1. Gas Laws: Introduction At the conclusion of our time together, you should be able to: 1. List 5 properties of gases 2. Identify the various parts of the kinetic molecular theory 3. Define pressure 4. Convert pressure into 3 different units 5. Define temperature 6. Convert a temperature to Kelvin

2. Importance of Gases Airbags fill with N 2 gas in an accident. Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3. Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 2 2 NaN 3 ---> 2 Na + 3 N 2

3. THREE STATES OF MATTER

4. General Properties of Gases There is a lot of “free” space in a gas. There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Gases diffuse and mix rapidly.

5. To Review  Gases expand to fill their containers  Gases are fluid – they flow  Gases have low density  1/1000 the density of the equivalent liquid or solid  Gases are compressible  Gases effuse and diffuse

6. Properties of Gases Gas properties can be modeled using math. This model depends on — V = volume of the gas (L) V = volume of the gas (L) T = temperature (K) T = temperature (K) – ALL temperatures in the entire unit MUST be in Kelvin!!! No Exceptions! n = amount (moles) n = amount (moles) P = pressure (atmospheres) P = pressure (atmospheres)

7. Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.  Gases consist of tiny particles that are far apart relative to their size.  Collisions between gas particles and between particles and the walls of the container are elastic collisions  No kinetic energy is lost in elastic collisions

8. Ideal Gases Ideal Gases (continued)  Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion  There are no forces of attraction between gas particles  The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

9. Pressure  Is caused by the collisions of molecules with the walls of a container  Is equal to force/unit area  SI units = Newton/meter 2 = 1 Pascal (Pa)  1 atmosphere = 101.325 kPa (kilopascal)  1 atmosphere = 1 atm = 760 mm Hg = 760 torr  1 atm = 29.92 in Hg = 14.7 psi = 0.987 bar = 10 m column of water.

10. Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

11. An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

12. PressureColumn height measures Pressure of atmosphere 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg * = 14.7 pounds/in2 (psi) = 101.325 kPa (SI unit is PASCAL)

13. 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 Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

14. Let’s Review: Standard Temperature and Pressure “STP” Allows us to compare amounts of different gases by converting them all to STP (the same temperature and pressure conditions) P = 1 atmosphere, 760 torr T =  C, 273 Kelvin The molar volume of an ideal gas is 22.42 liters at STP

15. Pressure Conversions A. What is 475 mm Hg expressed in atm? B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? = 1520 mm Hg = 0.625 atm 475 mm Hg x 29.4 psi x 1 atm 760 mm Hg 14.7 psi

16. 101.325 kPa 14.7 psi Pressure Conversions – Your Turn A. What is 2.00 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa? = 1520 torr 2.00 atm x 760 torr 1 atm = 221 kPa 32.0 psi x

17. Converting Celsius to Kelvin Gas law problems involving temperature require that the temperature be in KELVINS! Kelvins =  C + 273 °C = Kelvins - 273 Don’t use temp. when determining sig. figs. for your answer

18. Charles’ Law If n and P are constant, then V α T V and T are directly proportional. If one temperature goes up, the volume goes up! If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

19. Charles’ Law

20. Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. If one temperature goes up, the pressure goes up! If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850)

21. Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

22. Boyle’s Law Example A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

23. Now let’s put all 3 laws together into one big combined gas law …….

24. Combined Gas Law (for a fixed amount, moles, of gas) The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! (if a variable is not given in the problem, just leave it out)

25. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

26. Calculation P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 = T 1 T 2 Solving for T 2 = 604 K T 2 = 604 K - 273 = 331 °C

27. Gas Laws: Avogadro’s and Ideal At the conclusion of our time together, you should be able to: 1.Describe Law with a formula. 1.Describe Avogadro’s Law with a formula. 2.Use Law to determine either moles or volume 2.Use Avogadro’s Law to determine either moles or volume 3.Describe the Law with a formula. 3.Describe the Ideal Gas Law with a formula. 4.Use Law to determine either moles, pressure, temperature or volume 4.Use the Ideal Gas Law to determine either moles, pressure, temperature or volume 5.Explain the Kinetic Molecular Theory

28. Avogadro’s Law Equal volumes of gases at the same T and P have the same number of molecules. V and n are directly related. twice as many molecules

29. Avogadro’s Law Summary  For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V1V1 n1n1 V2V2 n2n2

30. Standard Molar Volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro

31. Avogadro’s Law Practice #1 V1V1 n1n1 V2V2 n2n2 4.00 L 0.21 mol 7.12 L n2n2 0.37 mol total 0.16 mol added

32. Ultimate Comined Gas Law (include Aavogadro’s law in the combined gas law) The relationship between T, P, n, and V for a gas is a constant which we call the ideal gas constant, R (=0.08206 L atm/ mole K). So we can can set one side as equal to R (the gas constant) and it is now called the ideal gas law.

33. IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION! P V = n R T

34. Ideal Gas Law PV = nRT  P = pressure in atm (you may need to convert)  V = volume in liters  n = moles  R = proportionality constant  = 0.08206 L atm/ mol·   T = temperature in Kelvin

35. R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. R = 0.08206 R = 0.08206 L atm mol K

36. Using PV = nRT How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 o C? Solution Solution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C + 273 = 298 K T = 25 o C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm And we always know R, 0.08206 L atm / mol K

37. How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? Solution Solution 2. Now plug in those values and solve for the unknown. PV = nRT RT RT RT RT Using PV = nRT n = 1.1 x 10 3 mol (or about 30 kg of gas)

38. Ideal Gas Law Problems #1 (5.6 atm)(12 L)(0.08206 atm*L / mol*K )(T) 200 K (4 mol)

39. Gas Laws: Dalton’s Law At the conclusion of our time together, you should be able to: 1.Explain Dalton’s Law 2.Use Dalton’s Law to solve a problem

40. Dalton’s Law John Dalton 1766-1844

41. Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... This is particularly useful in calculating the pressure of gases collected over water.

42. Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P H 2 O + P O 2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

43. Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mm Hg 20.95% O 2 159.2 mm Hg 0.94% Ar7.1 mm Hg 0.03% CO 2 0.2 mm Hg P AIR = P N + P O + P Ar + P CO = 760 mm Hg 2 2 2 Total Pressure = 760 mm Hg

44. Collecting a gas “over water” Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

45. Table of Vapor Pressures for Water

46. Solve This! A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H 2 gas? 768 torr – 17.5 torr = 750.5 torr

47. Dalton’s Law of Partial Pressures Also, for a mixture of gases in a container, because P, V and n are directly proportional if the other gas law variables are kept constant: n Total = n 1 + n 2 + n 3 +... V Total = V 1 + V 2 + V 3 +... This is useful in solving problems with differing numbers of moles or volumes of the gases that are mixed together.

48. Gas Laws: Gas Stoichiometry At the conclusion of our time together, you should be able to: 1.Use the Ideal Gas Law to solve a gas stoichiometry problem.

49. Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

50. Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O 2 22.4 L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 = 0.36 L O 2 at STP

51. Gas Stoichiometry: Practice! How many grams of He are present in 8.0 L of gas at STP? = 1.4 g He 8.0 L He x 1 mol He 22.4 L He x 4.00 g He 1 mol He

52. Gas Stoichiometry Trick If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H 2 (g) + N 2 (g)  2NH 3 (g) 3 moles H 2 + 1 mole N 2  2 moles NH 3 67.2 liters H 2 + 22.4 liter N 2  44.8 liters NH 3

53. Gas Stoichiometry Trick Example How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature? 3 H 2 (g) + N 2 (g)  2NH 3 (g) 12 L H 2 L H 2 = L NH 3 L NH 3 3 2 8.0

54. What if the problem is NOT at STP? 1. You will need to use PV = nRT

55. Gas Stoichiometry Example on HO How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 10.5 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 0.13 mol O 2

56. Gas Stoichiometry Example on HO How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 3.2 L O 2 (1.0 atm)(V)(0.08206 atm*L / mol*K )(298 K)(0.13 mol)

57. Gas Laws: Dalton, Density and Gas Stoichiometry At the conclusion of our time together, you should be able to: 1.Explain Dalton’s Law and use it to solve a problem. 2.Use the Ideal Gas Law to solve a gas density problem. 3.Use the Ideal Gas Law to solve a gas stoichiometry problem.

58. Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0  C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = “n” mol O 2 50.0 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 = 0.612 mol O 2 = 16.7 L

59. Try this one! How many L of O 2 are needed to react 28.0 g NH 3 at 24°C and 0.950 atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

60. Gas Laws: Finding “R” At the conclusion of our time together, you should be able to: 1.Use the Ideal Gas Law to find the value of “R”. See the problem on p. 50

61. Partial Pressure of CO 2 Carbon Dioxide gas over water at 17.0 degrees C and 97.932 kPa. What is the pressure of the CO 2 gas? 97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm 1.90 kPa

62. Determine the Moles of CO 2 Used: Mass of canister before-mass of canister after 17.099 g – 16.524 g= 0.575 g 0.575 g CO 2 1 mol CO 2 44.01 g CO 2 0.0131 mol CO 2

63. (0.948 atm)(0.347 L) (R)(290 K) 0.0868 atm*L / mol*K (0.0131 mol)

64. 0.08206 atm*L / mol*K (standard) 0.0868 atm*L / mol*K (experimental results) - 0.0048 atm*L / mol*K 0.08206 atm*L / mol*K x 100 - 5.85 % error - 0.00474 atm*L / mol*K (experimental error)