Gases CHAPTER 10

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Gases CHAPTER 10

• Powerpoint from Melissa Brophys website
  (http//teacherweb.com/TX/McNeilHS/brophy/photo7.a
  spx) and modified for our class

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Gases (Chapter 10)

• Resources and Activities
• Textbook - chapter 10 ppt file
• Online practice quiz from Pearson - chapter 10
• Lab activities
• (1) Molar volume of a gas (wet lab)
• (2) Virtual labs gas laws (next slide)
• POGIL activities
• Ideal gas law
• Partial Pressure
• Comprehensive tutorial and animations on
  Intermolecular forces
• http//www.chem.purdue.edu/gchelp/liquids/imf2.htm
  l

Chemtour videos from W.W. Norton (chapter 6
ideal gas law Daltons law molecular speed)
http//www.wwnorton.com/college/chemistry/gilber
t2/contents/ch06/studyplan.asp See NEXT SLIDE
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• Animation on gases to view in class
• http//glencoe.com/sites/common_assets/advanced_p
  lacement/chemistry_chang9e/animations/chang_7e_esp
  /gam2s2_6.swf
• Independent work - students to view animations
  interactive activities (4 in total) and write
  summary notes on each. These summaries are to be
  included in your portfolio
• http//glencoe.mcgraw-hill.com/sites/0035715985/s
  tudent_view0/chapter5/animations_center.html
• Virtual lab activity (To complete at home and to
  hand in for grading)
• Boyles Law Objective Explore the effects of V
  on P at constant T (for a fixed quantity of gas
  molecules ). Task Generate, record and graph
  data for three gases. Print and Discuss your
  results. Go to http//www.chem.iastate.edu/group
  /Greenbowe/sections/projectfolder/flashfiles/gasla
  w/boyles_law_graph.html
• Charles Law Objective Explore the effect of T on
  V at constant P (for a fixed quantity of gas
  molecules). Task generate, record and graph
  data. Print and discuss your results. Go to
• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/flashfiles/gaslaw/charles_law.htm
  l

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Activities and Problem set for chapter 10 (due
date_______)

• Complete the chapter 10 packet part (a)
  Characteristics of Gases ( sample, practice and
  integrative exercises - show work) and part (b)
  AP Chemistry Gas Law Practice
• Do all chapter 10 GIST and Visualizing concepts
  problems - write out questions and answers show
  work. (NB Photocopy of questions is also
  acceptable)

• TextBook ch. 10 all sections required for
  regents (in part), SAT II and AP exams
• Lab activities (wet lab and virtual lab)
• POGILS (2)
• Online practice quiz ch 10 due by_____

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Measurements on Gases

• Properties of gases
• Gases uniformly fill any container
• Gases are easily compressed
• Gases mix completely with any other gas
• Gases exert pressure on their surroundings
• Pressure Force
• Area

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Measurements on Gases

• Measuring pressure
• The barometer measures atmospheric pressure
• Inventor Evangelista Torricelli (1643)

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Measurements on Gases

• The manometer measures confined gas pressure

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Measurements on Gases

• Units
• mm Hg (torr)
• 760 torr standard pressure
• Kilopascal (kPa)
• 101.325 kPa standard pressure
• Atmospheres
• 1 atmosphere (atm) standard pressure
• STP 1 atm 760 torr 760 mmHg 101.325 kPa

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Measurements on Gases

• Example Convert 0.985 atm to torr and to kPa.

0.985 atm
760 torr
749 torr
1 atm
0.985 atm
101.325 kPa
98.3 kPa
1 atm
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The Gas Laws of Boyle, Charles and Avogadro

• Boyles Law (Robert Boyle, 1627-1691)
• The product of pressure times volume is a
  constant, provided the temperature remains the
  same
• PV k

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The Gas Laws of Boyle, Charles and Avogadro

• P is inversely related to V
• The graph of P versus V is hyperbolic
• Volume increases linearly as the pressure
  decreases
• As you squeeze a zip lock bag filled with air
  (reducing the volume), the pressure increases
  making it difficult to keep squeezing

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The Gas Laws of Boyle, Charles and Avogadro

• At constant temperature, Boyles law can be used
  to find a new volume or a new pressure
• P1V1 k P2V2
• or P1V1 P2V2
• Boyles law works best at low pressures
• Gases that obey Boyles law are called Ideal
  gases

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The Gas Laws of Boyle, Charles and Avogadro

• Example A gas which has a pressure of 1.3 atm
  occupies a volume of 27 L. What volume will the
  gas occupy if the pressure is increased to 3.9
  atm at constant temperature?
• P1 1.3 atm
• V1 27 L
• P2 3.9 atm
• V2 ?

