1. CHEMISTRY CHAPTERS 12 & 13 THE STATES OF MATTER AND GAS LAWS

2. CHAPTER 12 THE STATES OF MATTER

3. KINETIC-MOLECULAR THEORY OF GASES- VALID ONLY AT EXTREMELY LOW DENSITY 1. A gas is composed of particles, usually molecules or atoms. We treat the particles as… Hard spheres Insignificant volume Far from each other

4. 2.THE PARTICLES IN A GAS MOVE RAPIDLY IN CONSTANT RANDOM MOTION. 3. ALL COLLISIONS ARE PERFECTLY ELASTIC. The average speed of an O 2 molecule is 1656 km/hr!!!

5. KINETIC ENERGY (KE)- THE ENERGY AN OBJECT HAS BECAUSE OF ITS MOTION. WHEN A GAS IS HEATED, IT ABSORBS THERMAL ENERGY. SOME OF THIS ENERGY IS CONVERTED TO KE TO INCREASE THE MOTION OF PARTICLES.

6. * AS A SUBSTANCE MOVES FROM SOLID TO GAS THE KE INCREASES THE AVERAGE KE OF A GAS IS PROPORTIONAL TO THE KELVIN TEMPERATURE. *PARTICLES AT 200K HAVE TWICE THE KE OF PARTICLES AT 100K. *THE KELVIN TEMP SCALE IS USED BECAUSE 0K (ABSOLUTE ZERO) IS THE TEMP AT WHICH ALL MOTION CEASES.

7. GAS PRESSURE- THE RESULT OF THE COLLISIONS OF GAS PARTICLES WITH AN OBJECT Where are the pressurized areas in this picture?

8. ATMOSPHERIC PRESSURE- The pressure caused by the weight of the atmosphere. It decreases with an increase in elevation because the atmospheric gases are less dense

9. ARE ATMOSPHERIC PRESSURES HIGHER OR LOWER IN THE MOUNTAINS? WHY? The pressure is lower in the mountains because there are fewer air particles pressing down on you. This means that there are fewer particle collisions.

10. PHASE CHANGES The transition between the different phases of matter. Some phase changes require more energy to occur: Melting Vaporization sublimation Other release energy: Freezing Condensation Deposition The temperature of a substance does not change during a phase change.

11. VAPORIZATION- THE PROCESS BY WHICH A LIQUID CHANGES TO A GAS OR VAPOR CONDENSATION - when a gas becomes a liquid

12. MELTING- CHANGE OF A SOLID TO A LIQUID FREEZING PROCESS OF A LIQUID BECOMING A SOLID

13. SUBLIMATION- CHANGE OF A SOLID TO A GAS WITHOUT GOING THROUGH THE LIQUID PHASE DEPOSITION IS THE OPPOSITE EX. IODINE, ICE CUBE “SHRINKAGE”, FREEZE DRYING, FREEZER BURN

14. CHAPTER 13 THE BEHAVIOR OF GASES

15. KMT: A SUMMARY States that: Gases are composed of particles that are considered to be hard spheres with little volume. These particles are spaced far apart from one another and are in constant motion. They collide in a perfectly elastic manner so that energy is never lost.

16. PROPERTIES OF GASES Gas properties can be modeled using math. Model depends on— V = volume of the gas (L) T = temperature (K) ALL temperatures in the entire unit MUST be in Kelvin!!! No Exceptions! n = amount (moles) P = pressure (atmospheres)

17. AND NOW, WE PAUSE FOR THIS COMMERCIAL MESSAGE FROM STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure STP allows us to compare amounts of gases between different pressures and temperatures Standard Pressure Standard Temperature 1 atm 00C00C00C00C 273 K

18. WHAT HAPPENS WHEN YOU ADD OR REMOVE GAS FROM A CONTAINER? Adding gas to a container Increases the number of particles Increases the number of collisions And thus increases the pressure in the container Removing gas from a container Decreases the number of particles Decreases the number of collisions And thus decreases the pressure

19. Airing up a balloon illustrates this! The number of particles of gas and pressure are directly related. double # of particles = double pressure PRESSURE AND PARTICLES ARE RELATED

20. The Effect Of Changing The Size Of The Container Increase Volume  Decrease Pressure Decrease Volume  Increase Pressure Volume and Pressure are indirectly related. If you cut the volume in half, the pressure doubles.

