Learning Log b Why are you advised to open windows slightly if a tornado approaches?

Dalton’s Law Ideal Gas Law (p , ) Ch. 10 & 11 - Gases

A.Dalton’s Law of Partial Pressures b Total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. In the absence of a chemical reaction P T = P 1 + P 2 + P 3 +…

Practice Problem Dalton’s Law of Partial Pressures b Calculate the partial pressure in millimeters of mercury exerted by the four main gases in air at 760 mm Hg: nitrogen, oxygen, argon, and carbon dioxide. Their abundance by volume is %, 20.95%, %, and 0.035%, respectively.

Practice Problem b N: (760 mm Hg)(0.7808) = mm Hg b O: (760 mm Hg)(0.2095) = mm Hg b Ar: (760 mm Hg)( ) = mm Hg b C: (760 mm Hg)( ) = mm Hg mm Hg mm Hg mm Hg mm Hg = 760 mm Hg

Practice Problem A mixture of four gases exerts a total pressure of 1200 mm Hg. Gases A and B each exert 420 mm Hg. Gas C exerts 280 mm Hg. What pressure is exerted by gas D? P T = P 1 + P 2 + P 3 + P = 420 mm Hg mm Hg mm Hg + P = 1120 mm Hg + P 4 P 4 = 80 mm Hg

B. Vapor pressure of water b Gases are often collected in lab by water displacement and are mixed with water vapor b P atm = P gas + P H2O b To determine the pressure of the gas collected – subtract the vapor pressure of the water at that temperature from the current atmospheric pressure

Learning Log (before lab) b Oxygen gas from the decomposition of hydrogen peroxide was collected by water displacement. The barometric pressure and the temperature during the experiment were torr and 20.0 °C. What was the partial pressure of the oxygen collected?

PracticePractice b P atm = P gas + P H2O b P O2 = P atm - P H2O (20.0 °C) b vapor pressure of water at 20.0 °C is 17.5 torr (from TableA-8) b P O2 = torr – 17.5 torr b P O2 = torr

C. Ideal Gas Law b The mathematical relationship among pressure, volume, temperature and the number of moles of a gas. b Derived by combining the gas laws. PV=nRT

PV T VnVn PV nT D. Ideal Gas Constant = k IDEAL GAS CONSTANT R= L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm E. Ideal Gas Law Problems b Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.

GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm 3  kPa/mol  K E. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3

F.Finding Molar Mass from the Ideal Gas Law n = mass n = m Molar mass M PV = mRT OR M PV=nRT M = mRT PV

G.Finding Density from the Ideal Gas Law D = mass or D = m volume V M = DRT P M = mRT PV D = MP RT

PRACTICE PROBLEMS b At 28°C and at, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas? b P = atm V = 1.00 L b T = 28°C = 273 = 301 K m = 5.16 g b M = mRT PV = (5.16 g) ( L∙atm/mol∙K) (301K) (0.974 atm)(1.00 L) = 131g/mol

PRACTICE PROBLEMS b What is the density of a sample of ammonia gas, NH 3, if the pressure is atm and the temperature is 63.0°C? b P = atm T = 63.0°C = 336 K M = g/mol R = L∙atm/mol∙K b D = MP = ( g/mol)(0.928 atm) RT (0.0821L∙atm/mol∙K)((336 K) = g/L NH 3

Homework Assignment b Workbook. b Complete problems #1 and 2 on pp , 1-2 p. 178, 1-2 p. 180

Practice Test Part 2 b P. 181 – 182 #1-7, 12,

Practice test Part 2 b Workbook p. 182 #1-7