(1.3atm)(27L) (3.9atm)V2 V2 9.0 L
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The Gas Laws of Boyle, Charles and Avogadro

• Charles Law (Jacques Charles, 1746 1823)
• The volume of a gas increase linearly with
  temperature provided the pressure remains
  constant
• V bT V b V1 b V2
• T T1 T2
• or V1 V2
• T1 T2

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The Gas Laws of Boyle, Charles and Avogadro

• Temperature must be measured in Kelvin
• ( K C 273)
• 0 K is absolute zero

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The Gas Laws of Boyle, Charles and Avogadro

• Example A gas at 30C and 1.00 atm has a volume
  of 0.842L. What volume will the gas occupy at
  60C and 1.00 atm?
• V1 0.842L
• T1 30C (273 303K)
• V2 ?
• T2 60C (273 333K)

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• 0.842L V2
• 303K 333K
• V2 0.925L

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The Gas Laws of Boyle, Charles and Avogadro

• Avogadros Law (Amedeo Avogadro, 1811)
• For a gas at constant temperature and pressure,
  the volume is directly proportional to the number
  of moles, n
• V an V/n a
• V1 a V2 or V1 V2
  
• n1 n2 n1 n2

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The Gas Laws of Boyle, Charles and Avogadro

• Example A 5.20L sample at 18C and 2.00 atm
  contains 0.436 moles of a gas. If we add an
  additional 1.27 moles of the gas at the same
  temperature and pressure, what will the total
  volume occupied by the gas be?
• V1 5.20L
• n1 0.436 mol
• V2 ?
• n2 1.27 mol 0.436 mol 1.706 mol

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The Gas Laws of Boyle, Charles and Avogadro

• 5.20 L V2
• 0.436 mol 1.706 mol
• x 20.3L

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The Ideal Gas Law

• Derivation from existing laws
• Vk VbT Van
• P
• V kba(Tn)
• P

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The Ideal Gas Law

• Constants k, b, a are combined into the universal
  gas constant (R),
• V nRT or PV nRT
• P

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The Ideal Gas Law

• R PV using standard numbers will give you R
• nT
• STP
• P 1 atm
• V 22.4L
• n 1 mol
• T 273K
• Therefore if we solve for R, R 0.0821 L
  atm/mol K

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The Ideal Gas Law

• Example A sample containing 0.614 moles of a
  gas at 12C occupies a volume of 12.9L. What
  pressure does the gas exert?
• P ?
• V 12.9L
• n 0.614 mol
• R 0.0821 Latm/mol K
• T 12 273 285K

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The Ideal Gas Law

• (P)(12.9L) (0.614mol)(0.0821 Latm/mol
  K)(285K)
• Hint Rearrange and re-write (watch out for units
  in numerator and denominator)
• P 1.11 atm

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The Ideal Gas Law

• Solving for new volumes, temperature, or pressure
  (n remaining constant)
• Combined Law
• P1V1 nR P2V2 or P1V1 P2V2
• T1 T2 T1 T2

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The Ideal Gas Law

• Example A sample of methane gas at 0.848 atm
  and 4.0C occupies a volume of 7.0L. What volume
  will the gas occupy if the pressure is increased
  to 1.52 atm and the temperature increased to
  11.0C?
• P1 0.848 atm P2 1.52 atm
• V1 7.0L V2 ?
• T1 4273 277K T2 11273 284K

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The Ideal Gas Law

• (0.848atm)(7.0L) (1.52atm)(V2)
• 277K 284K
• V2 4.0L

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Gas Stoichiometry

• Molar volume
• One mole of an ideal gas occupies 22.4L of volume
  at STP
• Things to remember
• Density mass
• volume
• n grams of substance m
• molar mass M
• PV mRT
• M

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Gas Stoichiometry

• Example A sample containing 15.0g of dry ice
  (CO2(s)) is put into a balloon and allowed to
  sublime according to the following equation
  CO2(s) ? CO2(g)
• How big will the balloon be (i.e. what is the
  volume of the balloon) at 22C and 1.04 atm,
  after all of the dry ice has sublimed?

