1

Chapter 5: Gases and the Kinetic - Molecular Theory

5.1 An Overview of the Physical States of Matter5.2 Gas Pressure and its Measurement5.3 The Gas Laws and Their Experimental Foundations5.4 Further Applications of the Ideal Gas Law5.5 The Ideal Gas Law and Reaction Stoichiometry5.6 The Kinetic - Molecular Theory: A Model for Gas

Behavior5.7 Real Gases: Deviations from Ideal Behavior

Important Characteristics of Gases

1) Gases are highly compressibleAn external force compresses the gas sample and decreases itsvolume, removing the external force allows the gas volume toincrease.

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

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

4) Most gases have low densitiesGas densities are on the order of grams per liter whereas liquidsand 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 under Normal Conditions (300 K 1 atm)

Substance Formula MM(g/mol)

Important variable that describe a gas are pressure (P), temperature (T), and volume (V)

Temperature is a measure of the kinetic energy of the gas atoms or molecules

Pressure arises from collisions of the gas atoms or molecules with the container walls

Chapter 5: Gases and the Kinetic - Molecular Theory

5.2 The Gas Laws of Boyle, Charles, and Avogadro5.3 The Ideal Gas Law5.4 Gas Stoichiometry5.5 Dalton’s Law of Partial Pressures5.6 The Kinetic - Molecular Theory of Gases5.7 Effusion and Diffusion5.8 Collisions of Gas Particles with Container Walls5.9 Intermolecular Collisions5.10 Real Gases5.11 Chemistry in the Atmosphere

Gas pressure can be measured using a manometer

760 mm Hg = 1 atm

2

Magdeburg spheresBoyle’s Law: At constant T, the product PV for a

given amount of a gas is constant

CHARLES’S LAW: V/T = constantif P, n held constant

Basis for absolute temperature scale

T(K) = T(°C) +273.15

Higher T (a higher KE) yields higher gas velocity, raising Pgas initially until top wall moves, which increases V until Pgas = Patm.

AVOGADRO’SLAW:

V/n = constant

at constant T, P

Moles gas in B is double that in Aso V is double

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

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

Independent of substance! In the limit that P →0,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 J mol-1 K-1

• R = 0.08206 L atm mol-1 K-1

3

The Ideal Gas Law Subsumes the Other Gas Laws-- for a fixed amount at constant temperature

• PV = constant PV = nRTBoyle’s Law

– for a fixed amount at constant volume• P increases linearly with T P = (nR/V)TAmonton’s Law

– for a fixed amount at constant pressure• V increases linearly with T V = (nR/P)TCharles’s Law

– for a fixed volume and temperature• P increases linearly with n P = (RT/V)nAvogadro’s 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.

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 VT

P1 x V1 =P2 x V2

EP100: Change of Three Variables - I

A gas sample in the laboratory has a volume of 45.9 L at 25.0 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by compressing the gas to a new volume of 31.0 L what is the pressure in atm?

P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atmP2 = ?V1 = 45.9 L V2 = 31.0 LT1 = 25.0 oC + 273 = 298 KT2 = 155 oC + 273 = 428 K

4

1 1 2 2

1 2

PV PVT T

=

2

2

31.0 L0.978 atm × 45.9 L298 K 428 K

428 K 0.978 atm × 45.9 L298 K 31.0 L

P

P

×=

×= =

×2.08 atm

EP101: Gas LawCalculate the pressure (in atm) in a container filled with 5.038 kg 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.

T = 18.8 oC + 273.15 K = 292.0 K

Xe5038 g Xen = 38.37 mol Xe

131.3 g Xe / mol=

38.37 mol 0.0821 L atm 292.0 K87.5 L

nRTPV

× ×= = = 10.5 atm

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.

(1) 2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

mol NaN3 = 60.0 g NaN3 / 65.02 g NaN3 / mol == 0.9228 mol NaN3

mol N2= 0.9228 mol NaN3 x 3 mol N2/2 mol NaN3= 1.38 mol N2

T = 273.15 +21.0 = 294.2 KP = 823 mm x 1 atm/760. mm = 1.08 atm

-1 -11.38 mol×0.08206 L atm mol K 294.2 K1.08 atm

nRTVP

×= = = 30.8 L

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.

