gas behave exactly the same way.

Daltons Law of Partial Pressures - I

In a mixture of gases, each gas contributes to the

total pressure: the pressure it would exert if the gas

were present in the container by itself.

To obtain a total pressure, add all of the partial

pressures: Ptotal = P1+P2+P3+PN

Pressure exerted by an ideal gas mixture is determinedby the total number of moles:

P=(ntotal RT)/Vntotal = sum of the amounts of each gas pressure

the partial pressure is the pressure of gas if it was presentby itself.

P1 = (n1 RT)/V P2 = (n2 RT)/V P3 = (n3RT)/V

the total pressure is the sum of the partial pressures.P= (n1 RT)/V + (n2 RT)/V + (n3RT)/V= (n1+n2+ n3)(RT/V) =(ntotal RT)/V

Partial pressures in terms of mol fractions

1 1 1 1

1 2 3

P n= P = x PP n + n + n

EP104: Daltons Law of Partial PressuresA 2.00 L flask contains 3.00 g of CO2 and 0.100 g

of helium at a temperature of 17.0 oC.

What are the partial pressures of each gas, and

the total pressure?

T = 17.0 oC + 273.15 =

nCO2 =


6/21

EP105: Daltons Law using mole fractionsA mixture of gases contains 4.46 mol Ne, 0.740

mol Ar and 2.15 mol Xe. What are the partialpressures of the gases if the total pressure is2.00 atm ?

Strategy:(1) Calculate mol fraction each gas

(2) Partial pressure = mol fraction x total P

Total # moles =

XNe =

PNe =

Whats Behind the Ideal Gas Law?

A purely empirical law - solely the

consequence of experimental

observations

Explains the behavior of gases over a

limited range of conditions.

A macroscopic explanation. Says

nothing about the microscopic behavior

of the atoms or molecules that make up

the gas.

The Kinetic Theory of Gases

Starts with a set of assumptions about themicroscopic behavior of matter at the atomic level.

Supposes that the constituent particles (molecules)of the gas obey the laws of classical physics.

Accounts for the random behavior of the particleswith statistics, thereby establishing a new branchof physics - statistical mechanics.

Offers an explanation of the macroscopic behaviorof gases.

Predicts experimental phenomena that ideal gaslaw cant predict. (Maxwell-Boltzmann Speed

Distribution)

The Four Postulates of the Kinetic Theory

A pure gas consists of a large number ofidentical molecules separated by distancesthat are great compared with their size.

The gas molecules are constantly moving inrandom directions with a distribution of

speeds. The molecules exert no forces on one

another between collisions, so betweencollisions they move in straight lines withconstant velocities.

The collisions of molecules with the walls ofthe container are elastic; no energy is lostduring a collision. Collisions with the wall ofthe container are the cause of pressure.

An ideal gas

molecule in a cube

of sides L.


7/21

Cartesian Coordinate System for the Gas Particle

Cartesian Components of a

Particles Velocity

n.informatioldirectiona

noconveysandquantityscalaraisthatNote

axes.Cartesian

principalthreethealongvelocitytheofcomponents

areand,andparticletheofspeedtheisWhere

2222

u

uuuu

uuuu

zyx

zyx ++=

Consider the force exertedon the wall by the

component of velocityalong the x axis.

Calculation of the Force Exerted on a

Container by Collision of a Single Particle

( )

( )

2

22 22

2

length of boxspeed

22 and similarly for and

,

22 2 2

x x

x

x xx x y z

yx ztot

muuF ma m

t t

mu mu

Ltu

u muF mu F F

L L

Thus

mumu mu mF u

L L L L

= = =

=

= =

= =

= + + =

Calculation of the Pressure in Terms of

Microscopic Properties of the Gas Particles

2

,

2

Since the number of particles in a given gas sample can be expressed

average over

as , where is the number of moles and is Avogadro's

square of speed of all particles

!

num

tot one particle

A A

mF u

L

nN n N

=

2

2 2

2 3

2

ber,

2

2 /

6 3

1

2 2

3

tot A

TotA A

Tot

A

m

F nN uL

F mu L muP nN nN

Area L L

nN mu

PV

=

= = = =

= =

2

A

munN

3V

n Kinetic Energy/mole2

3 V


8/21

The Kinetic Theory Relates the

Average Kinetic Energy of theParticles to Temperature

from ideal gas law

Note that is independent of molecular mass

2 or

3

3

2

!

avg

avg

nKEP

V

RTPV

KEn

=

= =

avg

avg

3KE = RT

2

KE

Root Mean Square Speed and

Temperature

2

2

1 3

2 2A

rms

N mu RT

u u

=

= 3RT

M

whereR is the gas constant (8.314 J/mol K), Tis the

temperature in kelvin, andMis the molar mass expressed in

kg/mol (to make the speed come out in units of m/s).

