1. Chapter 14 Properties of Gases Section 14.1 The Behavior of Gases 1

2. The Properties of Gases Gas can expand to fill its container Gases are easily compressed, or squeezed into a smaller volume. Compressibility Compressibility – measure of how much the volume of matter decreases under pressure. 2

3. Compressibility of Gases What factors do you think affect the pressure of the air inside the soccer ball? 1. Temperature of the air inside the ball 2. Volume of the ball 3

4. Kinetic Theory & Gases Gases are easily compressed because of the space between the particles in a gas. Under pressure, the particles in a gas are forced closer together, or compressed. 4

5. Factors Affecting Gas Pressure Pressure (P) - kPa Volume (V) - liters Temperature (T) - Kelvin Number of moles (n) The amount of gas, volume, and temperature are factors that affect gas pressure 5

6. Amount of Gas & Gas Pressure Collisions of particles with the inside walls of the raft result in the pressure that is exerted by the gas. By adding gas, you increase the number of particles. Increasing the number of particles increases the number of collisions, which is why the gas pressure increases. 6

7. When a gas is put into a closed rigid container, the pressure increases as more particles are added Because the container is rigid, the volume of the gas is constant. Assume the temperature doesn't change Doubling the number of particles of gas, doubles the pressure. As gas is removed, the pressure inside the container is reduced. 7

8. Cause and Effect If the pressure of the gas in a sealed container is lower than the outside air pressure, air will rush into the container when the container is opened. When the pressure of the gas in a sealed container is higher than the outside air pressure, the gas will flow out of the container when the container is unsealed. 8

9. Volume & Gas Pressure You can raise the pressure exerted by a contained gas by reducing its volume. The more gas is compressed, the greater is the pressure the gas exerts inside the container. 9

10. When cylinder has a volume of 1 L, the pressure is 100 kPa If volume is halved to 0.5 L, the pressure doubles to 200kPa If volume is doubled to 2.0 L, the pressure of the volume is cut in half to 50 kPa. 10

11. Temperature & Gas Pressure A sealed bag of potato chips may bulge at the seams if placed in the sun. Bag bulges because an increase in the temperature of the gas inside the bag causes an increase in its pressure. As gas inside bag is heated, the temperature increases, increasing kinetic energy of the particles, and causing more collisions, thus more pressure. 11

12. Questions 1. What effect would tripling the number of particles of a gas in a closed container have on the pressure exerted? Gas pressure would triple 2. What effect would doubling the volume of an enclosed gas have on the pressure? Gas pressure would decrease by half 3. How does the pressure of an enclosed gas change with increasing temperature? The number and force of collisions increase with temperature, and the pressure increases 12

13. 4. Why is a gas easy to compress? Because of the space between particles in a gas 5. List three factors that can affect gas pressure? Temperature, volume, & amount of gas 6. Why does a collision with an air bag cause less damage than a collision with a steering wheel? Gas in the inflated airbag can be compressed, and absorbs some of the energy from the impact. The solid steering wheel cannot. 13

14. 7. If temperature is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one quarter of its original value? The volume would need to increase by a factor of four 8. How does a decrease in temperature affect the pressure of a contained gas If temperature decreases, the pressure will decrease 14

15. End of Section 14.1 15

16. Section 14.2 The Gas Laws 16

17. Boyle’s Law (Pressure & Volume) Conversely, if the temperature is constant, as the pressure of a gas decreases, the volume increases. Robert Boyle was the first person to study this pressure-volume relationship. In 1662, Boyle proposed a law to describe the relationship. If the temperature is constant, as the pressure of a gas increases, the volume decreases. 17

18. Boyle’s Law (Pressure & Volume) Boyle’s Law Boyle’s Law – states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. P 1 V 1 = P 2 V 2 18

19. Sample Problem Using Boyle’s Law A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume the temperature remains constant) P 1 = 103 kPa P 2 = 25.0 kPa V 1 = 30.0 L V 2 = ? L P 1 V 1 = P 2 V 2 103 X 30 = 25 X V 2 V 2 = (30.0 L) (103 kPa) 25.0 KPa V 2 = 124 L 19

20. Sample Problem Using Boyle’s Law A gas with a volume of 4.00 L at a pressure of 205 KPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? P 1 = 205 kPa P 2 = ? kPa V 1 = 4.00 LV 2 = 12.0 L P 1 V 1 = P 2 V 2 205 X 4 = P 2 X 12 P 2 = (4.00 L) (205 kPa) 12.0 L P 2 = 68.3 KPa 20

21. Sample Problem Using Boyle’s Law Nitrous oxide (N 2 O) is used as an anesthetic. The pressure on 2.50 L of N 2 O changes from 105 KPa to 40.5 KPa. If the temperature does not change, what will the new volume be? P 1 = 105 kPa P 2 = 40.5 kPa V 1 = 2.50 LV 2 = ? L P 1 V 1 = P 2 V 2 105 X 2.5 = 40.5 X V 2 V 2 = (2.50 L) (105 kPa) 40.5 KPa V 2 = 6.48 L 21

22. Sample Problem Using Boyle’s Law The volume of a gas at 99.6 kPa and 24ºC is 4.23L. What volume will it occupy at 93.3 kPa and 24ºC? P 1 = 99.6 kPa P 2 = 93.3 kPaT 1 = 24ºC V 1 = 4.23 L V 2 = ? L T 2 = 24ºC P 1 V 1 = P 2 V 2 99.6 X 4.23 = 93.3 X V 2 V 2 = (4.23 L) (99.6 kPa) 93.3 kPa V 2 = 4.52 L 22

23. Charles’s Law Temperature and Volume In 1787, French physicist Jacques Charles studies the effect of temperature on the volume of a gas at constant pressure. As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant. 23