21. The Effect of Heating or Cooling a Gas Compressing A Gas Increases Its Temperature. Expanding A Gas Decreases Its Temperature. Heating a Gas Increases KE Increases the number of collisions with the walls of the container Thus increasing the pressure Cooling a Gas Decreases KE Decreases the number of collisions with the walls of the container Thus decreasing the pressure

22. The Effect of Heating or Cooling a Gas Kelvin temperature and pressure are directly related. If you double the Kelvin temp  double pressure

23. BELLWORK TUESDAY, JANUARY 7 TH Three soda cans are placed into three different situations as shown below. Which soda can has particles with the highest kinetic energy and why? What does this do to the pressure inside the can?

24. THE GAS LAWS

25. 1. Boyle’s law for pressure-volume changes -for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. “ We Boyle Peas and Vegetables ” P 1 V 1 = P 2 V 2 or

26. BOYLE’S LAW Examples: Syringes Balloons Bicycle pumps

27. Example: A gas is collected in a 242 ml container. The pressure of the gas in the container is measured and determined to be 2.71 atm. What is the volume of this gas at 5.32 atm? Assume the temperature is constant. P 1 = 2.71 atm P 2 = 5.32 atm V 1 = 242 mL V 2 = ? P 1 V 1 = P 2 V 2 (2.71 atm)(242 mL) = (5.32 atm)V 2 V 2 = 123 mL

28. 2. Charles’ law for temperature-volume changes The volume of a fixed mass of gas is directly proportional to its kelvin temperature if the pressure is kept constant. Temperature must be in Kelvin! “Charlie Brown’s Christmas is on TV” V 1 = V 2 or V 1 T 2 = V 2 T 1 T 1 T 2

30. CHARLES’S LAW Examples: Hot air balloons

31. Example: A sample of gas at 15 o C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 o C and 1 atm? T 1 = 15 o C + 273 =288K V 1 = V 2 T 2 = 38 o C + 273 = 311K T 1 T 2 V 1 = 2.58 L V 2 = ? 2.58L = V 2 288K 311K V 2 = 2.79L

32. 3. Gay-lussac’s law for temperature-pressure changes The pressure of a gas is directly proportional to the kelvin temperature if the volume is kept constant. Temperature must be in kelvin! GayLe drives a PT Cruiser! P 1 = P 2 or P 1 T 2 = T 1 P 2 T 1 T 2

34. Example: A 2.00 L flask contains helium gas at a pressure of 2.5 atm and a temperature of 0 o C. What would be the pressure in a flask if the temperature is increased to 150. O C? P 1 = 2.5 atm P 2 =? T 1 = 0 o C + 273 = 273K T 2 = 150. o C + 273 = 423K P 1 = P 2 2.5 atm = P 2 T 1 T 2 273K 423K P 2 = 3.87 atm

35. BELLWORK WEDNESDAY, 1/9 Complete the following statements. When the pressure of a gas goes up, the volume… This means it is a(n) __________________ relationship. When the temperature of a gas goes up, the volume… This means it is a(n) __________________ relationship. When the temperature of a gas goes up, the pressure… This means it is a(n) __________________ relationship.

36. THE COMBINED GAS LAW P 1 V 1 = P 2 V 2 T 1 T 2 “PEAS AND VEGETABLES ON THE TABLE” BY CANCELING OUT TERMS REMAINING CONSTANT, WE CAN DERIVE BOYLE’S, CHARLES’, AND GAY-LUSSAC’S LAWS. P 1 V 1 = P 2 V 2 T 1 T 2

37. EX. IF A HELIUM-FILLED BALLOON HAS A VOLUME OF 3.40 L AT 25.0 O C AND 0.6875 ATM, WHAT IS ITS VOLUME AT STP? V 1 = 3.40L V 2 = ? P 1 = 0.6875 atm P 2 = 1 atm T 1 = 25.0 o C + 273 = 298K T 2 = 0 o C + 273 = 273K P 1 V 1 = P 2 V 2 (0.6875 atm)(3.40L) = (1 atm)V 2 T 1 T 2 298K 273K V 2 = 2.14 L

38. BELLWORK THURSDAY JAN. 9 A sample of air has a volume of 550.0mL at 106 o C. At what temperature will its volume be 700.0mL at constant pressure?

39. A flask contains 1.4 L of an ideal gas at 50.0 o C and 1.44 atm of pressure. If the gas is compressed to 0.7 L and the temperature is raised to 100.0 o C what will be the new pressure in the container? V 1 = 1.4 L V 2 = 0.7 L P 1 = 1.44 atmP 2 = ? T 1 = 50.0 o C + 273 = 323K T 2 = 100.0 o C + 273 = 373K P 1 V 1 = P 2 V 2 (1.44 atm)(1.4L) = (P 2 )(0.7 L) T 1 T 2 323K 373K P 2 = 3 atm

40. Ideal gas- follows the gas laws at all conditions of temperature and pressure. - Ideal gases don’t exist  Real gases can be liquefied and sometimes solidified; ideal gases can not. 12.4 Real vs. Ideal Gases

41. *Gases Behave Most Ideally At High Temperature And Low Pressure. (Just Like Students In The Summer!!!!!!)