Not all molecules have the same speed.

But what is the distribution of speeds in a

large number of molecules? The kinetic

theory predicts the distribution function

for the molecular speeds.

Molecular Speed Distribution The Mathematical Description of theMaxwell-Boltzmann Speed Distribution

Kinretemperatu

andJ/K,101.38066constantsBoltzman'

kginparticleofmass,/inpeed:where

24)(

23-

22 2

=

==

==

=

T

k

msmsu

euTk

muf

B

Tkmu

B

B

f(u)du gives the fraction of molecules that have

speeds between u and u + u.

The distribution of

molecular speeds for N2at three temperatures

Features of the Speed Distribution

The most probable speed is at the peak ofthe curve.

The most probable speed increases as the

temperature increases.

The distribution broadens as thetemperature increases.


9/21

The Speed Distribution Depends on

Molecular Mass

The most probable speed increases as the

molecular mass decreases.

The distribution broadens as the molecular

mass decreases.

Relationship between molar mass and

molecular speed

Calculation of Molecular Speeds

and Kinetic Energies at 300.0 KMolecule H2 CH4 CO2

MolecularMass (g/mol)

2.016 16.04 44.01

Kinetic Energy(J/molecule)

6.213 x 10- 21 6.213 x 10- 21 6.213 x 10- 21

Speed(m/s)

1,926 683.8 412.4

2

2

1 3

2 2A

rms

N mu RT

u u

=

= 3RTM

Molecular Mass and Molecular Speeds

Molecule H2 CH4 CO2

Molecular

Mass (g/mol)

Kinetic Energy

(J/molecule)

rms speed

(m/s)

2.016 16.04 44.01

6.213 x 10 - 21 6.213 x 10 - 21 6.213 x 10 - 21

1,926 683.8 412.4

Larger

Same Kinetic Energy

Slower

Various Ways to

designate

the averageSpeed.


10/21

The Three Measures of the Speed of a

Typical Particle

3

2

8

rms

mp

avg

RT

u M

RTu

M

RTu u

M

=

=

= =

EP106: Calculate the Kinetic Energy of aHydrogen Molecule traveling at 1.57 x 103

m/sec.

Strategy: (1) calculate mass of molecule. (2) Use KE=1/2mu2

Mass =

KE =

Higher T (a higher KE) yields higher gas velocity, raising

Pgas initially until top wall moves, which increases V until

Pgas = Patm.

The Process of Effusion

EffusionEffusion is the process whereby a gas escapes from its

container through a tiny hole into an evacuated space.

According to the kinetic theory a lighter gas effuses faster

because the most probable speed of its molecules is h igher.

Therefore more molecules escape through the tiny hole in unit

time.

This is made quantitative in Graham's Law of effusion: The rate

of effusion of a gas is inversely proportional to the square root

of its molar mass.

1Rate of effusion speed

M

Effusion CalculationEP107: Calculate the ratio of the effusion rates of heliumand hydrogen through a small hole.

Strategy:

(1)The effusion rate is inversely proportional to square root of

molecular mass.(2) find the ratio of the molecular masses and take the square

root.

(3)The inverse of the ratio of the square roots is the effusion rate

ratio.


11/21


12/21

The ideal gas law is accurate over only a rangeof conditions

The effect of intermolecular attractions on measured gas

pressure.

The effect of

molecular

volume on

measured gas

volume.

The van der Waals equation of state is valid over a

wider range of conditions than the ideal gas law:

( ) nRTnbVV

anP =

+

2

2

Where P is the measured pressure, Vis the

container volume, n is the number of moles of gas

and T is the temperature. a and b are the van der

Waals constants, specific for each gas.

EP109: van der Waals Calculation of a Real gasProblem: A tank of 20.00 liters contains chlorine gas at a temperature

of 20.000C at a pressure of 2.000 atm. The tank is pressurized to a new

volume of 1.000 L and a temperature of 150.000C. Calculate the final

pressure using the ideal gas equation, and the Van der Waals equation.


