Ideal Gases and Molar Volume

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Ideal Gases and Molar Volume
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Volume, Pressure, Temperature

• and now actual amount of gas (number of gas
  molecules contained in a sample)
• Amedeo Avogadro, 1800s, asked the question

If you have the same volumes of oxygen,
nitrogen, carbon dioxide in 3 different
balloons at the same temperature and pressure,
how do the numbers of gas molecules compare?
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• After careful observation and experimentation he
  proposed
• Equal volumes of all gases at the same
  temperature and pressure contain the same number
  of molecules.
• Now known as
• AVOGADROS LAW

Italian stamp from 1956, with Avogadros Law in
Italian
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Molar Volume

• volume occupied by one mole of a substance
• All gases have equal molar volume if measured at
  the same temp and pressure
• At conditions of 0ºC and 1 atm, the molar volume
  of any gas sample is 22.4 L (no similar simple
  relationship for solids and liquids)

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Simplifies chemical reactions involving gases

• Consider the following reaction
• N2(g) O2(g) ? 2 NO(g)
• Interpreted as
• 1 volume of N2(g) and 1 volume of O2(g) combine
  to form 2 volumes of NO(g)
• Sample problem
• What volume (in liters) of NO(g) is produces as 3
  L N2(g) and 3 L O2(g) react? (See equation
  above.)
• Because the reactants have tripled, the product
  is also tripled6 L NO(g)

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• Formation of water (vapor)
• 2 H2(g) O2(g) ? 2 H2O(g)
• Can be interpreted as
• 2 volumes H2(g) added to 1 volume O2(g) combine
  to form 2 volumes H2O(g)
• Volume can be either liters of cm3
• So 200.0 L H2(g) 100.0 L O2(g) would produce
  how many liters of water vapor?
• 100.0 L H2O(g)

Visible water vapor on Earth, NASA
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Sample problem 1

• In the following equation, what volume of C2H6(g)
  is needed to produce 12 L CO2(g)? All
  measurements are at the same temperature.
• 2 C2H6(g) 7 O2(g) ? 4 CO2(g) 6 H2O(g)
• Answer 2 L C2H6(g) to 4 L CO2(g) so,
• 6 L C2H6(g) to 12 L CO2(g)

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Sample Problem 2

• Using the same scenario (from 1), how many
  liters of O2(g) would be needed to produce 12 L
  CO2(g)? Think through this question and make a
  prediction.
• Answer 7 O2(g) ? 4 CO2(g), so
• ? L CO2(g) 12 L CO2(g) x 7 O2(g)
• 4
  CO2(g)
• 21 L CO2(g)

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Homework

• A.18 Molar Volume and Reactions of Gases p.329
  1-3