P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atmT = 55.0 oC + 273.15 = 328.2 K

-1 -11 mol×0.08206 L atm mol K ×328.2 K0.989 atm

nRTVP

= =

= 27.2 L

17.03 g27.2 L

massdensityV

= = = -10.626 g L

5

Calculation of Molar Mass

n = = P x VR x T

MassMolar Mass

Molar Mass = M = Mass x R x TP x V

m RTnRT mRTMVP P MPmRT RTMVP P

ρ

= = =

= =

EP103: A volatile liquid is placed in 590.0 ml flask and allowed to boil until only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the masses of the empty and gas filled flask were 148.375g and 149.457 g respectively, 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

Mass of gas = 149.457g - 148.375g = 1.082 g-1 -11.082 g ×0.08206 L atm mol K ×373K

0.968 atm 0.5900 L

mRTMPV

= =×

= -158.0 g mol

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.

Dalton’s 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

6

• Pressure exerted by an ideal gas mixture is determined by 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 present by 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 11 1

1 2 3

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

EP104: Dalton’s 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 = 290.2 K

nCO2 = 3.00 g CO2/ (44.01 g CO2 / mol CO2)= 0.0682 mol CO2

2

2

-1 -10.0682 mol ×0.08206 L atm mol K ×290.2 K2.00 L

COCO

n RTP

P= =

= 0.812 atm

nHe = 0.100 g He / (4.003 g He / mol He)= 0.0250 mol He

PTotal = PCO2 + PHe = 0.812 atm + 0.298 atmPTotal = 1.110 atm

-1 -10.0250 mol ×0.08206 L atm mol K ×290.2 K2.00 L

HeHe

n RTPP

= =

= 0.297 atm

EP105: Dalton’s Law using mole fractionsA mixture of gases contains 4.46 mol Ne, 0.740

mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ?

Strategy:(1) Calculate mol fraction each gas(2) Partial pressure = mol fraction x total P

Total # moles = 4.46 + 0.74 + 2.15 = 7.35 mol

XNe = 4.46 mol / 7.35 mol = 0.607PNe = XNe PTotal = 0.607 ( 2.00 atm) = 1.21 atm

XAr = 0.740 mol / 7.35 mol = 0.101PAr = XAr PTotal = 0.101 (2.00 atm) = 0.202 atm

XXe = 2.15 mol / 7.35 mol = 0.293 PXe = XXe PTotal = 0.293 (2.00 atm) = 0.586 atm

What’s 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.

7

The Kinetic Theory of Gases• Starts with a set of assumptions about the

microscopic 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 particles

with statistics, thereby establishing a new branch of physics - statistical mechanics.

• Offers an explanation of the macroscopic behavior of gases.

• Predicts experimental phenomena that ideal gas law can’t predict. (Maxwell-Boltzmann Speed Distribution)

The Four Postulates of the Kinetic Theory

• A pure gas consists of a large number of identical molecules separated by distances that are great compared with their size.

• The gas molecules are constantly moving in random directions with a distribution of speeds.

• The molecules exert no forces on one another between collisions, so between collisions they move in straight lines with constant velocities.

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

An ideal gas molecule in a cube of sides L.

Cartesian Coordinate System for the Gas Particle

Cartesian Components of a Particle’s Velocity

n.informatio ldirectionano conveys andquantity scalar a is that Note

axes.Cartesian principal three thealong velocity theof components

are and , and particle theof speed theis Where

2222

u

uuuu

uuuu

zyx

zyx ++=

8

Consider the force exerted on the wall by the component of velocity along the x axis.

Calculation of the Force Exerted on a Container by Collision of a Single Particle

( )

( )

2

22 22

2length of box

speed

22 and similarly for and

,22 2 2

x x

x

x xx x y z

yx ztot

muuF ma mt t

mu muLtu

u muF mu F FL L

Thusmumu mu mF u

L L L L

ΔΔ⎛ ⎞= = =⎜ ⎟Δ Δ⎝ ⎠Δ =

Δ = =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= + + =

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 uL

nN n N

=

2

2 2

2 3

2

ber, 2

2 /6 3

12 23

tot A

TotA A

Tot

A

mF nN uL

F mu L muP nN nNArea L L

nN muP

V⎡ ⎤⎢ ⎥⎣

=

= = = =

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥= =

⎢ ⎥⎢ ⎥⎦

⎦⎣

2

AmunN3V

n Kinetic Energy/mole23 V

The Kinetic Theory Relates the Average Kinetic Energy of the

Particles 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 = RT2

KE

Root Mean Square Speed and Temperature

2

2

1 32 2A

rms

N mu RT

u u

⎛ ⎞ =⎜ ⎟⎝ ⎠

≡ =3RTM

where R is the gas constant (8.314 J/mol K), T is the temperature in kelvin, and M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).