13/21

2

2vdW

nRT n aP

V nb V =

=

57.75 atmidealP =

If attractive interactions dominateideal vdW P P>

EP110: Gas Law Stoichiometry

Problem: A 2.79 L container of ammonia gas for which P= 0.776 atm

and T=18.7 oC is connected to a 1.16 L container of HCl gas for which

P= 0.932 atm and T= 18.7 oC. What mass of solid ammonium chloride

will be formed? What gas is left in the combined volume, and what

is its pressure?

Strategy:1) Calculate the moles of each reactant and determine limiting reactant.

2) Calculate the mass of product from balanced equation

3) Determine which reactant is left and its pressure

Solution:

Energy used to create wealth and demand will grow

All aspect of economy depend on energy

U.S.A.Canada

Takenfrom R. G. Watts, EngineeringResponse to Global Climate Change, Lewis

Publishers, NewYork, 1997.100

500

1000

5000

10,000

$ 20,000

50,000

$ 200

2000

0.01 0.10 1.0 10 100

POWER CONSUMPTION

MeanGrossDomesticProduct

($perPersonperYear)

Bangladesh

China

MexicoPoland

South Korea(Former

U.S.S.R.)

France

Japan

U.K.

SLOPE = 23/kWhr

AFFLUENCE

POVERTY

kW/person

Mean Global Energy

Consumption, 1998

4.52

2.72.96

0.286

1.21

0.286

0.828

0

1

2

3

4

5

TW

Oil Coal Biomass Nuclear Gas Hydro Renew

Total: 12.8 TW U.S.: 3.3 TW (99 Quads)

80% from fossils fuels

2004: 80 millions barrels oil per day

The Trouble with Using

HydrocarbonsAll those carbon atoms react to form CO2 molecules.

CO2 is a greenhouse gas.It traps infrared radiation in the

troposphere, heating lower atmosphere.

Earths Surface absorbs visible light.

emits thermal radiation in infrared.

CO2CO2CO2

CO2

CO2CO2

CO2CO2

CO2

CO2


14/21

Combustion produces other

greenhouse gases

Combustion of petroleum produces CO, CO2, NO and

NO2 together with unburned petroleum.

N2 + O2 2 NO; 2 NO + O2 2 NO2

NO2 NO + O (reactive) : O + O2 O3 (ozone)

This net production of ozone then produces other

pollutants.

World World Energy EnergyMillions of Barrels per Day (Oil Equivalent)

300

200

100

0

1860 1900 1940 1980 2020 2060 2100

Source: John F. Bookout (President of Shell USA) ,Two Centuries of Fossil Fuel Energy

International Geological Congress, Washington DC; July 10,1985.

Episodes, vol 12, 257-262 (1989).

1860 1900 1940 1980 2020 2060 21001860 1900 1940 1980 2020 2060 2100

0

5

10

15

20

25

30

35

40

45

50

Oil

Coal

Gas

Fissio

n

Biom

ass

Hydroelec

tric

Solar

,win

d,geothe

rmal

0.5%

Source:BP &IEA

2004

0

5

10

15

20

25

30

35

40

45

50

Oil

Coal

Gas

Fusio

n/Fiss

ion

Biom

ass

Hydroelec

tric

Solar

,win

d,geothe

rmal

2050

The ENERGY REVOLUTION(The Terawatt Challenge 1TW = 1012 W)

14.5 Terawatts

220 M BOE/day30 -- 60 Terawatts

450 900 MBOE/day

The Basis of Prosperity

20st

Century = OIL21st Century = ??

Consequences of Increasing

World Energy Demand

Ten billion people consuming energyas North Americans and Europeans do todaywould raise world energy demand tenfold

6 billion tons of carbon released intoatmosphere per year today

More than 6 tons per capit a in US

60 bill ion tons per year in 2100 ???

83 billion tons CO2 per year in 2100 ???

Is Greenhouse Effect Bad?

Lets compare Mars, Earth, Venus.

Altitude(km

)

50

100

Temperature (Kelvin)100 400 800

A little greenhouse effect

is good. s show surfacetemperature without the

greenhouse effect.

A lot of greenhouse effect isvery bad. Example: Venus.

Temperature over the last 420,000 yearsIntergovernmental Panel on Climate Change

We are

here

CO2


15/21

Unstable

Glaciers

Surface melt onGreenland ice sheet

descending intomoulin, a vertical shaftcarrying the water to

base of ice sheet.

Source: Roger Braithwaite

If CO2 can be captured from power plants , growth

in greenhouse gases can be avoided.

Buzzword: carbon sequestration

Current technologies use cryogenic capture or amine

absorbers. 10 times more expensive than acceptable

Natural capture by forests and oceans

Capture and injection into geological formations?

