Notes Solutions Chapter 05 Large

8/11/2019 Notes Solutions Chapter 05 Large

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Chapter 5: Gases and theKinetic - Molecular Theory

5.1 An Overview of the Physical States of Matter 5.2 Gas Pressure and its Measurement5.3 The Gas Laws and Their Experimental Foundations

5.4 Further Applications of the Ideal Gas Law5.5 The Ideal Gas Law and Reaction Stoichiometry5.6 The Kinetic - Molecular Theory: A Model for Gas

Behavior 5.7 Real Gases: Deviations from Ideal Behavior


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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 is

cooled 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 miscible

Gases mix in any proportion such as in air, a mixture of many gases.


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Helium He 4.0 Neon Ne 20.2 Argon Ar 39.9

Hydrogen H 2 2.0 Nitrogen N 2 28.0 Nitrogen Monoxide NO 30.0

Oxygen O 2 32.0 Hydrogen Chloride HCl 36.5 Ozone O 3 48.0 Ammonia NH

317.0

Methane CH 4 16.0

Substances that are Gases under

Normal Conditions (300 K 1 atm)Substance Formula MM(g/mol)


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

Temperature is a measure of the kinetic energy of the gasatoms or molecules

Pressure arises from collisions of the gas atoms or moleculeswith the container walls


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Chapter 5: Gases and theKinetic - Molecular Theory

5.2 The Gas Laws of Boyle, Charles, and Avogadro5.3 The Ideal Gas Law5.4 Gas Stoichiometry

5.5 Daltons Law of Partial Pressures5.6 The Kinetic - Molecular Theory of Gases5.7 Effusion and Diffusion

5.8 Collisions of Gas Particles with Container Walls5.9 Intermolecular Collisions5.10 Real Gases5.11 Chemistry in the Atmosphere

Gas pressure can

be measuredusing amanometer

760 mm Hg =1 atm


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Magdeburg spheres


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Boyles Law: At constant T, the product PV for agiven amount of a gas is constant


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CHARLESS LAW: V/T = constantif P, n held constant

Basis for absolutetemperature scale

T(K) = T(C) +273.15


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Higher T (a higher KE) yields higher gas velocity, raisingP gas initially until top wall moves, which increases V untilP gas = P atm .


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AVOGADROSLAW:

V/n = constant

at constant T, P

Moles gas in B isdouble that in A

so V is double


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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 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 Jmol -1 K-1

R = 0.08206 L atm mol -1 K-1


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

PV = constant PV = nRTBoyles Law

for a fixed amount at constant volume P increases linearly with T P = (nR/V)TAmontons Law

for a fixed amount at constant pressure V increases linearly with T V = (nR/P)TCharless Law

for a fixed volume and temperature P increases linearly with n P = (RT/V)nAvogadros Law


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Standard Temperature and Pressure (STP)A set of Standard conditions have been chosen to make it easier tounderstand the gas laws, and gas behavior.

Standard Temperature = 0 0 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 .


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Many gas law problems involve a change ofconditions, 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


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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 155oC bycompressing the gas to a new volume of 31.0 L

what is the pressure in atm?

P 1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atmP 2 = ?

V1 = 45.9 L V 2 = 31.0 LT1 = 25.0 oC + 273 = 298 KT2 = 155 oC + 273 = 428 K


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1 1 2 2

1 2

PV PV T T

=

2

2

31.0 L0.978 atm 45.9 L298 K 428 K

428 K 0.978 atm 45.9 L

298 K 31.0 L

P

P

=

= =

2.08 atm


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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 Xe

n = 38.37 mol Xe

131.3 g Xe / mol

=

38.37 mol 0.0821 L atm 292.0 K 87.5 L

nRT P

V = = =10.5 atm


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EP102: Sodium Azide Decomposition - ISodium azide (NaN 3) is used in air bags in automobiles. Write

a balanced chemical equation for the decomposition ofNaN 3 into N 2(g) and Na(s). Calculate the volume in liters ofnitrogen gas generated at 21.0 oC and 823 mm Hg by thedecomposition of 60.0 g of NaN 3.

Strategy:

(1) Write balanced equation.(2) Use stoichiometric coefficients to calculate moles of N 2.(3)Convert given parameters into appropriate units.

