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GASES and theMOLE. Marti Andreski. Introduction. the # of gas molec\'s in a container determines the P at a given T Avogadro\'s Principle: "At equal T\'s and P\'s, equal V\'s of gases contain the same # of molecules"

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marti andreski

GASES and theMOLE

Marti Andreski

introduction
Introduction
  • the # of gas molec\'s in a container determines the P at a given T
  • Avogadro\'s Principle: "At equal T\'s and P\'s, equal V\'s of gases contain the same # of molecules"
  • "n" relates to the number of moles of gas at constant T and P, if V1=V2, then n1 = n2
molar volume @stp
Molar Volume @STP
  • 1 mol of any gas occupies the same V as 1 mol of any other gas
  • molar volume = the V of 1 mole of a gas at STP = 22.4 L
ideal gas equation
Ideal Gas Equation
  • P V = n R T
  • comes from Boyle\'s Law, Charles\'s Law, & molar volume
  • if we substitute in all standard conditions, we can solve for "R"
solving for r
Solving for R
  • P V = n R T
  • Rearrange to solve for R
  • R = PV / nT

Memorize These

R = (1 atm)(22.4 L) / (1 mol)(273 K) =

0.0821

____ L. atm/mol K

R = (101.3 kPa)(22.4 L) / (1 mol)(273 K) =

___ L. kPa/mol K

8.31

R = (760 torr)(22.4 L) / (1 mol)(273 K) =

62.4

___ L. torr/mol K

repeat memorize all of those values for r
Repeat: Memorize all of those values for R!
  • this crazy unit doesn\'t really exist as a measurable entity but is used for cancellation purposes
molecular mass
Molecular Mass
  • what are the units for MM? How can you use PV = nRT?
  • can be found from lab measurements in this way:
  • MM = m R T
  • P V
  • ie: What is the molecular mass of a gas if 150.0 cm3 of it has a mass of 0.922 g at 99.0oC and 107.0 kPa?
molecular mass problem
Molecular Mass Problem
  • ie: What is the molecular mass of a gas if 150.0 cm3 of it has a mass of 0.922 g at 99.0oC and 107.0 kPa?
  • MM = m R T
  • P V
and another formula
And another formula...
  • Since…
  • MM = m R T
  • P V
  • Then…
  • MM = DRT
  • P
gas stoichiometry at standard conditions
Gas Stoichiometry at Standard Conditions
  • main point: at STP, 1 mole of any gas occupies a V of 22.4 L
  • ie: what V of H2 at STP can be produced from the reaction of 6.54 g of Zn with HCl?
  • ie: how many g NaCl can be produced by the reaction of 112 cm3 of chlorine at STP with an excess of sodium?
gas stoichiometry at non standard conditions
Gas Stoichiometry at Non-Standard Conditions
  • Four types of problems to expect
  • Each type requires a slightly different approach
a mass volume non
a) Mass - Volume (Non)
  • what V of chlorine gas at 24.0oC and 99.2 kPa would be required to react with 2.51 g Ag according to:
  • 2Ag + Cl2 ---> 2AgCl
  • i) Stoich “pretending” standard – this would be your answer if conditions were STP
  • ii) VPT the answer
b volume non mass
b. Volume (Non) – Mass
  • What mass of mercury (II) chloride will react with 0.567 dm3 of ammonia at 27oC and 102.7 kPa?
  • HgCl2 + 2NH3 ---> Hg(NH2)Cl + NH4Cl
  • i) VPT first
  • ii) Stoich
c volume non volume same
c. Volume (Non) - Volume (same)
  • what volume of oxygen at 100oC and 105.5 kPa is required to burn 684 cm3 of methane at the same T and P?
  • CH4+ 2O2 --> CO2 + 2H20
  • i) if T & P do not change, no need to convert; just do Stoich
d volume non volume diff
d. Volume (Non) - Volume (Diff)
  • what volume of oxygen at 26.0oC and 102.5 kPa is required to burn 684 cm3 of methane at 101oC and 107.5 kPa?
  • CH4+ 2O2 --> CO2 + 2H20
  • i) VPT to standard
  • ii) Stoich
  • iii) VPT answer to other non-std
graham s law
Graham’s Law
  • "the relative rates at which two gases under identical condit\'s of T & P will pass thru a small hole vary inversely with the square roots of their molecular masses"
  • we know: KE1 = KE2 (at same T)
  • m1v12 = m2v22
  • 2 2
  • ...and V1 = m2
  • V2 m1
  • ie: what is the ratio of the speed of H2 to O2 when T for both is equal?
  • ...so, H2 is moving 3.98 times faster than O2
dalton s law of partial pressures
Dalton’s Law of Partial Pressures
  • the total P in a container is the sum of the partial P\'s of the gases in the container
  • see Water Vapor Pressure Table
  • it is easy to collect gases as they bubble through water. However, if a gas is collected "over water" a correction must be made which accounts for the amount of water vapor present at that T
partial pressure problem
Partial Pressure Problem
  • ie: a qty of gas is collected over water at 8.00oC in a 353 cm3 vessel. The manometer indicates a P of 84.5 kPa. What V would the dry gas occupy at standard P and 8.00oC?
  • i) subtract water vapor 84.5 - 1.1 = 83.4 kPa
  • ii) combined gas law
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