Capture and injection into deep oceans?

Mineralization; react with MgO to form MgCO3?

Microbial conversion into CH4 and/or acetates?


16/21

Alternate Forms of Energy

Fossil fuels will be hard to replace.

Small volume large energy release.

Atomic nuclei? Most are very stable.A few large ones can be induced to fission.

The

seneutr

onshi

t

Oth

ernucle

icau

sing

Cha

inrea

ctio

n.

Nuclear Power Excellent energy

output: 1014 J/kg,

= 10,000 gallons of

gasoline.

Need good, solid

containment

vessels.

Final products

are still radioactive,

(alpha, beta decay).

Need long termdisposal solution.

94

Nuclear Reactors in the World

By the end of 2002:30 countries441 reactors

359 GWe

16% of electricity production

7% of primary energy

Source : AIEASource : AIEA

Wind Power

Turbines can provide

~ MWatt when running.

Wind farms can have up to

200 turbines. over 500 gallons of gas/day.

Problem: calm. Answer: storage.

fan gearbox dynamo

Electric Potential of Wind

http://www.nrel.gov/wind/potential.html

In 1999, U.S

consumed

3.45 trillion

kW-hr of

Electricity =0.39 TW


17/21

Wave and Geothermal Power

Wave farms: Convert wave motion to circular drive

turbines: ~50 kWatts/mAt tectonic plate boundaries, geothermal plants can tap the heat

of the earths interior

Geothermal Energy

Potential

Renewable Energy Cost TrendsLevelized cents/kWh in constant $20001

Wind

1980 1990 2000 2010 2020

PV

COEcents/kWh

1980 1990 2000 2010 2020

40

30

20

10

0

100

80

60

40

20

0

BiomassGeothermal Solar thermal

1980 1990 2000 2010 2020 1980 1990 2000 2010 2020 1980 1990 2000 2010 2020

COEcents/kWh

10

8

6

4

2

0

70

60

50

40

30

20

10

0

15

12

9

6

3

0

Source: NREL Energy Analysis Office (www.nrel.gov/analysis/docs/cost_curves_2002.ppt)1These graphs are reflections of historical cost trends NOT precise annual historical data.

Updated: October 2002

Problem with Alternatives to

Hydrocarbons

Hydrocarbons:

1) store a lot of energy compactly.

2) are currently cheap.

Alternatives:

1) have large footprint.

2) may generate enough total energy, but atlow power rates.

3) solar and wind have low duty cycle.

165,000 TWof sunlighthit the earth

Solar Power

Suns main process: Turning H to He (fusion).

Suns output 4 x 1026 Watts (or Joules/sec).

We see ~200 W/m2 (in the US).So at 15% efficiency, 1 m2, 10 hrs of sunlight 1 MJoule/day.

Problem:Night, clouds.

Answer:

Storage.(batteries, fuel cells).


18/21

6 Boxes at 3.3 TW Each = 206 Boxes at 3.3 TW Each = 20 TWeTWe

From D. Kamen, UC Berkeley Solar Land Area Requirements

3 TW

Solar Across Scales

Moscone Center, SF: 675,000 W

Kenyan PV market:

Average system: 18W

US home: 2400 W

From D. Kamen, UC Berkeley

Learning Curve for PV Modules

(crystalline sili con)

2000

1980

1990

1

10

100

1 1010-110-210-310-4 102 103

Cumulative installed PV Peak Power [GWp]

[/Wp]

20202010

Today PV electricity costs about $0.20 - 0.25/kWh,Which can be compared with $0.32/kWh PG&E charges for TOU

customers during peak time (noon-6pm)


Efficiency(%)

Cu(In,Ga)Se2Amorphous Si:H(stabilized)CdTe

Universityof Maine

Boeing

Boeing

Boeing

Boeing

ARCO

NREL

Boeing

Euro-

CIS

12

8

4

0200019951990198519801975

United Solar

16

20

24

28

32

Three-junction (2-terminal, monolithic)

Two-junction (2-terminal, monolithic)

NREL/Spectrolab

NREL

NREL

Japan

Energy

Spire

NorthCarolinaState University

Multijunction Concentrators

Thin Film Technologies

Best Research-Cell Efficiencies

Varian

RCA

Solarex

UNSW

UNSW

ARCO

UNSWUNSW

UNSWSpire

Stanford

Westing-house

Crystalline Si CellsSingle Crystal

Multicrystalline

Thin Si

UNSWGeorgia Tech

Georgia Tech

Sharp

Solarex Astro-Power

NREL

AstroPower

Spectrolab

NREL

Emerging PV

Masushita

Monosolar

Kodak

Kodak

AMETEK

PhotonEnergy

Univ.S.Florida

NREL

NREL

NREL

Princeton

U. Konstanz

U.CaliforniaBerkeley

Organic Cells

NREL

NRELCu(In,Ga)Se214x Concentration

Source: T. J. Berniard, NREL


Thin film-based PV modules could be printed on flexibleplastic backing using a printing press