(4) Calculate V using the ideal gas law.

(1) 2 NaN 3 (s ) 2 Na ( s ) + 3 N 2 (g)


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mol NaN3

= 60.0 g NaN3

/ 65.02 g NaN3

/ mol == 0.9228 mol NaN 3

mol N 2= 0.9228 mol NaN 3 x 3 mol N 2/2 mol NaN 3= 1.38 mol N 2

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

-1 -1

1.38 mol0.08206 L atm mol K 294.2 K 1.08 atm

nRT V P

= = =30.8 L


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EP102a: Calculate the density of ammonia gas(NH 3) in grams per liter at 752 mm Hg and55.0 oC.

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

(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


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-1 -11 mol0.08206 L atm mol K 328.2 K 0.989 atm

nRT V

P= =

= 27.2 L

17.03 g27.2 L

massdensity

V = = = -10.626 g L


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Calculation of Molar Mass

n = =P x VR x T

MassMolar Mass

Molar Mass = M =Mass x R x T

P x V


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m RT nRT mRT M V P P MP

mRT RT M

VP P

= = =

= =


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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 and736 mm Hg pressure. If the masses of the empty and gas filled flask were148.375g 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 -1

1.082 g 0.08206 L atm mol K 373K 0.968 atm 0.5900 L

mRT M PV

= =

= -158.0 g mol


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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 asa whole.

All molecules in a sample of an idealgas behave exactly the same way.


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Daltons Law of Partial Pressures - I

In a mixture of gases, each gas contributes to thetotal 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: P total = P 1+P 2+P 3+P N


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Pressure exerted by an ideal gas mixture is determined

by the total number of moles:P=(n total RT)/Vn total = sum of the amounts of each gas pressure

the partial pressure is the pressure of gas if it was presentby itself.P 1 = (n 1 RT)/V P 2 = (n 2 RT)/V P 3 = (n 3RT)/V

the total pressure is the sum of the partial pressures.P= (n 1 RT)/V + (n 2 RT)/V + (n 3RT)/V

= (n 1+n 2+ n 3)(RT/V) =(n total RT)/V

Partial pressures in terms of mol fractions

1 11 1

1 2 3

P n= P = x P

P n + n + n


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EP104: Daltons Law of Partial Pressures

A 2.00 L flask contains 3.00 g of CO 2 and 0.100 gof 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 CO 2/ (44.01 g CO 2 / mol CO 2)= 0.0682 mol CO 2

22

-1 -10.0682 mol 0.08206 L atm mol K 290.2 K 2.00 L

COCO

n RT P P= == 0.812 atm


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

= 0.0250 mol He

P Total = P CO2 + P He = 0.812 atm + 0.298 atmP Total = 1.110 atm

-1 -10.0250 mol 0.08206 L atm mol K 290.2 K 2.00 L

He He

n RT P P= == 0.297 atm


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X Ar = 0.740 mol / 7.35 mol = 0.101P Ar = X Ar P Total = 0.101 (2.00 atm) = 0.202 atm

XXe = 2.15 mol / 7.35 mol = 0.293P Xe = XXe P Total = 0.293 (2.00 atm) = 0.586 atm


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Whats Behind the Ideal Gas Law?

A purely empirical law - solely theconsequence of experimentalobservations

Explains the behavior of gases over alimited range of conditions. A macroscopic explanation. Says

nothing about the microscopic behaviorof the atoms or molecules that make upthe gas.


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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 branchof physics - statistical mechanics. Offers an explanation of the macroscopic behavior

of gases. Predicts experimental phenomena that ideal gas

law cant predict. (Maxwell-Boltzmann SpeedDistribution)


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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 ofspeeds.

The molecules exert no forces on oneanother 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.


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An ideal gasmolecule in a cubeof sides L.


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Cartesian Coordinate System for the Gas Particle


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Cartesian Components of aParticles Velocity

n.informatioldirectionanoconveysand quantityscalaraisthat Note

axes.Cartesian principalthreethealongvelocitytheof components

areand ,and particletheof speed theisWhere

2222

u

uuuu

uuuu

z y x

z y x ++=


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Consider the force exerted

on the wall by thecomponent of velocityalong the x axis.