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Gas Stoichiometry

• Moles of CO2(s)
• 15g 1 mol 0.341 mol CO2(s)
• 44 g
• 0.341 mol CO2(s) 1 CO2(g) 0.341 mol CO2(g)
• 1 CO2(s) n

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Gas Stoichiometry

• P 1.04 atm
• V ?
• n 0.341 mol
• R 0.0821 Latm/mol K
• T 22C 273 295K
• V 7.94L

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Gas Stoichiometry

• Limiting Reactant Example 0.500L of H2(g) are
  reacted with 0.600L of O2(g) at STP according to
  the equation 2H2(g) O2(g) ? 2H2O(g)
• What volume will the H2O occupy at 1.00 atm and
  350C?

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Gas Stoichiometry

• H2 0.5L 1 mol 2 H2O 0.0223 mol H2O
• 22.4L 2 H2
• O2 0.6L 1 mol 2 H2O 0.0536 mol H2O
• 22.4L 1 O2

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Gas Stoichiometry

• H2
• (1 atm)(V) (0.0223mol)(0.0821 Latm/mol
  K)(623K)
• V 1.14L
• O2
• (1atm)(V) (0.0536mol)(0.0821 Latm/mol
  K)(623K)
• V 2.74L

Limiting reactant and how much volume is produced!
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Gas Stoichiometry

• Density/Molar Mass example A gas at 34C and
  1.75 atm has a density of 3.40g/L. Calculate the
  molar mass of the gas.
• D mass 3.40 3.40g
• V 1L
• P 1.75 atm R 0.0821 Latm/mol K
• V 1L T 34 273 307K
• m 3.40 g M x

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Gas Stoichiometry

• (1.75atm)(1L) (3.4g)(0.0821 Latm/mol
  K)(307K)
• M
• M molar mass
• M 48.97 g/mol

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Daltons Law of Partial Pressures (John Dalton,
1803)

• Statement of law
• for a mixture of gases in a container, the total
  pressure exerted is the sum of the pressures each
  gas would exert if it were alone
• It is the total number of moles of particles that
  is important, not the identity or composition of
  the gas particles.

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Daltons Law of Partial Pressures

• Derivation
• Ptotal P1 P2 P3
• P1 n1RT P2 n2RT P3 n3RT .
• V V V
• Ptotal n1RT n2RT n3RT
• V V V
• Ptotal (n1n2n3) (RT)
• V
• Ptotal ntotal (RT)
• V

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Daltons Law of Partial Pressures

• Example Oxygen gas is collected over water at
  28C. The total pressure of the sample is 5.5
  atm. At 28C, the vapor pressure of water is 1.2
  atm. What pressure is the oxygen gas exerting?
• Ptotal PO2 PH2O
• 5.5 x 1.2
• X 4.3 atm

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Daltons Law of Partial Pressures

• Mole Fraction
• The ratio of the number of moles of a given
  component in a mixture to the total number of
  moles in the mixture
• For an ideal gas, the mole fraction (x)
• x1 n1 P1
• ntotal Ptotal

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Daltons Law of Partial Pressures

• Example The vapor pressure of water in air at
  28C is 28.3 torr. Calculate the mole fraction
  of water in a sample of air at 28C and 1.03 atm.
  
• XH2O PH2O 28.3 torr
• Pair 783 torr
• 0.036

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Daltons Law of Partial Pressures

• Example A mixture of gases contains 1.5 moles
  of oxygen, 7.5 moles of nitrogen and 0.5 moles of
  carbon dioxide. If the total pressure exerted is
  800 mmHg, what are the partial pressures of each
  gas in the mixture?

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Daltons Law of Partial Pressures

• Calculate the total number of moles
• 1.5 7.5 0.5 9.5 moles
• Use mole fractions for each individual gas
• PO2 800mmHg x 1.5 mol 126 mmHg
• 9.5 mol
• PN2 800mmHg x 7.5 mol 632 mmHg
• 9.5 mol
• PCO2 800mmHg x 0.5 mol 42 mmHg
• 9.5 mol

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Kinetic Molecular Theory of Gases

• Postulates of the KMT Related to Ideal Gases
• The particles are so small compared with the
  distances between them that the volume of the
  individual particles can be assumed to be zero
• The particles are in constant motion. Collisions
  of the particles with the walls of the container
  cause pressure.
• Assume that the particles exert no forces on each
  other
• The average kinetic energy of a collection of gas
  particles is assumed to be directly proportional
  to the Kelvin temperature of the gas