19/21

Biodiesel ProductionBiodiesel Production

100 lbs. of soybean oil

+

10 lbs. methanol

=

100 lbs. soy biodiesel

(B100)

+

10 lbs. of glycerin

~ 100,000 TW of energy is receivedfrom the sun

Amount of land needed to capture 13 TW:

20% efficiency (photovoltaic) = 0.23%

1% efficiency (bio-mass) = 4.6%

Steven Chu Berkeley

> 1% conversion

efficiency may be

feasible.

Miscanthus yields: 30 dry tons/acre

100 gallons of ethanol / dry ton possible 3,000 gal/acre.

100 M out of 450 M acres ~300 B gal / year of ethanol

US consumption (2004) = 141 B gal of gasoline

~ 200 B gal of ethanol / year

US also consumes 63 B gal diesel

Steven Chu Berkeley

Greenhouse Gases

-

25

50

75

100

125

-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24

Net Energy (MJ / L)

NetGHG(

gCO2e/

MJ-ethanol)

Pimentel

Patzek

Shapouri

Wang

Graboski

Gasoline

de Oliviera

Today

CO2Intensive

Cellulosic

Original data

Commensurate values

Gasoline

EBAMM cases

Dan Kammen, et al. (2006)

Sunlight

CO2, H20,

NutrientsBiomass

Chemicalenergy

Self-fertilizing,

drought and pest

resistant

Improved

conversion of

cellulose into

chemical fuel

Cellulose 40-60% Percent Dry Weight

Hemicellulose 20-40%

Lignin 10-25%

The majority of a plant is structural material

Steven Chu Berkeley


20/21

40% of Oil used for Vehicles:

Do we need bigger SUVs..

The Kenworth Grand Dominator- Extra high roof/cathedral ceilings- Power expandable sides- Full lavatory

The Peterbuilt Crusader

All Sport DenaliThe worlds first two story high

performance sport brute

Crusader-E Edition: includes elevator

Source: http://poseur.4x4.org/futuresuv.html


or more fuel efficient cars?

Fuel Cell Detail1. Hydrogen goes to Anode,

(can use hydrocarbon fuel)

Oxygen goes to Cathode.

2. Anode strips electrons from

hydrogen, H+ ions enter the

membrane

3. Since electrons cannot

enter the membrane they

go through the external circuit.

4. When electrons get back to the

Cathode, they combine with H+

and O to form water.

membrane

OfficeofFossil Energy, U.S.Department ofEnergyOffice of Fossil Energy, U.S. Department of Energy

Fuel Cell Points

Each individual cell provides ~0.7 V.

Use many in a stack.

Where do you get Hydrogen?

can use hydrocarbons,

wastewater digesters, landfills,

biomass.

can also run the fuel cell backward

(use solar, wind, etc. power to convert

water to H2 and O2).

Greenhouse Comparison:

FCVs, ICEs and Hybrid Vehicles

Source: Bevilacqua-Knight, 2001



21/21

Generating H2 by

splitting water using

visible light

From M. Graetzel, ETH Lausanne

Chemists vital to solution Availability of fossil fuels has enabled mass of humanity

to move beyond subsistence living.

But we really need to figure out how to live without

them.

Carbon loading of the atmosphere is reaching alarming

levels.

Scientific consensus on global warming.

Clean alternatives like solar, wind, etc. have problems of

rate, efficiency. Need to solve them. Nuclear will play a

role.

Answers to Numerical Problems

EP100: 2.08 atm EP101: 10.5 atm

EP102 30.8 L EP102a: 0.626 g L-1

EP103: 58.0 g mol-1

EP104: CO2 0.812 atm He 0.298 atm total 1.110 atm

EP105: Ne 1.21 atm Ar 0.202 atm Xe 0.586 atm

EP 106: 4.13 x 10-21 J EP 107: 0.7097

EP 108: 1 .463 EP109: ideal 57.75 atm vdW 45.75 a tm

EP 110: 2.41 g NH4Cl PNH3 = 0.274 atm

Documents

Notes Chapter 05