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# Law of Gases

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Dalton's Law, Avogadro's Law, Molar Volume, and Ideal Gas Law

Dalton's Law, Avogadro's Law, Molar Volume, and Ideal Gas Law

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• 1. Project in Chemistry SY: 2012-2013 Submitted by: Axl Merk Mindajao Mary Rose Baccay III-OBB Submitted to: Mrs. Panlasiqui
• 2. Recall 1. If the pressure exerted on a gas is tripled, what will happen to the volume (assuming the temperature and amount of gas remains constant)? A. x2 C. x 1/2 B. X3 D. x 1/3
• 3. 2. Carbon dioxide occupies a 2.54 L container at STP. What will be the volume when the pressure is 150 KPa and 26 C? A. 1.89 L C. 0.163 L B. 42300 L D. 14.1 L
• 4. 3. In which of the following sets of conditions would a gas most act as a real gas? A. high temperature and low pressure B. high temperature and high pressure C. low temperature and high pressure D. low temperature and low pressure
• 5. 4. Which of the following terms is used to describe the average kinetic energy of the molecules in a sample? A. manometer C. pressure B. temperature D. volume
• 6. 5. Which of the following is not a standard pressure measurement? A. 101.3 kPa C. 760 mm Hg B. 1 atm D. 273 K
• 7. 6. The volume of a gas is increased from 150.0 mL to 350.0 mL by heating it. If the original temperature of the gas was 25.0 °C, what is the final temperature? A. -146 °C C. 695 °C B. 10.7 °C D. 150 °C
• 8. 7. What is the name of the phase change that exists when a solid turns into a liquid? A. depositon C. sublimation B. melting D. freezing
• 9. 8. A balloon is inflated to a volume of 130 ml at a pressure of 690 mmHg. If the pressure is increased to 1000 mmHg, what will the new volume be? A. 188 ml C. 89.7 ml B. 98.8 ml D. 40300 ml
• 10. 9. What type of relationship is shown between temperature and pressure? A. direct C. exponential B. inverse D. logarithmic
• 11. 10. How many mm Hg is equal to 121 kPa? A. 1.19 C. 636 B. 16.1 D. 908
• 12. Answers 1.d 6. c 2.a 7. b 3.c 8. c 4.b 9. a 5.d 10. d
• 13. Title Dalton’s Law of Partial Pressure
• 14. Objectives To demonstrate the ability to use Dalton’s Law of Partial Pressure in calculations. To calculate the total pressure of a mixture of gases or the partial pressure of a gas in a mixture of gases. To employ Dalton’s Law of Partial Pressure to predict the pressure of a gas mixture.
• 15. Unlocking of Words • Pressure- the force exerted by gas particles that hit the walls of a container; a force applied per unit area. • Partial pressure- the individual pressure of each gas in a mixture. • Water vapor- gaseous water found in the air that is below the boiling point of water.
• 16. Lesson • The English chemist John Dalton investigated pressures in mixture of gases. Dalton’s Law of Partial Pressure states that: “At constant volume and temperature, the total pressure in a mixture of gases is equal to the sum of the partial pressures of the component gases.”
• 17. • In equation form, PT=P1+P2+.... where PT= total pressure P1= partial pressure of gas 1 P2= partial pressure of gas 2
• 18. Activity Problem 1: Oxygen gas occupies 500 ml at 20°C and 760 mmHg. What volume will it occupy if it is collected over water at 30°C and 750 mmHg? (Water vapor pressure at 30°C 31.8 mmHg)
• 19. • Solution: Step 1 Given: P1= 760 mmHg V1= 500 ml T1= 20°+273 = 293 K P2 = 750 mmHg-31.8 mmHg = 718.2 mmHg T2 = 30°C+273 = 303 K
• 20. Step 2: The combined gas formula P1V1= P2V2 V2= P1 T2 T1 T2 P2 T1 Step 3: V2= 500 ml 760 mmHg 303 K 718.2 mmHg 93 K V2= 547 ml
• 21. Problem 2: Calculate the mass of 400. mL of carbon dioxide collected over water at 30° C and 749 mm Hg. Solution: PT = Pgas + Pwater= 749 mmHg R = 0.0821 L·atm/mol·K Pgas = 749 mmHg–31.8 mmHg = 717 mmHg V = 400.0 L
• 22. T = 30° C + 273 = 303 K PV = nRT n = PV/RT n = 717 mm x 1 atm/760 mm x 400.0 mL x 1L/10 ³mL/(0.0821 L·atm/mol·K x 303 K) n = 0.0152 mol CO2 n = 0.0152 mol CO2 x 44.01 g CO2/1 mol CO2 = 0.669 g CO2
• 23. Exercise 1. Determine the total pressure of a gas mixture that contains oxygen, nitrogen, and helium if the partial pressures of the gases are: PO2 = 20.2 kPa PN2 = 46.7 kPa PHe = 26.7 kPa
• 24. 2. A 10 L flask at 298 K contains a gaseous mixture of CO and CO2 at a total pressure of 1520 mm of Hg. If 0.20 mole of CO is present, find the partial pressure of CO and that of CO2?