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Kinetic Molecular Theory of Gases

• Explaining Observed Behavior with KMT
• P and V (T constant)
• As V is decreased, P increases
• V decrease causes a decrease in the surface area.
  Since P is force/area, the decrease in V causes
  the area to decrease, thus increasing the P

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Kinetic Molecular Theory of Gases

• P and T (V constant)
• As T increase, P increases
• The increase in T causes an increase in average
  kinetic energy. Molecules moving faster collide
  with the walls of the container more frequently,
  and with greater force.

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Kinetic Molecular Theory of Gases

• V and T (P constant)
• As T increases, V also increases
• Increased T creates more frequent, more forceful
  collisions. V must increase proportionally to
  increase the surface area, and maintain P

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Kinetic Molecular Theory of Gases

• V and n (T and P constant)
• As n increases, V must increase
• Increasing the number of particles increases the
  number of collisions. This can be balanced by an
  increase in V to maintain constant P

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Kinetic Molecular Theory of Gases

• Daltons Law of partial pressures
• P is independent of the type of gas molecule
• KMT states that particles are independent, and V
  is assumed to be zero. The identity of the
  molecule is therefore unimportant

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Kinetic Molecular Theory of Gases

• Root Mean Square Velocity
• Velocity of a gas is dependent on mass and
  temperature
• Velocity of gases is determined as an average
• M mass of one mole of gas particles in kg
• R 8.3145 J/Kmol
• joule kgm2/s2
• urms 3RT
• M

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Kinetic Molecular Theory of Gases

• Example Calculate the root mean square velocity
  for the atoms in a sample of oxygen gas at 0C.
• MO2 32g/mol
• 32g 1 kg 0.032 kg
• 1000g
• T 0 273 273K
• R 8.3145 J/Kmol

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Kinetic Molecular Theory of Gases

• Urms 3(8.3145 kgm2 s-2K-1 mol-1)(273 K)
  461ms-1
• 0.032 kgmol-1
• Answer 461ms-1 or 461 m/s
• Recall 1 Joule 1 J 1 kg m2/s2 1 kg m2 s-2

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Kinetic Molecular Theory of Gases

• Mean Free Path distance between collisions
• Average distance a molecule travels between
  collisions
• 1 x 10 -7 m for O2 at STP

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Effusion and Diffusion

• Effusion
• Movement of a gas through a small opening into an
  evacuated container (vacuum)
• Grahams law of effusion
• Rate of effusion for gas 1 vM2
• Rate of effusion for gas 2 vM1

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Effusion and Diffusion

• Example How many times faster than He would NO2
  gas effuse?
• MNO2 46.01 g/mol larger mass
• MHe 4.003 g/mol effuses slower
• Rate He 46.01 He effuses 3.39
• Rate NO2 4.003 times faster

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Real Gases and van der Waals Equation (Johannes
van der Waals, 1873)

• Volume
• Real gas molecules do have volume
• Volume available is not 100 of the container
  volume
• n number of moles
• b is an empirical constant, derived from
  experimental results

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Real Gases and van der Waals Equation

• Ideal P nRT
• V
• Real P nRT
• V-nb

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Real Gases and van der Waals Equation

• Pressure
• Molecules of real gases do experience attractive
  forces
• a proportionality constant determined by
  observation of the gas
• Pobs Pideal a(n/V)2

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Real Gases and van der Waals Equation

• Combining to derive van der Waals equation
• Pobs nRT - n2a
• V-nb V2
• And then rearranging
• (Pobs n2a/V2)(V-nb) nRT

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Chemistry in the Atmosphere

• Composition of the Troposphere
• The atmosphere is composed of 78 N2, 21 O2,
  0.9Ar, and 0.03 CO2 along with trace gases.
• The composition of the atmosphere varies as a
  function of distance from the earths surface.
  Heavier molecules tend to be near the surface due
  to gravity.

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Chemistry in the Atmosphere

• Upper atmospheric chemistry is largely affected
  by ultraviolet, x ray, and cosmic radiation
  emanating from space. The ozone layer is
  especially reactive to ultraviolet radiation
• Manufacturing and other processes of our modern
  society affect the chemistry of our atmosphere.
  Air pollution is a direct result of such
  processes.