avogadro’s principle

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Avogadro’s Principle • Gas particles = big, little, heavy, light • Doesn’t matter = so far apart • Therefore, a 1000 krypton (big) atoms occupy the same space as 1000 helium (little) atoms • Or as Avogadro puts it: “equal volumes of gases at the same temperature and pressure contain equal number of particles”….which means equal volume = equal particles

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Avogadro’s Principle. Gas particles = big, little, heavy, light Doesn’t matter = so far apart Therefore, a 1000 krypton (big) atoms occupy the same space as 1000 helium (little) atoms Or as Avogadro puts it: - PowerPoint PPT Presentation

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Page 1: Avogadro’s Principle

Avogadro’s Principle• Gas particles = big, little, heavy, light

• Doesn’t matter = so far apart

• Therefore, a 1000 krypton (big) atoms occupy the same space as 1000 helium (little) atoms

• Or as Avogadro puts it:

“equal volumes of gases at the same temperature and pressure contain equal number of particles”….which means equal volume = equal particles

Page 2: Avogadro’s Principle

Molar Volume of Gases

• The volume for one mole of gas at STP

• STP = standard temperature & pressure

= 0.00o C and 1.00 atm

= 22.4 L for any gas

(remember Avogrado)

• Can be used to calculate number of particles, moles or mass of a gas.

Page 3: Avogadro’s Principle

Try It• How many moles of gas would be found in

34.36 L of nitrogen? How many molecules of nitrogen would this be?

• Calculate the volume that .881 mol of oxygen will occupy at STP

• Calculate the volume of gas for 20.00 x 1023 molecules of carbon dioxide

Page 4: Avogadro’s Principle

Answers

• 34.36 L x 1 mole / 22.4 L = 1.534 mol

• 1.534 mol x (6.02 x 1023 molecules / 1 mole) = 9.235 x 1023 molecules

• .881 mol x 22.4 L / 1 mole = 19.7 L

• (20.00 x 1023 molecules) x (1 mole / 6.02 x 1023 molecules) = 3.32 moles

3.32 moles x 22.4 L / 1 mole = 74.4 L

Page 5: Avogadro’s Principle

Try It

• Calculate the volume that 4.5 kg of ethylene gas (C2H4) will occupy at STP.

• How many grams of methane are (CH4) in 3,000 L?

Page 6: Avogadro’s Principle

Answers• 4.5 kg x 1000 g / 1 kg = 4500 g C2H4

4500 g x 1 mole / 28.05 g = 160.4 mol

160.4 mol x 22.4 L / 1 mol = 3, 593 L

• 3,000 L CH4 x 1 mole / 22.4 L = 134 mol

134 mol x 16.04 g / 1 mol = 2150 g or 2.15 kg

Page 7: Avogadro’s Principle

Ideal Gases• Gases whose particles take up no space

and have no intermolecular interactions, regardless of the pressure and temperature of the gas.

• No such thing as an ideal gas since all gases have some volume & some intermolecular force

Page 8: Avogadro’s Principle

Ideal Gas Deviations

• Real gases behave very close to what the ideal gas law predicts

• Except for two extreme conditions: really high pressure or really low temperature

• And 2 types of molecules: polar and large molecules

Page 9: Avogadro’s Principle

Ideal Gas Law• Includes moles or number of gas particles

in the combined gas law

• An increase in moles or particles leads to a change in at least one of the other 3 variables (temperature, pressure, volume)

• So PV / T = constant that depends on the number of moles of gas involved thus….

Page 10: Avogadro’s Principle

Ideal Gas Law• Ideal Gas Law: PV = nRT or PV = mRT/M

n = # of moles R = gas constant

• Value for R depends on unit used for pressure (if atm, R = .0821 L.atm / mol.K)

• n can also = m / M

m = mass present M = molar mass

Page 11: Avogadro’s Principle

Try It

• Calculate the number of moles of gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm.

• Determine the kelvin temperature required for .0470 mol of gas to fill a balloon to 1.20 L under .988 atm pressure.

Page 12: Avogadro’s Principle

Answers

• PV = nRT so n = PV / RT1.50 atm x 3.0 L / .0821 L.atm / mol.K x 300 K

= .18 mol

• PV = nRT so T = PV / nR

.988 atm x 1.20 L / .0470 mol x .0821 L.atm / mol.K

= 307 K or 34o C

Page 13: Avogadro’s Principle

Density & Ideal Gases

• Since PV = mRT / M where

m = mass present M = molar mass

• Then M = mRT /PV or M = DRT/P where D = density

– This means the ideal gas law can be used to calculate a gas’ density

Page 14: Avogadro’s Principle

Try It

• What is the molar mass of a pure gas that has a density of 1.40 g/L at STP?

• Calculate the density a gas will have at STP if it molar mass is 44.09 g/mol.

Page 15: Avogadro’s Principle

Answers

• M = DRT / P = 1.40g/L x .0821 L.atm / mol.K x 273 K / 1.00 atm

= 31.4 g/mol

• D = MP / RT

= 44.09 g/mol x 1.00atm / .0821L.atm / mol.K x 273K

= 1.97 g/L or .00197 g/ml