1. Gases Gases

2. Kinetic Theory of Gases A gas consists of small particles that move rapidly in straight lines. have essentially no attractive (or repulsive) forces. are very far apart. have very small volumes compared to the volume of the container they occupy. have kinetic energies that increase with an increase in temperature.

3. Properties That Describe a Gas Gases are described in terms of four properties: pressure (P), volume(V), temperature(T), and amount(n).

4. Gas Pressure Gas pressure is described as a force acting on a specific area. Pressure (P) = Force Area

5. NEWTON The SI unit for force is the newton, (N), the force that will increase the speed of a one-kilogram mass by one meter per second each second that the force is applied. example: consider a person with a mass of 51 kg (about 112 lbs). At Earth’s surface, gravity has an acceleration of 9.8 m/s 2. The force the person exerts on the ground is therefore 51 kg  9.8 m/s 2 = 500 kg m/s 2 = 500 N Pressure and Force Fig Newton Demonstration

6. Which situation results in the greatest pressure? Which situation results in the least pressure? ◦A 120 lb person standing on her tip-toes ◦A 120 lb person standing normally (flat feet) ◦A 120 lb person lying on her back on the floor

7. Relationship Between Pressure, Force, and Area

8. Gas Pressure The greater the force on a given area, the greater the pressure. (directly proportional) The smaller the area is on which a given force acts, the greater the pressure. (inversely proportional) Pressure (P) = Force Area Compare to pizza graphs

9. Units of Pressure

10. Units of pressure 1 atm = 760 mm Hg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in. 2 1 atm = 101.325 kPa 1 atm = 101,325 N/m 2 1atm = 1.01325 bars

11. Units of Pressure Average atmospheric pressure at sea level at 0°C = 760 mm Hg One atmosphere of pressure (atm) is defined as being exactly equivalent to 760 mm Hg. 1 mm Hg is also called 1 torr in honor of Torricelli for his invention of the barometer (device used to measure pressure).

12. SI Units of Pressure One pascal (Pa) is defined as the pressure exerted by a force of one newton (1 N) acting on an area of one square meter. 1 Pa = 1 N/m 2 (1 Pa is small; 6 Fig Newton cookies on 1 m 2 of fabric) ◦The unit is named for Blaise Pascal, a French mathematician and philosopher who studied pressure during the seventeenth century. 1 atm = 101,325 PA or 101.325 kPa.

13. Pressure Conversions Sample Problem A The average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in a. millimeters of mercury (mm Hg) and b. kilopascals (kPa)

14. Sample Problem A Solution Given: atmospheric pressure = 0.830 atm Unknown: a. pressure in mm Hg b. pressure in kPa Solution:conversion factor a. b.

15. Learning Check The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 2.00 atm x 760 mm Hg = 1 atm 1520 mm Hg

16. Closed End Manometer Reading Closed End Manometer Direct read instrument Height(h) is the mmHg pressure of the gas in the bulb (ALWAYS).

17. Open End Manometer Reading Open End Manometer Atmosphere vs. gas Higher pressure is down Distance between Hg levels is the difference between the the pressure of the two sides “There is a Battle going on”

18. Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth.

19. Altitude and Atmospheric Pressure Atmospheric pressure is about 1 atm(atmosphere) at sea level. depends on the altitude and the weather. is lower at high altitudes where the density of air is less. is high for sunny and low for cloudy and rainy conditions ◦Higher pressure = stable air

20. Barometer & Barometric pressure A barometer measures the pressure exerted by the gases in the atmosphere. indicates atmospheric pressure as the height in mm of the mercury column.

21. Dalton’s Law of Partial Pressures P t = p x + p y + p z …. Pressure total is the sum of all of the individual gas pressures (partial pressure) in a container The pressure of one gas does not affect the pressure of other gases. ◦(proposed by John Dalton- atomic theory) Could be more than 3 gases in the mixture.