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Calculation of the Force Exerted on a

Container by Collision of a Single Particle( )

( )

2

22 22

2

length of box

speed2

2 and similarly for and

,

22 2 2

x x

x

x x x x y z

y x ztot

muuF ma m

t t

mu mu

Lt

uu mu

F mu F F L L

Thus

mumu mu mF u

L L L L

= = = =

= =

= =

= + + =

C l l i f h P i T f


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Calculation of the Pressure in Terms of

Microscopic Properties of the Gas Particles2,

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

12 23

tot A

Tot A A

Tot

A

mF nN u

L

F mu L muP nN nN

Area L L

nN muP

V

=

= = = =

= =

2

A

mu nN

3V

n Kinetic Energy/mole 2 3 V

The Kinetic Theory Relates the


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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

nKE P V

RT PV

KE n

=

= =

avg

avg

3 KE = RT

2 KE


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Root Mean Square Speed andTemperature

2

2

1 32 2 A

rms

N mu RT

u u

=

= 3RT

M

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

2,

2

independent of !

tot one particle

mF u

L

m

=


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Not all molecules have the same speed.

But what is the distribution of speeds in alarge number of molecules? The kinetictheory predicts the distribution function

for the molecular speeds.

Molecular Speed Distribution


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The Mathematical Description of theMaxwell-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

m f u u e

k T u s m s m

k R N T

=

= =

= = =

f(u)du is the fraction of molecules that havespeeds between u and u + u.

h d b f


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The distribution of

molecular speeds for N 2at three temperatures

F f h S d Di ib i


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

the curve.

The most probable speed increases as thetemperature increases.

The distribution broadens as thetemperature increases.


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Relationship between molar mass and

molecular speed at fixed T

The Speed Distribution Depends on


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The Speed Distribution Depends on

Molecular MassThe most probable speed increases as themolecular mass decreases.

The distribution broadens as the molecularmass decreases.

Calculation of Molecular Speeds


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Calculation of Molecular Speeds

and Kinetic Energies at 300.0 KMolecule H 2 CH 4 CO 2 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 2 A

rms

N mu RT

u u

=

= 3RT

M

Various Ways to


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Various Ways todesignate

the averageSpeed.

Molecular Mass and Molecular Speeds


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Molecule H 2 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

Th Th M f th S d f


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The Three Measures of the Speed of aTypical Particle

3

2

8

rms

mp

avg

RT u M

RT u M

RT u u

=

=

= =

EP106: Calculate the Kinetic Energy of a


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EP106: Calculate the Kinetic Energy of a

Hydrogen Molecule traveling at 1.57 x 10 3m/sec.

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

Mass = 2.016 g/mol x 10 -3 kg/g

6.022 x 1023

molecules / mol= 3.348 x 10 - 27 kg / H 2 molecule

KE = 1/2 mu 2 = 1/2 (3.348 x 10 - 27 kg/ molecule)x ( 1.57 x 10 3 m/sec) 2

KE = 4.13 x 10 - 21 kg m 2/ sec 2 moleculeKE = 4.13 x 10 - 21 J / molecule

The Process of Effusion


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The Process of Effusion

Effusion


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EffusionEffusion is the process whereby a gas escapes from itscontainer 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 unittime.

This is made quantitative in Graham's Law of effusion: The rateof effusion of a gas is inversely proportional to the square rootof its molar mass.

1Rate of effusion speed


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Effusion Calculation

2

mass of He 4.0031.409

mass of H 2.016= =

EP107: 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 squareroot.(3)The inverse of the ratio of the square roots is the effusion rateratio.

2

effusion rate He 10.7097

effusion rate H 1.409= =


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Path of one

particle indiffusingthrough thegas.

Diffusion


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Diffusion

The movement of one gas through another bythermal random motion.

Diffusion is a very slow process in airbecause the mean free path is very short (forN2 at STP it is 6.6x10 -8 m). Given the nitrogenmolecules high velocity, the collisionfrequency is also very high(7.7x10 9 collisions/s).

