1. GASES Chapter 10

2. The Atmosphere The atmosphere is a gaseous solution of nitrogen, N 2, and oxygen, O 2. The atmosphere both supports life and acts as a waste receptacle for the exhaust gases that accompany many industrial processes leading to many types of pollution, including smog and acid rain.

3. A Gas -Uniformly fills any container. -Mixes completely with any other gas -Exerts pressure on its surroundings. -Is easily compressed.

4. Simple barometer invented by Evangelista Torricelli. A barometer is used to measure atmospheric pressure.

5. Atmospheric Pressure Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity--the weight of the air. At sea level, atmospheric pressure is 760mm. Atmospheric pressure varies with altitude-- at Breckenridge, CO (elevation 9600 ft), the pressure is 520 mm.

6. Figure 12.1: The pressure exerted by the gases in the atmosphere can demonstrated by boiling water in a can

7. Pressure -is equal to force/unit area -SI units = Newton/meter 2 = 1 Pascal (Pa) -1 standard atmosphere = 101,325 Pa -1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

8. A simple manometer s a device for measuring the pressure of a gas in a container.

9. Pressure Unit Conversions The pressure of a tire is measured to be 28 psi. What would the pressure be in atmospheres, torr, and pascals. (28 psi)(1.000 atm/14.69 psi) = 1.9 atm (28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 10 3 torr (28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 10 5 Pa

10. Volume of a gas decreases as pressure increases at constant temperature

11. BOYLE’S LAW DATA P vs VV vs 1/P

12. Figure 12.6: Illustration of Boyle’s law

13. Boyle’s Law * Robert Boyle -- Irish scientist (1627-1691) (Pressure)( Volume) = Constant (T = constant) P 1 V 1 = P 2 V 2 (T = constant) V  1/P (T = constant) ( * Holds precisely only at very low pressures.) The temperature and amount of gas must remain unchanged.

14. A gas that strictly obeys Boyle’s Law is called an ideal gas.

15. Boyle’s Law Calculations A 1.5-L sample of gaseous CCl 2 F 2 has a pressure of 56 torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be? Decrease P 1 = 56 torr P 2 = 150 torr V 1 = 1.5 L V 2 = ? V 1 P 1 = V 2 P 2 V 2 = V 1 P 1 /P 2 V 2 = (1.5 L)(56 torr)/(150 torr) V 2 = 0.56 L

16. Boyle’s Law Calculations In an automobile engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure? P 1 = 1.00 atm P 2 = ? V 1 = 0.725 L V 2 = 0.075 L V 1 P 1 = V 2 P 2 P 2 = V 1 P 1 /V 2 P 2 = (0.725 L)(1.00 atm)/(0.075 L) P 2 = 9.7 atm Is this answer reasonable?

17. Plot of V vs. T( o C) for several gases

18. Volume of a gas increases as heat is added when pressure is held constant.

19. Charles’s Law Jacques Charles -- French physicist (1746- 1823) The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bT (P = constant) y = mx + b b = a proportionality constant

20. Charles’s Law Temperature must be in Kelvin!!

21. Charles’s Law Calculations A 2.0-L sample of air is collected at 298 K and then cooled to 278 K. The pressure is held constant. Will the volume increase or decrease? What will the new volume be? V 1 = 2.0 L V 2 = ? T 1 = 298 K T 2 = 278 K V 1 /V 2 = T 1 /T 2 V 2 = V 1 T 2 /T 1 V 2 = (2.0 L)(278 K)/(298 K) V 2 = 1.9 L Decrease

22. Charles’s Law Calculations Consider a gas with a a volume of 0.675 L at 35 o C and 1 atm pressure. What is the temperature (in C o ) of the gas when its volume is 0.535 L at 1 atm pressure? V 1 = 0.675 L V 2 = 0.535 L T 1 = 35 o C + 273 = 308 K T 2 = ? V 1 /V 2 = T 1 /T 2 T 2 = T 1 V 2 /V 1 T 2 = (308 K)(0.535 L)/(0.675 L) T 2 = 244 K -273 T 2 = - 29 o C

23. At constant temperature and pressure, increasing the moles of a gas increases its volume.

24. Figure 12.11: The total pressure of a mixture of gases depends on the number of moles of gas particles (atoms or molecules) present

25. Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an y = mx + b a = proportionality constant V = volume of the gas n = number of moles of gas

26. AVOGADRO’S LAW Temperature and pressure must remain constant!!

27. AVOGADRO’S LAW A 12.2 L sample containing 0.50 mol of oxygen gas, O 2, at a pressure of 1.00 atm and a temperature of 25 o C is converted to ozone, O 3, at the same temperature and pressure, what will be the volume of the ozone? 3 O 2(g) ---> 2 O 3(g) (0.50 mol O 2 )(2 mol O 3 /3 mol O 2 ) = 0.33 mol O 3 V 1 = 12.2 L V 2 = ? n 1 = 0.50 mol n 2 = 0.33 mol V 1 /V 2 = n 1 /n 2 V 2 = V 1 n 2 /n 1 V 2 = (12.2 L)(0.33 mol)/(0.50 mol) V 2 = 8.1 L

28. COMBINED GAS LAW V 1 / V 2 = P 2 T 1 / P 1 T 2 V 2 = V 1 P 1 T 2 /P 2 T 1 V 2 = (309 K)(0.454 atm)(3.48 L) (258 K)(0.616 atm) V 2 = 3.07 L What will be the new volume of a gas under the following conditions? V 1 = 3.48 L V 2 = ? P 1 = 0.454 atm P 2 = 0.616 atm T 1 = - 15 o C + 273 = 258 K T 2 = 36 o C + 273 T 2 = 309 K

29. Pressure exerted by a gas increases as temperature increases provided volume remains constant.

30. If the volume of a gas is held constant, then V 1 / V 2 = 1. Therefore: P 1 / P 2 = T 1 / T 2 Pressure and Kelvin temperature are directly proportional.

