Marti Andreski

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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: &quot;At equal T's and P's, equal V's of gases contain the same # of molecules&quot;

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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"
• "n" relates to the number of moles of gas at constant T and P, if V1=V2, then n1 = n2
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
• 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
• 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!
• this crazy unit doesn't really exist as a measurable entity but is used for cancellation purposes
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
• 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...
• Since…
• MM = m R T
• P V
• Then…
• MM = DRT
• P
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
• Four types of problems to expect
• Each type requires a slightly different approach
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
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)
• 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)
• 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
• "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
• 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
• 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