Diffusion also follows Graham's law:

M

1diffusionof Rate


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Diffusion of agas particlethrough a

space filled with other

particles

Relative Diffusion Rates of NH 3 and HCl


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Relative Diffusion Rate of NH 3 compared to HCl inthe reaction

NH3(g ) + HCl( g ) = NH 4Cl(s )

HCl = 36.46 g/mol NH 3 = 17.03 g/mol

Rate NH3 = Rate HCl x ( 36.46 / 17.03 ) 1/ 2

Rate NH3 = 1.463 Rate HCl

Gaseous Diffusion Separation of


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Gaseous Diffusion Separation ofUranium 235 / 238

235 UF 6 vs 238 UF 6

Separation Factor = S =

after Two runs S = 1.0086after approximately 2000 runs

235 UF 6 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


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of conditions

The effect of intermolecular attractions on measured gaspressure


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pressure.


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The effect of molecular volume onmeasured gasvolume.

The van der Waals equation of state is valid over awider range of conditions than the ideal gas law:


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wider range of conditions than the ideal gas law:

( ) nRT nbV V anP =

+ 2

2

Where P is the measured pressure, V is thecontainer volume, n is the number of moles of gasand T is the temperature. a and b are the van der Waals constants, specific for each gas.


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EP109: van der Waals Calculation of a Real gas


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Problem: A tank of 20.00 liters contains chlorine gas at a temperatureof 20.00 0C at a pressure of 2.000 atm. The tank is pressurized to a newvolume of 1.000 L and a temperature of 150.00 0C. Calculate the final pressure using the ideal gas equation, and the Van der Waals equation.

-1 -1

2.000atm20.00L1.663 mol

0.8206 L atm mol K 273.15 K PV

n RT

= = =

-1 -11.663 mol 0.08206 L atm mol K 423.15 K 57.75 atm

1.000LidealnRT

PV

= = =

=


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( )( )

2

2

-1 -1

-1

2 2 -2

2

1.663 mol 0.8206 L atm mol K 423.15 K 1.000L 1.663 mol 0.0562 mol L

1.663 mol 6.49atm L mol 63.7 19.91.00

45.8 a0 L

tm

vdW

nRT n aP

V nb V =

=

= =

57.75 atmidealP

If attractive interactions dominateideal vdW P P>

EP110: Gas Law Stoichiometry


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Problem: 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 whichP= 0.932 atm and T= 18.7 oC. What mass of solid ammonium chloridewill be formed? What gas is left in the combined volume, and whatis 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: NH 3 ( g) + HCl ( g) NH 4Cl ( s)

T NH3 = 18.7 oC + 273.15 = 291.9 K

3 -1 -1

0.776 atm 2.79 L0.0903 mol NH

PV n

= = =


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3 1 1

0.08206 L atm mol K 291.9 K RT

-1 -1

0.932 atm 1.16 L0.0451 mol

0.08206 L atm mol K 291.9 K

HCl

PV n

RT

= = =

HCl is the limiting reactant

453.5 g

mass product 0.0451 mol 2.41 g NH Clmol

= =

3 NH remaining 0.0903 mol - 0.0451 mol 0.0452 mol

V= 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 atm

3.95 L NH nRT

PV

= = =

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


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All aspect of economy depend on energy

U.S.A.Canada

Taken from R. G. Watts, Engineering

Response to Global Climate Change, LewisPublishers, 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

M

e a n

G r o s s

D o m e s

t i c

P r o

d u c

t

( $ p e r P e r s o n p e r Y e a r )

Bangladesh

China

MexicoPoland

South Korea(Former U.S.S.R.)

FranceJapan

U.K.

SLOPE = 23/kWhr

AFFLUENCE

POVERTY

kW/person

Mean Global Energy


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Consumption, 19984.52

2.7 2.96

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


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HydrocarbonsAll those carbon atoms react to form CO 2 molecules.CO 2 is a greenhouse gas.

It traps infrared radiation in thetroposphere, heating lower atmosphere.

Earths Surface absorbs visible light.

emits thermal radiation in infrared.

CO2CO2

CO2

CO2

CO2CO2

CO2CO2

CO2

CO2

Combustion produces other


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greenhouse gasesCombustion of petroleum produces CO, CO 2, NO and

NO 2 together with unburned petroleum.

N2 + O2 2 NO; 2 NO + O 2 2 NO 2

NO 2 NO + O (reactive) : O + O 2 O3 (ozone)

This net production of ozone then produces other pollutants.