2,

2

independent of !

tot one particlemF uLm

=

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

9

The Mathematical Description of the Maxwell-Boltzmann Speed Distribution

2 22

-23

( ) 42

where: peed in , mass of particle in kg Boltzman's constant = 1.38066 10 J/K,

and temperature in K

Bmu k T

B

B A

mf u u ek T

u s m s mk R N

T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠= =

= = ×=

f(u)du is the fraction of molecules that have speeds between u and u + Δu.

The distribution ofmolecular speeds for N2at three temperatures

Features of the Speed DistributionThe most probable speed is at the peak of

the curve.

The most probable speed increases as the temperature increases.

The distribution broadens as the temperature increases.

Relationship between molar mass andmolecular speed at fixed T

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.

Calculation of Molecular Speeds and Kinetic Energies at 300.0 K

Molecule H2 CH4 CO2 Molecular Mass (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 32 2A

rms

N mu RT

u u

⎛ ⎞ =⎜ ⎟⎝ ⎠

≡ =3RTM

10

Various Ways to designate

the ‘average’Speed.

Molecular Mass and Molecular Speeds

Molecule H2 CH4 CO2

MolecularMass (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

The Three Measures of the Speed of a Typical Particle

3

2

8

rms

mp

avg

RTuMRTuM

RTu uMπ

=

=

= =

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

m/sec.

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

Mass = 2.016 g/mol x 10-3 kg/g6.022 x 1023 molecules / mol

= 3.348 x 10- 27 kg / H2 molecule

KE = 1/2 mu2 = 1/2 (3.348 x 10- 27kg/ molecule)x ( 1.57 x 103 m/sec)2

KE = 4.13 x 10- 21 kg m2/ sec2 moleculeKE = 4.13 x 10- 21 J / molecule

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 higher. 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 speedM

∝ ∝

11

Effusion Calculation

2

mass of He 4.003 1.409mass of H 2.016

= =

EP107: Calculate the ratio of the effusion rates of helium and hydrogen through a small hole.

Strategy:(1)The effusion rate is inversely proportional to square root ofmolecular 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.

2

effusion rate He 1 0.7097effusion rate H 1.409

= =

Path of one particle in diffusing through the gas.

Diffusion• The movement of one gas through another by

thermal random motion. • Diffusion is a very slow process in air

because the mean free path is very short (for N2 at STP it is 6.6x10-8 m). Given the nitrogen molecule’s high velocity, the collision frequency is also very high (7.7x109 collisions/s).

• Diffusion also follows Graham's law:

M1diffusionofRate ∝

Diffusion of agas particlethrough aspace filledwith otherparticles

Relative Diffusion Rates of NH3 and HCl

Relative Diffusion Rate of NH3 compared to HCl in the reaction

NH3(g) + HCl(g) = NH4Cl(s)

HCl = 36.46 g/mol NH3 = 17.03 g/mol

RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2

RateNH3 = 1.463 RateHCl

12

Gaseous Diffusion Separation of Uranium 235 / 238

235UF6 vs 238UF6

Separation Factor = S =

after Two runs S = 1.0086after approximately 2000 runs

235UF6 is > 99% Purity !Y - 12 Plant at Oak Ridge National Lab

(238.05 + (6 x 19))0.5

(235.04 + (6 x 19))0.5

The ideal gas law is accurate over only a range of conditions

The effect of intermolecular attractions on measured gas pressure.

The effect ofmolecularvolume onmeasured gasvolume.

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, V is the container volume, n is the number of moles of gas and T is the temperature. a and b are the van derWaals constants, specific for each gas.

13

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

-1 -1

2.000atm×20.00L 1.663 mol0.8206 L atm mol K ×273.15 K

PVnRT

= = =

-1 -11.663 mol 0.08206 L atm mol K ×423.15 K 57.75 atm1.000Lideal

nRTPV

×= = =

( )( )

2

2

-1 -1

-1

2 2 -2

2

1.663 mol 0.8206 L atm mol K ×423.15 K1.000L 1.663 mol 0.0562 mol L

1.663 mol 6.49atm L mol63.7 19.9

1.0045.8 a

0 Ltm

vdWnRT n aP

V nb V= −

−×

=− ×

×− = − =

57.75 atmidealP =

If attractive interactions dominateideal vdWP P>

EP110: Gas Law StoichiometryProblem: A 2.79 L container of ammonia gas for which P= 0.776 atmand 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 equation3) Determine which reactant is left and its pressure

