1. Chapter 14 The Behavior of Gases

2. Section 14.2 The Gas Laws l\

3. Converting Pressure 1 atmosphere = 760 mmHg 1 atmosphere = 760 mmHg 1 atmosphere = 101 325 Pa 1 atmosphere = 101 325 Pa 1 atmosphere = 101.325 kPa 1 atmosphere = 101.325 kPa 1 atmosphere = 760 torrs 1 atmosphere = 760 torrs 1 atmosphere = 14.7 psi 1 atmosphere = 14.7 psi 1 atmosphere = 29.92 in Hg 1 atmosphere = 29.92 in Hg

4. STP Standard temperature and pressure Standard temperature and pressure 1atmosphere 1atmosphere 0 degrees Celsius 0 degrees Celsius STP is also a motor oil. That’s cool, if irrelevant. STP is also a motor oil. That’s cool, if irrelevant.

5. The Gas Laws are mathematical. The gas laws will describe HOW gases behave.

6. Four Variables Describe a Gas 1. 1. pressure (P) in atm 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles

7. Four Variables Describe a Gas 1. 1. pressure (P) in atm 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles Held constant in Section 14.2

8. Robert Boyle (1627-1691) Boyle was born into an aristocratic Irish family

9. #1. Boyle’s Law - 1662 Pressure x Volume = a constant Equation: P 1 V 1 = P 2 V 2 (T = constant) Gas pressure is inversely proportional to the volume, when temperature is held constant.

10. Graph of Boyle’s Law – page 418 Boyle’s Law says the pressure is inverse to the volume. Note that when the volume goes up, the pressure goes down

11. http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/frglab2.html

13. Boyle’s law

15. Key mathematical points: The product of corresponding P and V values is a constant PV = constant The product of corresponding P and V values is a constant PV = constant P 1 V 1 =P 2 V 2 P 1 V 1 =P 2 V 2 V is inversely proportional to P V is inversely proportional to P V is directly proportional to 1/P: V~1/P V is directly proportional to 1/P: V~1/P

16. A sample problem on volume-pressure relationship. A helium balloon contains 30.0 L of helium gas sat 103 kPa. 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. Answer: 124 L

17. Jacques Charles (1746-1823) French Physicist

18. #2. Charles’s Law - 1787 The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant. This extrapolates to zero volume at a temperature of zero Kelvin.

19. Converting Celsius to Kelvin Always use the temperature in Kelvin. Kelvin =  C + 273.15 °C = Kelvin – 273.15

20. Joseph Louis Gay-Lussac (1778 – 1850)  French chemist and physicist

21. #3. Gay-Lussac’s Law - 1802 The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant.

22. http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/frglab2.html

23. Charles’s law

25. A sample problem on volume-temperature relationship. A balloon inflated in a room at 24 ˚C has a volume of 4.00 L. The balloon is then heated to a temperature of 58 ˚C. What is the new volume if the pressure remains constant? Answer: 4.46 L

26. A sample problem on pressure-temperature relationship. 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? Answer: 2.58 kPa

27. Given: A 58 L sample of dry air is cooled from 127°C to -23°C while the pressure is maintained at 2.85 atm. What is the final volume? Given: A 58 L sample of dry air is cooled from 127°C to -23°C while the pressure is maintained at 2.85 atm. What is the final volume? Solution: V 2 =V 1 T 2 /T 1 V 2 =58L*250K/400K=36L - Looks realistic. Solution: V 2 =V 1 T 2 /T 1 V 2 =58L*250K/400K=36L - Looks realistic.

28. #4. The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

29. The combined gas law contains all the other gas laws! If the temperature remains constant... P1P1 V1V1 T1T1 x = P2P2 V2V2 T2T2 x Boyle’s Law

30. The combined gas law contains all the other gas laws! If the pressure remains constant... P1P1 V1V1 T1T1 x = P2P2 V2V2 T2T2 x Charles’s Law

31. u The combined gas law contains all the other gas laws! u If the volume remains constant... P1P1 V1V1 T1T1 x = P2P2 V2V2 T2T2 x Gay-Lussac’s Law

32. A sample problem on pressure -volume-temperature relationship. A gas at 155 kPa and 25˚C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125 ˚C. What is the new volume? Answer: 0.342 L

33. Section 14.3 Ideal Gases OBJECTIVES: OBJECTIVES: Compute the value of an unknown using the ideal gas law.

34. Section 14.3 Ideal Gases OBJECTIVES: OBJECTIVES: Compare and contrast real an ideal gases.

35. 5. The Ideal Gas Law #1 Equation: P x V = n x R x T Pressure times Volume equals the number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin. R = 8.31 (L x kPa) / (mol x K) The other units must match the value of the constant, in order to cancel out. The value of R could change, if other units of measurement are used for the other values (namely pressure changes)

36. We now have a new way to count moles (the amount of matter), by measuring T, P, and V. We aren’t restricted to only STP conditions: P x V R x T The Ideal Gas Law n =

37. Ideal Gases We are going to assume the gases behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure An ideal gas does not really exist, but it makes the math easier and is a close approximation. Particles have no volume? Wrong! No attractive forces? Wrong!

38. Ideal Gases There are no gases for which this is true (acting “ideal”); however, Real gases behave this way at a) high temperature, and b) low pressure. Because at these conditions, a gas will stay a gas Sample Problem

39. #6. Ideal Gas Law 2 Equation: P x V = m x R x T M Allows LOTS of calculations, and some new items are: m = mass, in grams M = molar mass, in g/mol Molar mass = m R T P V

40. Density Density is mass divided by volume m V so, m M P V R T D = =

42. Ideal Gases don’t exist, because: 1. 1. Molecules do take up space 2. 2. There are attractive forces between particles - otherwise there would be no liquids formed

43. Real Gases behave like Ideal Gases... When the molecules are far apart. The molecules do not take up as big a percentage of the space We can ignore the particle volume. This is at low pressure

44. Real Gases behave like Ideal Gases… When molecules are moving fast This is at high temperature Collisions are harder and faster. Molecules are not next to each other very long. Attractive forces can’t play a role.

45. Section 14.4 Gases: Mixtures and Movements OBJECTIVES: Relate the total pressure of a mixture of gases to the partial pressures of the component gases.

46. Section 14.4 Gases: Mixtures and Movements OBJECTIVES: Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses.

47. #7 Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... P 1 represents the “partial pressure”, or the contribution by that gas. Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.

48. Collecting a gas over water Connected to gas generator

49. If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3: 2 atm + 1 atm + 3 atm = 6 atm 1 2 3 4

50. Diffusion is: Effusion: Gas escaping through a tiny hole in a container. Both of these depend on the molar mass of the particle, which determines the speed. u Molecules moving from areas of high concentration to low concentration. u Example: perfume molecules spreading across the room.

51. Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Molecules move from areas of high concentration to low concentration.

52. Effusion: a gas escapes through a tiny hole in its container -Think of a nail in your car tire… Diffusion and effusion are explained by the next gas law: Graham’s

53. 8. Graham’s Law The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules. Derived from: Kinetic energy = 1/2 mv 2 m = the molar mass, and v = the velocity. Rate A  Mass B Rate B  Mass A =

54. With effusion and diffusion, the type of particle is important: Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass. Helium effuses and diffuses faster than nitrogen – thus, helium escapes from a balloon quicker than many other gases Graham’s Law