World World Energy Energy


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Millions 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 EnergyInternational 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

The ENERGY REVOLUTION(The Terawatt Challenge 1TW = 10 12 W)


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0

5

10

15

20

25

30

35

40

45

50

O i l

C o a l G a

s

F i s s i o

n

B i o m a

s s

H y d r o

e l e c t r

i c

S o l a r

, w i n d

, g e o t

h e r m a

l

0.5%

Source: BP & IEA

2004

0

5

10

15

20

25

30

35

40

45

50

O i l

C o a l G a

s

F u s i o n

/ F i s s

i o n

B i o m a

s s

H y d r o e l

e c t r i c

S o l a r

, w i n d , g e

o t h e r m

a l

2050

14.5 Terawatts

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

The Basis of Prosperity20 st Century = OIL21 st Century = ??

Consequences of Increasing


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World Energy DemandTen billion people consuming energy

as 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 CO 2 per year in 2100 ???

Is Greenhouse Effect Bad?


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Lets compare Mars , Earth , Venus.

A l t i t u d e

( k m

)

50

100

Temperature (Kelvin)100 400 800

A little greenhouse effectis good. s show surfacetemperature without thegreenhouse effect.

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

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


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We arehere

CO 2


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Unstable


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Glaciers

Surface melt onGreenland ice sheetdescending into

moulin , a vertical shaft

carrying the water tobase of ice sheet.

Source: Roger Braithwaite


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If CO 2 can be captured from power plants , growth

i h b id d


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in greenhouse gases can be avoided.Buzzword: carbon sequestration

Current technologies use cryogenic capture or amineabsorbers. 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 MgCO 3?

Microbial conversion into CH 4 and/or acetates?


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Alternate Forms of Energy


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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.

T h e s e n

e u t r o n

s h i t

O t h

e r n u c l e i

c a u s i n

g

C h a i n r e a c t i o

n .

N l P


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Nuclear Power Excellent energyoutput: 10 14 J/kg,

= 10,000 gallons ofgasoline. Need good, solid

containmentvessels.

Final productsare still radioactive,

(alpha, beta decay).Need long termdisposal solution.

Nuclear Reactors in the World

Source : AIEASource : AIEA


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95

By the end of 2002:30 countries441 reactors

359 GWe

16% of electricity production

7% of primary energy

Wi d P


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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 gearboxdynamo

Electric Potential of Wind


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

In 1999, U.Sconsumed

3.45 trillionkW-hr of Electricity =0.39 TW

Wave and Geothermal Power


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Wave farms: Convert wave motion to circular driveturbines: ~50 kWatts/mAt tectonic plate boundaries, geothermal plants can tap the heatof the earths interior

Geothermal EnergyPotential


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Renewable Energy Cost TrendsLevelized cents/kWh in constant $2000 1

Wi d PV40 100


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Wind

1980 1990 2000 2010 2020

PV

C

O E c e n t s /

k W h

1980 1990 2000 2010 2020

30

20

10

0

80

60

40

200

BiomassGeothermal Solar thermal

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

C O E c e n t s / k

W h

10

8

6

4

2

0

7060

5040302010

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

45

55

Space HeatingSpace CoolingWater HeatingLightingRefrigeratorsPCTV

h

Business-As-Usual

2700

3300A

Energy

Conclusions:

A) Moderate path :l


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25

35

1990 1995 2000 2005 2010 2015 2020 2025Year

Other

Moderate Scenario

1990 Emissions Levels

1500

2100

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

) p :1.5% annualimprovementEmissions growthhalted at a savings

B) Aggressivepath :2.9% annualimprovementMeet Kyoto levels

by efficiency alone& facilitate cleanenergy marketdevelopment

Problem with Alternatives to

Hydrocarbons


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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.


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165,000 TWof sunlighthit the earth

Solar Power


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Suns main process: Turning H to He (fusion).Suns output 4 x 10 26 Watts (or Joules/sec).

We see ~200 W/m2

(in the US).So at 15% efficiency, 1 m 2, 10 hrs of sunlight 1 MJoule/day .

Problem:

Night, clouds. Answer:Storage.

