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: "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


Summary gas stoichiometry

Summary – Gas Stoichiometry


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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