Solution:Balanced equation: NH3 (g) + HCl (g) → NH4Cl (s)

TNH3 = 18.7 oC + 273.15 = 291.9 K

3 -1 -1

0.776 atm 2.79 L 0.0903 mol0.08206 L atm mol K 291.9 KNH

PVnRT

×= = =

×

-1 -1

0.932 atm 1.16 L 0.0451 mol0.08206 L atm mol K 291.9 K

HClPVnRT

×= = =

×

HCl is the limiting reactant

453.5 gmass product 0.0451 mol 2.41 g NH Cl

mol= × =

3NH remaining 0.0903 mol - 0.0451 mol 0.0452 molV= 1.16 L + 2.79 L = 3.95 L

= =

3

-1 -10.0452 mol 0.08206 L atm mol K 291.9 K 0.274 atm3.95 LNH

nRTPV

× ×= = =

Energy used to create wealth and demand will growAll aspect of economy depend on energy

U.S.A.Canada

Taken from R. G. Watts, Engineering Response to Global Climate Change, Lewis Publishers, New York, 1997.

100

500

1000

5000

10,000

$ 20,000

50,000

$ 200

2000

0.01 0.10 1.0 10 100

POWER CONSUMPTION

Mea

n G

ross

Dom

estic

Pro

duct

($

per

Per

son

per Y

ear)

Bangladesh

China

MexicoPoland

South Korea(FormerU.S.S.R.)

FranceJapan

U.K.

SLOPE = 23¢/kW•hr

AFFLUENCE

POVERTY

kW/person

Mean Global Energy Consumption, 1998

4.52

2.7 2.96

0.286

1.21

0.2860.828

0

1

2

3

4

5

TW

Oil Coal Biomass NuclearGas Hydro Renew

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

80% from fossils fuels

2004: 80 millions barrels oil per day

14

The Trouble with Using Hydrocarbons

All those carbon atoms react to form CO2 molecules. CO2 is a greenhouse gas.

It traps infrared radiation in thetroposphere, heating lower atmosphere.

Earth’s Surface absorbs visible light.emits thermal radiation in infrared.

CO2CO2CO2

CO2

CO2CO2

CO2 CO2CO2

CO2

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

01860 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

05

101520253035404550

OilCoa

lGas

Fission

Biomass

Hydroe

lectric

Solar, w

ind, g

eothe

rmal

0.5%

Source: BP & IEA

2004

0

5

10

15

20

25

30

35

40

45

50

OilCoa

lGas

Fusion

/ Fiss

ion

Biomass

Hydroe

lectric

Solar, w

ind, g

eothe

rmal

2050

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

14.5 Terawatts

220 M BOE/day30 -- 60 Terawatts450 – 900 MBOE/day

The Basis of Prosperity20st Century = OIL21st Century = ??

Consequences of IncreasingWorld Energy Demand…

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

6 billion tons of carbon released into atmosphere per year today

More than 6 tons per capita in US

…60 billion tons per year in 2100 ???

…83 billion tons CO2 per year in 2100 ???

Is Greenhouse Effect Bad?Let’s compare Mars, Earth, Venus.

Alti

tude

(km

)

50

100

Temperature (Kelvin)100 400 800

A little greenhouse effectis good. ’s show surfacetemperature without thegreenhouse effect.A lot of greenhouse effect is very bad. Example: Venus.

15

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

We are here

CO2

Unstable Glaciers

Surface melt on Greenland ice sheet

descending into moulin, a vertical shaft carrying 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

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.

Thes

e neu

tron

s hit

Oth

er n

uclei

caus

ing

Chain

reac

tion.

Nuclear Power• Excellent energy

output: 1014 J/kg,= 10,000 gallons of

gasoline.• Need good, solid

containmentvessels.

• Final productsare still radioactive,(alpha, beta decay).Need long term disposal solution.

95

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

~ ½ MWatt when running. • Wind farms can have up to

200 turbines. over 500 gallons of gas/day.