(batteries, fuel cells).

From D. Kamen, UC Berkeley


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

Solar Land Area Requirements


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3 TW

Solar Across Scales


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Moscone Center, SF: 675,000 W

Kenyan PV market:Average system: 18W

US home: 2400 W


Learning Curve for PV Modules

(crystalline silicon)


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(crystalline silicon)

2000

1980

1990

1

10

100

1 1010 -110 -210 -310 -4 10 2 10 3

Cumulative installed PV Peak Power [GW p]

[/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)


28

32

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

NREL/Spectrolab

NRELNREL

JapanEnergy

Multijunction Concentrators

Best Research-Cell Efficiencies

Spectrolab



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E f f i c i e n c y

( % )

Cu(In,Ga)Se 2 Amorphous Si:H(stabilized)CdTe

Universityof Maine

Boeing

Boeing

Boeing

Boeing ARCO

NREL

Boeing

Euro-CIS

12

8

4

0200019951990198519801975

United Solar

16

20

24

8

Spire

NorthCarolina

State University

Thin Film Technologies

Varian

RCA

Solarex

UNSW

UNSW

ARCO

UNSWUNSW

UNSWSpire

Stanford

Westing-house

Crystalline Si CellsSingle CrystalMulticrystallineThin Si

UNSWGeorgia Tech

Georgia Tech

Sharp

Solarex Astro-Power

NREL

AstroPower

NREL

Emerging PV

Masushita

Monosolar

Kodak

Kodak

AMETEK

PhotonEnergy

Univ.S.Florida

NREL

NREL

NREL

Princeton

U. Konstanz

U.CaliforniaBerkeleyOrganic Cells

NREL

NRELCu(In,Ga)Se 214x Concentration

Source: T. J. Berniard, NREL

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


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Biodiesel ProductionBiodiesel Production100 lbs. of soybean oil


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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


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Amount of land needed to capture 13 TW:

20% efficiency (photovoltaic) = 0.23%1% efficiency (bio-mass) = 4.6%

Steven Chu Berkeley

Miscanthus yields : 30 dry tons/acre

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

100 M t f 450 M 300 B g l / f th l


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> 1% conversionefficiency may be

feasible.

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


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

The majority of a plant is structural material


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SunlightCO2, H20,

NutrientsBiomass Chemical

energy

Self-fertilizing,drought and pestresistant

Improvedconversion ofcellulose intochemical fuel

Lignin 10-25%

Steven Chu Berkeley

Greenhouse Gases

125

CO I t i


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-

25

50

75

100

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

Net Energy (MJ / L)

N e

t G H G

( g

C O 2 e

/ M J - e t h

a n o

l )

Pimentel

Patze k

Shapouri

Wang

Graboski

Gasoline

de Oliviera

Today

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..


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ggThe 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 bruteCrusader-E Edition: includes elevator

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


or more fuel efficient cars?


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Fuel Cell Detail1. Hydrogen goes to Anode,

(can use hydrocarbon fuel)


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( y )Oxygen goes to Cathode .

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

3. Since electrons cannotenter the membrane theygo through the external circuit .

4. When electrons get back to theCathode, they combine with H+and O to form water.

m em b r a n e

Office of Fossil Energy, U.S. Department of EnergyOffice of Fossil Energy, U.S. Department of Energy

Fuel Cell Points


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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 H 2 and O 2).

Greenhouse Comparison:FCVs, ICEs and Hybrid Vehicles



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Source: Bevilacqua-Knight, 2001


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Generating H2

bysplitting water using

i ibl li h


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visible light

From M. Graetzel, ETH Lausanne

Chemists vital to solution Availability of fossil fuels has enabled mass of humanityto move beyond subsistence living.


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But we really need to figure out how to live without

them.

Carbon loading of the atmosphere is reaching alarminglevels.

Scientific consensus on global warming.

Clean alternatives like solar, wind, etc. have problems ofrate, efficiency. Need to solve them. Nuclear will play arole.

Answers to Numerical ProblemsEP100: 2.08 atm EP101: 10.5 atm

EP102 30.8 L EP102a: 0.626 g L -1


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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 NH 4Cl PNH 3 = 0.274 atm

Notes Solutions Chapter 05 Large

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