42. The Ideal Gas Law -Allows Us To Include the Amount Of Gas (Moles) In Our Calculations. PV = Nrt “Puv Nert” P = Pressure In Atm V = Volume In L N = # Of Moles R = Ideal Gas Constant = 0.0821 (L. Atm)/(Mol. K) T = Temperature In Kelvin

43. Example A 5.0 L flask contains 0.60 g O 2 at a temperature of 22 o C. What is the pressure (in atm) inside the flask? P=? V= 5.0L R = 0.0821 (L. atm/mol. K) T = 22°C +273= 295K n = 0.60g O 2 1 mol O 2 = 0.01875 mol 32.0g O 2 PV = nRT P(5.0L) = (0.01875mol)(0.0821Latm/molK)(295K) P = 0.091 atm

44. Example How many grams of krypton are present in a 600. mL container at 1010 o C in which the pressure of krypton is 10.0 atm? P = 10.0 atm V= 600.mL = 0.600 L n = ? R = 0.0821 (L. atm)/(mol. K) T = 1010 o C + 273 = 1283K PV = nRT (10.0atm)(0.600L)= n(0.0821 L. atm)/(mol. K)(1283K) n = 0.05696 mol Kr 0.05696mol Kr 83.8g Kr = 4.77g Kr 1 mol Kr

45. GAS MOLECULES: MIXTURES AND MOVEMENTS

46. DALTON’S LAW OF PARTIAL PRESSURE -At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures. P total = P 1 + P 2 + P 3 …

47. Example: determine the total pressure of a gas mixture that contains nitrogen and oxygen if the partial pressure of the nitrogen is 7.25 atm and the partial pressure of the oxygen is 4.26 atm. P 1 + P 2 = P total 7.25 atm+ 4.26 atm = 11.51 mm Hg

48. Gases are often collected by water displacement. The total of the gas pressure plus the water vapor pressure is equal to the atmospheric pressure. When we work a problem like this we must always look up and subtract the water vapor pressure to get the gas pressure.

49. Example: A sample of N 2 gas is collected by the downward displacement of water from an inverted bottle. What is the partial pressure of the N 2 gas at 20.0 o C, if the atmospheric pressure is 7.52 atm? The water vapor pressure is 0.175 atm at 20.0 o C. P H2O + P N2 = P total 0.175 atm+ P N 2 = 7.52 atm P N2 = 7.35 atm

50. Avogadro’s Hypothesis- Equal volumes of gases at the same temperature and pressure have equal numbers of particles. STP = 1 ATM AND 0 O C

51. AVOGADRO’S LAW If two gases have the same temperature, pressure and volume, they will have the same number of particles or moles. one mole of gas at STP has a molar volume of 22.4L 1 mole = 22.4 L

52. Ex. How many liters will 2.0 moles N 2 O occupy at STP? 2.0 mol N 2 O 22.4 L N 2 O = 44.8 L N 2 O 1 mol N 2 O

53. Ex. How many moles of hydrogen gas are there in 15.8L? 15.8 L H 2 1 mol H 2 = 0.705 mol H 2 22.4 L H 2

54. GAS STOICHIOMETRY the same as regular stoichiometry, but using molar volume (22.4 L = 1 mole) 1 mole = 22.4 L 1 mole = 6.02 x 10 23 representative particles 1 mole = the molar mass of a substance

55. At STP, how many liters of O 2 will be produced by the decomposition of 25g of KClO 3 ? 2 KClO 3  2KCl + 3O 2 25 g KClO 3 1 mole KClO 3 3 mol O 2 22.4 L O 2 122.548 g KClO 3 2 mol KClO 3 1 mol O 2 6.85 L O 2

56. Ex. How many grams of H 2 O will be produced by the addition of 5L H 2 to excess O 2 at STP? First, write a balanced chemical equation 2 H 2 + O 2  2 H 2 O Now use stoichiometry 5 L H 2 1 mol H 2 2 mol H 2 O 18.015 g H 2 O 22.4 L H 2 2 mol H 2 1 mol H 2 O 4.02 g H 2 O

57. If 100mL of hydrogen are produced from zinc reacting with HCl at STP, what mass of zinc is required? First, write a balanced chemical equation Zn + 2HCl  H 2 + ZnCl 2 Now use stoichiometry 100mL H 2 1 L H 2 1 mol H 2 1 mol Zn65.38 g Zn 1000 mLH 2 22.4 L H 2 1 mol H 2 1 mol Zn 0.292 g Zn