24. Charles’s Law Charles’s Law – states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. Charles’s Law Temperature and Volume V 1 T 1 = V 2 T 2 24

25. Sample Problem Using Charles’s Law A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant? When using gas laws always express the temperatures in kelvins! T 1 = 24ºC + 273 = 297 K V 1 = 4.00 L T 2 = 58ºC + 273 = 331 K V 2 = ? L V 1 T 1 = V 2 T 2 4 297 = x 331 V 2 = (4.00 L) (331 K) = 4.46 L 297 K 25

26. Sample Problem Using Charles’s Law If a sample of gas occupies 6.80 L at 325ºC, what will its volume be at 25ºC if the pressure does not change? T 1 = 325ºC or 598 K V 1 = 6.80 L T 2 = 25ºC or 298 K V 2 = ? L V 2 = (6.80 L) (298 K) = 3.39 L 598 K V 1 T 1 = V 2 T 2 6.8 598 = x 298 26

27. Gay-Lussac’s Law Pressure and Temperature As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant. Joseph Gay-Lussac discovered the relationship between the pressure and the temperature of gas in 1802. 27

28. Gay-Lussac’s Law – states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant. Gay-Lussac’s Law Pressure and Temperature P 1 T 1 = P 2 T 2 28

29. Sample Problem Using Gay-Lusaac’s Law A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K? P 1 = 6.58 kPaT 1 = 539 K P 2 = ? kPaT 2 = 211 K P 2 = (6.58 kPa) (211 K) = 2.58 kPa 539 K P 1 T 1 = P 2 T 2 6.58 539 = x 211 29

30. Sample Problem Using Gay-Lusaac’s Law The pressure in a car tire is 198 kPa at 27ºC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. P 1 = 198 kPa T 1 = 300 K P 2 = 225 kPa T 2 = ? K T 2 = (225 kPa) (300 K) = 341 K 198 kPa P 1 T 1 = P 2 T 2 198 300 = 225 x 30

31. Gases If the gas is heated (T2 > T1), the new pressure is greater. (volume constant) If the gas is heated (T2 > T1), the new volume is greater because the gas expands. (pressure constant) If the gas is cooled (T2 < T1), the new pressure is less. (volume constant) If the gas is cooled (T2 < T1), the new volume is smaller because the gas contracts. (pressure constant) 31

32. Combined Gas Law There is a single expression that combines Boyle’s, Charles’s and Gay-Lusaac’s Law. The combined gas law describes the relationship among the pressure, temperature, and volume of an enclosed gas. The combined gas law allows you to do calculation for situations in which only the amount of gas is constant P 1 V 1 T 1 = P 2 V 2 T 2 32

33. Sample Problem Using Combined Gas Law A gas at 155 kPa and 25º C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125º C. What is the new volume? P 1 = 155 kPa T 1 = 298 KV 1 = 1.00 L P 2 = 605 kPa T 2 = 398 KV 2 = ? V 2 = (155)(1.00)(398) = 0.342 L (298)(605) P 1 V 1 T 1 = P 2 V 2 T 2 155 x 1 298 = 605 x V 2 398 33

34. Sample Problem Using Combined Gas Law A 5.00 L air sample has a pressure of 107 kPa at a temp of -50 º C. If the temperature is raised to 102 º C and the volume expands to 7.00 L, what will the new pressure be? P 1 = 107 kPa T 1 = 223 KV 1 = 5.00 L P 2 = ? kPa T 2 = 375 KV 2 = 7.00 L P 2 = (107)(5.00)(375) = 128 kPa (223)(7.00) P 1 V 1 T 1 = P 2 V 2 T 2 107 x 5 223 = P 2 x 7 375 34

35. End of Section 14.2 35

36. Section 14.3 Ideal Gases 36

37. Ideal Gas Law With the combined gas law, you can solve problems with three variables: temperature, volume & pressure The combined gas law assumes that the amount of gas does not vary. The number of moles of gas is directly proportional to the number of particles. An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces. 37

38. Ideal Gas Law P V = n R T Pressure Volume moles ideal gas Temperature (kPa) (L)(mol)constant (K) 8.31 L · kPa / mol · K 38

39. Sample Problem Using Ideal Gas Law When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10 3 kPa. How many moles of helium does the sphere contain? P = 1.89 x 10 3 kPa V = 685 LT = 621 K PV = nRT n = PV / RT n = (1.89 x 10 3 ) (685) (8.31) (621) n = 251 mol He 39

40. Sample Problem Using Ideal Gas Law A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37ºC? Use a molar mass of 29 g for air. P = 102 kPaV = 2.20 LT = 310 K PV = nRT or PV / RT = n n = (102) (2.20) (8.31) (310) n = 0.087 mol air 0.087 mol air x 29g air / mol air = 2.5 g air 40

41. Ideal Gases & Real Gases Ideal gas Ideal gas – one that follows the gas laws at all conditions of pressure and temperature. Such a gas would have to conform precisely to the assumptions of kinetic theory. Its particles could have no volume, and there could be no attraction between particles in the gas. There is no gas for which these assumptions are true. 41

42. Question What pressure is exerted by 0.450 mol of gas at 25 C if the gas is in a 0.650 L container? V = 0.650 Ln = 0.450 molT = 298 K PV = nRT or P = nRT / V P = (0.450 )(8.31)(298) (0.650) P = 1.71 x 10 3 kPa 42

43. Question Determine the volume occupied by 0.582 mol of a gas at 15ºC if the pressure is 0.808 atm? P = 0.808 x 101.3n = 0.582 molT = 288 K = 81.8 kPa PV = nRT or V = nRT / P V = (0.582 )(8.31)(288) (81.8) V = 17.0 L 43

44. End of Chapter 14 44