notes solutions chapter 05 large
TRANSCRIPT
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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).
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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
From D. Kamen, UC Berkeley
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)
From D. Kamen, UC Berkeley
28
32
Three-junction (2-terminal, monolithic)Two-junction (2-terminal, monolithic)
NREL/Spectrolab
NRELNREL
JapanEnergy
Multijunction Concentrators
Best Research-Cell Efficiencies
Spectrolab
From D. Kamen, UC Berkeley
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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
From D. Kamen, UC Berkeley
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