Chemistry Chapter 11

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Chemistry Chapter 11

• Molecular Composition of Gases

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Volume and mass

• Gay-Lussac examined gas volume in reactions
• Noted 2 L H2 and 1 L O2 can form 2 L water vapor
• 212 volume relationship of HO water
• Simple definite proportions hold true for other
  gases in reactions
• this lead to .

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Gay-Lussacs Law of Combining Gas Volumes

• At constant temperature and pressure the volumes
  of gaseous reactants and products can be
  expressed as ratios of small whole numbers

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Avogadro (again)

• Combining volumes seemed to challenge the
  indivisibility of the atom
• Avogadro posited that some molecules might
  contain more than one atom (ex. O2 explains the
  212 HOwater ratio)
• Avogadro's Law Equal volumes of gases at the
  same temperature and pressure contain equal
  numbers of molecules

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Implications of Avogadro

• At the same temperature and pressure the volume
  of a gas varies directly with the number of
  molecules
• Avogadro believed that some elements must exist
  in diatomic form (H2, O2, N2)

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Avogadro continued

• H, O and water illustrate this well
• 2 volumes H2 1 volume O2 2 volumes H2O
• Leads to the balanced equation
• 2H2 O2 ? 2H2O
• Which confirms the diatomic molecule hunch nicely!

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Avogadro in algebra

• Gas volume is directly proportional to the amount
  of gas (number of particles) at a given
  temperature and pressure give us
• V kn
• Where
• V is volume
• n is the amount of gas (in moles)
• k is a constant

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Molar Volumes

• One mole of gas at STP will occupy 22.4 L
• 1 mole/ 22.4 L of ________gas can be used as a
  conversion factor to find number of particles,
  mass, or volume of a gas at STP
• Practice!
• Problems 1-3 page 337

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Ideal Gas Law
11-2

• A mathematical relationship among pressure,
  volume, temperature and number of moles
• To derive (see p. 341)
• Ideal gas law
• V nRT/P OR PV nRT
• V is volume, P is pressure, T is temperature, n
  is number of moles, and R is a constant

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The Ideal Gas Constant

• R (1 atm) (22.4 L)/(1 mol) (273.15 K)
• Or
• R 0.08205784 L x atm / mol x K
• (round to .0821)
• USE ONLY when units are appropriate!
• For any other units see chart on p. 342
• Practice Problems! p. 345

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Finding Molar Mass or Density

• Use V nRT/P but remember that n (number of
  moles) is equal to mass (m) /molar mass (M)
• Substituting gives
• PV mRT/M or M mRT/PV
• Density is just mass (m) per unit volume (V)
• Substituting gives
• M DRT/P or D MP/RT
• Practice problems! 1-4 page 346

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Stoichiometry of gases
11-3

• Volume ratios of gases in reactions can be used
  exactly as mole ratios are in standard mass-mole,
  mass-mass, mole mole etc. problems
• Practice! 1-2 pg. 348

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More Stoich!

• When given a volume for a reactant and a mass
  for a reactant
• Go through moles
• Need conditions (temp, pressure, etc) for each
  gas
• Ideal gas law works well for this
• Practice problems! 1-2, p.349 1-2 p.350

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Effusion and Diffusion

• Rates of either can be calculated!
• Remember! KE ½ m v2
• And for any two gases (A B, lets say) at the
  same temp KEA KEB so
• ½ MA vA2 ½ MBvB2
• Where M is molar mass and v is molecular velocity
• You can multiply by 2 to clean up and get
• MA vA2 MBvB2

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Effusion and diffusion (cont)

• Recall MA vA2 MB vB2
• Rearrange vA2 / vB2 MB / MA
• Take square roots vA / vB vMB / vMA
• Because rate of effusion is directly proportional
  to molecular velocity we can say that
• rate of effusionA / rate of effusionB vMB / vMA

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Grahams Law of effusion

• Rates of effusion of gases at the same
  temperature and pressure are inversely
  proportional to the square roots of their molar
  masses
• So? Density varies directly with molar mass so
• rate of effusionA / rate of effusionB vDB / vDA
• (Where D is density)

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Grahams

• Can be used to find density or molar mass of
  gases effusing.
• Practice problems! 1-3 p.355

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Thats all folks!