31. Ideal Gas Law -An equation of state for a gas. -“state” is the condition of the gas at a given time. PV = nRT

32. IDEAL GAS 1. Molecules are infinitely far apart. 2. Zero attractive forces exist between the molecules. 3. Molecules are infinitely small--zero molecular volume. What is an example of an ideal gas?

33. REAL GAS 1. Molecules are relatively far apart compared to their size. 2. Very small attractive forces exist between molecules. 3. The volume of the molecule is small compared to the distance between molecules. What is an example of a real gas?

34. Real vs. Ideal Gases When p  1 atm and T  O o C, real gases approximate ideal gases.

35. Ideal Gas Law PV = nRT R = proportionality constant = 0.08206 L atm   mol  P = pressure in atm V = volume in liters n = moles T = temperature in Kelvins Holds closely at P  1 atm

36. Ideal Gas Law Calculations A sample of hydrogen gas, H 2, has a volume of 8.56 L at a temperature of O o C and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present. p = 1.5 atm V = 8.56 L R = 0.08206 Latm/molK n = ? T = O o C + 273 T = 273K pV = nRT n = pV/RT n = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K) n = 0.57 mol

37. Ideal Gas Law Calculations A 1.5 mol sample of radon gas has a volume of 21.0 L at 33 o C. What is the pressure of the gas? p = ? V = 21.0 L n = 1.5 mol T = 33 o C + 273 T = 306 K R = 0.08206 Latm/molK pV = nRT p = nRT/V p = (1.5mol)(0.08206Latm/molK)(306K) (21.0L) p = 1.8 atm

38. Dalton’s Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present. P Total = P 1 + P 2 + P 3 +...

39. Dalton’s Law of Partial Pressures Calculations A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at.33 atm would have what total pressure? P Total = P 1 + P 2 + P 3 P Total = 1.25 atm + 2.55 atm +.33 atm P total = 4.13 atm

40. Water Vapor Pressure 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) When a sample of potassium chlorate is decomposed and the oxygen produced collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 o C. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 o C is 21 torr). How many moles of oxygen are produced? P ox = P total - P HOH P ox = 754 torr - 21 torr p ox = 733 torr

41. Water Vapor Pressure Continued p = (733 torr)(1 atm/760 torr) p = 0.964 atm V = 0.650 L n = ? T = 22 o C + 273 T = 295 K R = 0.08206 Latm/molK pV = nRT n = pV/RT n = (0.964 atm)(0.650 L) (0.08206 Latm/molK)(295K) n = 2.59 x 10 -2 mol

42. Kinetic Molecular Theory 1.Volume of individual particles is  zero. 2.Collisions of particles with container walls cause pressure exerted by gas. 3.Particles exert no forces on each other. 4.Average kinetic energy  Kelvin temperature of a gas. 5. Gases consist of tiny particles (atoms or molecules.

43. Plot of relative number of oxygen molecules with a given velocity at STP (Boltzmann Distribution).

44. Plot of relative number of nitrogen molecules with a given velocity at three different temperatures.

45. Kinetic Molecular Theory As the temperature of a gas increases, the velocity of the molecules increases. Molecules at a higher temperature have greater KE and hit the walls of the container harder and more often exerting more pressure.

46. Kinetic Molecular Theory As the temperature of a gas increases, the velocity of the molecules increases. Molecules at a higher temperature have greater KE and hit the walls of the container harder and more often exerting more pressure. If the container is not rigid, the container and volume of the gas will expand.

47. Standard Temperature and Pressure “STP” P = 1 atmosphere T =  C The molar volume of an ideal gas is 22.42 liters at STP

48. Molar Volume pV = nRT V = nRT/p V = (1.00 mol)(0.08206 Latm/molK)(273K) (1.00 atm) V = 22.4 L

49. Figure 12.12: The production of oxygen by thermal decomposition of KClO 3

50. GAS STOICHIOMETRY Not at STP Calculate the volume of oxygen gas produced at 1.00 atm and 25 o C by the complete decomposition of 10.5 g of potassium chlorate. 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) (10.5g KClO 3 )(1mol/122.6 g)(3 mol O 2 /2mol KClO 3 ) = 1.28 x 10 -1 mol O 2

51. Gas Stoichiometry Not at STP (Continued) p = 1.00 atm V = ? n = 1.28 x 10 -1 mol R = 0.08206 Latm/molK T = 25 o C + 273 = 298 K pV = nRT V = nRT/p V = (1.28 x 10 -1 mol)(0.08206Latm/molK)(298K) (1.00 atm) V = 3.13 L O 2

52. Gases at STP A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present? (1.75L N 2 )(1.000 mol/22.4 L) = 7.81 x 10 -2 mol N 2

53. Gas Stoichiometry at STP Quicklime, CaO, is produced by heating calcium carbonate, CaCO 3. Calculate the volume of CO 2 produced at STP from the decomposition of 152 g of CaCO 3. CaCO 3(s) ---> CaO (s) + CO 2(g) (152g CaCO 3 )(1 mol/100.1g)(1mol CO 2 /1mol CaCO 3 ) (22.4L/1mol) = 34.1L CO 2 Note: This method only works when the gas is at STP!!!!!

54. Chemistry is a gas!!