• Problem: calm.• Answer: storage.

fan gearbox dynamo

17

Electric Potential of Wind

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

In 1999, U.S consumed3.45 trillion kW-hr ofElectricity =0.39 TW

Wave and Geothermal Power

• Wave farms: Convert wave motion to circular drive turbines: ~50 kWatts/m•At tectonic plate boundaries, geothermal plants can tap the heatof the earth’s interior

Geothermal Energy Potential

Renewable Energy Cost TrendsLevelized cents/kWh in constant $20001

Wind

1980 1990 2000 2010 2020

PV

CO

E ce

nts/

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

CO

E ce

nts/

kWh

10

8

6

4

2

0

706050403020100

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

25

35

45

55

1990 1995 2000 2005 2010 2015 2020 2025Year

Space HeatingSpace CoolingWater HeatingLightingRefrigeratorsPCTVOther

Business-As-Usual

Moderate Scenario

1990 Emissions Levels

1500

2100

2700

3300A

25

35

45

55

1990 1995 2000 2005 2010 2015 2020 2025Year

Space HeatingSpace CoolingWater HeatingLightingRefrigeratorsPCTVOther

Business-As-Usual

Aggressive Scenario

1990 Emissions Levels

1500

2100

2700

3300B

Source: Kammen & Ling, in press

EnergyEfficiencyFutures

Conclusions:

A) Moderate path:1.5% annual improvementEmissions growth halted at a savings

B) Aggressive path:2.9% annual improvement Meet Kyoto levels by efficiency alone & facilitate clean energy market development

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 at low power rates.3) solar and wind have low duty cycle.

18

165,000 TWof sunlighthit the earth

Solar Power•Sun’s main process: Turning H to He (fusion).•Sun’s 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).

6 Boxes at 3.3 TW Each = 20 6 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 silicon)

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)

From D. Kamen, UC Berkeley

19

Effic

ienc

y (%

)

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

Universityof Maine

Boeing

Boeing

Boeing

BoeingARCO

NREL

Boeing

Euro-CIS

12

8

4

0200019951990198519801975

United Solar

16

20

24

28

32Three-junction (2-terminal, monolithic)Two-junction (2-terminal, monolithic)

NREL/Spectrolab

NRELNREL

JapanEnergy

Spire

NorthCarolinaState University

Multijunction Concentrators

Thin Film Technologies

Best Research-Cell Efficiencies

Varian

RCA

Solarex

UNSW

UNSW

ARCO

UNSWUNSW

UNSWSpireStanford

Westing-house

Crystalline Si CellsSingle CrystalMulticrystallineThin Si

UNSWGeorgia Tech

Georgia Tech

Sharp

Solarex Astro-Power

NREL

AstroPower

Spectrolab

NREL

•

Emerging PV

• ••• •

••• •

••

•••

•

Masushita

MonosolarKodak

Kodak

AMETEK

PhotonEnergy

Univ.S.Florida

NRELNREL

NREL

Princeton

U. Konstanz

U.CaliforniaBerkeleyOrganic Cells

NREL

NRELCu(In,Ga)Se214x Concentration

Source: T. J. Berniard, NREL

From D. Kamen, UC BerkeleyThin film-based PV modules could be printed on flexible

plastic backing using a printing press

Biodiesel ProductionBiodiesel Production100 lbs. of soybean oil

+10 lbs. methanol

=100 lbs. soy biodiesel

(B100)+

10 lbs. of glycerin

~ 100,000 TW of energy is received from 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

20

Sunlight

CO2, H20,

NutrientsBiomass Chemical

energy

Self-fertilizing,drought and pest

resistant

Improved conversion of cellulose into chemical fuel

Cellulose 40-60% Percent Dry WeightHemicellulose 20-40%

Lignin 10-25%

The majority of a plant is structural material

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 24Net Energy (MJ / L)

Net

GH

G (g

CO

2e /

MJ-

etha

nol)

Pimentel

Patzek

Shapouri

Wang

Graboski

Gasoline

de OlivieraToday

CO 2 Intensive

Cellulosic

Original data Commensurate values Gasoline EBAMM cases

Dan Kammen, et al. (2006)

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 CrusaderAll Sport Denali

The worlds first two story high performance sport brute

Crusader-E Edition: includes elevator

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

From D. Kamen, UC Berkeley

…or more fuel efficient cars?

Fuel Cell Detail1. Hydrogen goes to Anode,

(can use hydrocarbon fuel)Oxygen goes to Cathode.

2. Anode strips electrons fromhydrogen, H+ ions enter themembrane

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

Office of Fossil Energy, U.S. Department of EnergyOffice 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 convertwater to H2 and O2).

21

Greenhouse Comparison:FCVs, ICEs and Hybrid Vehicles

Source: Bevilacqua-Knight, 2001

From D. Kamen, UC Berkeley

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.

22

Answers to Numerical ProblemsEP100: 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 atm

EP 110: 2.41 g NH4Cl PNH3 = 0.274 atm