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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
- ii) VPT the answer

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

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