1. Chapter 11 Gases

2. 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

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

4. Units of Pressure Millimeters of Mercury and Torr: 1 mm Hg = 1 torr Atmosphere: 1 atm = 760 mmHg 1 atm = 760 torr 1 atm = 101.3 kPa 760 mmHg= 101.3 kPa 11-3

5. Units of Volume 1000mL = 1L Units of Temperature ⁰ C + 273 = K

6. 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

7. 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

8. 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 0.0685 mol of gas occupy at STP? 0.0685 mol x 22.4 L = 1.53 L 1 mol

9. Practice Ex. What quantity of gas, in moles, is contained in 2.21 L at STP? 2.21L x 1 mol = 0.0987 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

10. 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 = 0.970 kPa 11-7

11. 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 0.980 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

12. 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

13. 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

14. 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

15. 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

16. 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 0.855 atm and 10.0 o C (1.08 atm)(50.0L) = (0.855 atm) V 2 298 K 283 K V 2 = 60. L

17. 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) = 0.0821atmL/molK (1 mol)(273K) 11-19

18. Ideal Gas Law Ex. If the pressure exerted by a gas at 0 o C in a volume of 0.0010L is 5.00atm, how many moles of gas are present? (5.00atm)(0.0010L) = n(0.0821atmL/molK)(273K) n = 2 x 10 -4 mol 11-20

19. 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 2 32.00g (1.50atm)V = (3.13mol)(0.0821atmL/molK)(298K) V = 51 L 11-21

20. 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

21. 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 2 11-15

22. 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

23. Gas Stoich Ex. How many liters of CO 2 are produced at STP when 400.00g 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 400.00g CaCO 3 x 1mol CaCO 3 x 1mol CO 2 x 22.4L 100.09g CaCO 3 1mol CaCO 3 1mol CO 2 = 89.519 L CO 2 11-17

24. 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 2 1.82L CO 2 x 1mol CO 2 x 1molC 6 H 12 O 6 x 180.18gC 6 H 12 O 6 22.4L 2mol CO 2 1 mol = 7.32 g of C 6 H 12 O 6 11-18

25. 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. 11-22

26. 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

27. Ch. 11 The End! 11-26