22. Vapor Pressure of Water Vapor pressure of water is the pressure of the gas molecules above the liquid surface at a given temperature. Higher temperature means more molecules leave the surface and HIGHER pressure.

23. Dalton’s Law of Partial Pressures, continued Gases Collected by Water Displacement Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water in the reaction bottle. Dalton’s law of partial pressures can be applied to calculate the pressures of gases collected in this way. Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure known as vapor pressure.

24. If you raise the bottle until the water levels inside and outside the bottle are the same, the total pressure outside and inside the bottle will be the same. Reading the atmospheric pressure from a barometer and looking up the value of P H2O at the temperature of the experiment in a table, you can calculate P gas. To determine the pressure of a gas inside a collection bottle, you would use the following equation, which is an instance of Dalton’s law of partial pressures. P atm = P gas + P H2O Dalton’s Law of Partial Pressures, continued Gases Collected by Water Displacement, continued

25. Particle Model for a Gas Collected Over Water

26. Dalton’s Law of Partial Pressures, continued Sample Problem B Oxygen gas from the decomposition of potassium chlorate, KClO 3, was collected by water displacement. The barometric pressure (total pressure) and the temperature during the experiment were 731.0 torr and 20.0°C. respectively. What was the partial pressure of the oxygen collected?

27. Sample Problem B Solution Given: P Total = P atm = 731.0 torr P H 2 O = 17.5 torr (vapor pressure of water at 20.0°C, from Water Vapor Pressure table on Gas Laws Data and Formulas) P atm = Unknown: Solution: start with the equation: rearrange algebraically to :

28. Sample Problem B Solution, continued Substitute the given values of P atm and P H 2 O into the equation:

29. What is the Nitrogen pressure when caught over water at 22°C with a total pressure of 0.5atm. At 22°C, Vapor pressure (H2O) = 19.8mmHg P total = 0.5 atm = 380 mmHg Using Dalton’s formula P t = pN 2 + pH 2 0 380mmHg = pN 2 + 19.8mmHg pN 2 = 360.2mmHg Example with partial pressure of two gases given mass of each STOP!!!

30. Pressure and Volume (Boyle’s Law)

31. Boyle’s Law: Pressure-Volume Relationship Robert Boyle discovered that doubling the pressure on a sample of gas at constant temperature reduces its volume by one-half. This is explained by the kinetic-molecular theory: The pressure of a gas is caused by moving molecules hitting the container walls. If the volume of a gas is decreased, more collisions will occur, and the pressure will therefore increase. Likewise, if the volume of a gas is increased, less collisions will occur, and the pressure will decrease.

32. Boyle’s Law Boyle’s Law states that the pressure of a gas is INVERSELY related to its volume when T and n are constant. if volume decreases, the pressure increases P 1 *V 1 =P 2 *V 2

33. Learning Check For a cylinder containing helium gas indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant). 1) pressure decreases B (volume increases) 2) pressure increases A (volume decreases) Marshmallow demo

34. Learning Check If a sample of helium gas has a volume of 120 mL and a pressure of 850 mm Hg, what is the new volume if the pressure is changed to 425 mm Hg ? Known:P 1 = 850 mm HgP 2 = 425 mm Hg V 1 = 120 mLV 2 = ?? P 1 V 1 =P 2 V 2 Pressure ratio V 2 = V 1 x P 1 = 120 mL x 850 mm Hg = 240 mL P 2 425 mm Hg Decrease in pressure = increase in volume (inversely proportional)

35. Temperature and Volume (Charles’ Law) Balloon in Liquid Nitrogen Video (demo) http://www.youtube.com/watch?v=ZvrJgGhnmJo http://agpa.uakron.edu/p16/video.php?id=liquidnitro genballoon http://www.youtube.com/watch?v=ZvrJgGhnmJo Relationship discovered by the French scientist Jacques Charles in 1787. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

36. Charles’s Law: Volume- Temperature Relationship If pressure is constant, gases contract when cooled. ◦When the temperature decreases, the volume of a fixed number of gas molecules must decrease if the pressure is to stay constant.  At the lower temperature, the gas molecules move slower. They collide with the walls of the container less frequently and with lesser force.  The volume of a flexible container must then decrease in order for the pressure to remain the same.

