avogadro’s principle
DESCRIPTION
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 PresentationTRANSCRIPT
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
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.
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
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
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?
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
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
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
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….
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
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.
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
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
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.
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