Chapter 11 Gases

VARIABLES WE WILL SEE! Pressure (P): force that a gas exerts on a given area Volume (V): space occupied by gas Temperature (T): MUST be in Kelvin! Number of moles (n): quantity of gas molecules

STP STP: standard temperature and pressure, 0 o C and 1 atm. 11-2

Units of Pressure Millimeters of Mercury and Torr: 1 mm Hg = 1 torr Atmosphere: 1 atm = 760 mmHg 1 atm = 760 torr 1 atm = kPa 760 mmHg= kPa 11-3

Units of Volume 1000mL = 1L Units of Temperature ⁰ C = K

Practice Ex. The barometer reads 758 mm Hg. What is the atmospheric pressure in kPa? WHAT DO WE NEED?! THE PICKET FENCE!!! 1) The air pressure in a tire is 109 kPa. What is the pressure in atm? 11-4

Practice Ex. The thermometer reads 30 ⁰ C. What is that in kelvin? Ex) You read a graduated cylinder and the volume reads 56.5 mL. What is that in Liters (L)? 11-4

Gas Volumes Standard Molar Volume: the volume occupied by one mole of a gas at STP is 22.4L. 1 mole of gas = 22.4L of gas AT STP! Ex. What volume does mol of gas occupy at STP? mol x 22.4 L = 1.53 L 1 mol

Practice Ex. What quantity of gas, in moles, is contained in 2.21 L at STP? 2.21L x 1 mol = mol 22.4 L 5)At STP, what is the volume of 7.08 mol of nitrogen gas? 7.08 mol x 22.4 L = 159 L 1 mol 11-14

Boyle’s Law Boyle’s Law: the volume of a fixed mass of gas varies inversely with the pressure at constant temperature. P 1 V 1 = P 2 V 2 Ex. A gas occupies a volume of 458 mL at a pressure of 1.01 kPa. When the pressure is changed, the volume becomes 477 mL. What is the new pressure? (1.01 KPa) (458 ml) = P 2 (477 ml) P 2 = kPa 11-7

Practice P 1 V 1 = P 2 V 2 2) A gas occupies a volume of 2.45L at a pressure of 1.03 atm and temp. 293 K. What volume will the gas occupy if the pressure changes to atm if the temp. remains the same? (1.03 atm) (2.45 L) = (0.980 atm) V 2 V 2 = 2.58 L 11-7

Charles’s Law Charles’s Law: the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temp. V 1 = V 2 T 1 T 2 Ex. What will be the volume of a gas sample at 309K if it’s volume at 215K is 3.42L? Assume that the pressure is constant. 3.42L = V 2 215K 309K V 2 = 4.92L 11-8

Practice 3) A gas sample at 83 o C occupies a volume of 1400m 3. At what temperature will it occupy 1200m 3 ? Assume that the pressure is constant. 1400m 3 = 1200m 3 356K T 2 T 2 = 310 K or 32 o C 11-9

Gay-Lussac’s Law Gay-Lussac’s Law: the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temp. P 1 = P 2 T 1 T 2 Ex. The gas in a container is at a pressure of 3.00 atm at 25 o C. Directions on the container warn the user not to keep it in a place where the temp exceeds 52 o C. What would the pressure in the container be at 52 o C? 3.00 atm = P 2 298K 325K P 2 = 3.3 atm 11-10

Practice P 1 = P 2 T 1 T 2 4) At 120 o C the pressure of a sample of nitrogen is 1.07 atm, What will the pressure be at 205 o C, assuming constant volume? 1.07 atm = P 2 393K 478K P 2 = 1.3 atm 11-11

Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law: expresses the relationship between pressure, volume and temp (K) of a fixed amount of gas. Ex. A helium balloon has a volume of 50.0L at 25 o C and 1.08 atm. What volume will it have at atm and 10.0 o C (1.08 atm)(50.0L) = (0.855 atm) V K 283 K V 2 = 60. L

Ideal Gas Law Ideal Gas Law: the mathematical relationship among pressure, volume, temperature and the number of moles of a gas. Ideal Gas Law: PV = nRT How R (the ideal gas law constant) is derived: R = (1 atm)(22.4L) = atmL/molK (1 mol)(273K) 11-19

Ideal Gas Law Ex. If the pressure exerted by a gas at 0 o C in a volume of L is 5.00atm, how many moles of gas are present? (5.00atm)(0.0010L) = n(0.0821atmL/molK)(273K) n = 2 x mol 11-20

Practice 7) What volume would be occupied by 100. g of oxygen gas at a pressure of 1.50 atm and a temp. of 25 o C? 100.g O 2 x 1mol O 2 = 3.13 mol O g (1.50atm)V = (3.13mol)(0.0821atmL/molK)(298K) V = 51 L 11-21

Gas Stoich Gay-Lussac’s Law of Combining Volumes of Gases: at a constant temp and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers. (coefficients in balanced eq. can represent volumes!) 11-15

Gas Stoich Given a volume, find a volume: Ex. When 0.75 L of hydrogen reacts with bromine, what volume of HBr is produced? H 2 + Br 2 → 2HBr 0.75L H 2 x 2L HBr =1.5L HBr 1L H

Practice 6) What volume of sulfur dioxide gase is necessary to produce 11.4 L of water vapor? Sulfur is also a product. SO 2 + H 2 S → S + H 2 O SO 2 + 2H 2 S → 3S + 2H 2 O 11.4 L H 2 O x 1L SO 2 = 5.70 L SO 2 2L H 2 O 11-16

Gas Stoich Ex. How many liters of CO 2 are produced at STP when g of CaCO 3 react with HCl? CaCO 3 + HCl → CaCl 2 + CO 2 + H 2 O CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O g CaCO 3 x 1mol CaCO 3 x 1mol CO 2 x 22.4L g CaCO 3 1mol CaCO 3 1mol CO 2 = L CO

Gas Stoich Ex. Find the mass of sugar required to produce 1.82L of CO 2 at STP in the rxn: C 6 H 12 O 6 → 2C 2 H 6 O + 2 CO L CO 2 x 1mol CO 2 x 1molC 6 H 12 O 6 x gC 6 H 12 O L 2mol CO 2 1 mol = 7.32 g of C 6 H 12 O

Diffusion and Effusion Graham’s 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. rate of effusion of A = √M B rate of effusion of B √M A Diffusion: spontaneous mixing of the particles of two substances due to their random motion. Effusion: a process by which gas particles pass through a tiny opening

Diffusion and Effusion rate of effusion of A = √M B rate of effusion of B √M A Ex. Compare the rate of effusion of hydrogen and oxygen at the same temp. and pressure. rate of effusion of H 2 = √32.00 g/mol = 3.98 rate of effusion of O 2 √2.02 g/mol 11-23

Ch. 11 The End! 11-26