37. Charles’s Law: Volume- Temperature Relationship If pressure is constant, gases expand when heated. ◦When the temperature increases, the volume of a fixed number of gas molecules must increase if the pressure is to stay constant.  At the higher temperature, the gas molecules move faster. They collide with the walls of the container more frequently and with more force.  The volume of a flexible container must then increase in order for the pressure to remain the same.

38. Absolute Zero Absolute Zero = –273.15°C = 0 K Kinetic Energy = 0, so Temperature = 0 K Conversion K = 273.15 + °C Problem: What is 100°C in Kelvin? K = 273.15 + 100°C~ 373 K

39. Charles’ Law In Charles’ Law, the Kelvin temperature scale MUST be used in Gas Law relationships K = °C + 273 P and n are constant. Direct relationship: when the temperature of a gas increases, its volume increases. V 1 /T 1 =V 2 /T 2 This equation reflects the fact that volume and temperature are directly proportional to each other at constant pressure.

40. Calculations: Charles’ Law A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0°C, what is the new volume of the balloon (P constant)? Be sure to use the Kelvin (K) temperature in gas calculations. 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21°C = 294 KT 2 = 0°C = 273 K

41. Calculations: Charles’ Law 2. Solve Charles’ law for V 2 : V 1 = V 2 T 1 T 2 V 2 = V 1 x T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K STOP!!!

42. Gay-Lussac’s Law

43. Gay-Lussac’s Law Problem

45. Combined Gas Law

47. You have learned about equations describing the relationships between two or three of the four variables—pressure, volume, temperature, and moles—needed to describe a gas sample at a time. All of the gas laws you have learned thus far can be combined into a single equation, the ideal gas law: the mathematical relationship among pressure, volume, temperature, and number of moles of a gas. It is stated as shown below, where R is a constant: PV = nRT The Ideal Gas Law

48. Ideal Gas Law

49. Ideal Gas Law Example

51. Standard Temperature and Pressure Standard Temperature and Pressure (STP) Standard set of conditions so sets of data can be compared 1 atm and 0°C (273K)

52. Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules 1.0 L of nitrogen at STP and 1.0 L of oxygen at STP will have the same number of molecules (and therefore, the same number of moles) 3.0 moles of nitrogen and 3.0 moles of oxygen at the same temperature and pressure will have the same volume

53. Molar Volume of a Gas

54. Avogadro’s Law ◦At STP the molar volume of ANY gas is ◦22.4 L/mol ◦In other words, the volume of 1 mol of gas at STP is 22.4 L ◦How many moles of N 2 occupy 4.3 L at STP? ◦What is the volume (in liters) of 3.9 moles of O 2 gas at STP?

55. You can also use the molar volume of a gas to find the volume, at STP, of a known number of moles or a known mass of gas. example: at STP, Knowing the volume of a gas, you can use the conversion factor 1 mol/22.4 L to find the moles (and therefore also mass) of a given volume of gas at STP. example: at STP, Molar Volume of a Gas, continued

56. Sample Problem G a. What volume does 0.0685 mol of gas occupy at STP? b. What quantity of gas, in moles, is contained in 2.21 L at STP?

57. Molar Volume of a Gas, continued Sample Problem G Solution a. Given: 0.0685 mol of gas at STP Unknown: volume of gas Solution: Multiply the amount in moles by the conversion factor,.

58. Molar Volume of a Gas, continued Sample Problem G Solution, continued b. Given: 2.21 L of gas at STP Unknown: moles of gas Solution: Multiply the volume in liters by the conversion factor,.