Kinetic Molecular Theory Postulates of the Kinetic Molecular Theory of Gases Gases consist of particles (atoms or molecules) in constant, straight-line motion. Gas particles do not attract or repel each other (no interactions). Particles collide with each other and surfaces elastically. Collisions with walls of container define pressure (P = F/A). Gas particles are small, compared with the distances between them. Hence, the volume (size) of the individual particles can be assumed to be negligible (zero). 4. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas

Properties of Gases Gases expand to fill any container. random motion, no attraction Gases are fluids (like liquids). particles flow easily Gases have very low densities. lots of empty space; particles spaced far apart Gases are easily compressible. empty space reduced to smaller volume Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Collisions of Gas Particles Pressure = collisions on container walls

Changing the Size of the Container In a smaller container - particles have less room to move. Particles hit the sides of the container more often. This causes an increase in pressure. As volume decreases: pressure increases.

Pressure = Force/Area KEY UNITS AT SEA LEVEL (also known as standard pressure) 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi Sea level Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Barometers Mount Everest Sea level Sea level On top of Mount Everest

Temperature K = ºC + 273 ºF ºC K Always use temperature in Kelvin when working with gases. Std temperature = 273 K ºF -459 32 212 ºC -273 100 K 273 373 K = ºC + 273 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Standard Temperature & Pressure STP STP Standard Temperature & Pressure 0°C 273 K 1 atm 101.325 kPa - OR - Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Boyle’s Law As the pressure on a gas increases - the volume decreases 1 atm As the pressure on a gas increases - the volume decreases Pressure and volume are inversely related As the pressure on a gas increases 2 atm 4 Liters 2 Liters

Boyle’s Law Illustrated Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 404

PV = k Boyle’s Law P1 x V1 = P2 x V2 Volume (mL) Pressure (torr) P.V (mL.torr) 10.0 20.0 30.0 40.0 760.0 379.6 253.2 191.0 7.60 x 103 7.59 x 103 7.64 x 103 The pressure and volume of a gas are inversely related at constant mass & temp PV = k P1 x V1 = P2 x V2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Boyle’s Law example A quantity of gas under a pressure of 106.6 kPa has a volume of 380 cm3. What is the volume of the gas at standard pressure, if the temperature is held constant? P1 x V1 = P2 x V2 (106.6 kPa) x (380 cm3) = (101.3 kPa) x (V2) V2 = 400 cm3

Charles’s Law Timberlake, Chemistry 7th Edition, page 259

If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K one of 2 things happens

Either the volume will increase to 2 liters at 1 atm 600 K 300 K Either the volume will increase to 2 liters at 1 atm

300 K 600 K Or the pressure will increase to 2 atm.

Charles’ Law Volume (mL) Temperature (K) V / T (mL / K) 40.0 44.0 47.7 51.3 273.2 298.2 323.2 348.2 0.146 0.148 0.147 The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V1 / T1 = V2 / T2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gay-Lussac’s Law P1 / T1 = P2 / T2 The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume Temperature (K) Pressure (torr) P/T (torr/K) 248 691.6 2.79 273 760.0 2.78 298 828.4 373 1,041.2 Joseph Louis Gay-Lussac (1788-1850, France) His experiments led him to propose in 1808 the Law of Combining Volumes, which states that the volume of gases involved in a chemical reaction are in a small whole number ratio. P1 / T1 = P2 / T2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

V T PV T P T PV = k P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1 Combined Gas Law V T PV T P T PV = k P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

The Combined Gas Law A quantity of gas has a volume of 400 cm3 at STP. What volume will it occupy at 35oC and 83.3 kPa? (101.325 kPa) x (400 cm3) = (83.3 kPa) x (V2) 273 K 308 K P1 = 101.325 kPa T1 = 273 K V1 = 400 cm3 P2 = 83.3 kPa T2 = 35oC + 273 = 308 K V2 = ? cm3 V2 = 548.9 cm3

The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 cm3. What volume will it occupy at 20oC and 93.3 kPa? (101.325 kPa) x (500 cm3) = (93.3 kPa) x (V2) 273 K 293 K P1 = 101.325 kPa T1 = 273 K V1 = 500 cm3 P2 = 93.3 kPa T2 = 20oC + 273 = 293 K V2 = ? cm3 V2 = 582.8 cm3

Molar Volume (Avogadro) 1 mol of all gases @ STP have a volume of 22.4 L Avogadro’s Law V1/n1 = V2/n2 MOLAR VOLUME One mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Timberlake, Chemistry 7th Edition, page 268

PV = nRT Ideal Gas Law Brings together all gas properties. P = pressure V = volume (must be in liters) n = moles R = universal gas constant (0.082 or 8.314) T = temperature (must be in Kelvin) Can be derived from experiment and theory.

