Chemistry Chapter 11

Molecular Composition of Gases

Volume and mass

• Gay-Lussac examined gas volume in reactions• Noted: 2 L H2 and 1 L O2 can form 2 L water

vapor• 2:1:2 volume relationship of H:O: water• Simple definite proportions hold true for

other gases in reactions • this lead to ….

Gay-Lussac’s Law of Combining Gas Volumes

• At constant temperature and pressure the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Avogadro (again)

• Combining volumes seemed to challenge the indivisibility of the atom…

• Avogadro posited that some molecules might contain more than one atom (ex. O2 explains the 2:1:2 H:O:water ratio)

• Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

Implications of Avogadro

• At the same temperature and pressure the volume of a gas varies directly with the number of molecules

• Avogadro believed that some elements must exist in diatomic form (H2, O2, N2)

Avogadro continued

• H, O and water illustrate this well• 2 volumes H2 + 1 volume O2 = 2 volumes H2O

Leads to the balanced equation…2H2 + O2 2H2O

• Which confirms the diatomic molecule hunch nicely!

Avogadro in algebra

• Gas volume is directly proportional to the amount of gas (number of particles) at a given temperature and pressure give us:

V = kn• Where – V is volume– n is the amount of gas (in moles)– k is a constant

Molar Volumes

• One mole of gas at STP will occupy 22.4 L• 1 mole/ 22.4 L of ________gas can be used as

a conversion factor to find number of particles, mass, or volume of a gas at STP

Practice! • Problems 1-3 page 337

Ideal Gas Law

• A mathematical relationship among pressure, volume, temperature and number of moles

• To derive (see p. 341)• Ideal gas law:• V= nRT/P OR PV = nRT• V is volume, P is pressure, T is temperature, n

is number of moles, and R is a constant

11-2

The Ideal Gas Constant

• R= (1 atm) (22.4 L)/(1 mol) (273.15 K)Or

• R= 0.08205784 L x atm / mol x K• (round to .0821) • USE ONLY when units are appropriate!• For any other units see chart on p. 342• Practice Problems! p. 345

Finding Molar Mass or Density

• Use V= nRT/P but remember that n (number of moles) is equal to mass (m) /molar mass (M)

Substituting gives: PV= mRT/M or M= mRT/PV

• Density is just mass (m) per unit volume (V)Substituting gives: M= DRT/P or D= MP/RTPractice problems! 1-4 page 346

Stoichiometry of gases

• Volume ratios of gases in reactions can be used exactly as mole ratios are in standard mass-mole, mass-mass, mole mole etc. problems

• Practice! 1-2 pg. 348

11-3

More Stoich!

• When given a volume for a reactant and a mass for a reactant…

• Go through moles• Need conditions (temp, pressure, etc) for each

gas• Ideal gas law works well for this • Practice problems! #1-2, p.349 # 1-2 p.350

Effusion and Diffusion

• Rates of either can be calculated!• Remember! KE= ½ m v2

• And… for any two gases (A & B, lets say) at the same temp KEA = KEB so…

½ MA vA2 = ½ MBvB

2

• Where M is molar mass and v is molecular velocity• You can multiply by 2 to clean up and get

MA vA2 = MBvB

2

Effusion and diffusion (con’t)

• Recall: MA vA2 = MB vB

2

• Rearrange: vA2 / vB

2 = MB / MA

• Take square roots: vA / vB = √MB / √MA

• Because rate of effusion is directly proportional to molecular velocity we can say that:

rate of effusionA / rate of effusionB = √MB / √MA

Grahams Law of effusion

• Rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses

• So? Density varies directly with molar mass so…

rate of effusionA / rate of effusionB = √DB / √DA

(Where D is density)

Grahams…

• Can be used to find density or molar mass of gases effusing.

• Practice problems! #1-3 p.355

That’s all folks!