Ideal Gas Law What is the pressure of 0.18 mol of a gas in a 1.2 L flask at 298 K? PV = nRT P x (1.2 L) = (0.18 mol) x (.082) x (298 K) P = ? atm n = 0.18 mol T = 298 K V = 1.2 L R = .082 (L x atm)/(mol x K) P = 3.7 atm

D = (MM)P/RT Gas Density Larger particles are more dense. Gases are more dense at higher pressures and lower temperatures D = density (in g/L) P = pressure MM = molar mass (g/mol) R = universal gas constant T = temperature (must be in Kelvin) Can be derived from experiment and theory.

Gas Problems 1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.00C and 3.25 atm? Use proper sig figs. g/mol = (0.010g/ml) x (.082atm L/mol K) x (262 K) x (1/3.25 atm) x (1000ml/1 L) Molar mass = ? g/mol D = 0.010 g/ml T = 262 K P = 3.25 atm R = .082 (L atm)/(mol K) P = 66 g/mol

Gas Problems 2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.00C and a pressure of 185 kPa? Use proper sig figs. (185 kPa) x (V) = (3.35 mol) x (8.314 L kPa/mol K) x (320 K) PV = nRT V = ? L n = 3.35 mol T = 320 K P = 185 kPa R = 8.314 (L kPa)/(mol K) V = 48.2 L

Ptotal = P1 + P2 + ... Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. Ptotal = P1 + P2 + ... In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container. The mole ratio in a mixture of gases determines each gas’s partial pressure. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Mixtures and Dalton’s Law

Gas Collected Over Water When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

Look up water-vapor pressure on p.10 for 22°C. Dalton’s Law Hydrogen gas is collected over water at 22°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.6 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH2 + 2.6 kPa PH2 = 91.8 kPa Look up water-vapor pressure on p.10 for 22°C. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Dalton’s Law The total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa. What is the partial pressure of each gas. 3 mol He 7 mol gas PHe = (97.4 kPa) = 41.7 kPa 4 mol Ne 7 mol gas PNe = (97.4 kPa) = 55.7 kPa

Dalton’s Law Suppose you are given four containers – three filled with noble gases. The first 1 L container is filled with argon and exerts a pressure of 2 atm. The second 3 liter container is filled with krypton and has a pressure of 380 mm Hg. The third 0.5 L container is filled with xenon and has a pressure of 607.8 kPa. If all these gases were transferred into an empty 2 L container…what would be the pressure in the “new” container? PKr = 380 mm Hg Ptotal = ? PAr = 2 atm Pxe 607.8 kPa V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters

“Total Pressure = Sum of the Partial Pressures” …just add them up PKr = 380 mm Hg Ptotal = ? PAr = 2 atm Pxe 607.8 kPa V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters Dalton’s Law of Partial Pressures “Total Pressure = Sum of the Partial Pressures” PT = PAr + PKr + PXe + … P1 x V1 = P2 x V2 P1 x V1 = P2 x V2 (0.5 atm) (3L) = (X atm) (2L) (6 atm) (0.5 L) = (X atm) (2L) PKr = 0.75 atm Pxe = 1.5 atm PT = 1 atm + 0.75 atm + 1.5 atm PT = 3.25 atm

Partial Pressure A gas is collected over water at 649 torr and 26.00C. If its volume when collected is 2.99 L, what is its volume at STP? Use proper sig figs. (83.1 x 2.99) / 299 = (101.325 x V2) / 273 P1V1/T1 = P2V2/T2 PT = PG + Pw V2 = ? L V1 = 2.99 L T1 = 299 K T2 = 273 K PT = 649 torr P1 = 86.5 kPa – 3.4 kPa = 83.1 kPa P2 = 101.325 kPa V2 = 2.24 L

Gas Stoichiometry Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88oC. Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)

Gas Stoichiometry Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88oC. Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) 38.2 g excess V = ? L H2 P = 107.3 kPa T = 88oC (361 K) At STP, we’d use 22.4 L per mol, but we aren’t at STP.

Pressure and Balloons B When balloon is being filled: PA > PB A When balloon is filled and tied: PA = PB When balloon deflates: PA < PB A = pressure exerted BY balloon B = pressure exerted ON balloon

Balloon Riddle A B C When the balloons are untied, will the large balloon (A) inflate the small balloon (B); will they end up the same size or will the small balloon inflate the large balloon? Why? B C