• 25. • Solution: 1. Ptotal = PO2 + PN2 + PHe Ptotal = 20.2 + 46.7 + 26.7 Ptotal = 93.6 kPa 2. PCO + PCO2 = P = 1520 mm of Hg
• 26. = 0.49 atm. Partial pressure of CO2, pCO2= P - (pCO) = 2.0 - 0.49 = 1147.6 mm of Hg
• 28. Objectives • Determine how the amount of gas in a fixed volume at a fixed pressure and temperature depends upon the identity of the gas. • To do calculations involving Avogadro’s Law.
• 29. Unlocking of Words • Moles- another name for grams. • Temperature- a measure of the average kinetic energy of the particles of a substance. • Volume- the amount of space occupied by something or within a container.
• 30. Lesson • Amedeo Avogadro stated that the volume of a gas is directly related to its number of moles when temperature and pressure remain unchanged.
• 31. • If the moles of a gas are doubled, then the volume will double as long as the pressure and temperature remain the same. • To illustrate these two conditions, you may write: V1=V2 n1 n2
• 32. Activity Problem 1: A balloon containing 2 moles of helium has a volume of 880 ml. What is the new volume after 4 more moles of helium are added to the balloon at the same temperature and pressure?
• 33. Solution: Step 1 Initial condition Final condition V1= 880 ml V2=? n1= 2 moles n2= 6 moles Step 2 V2= 880 ml x 6 moles =2640 ml 2 moles
• 34. Problem 2: What volume of O2 measured at 1.86 x 106 Pa and 375 K, is needed to react completely with 1.78 L of H2, measured at the same pressure and temperature, to give H2O?
• 35. Solution: 2H₂ O₂ → 2H₂O The volume of O₂ needed is 1.78 L H₂ (1L O₂ · 2L H₂) = 0.890 L O₂
• 36. Exercise 1. How many molecules of O2 are present in 1.00L of O2 at STP? 2. Calculate the number of moles of ammonia gas, NH3, in a volume of 80 L of the gas measured at STP.
• 37. Solution: 1. ___1L STP____ = 0.0446 mol O2 22.4 L mol⁻ⁱ STP (6.02 × 1023 molecules mol- 1)(0.0446 mol) = 2.69 × 1022molecules
• 38. 2. V= number of moles × 22.414 L/mol no. of moles= volume of gas 22.414 = 80__ 22.414 = 3.569 L
• 39. Title Molar Volume
• 40. Objectives • To experimentally determine the volume of one mole of a gas at standard temperature and pressure. • To experimentally determine the value of the gas constant, R. • To calculate the standard molar volume of a gas from accumulated data.
• 41. Unlocking of Words • Molar mass- the mass of one mole of an element or compound equal to the atomic or formula weight. • Molar volume- the volume in liters of one mole of a gas at STP. • STP- the standard gas conditions of 1 atm of pressure and 273 K.
• 42. Lesson • At STP conditions (1atm and 273K), one mole of a gas occupies 22.4 liters. • Molar volume is an extension of Avogadro’s Law, w/c state that,”At any given temp. and pressure equal volume of any gases would contain equal no. of moles.”
• 43. Activity Problem 1: What is the volume of 56 g N₂ gas at STP? Solution: V= 56g N₂ x 1 mole N2 x 22.4 L N₂ 28 g N₂ 1 mole N₂ =44.8 L N₂
• 44. Problem 2: How many liters of 0.250 moles of HCl will occupy at STP? Solution: (x/0.250 mol)=(22.414L/1 mol) 0.250 mol x 22.414 L mol X= 5.60 L
• 45. Exercise 1. Find out the volume contained by 6.8 g of ammonia at STP? 2. A balloon contains 0.5 moles of pure helium gas at standard temperature and pressure. What is the volume of the balloon?
• 46. Solution: 1. Gram molecular mass of NH₃=17 g Molar volume= 22.4 liters 17g : 22.4 L 6.8 g : x x= (6.8 X 22.4)/17 x= 8.96 Liters
• 47. 2. Extract the data from the question: n(He) = 0.5 mol Vm = 22.71 L mol-1 (at STP 1 mole of gas occupies 22.71 L) V(He) = ? L Write the equation: V(He) = n(He) x Vm Substitute in the values and solve: V(He) = 0.5 x 22.71 = 11.4 L
• 48. Title Ideal Gas Law
• 49. Objectives • Demonstrate the ability to use the ideal gas content to a basic calculation and those that involve density and molecular mass. • Calculate the amount of gas at any specified conditions of pressure, volume and temperature.
• 50. Unlocking of Words • Ideal gas- an imaginary gas whose behavior is described by the gas laws. • Ideal gas law- a law that combines the 4 meausrable properties of a gas in the equation PV= nRT
• 51. Lesson • When the temperature, pressure and volume of a gas areknown, ideal gas equation can be use. • The ideal gas equation, PV=nRT, illustrates a direct relationships between volume, temperature and the no. of moles of a gas.
• 52. • Volume and pressure are inversely related. P= pressure in atom V= volume in liter n= no. of moles of gas T= kelvin temperature R= 0.082 L atm/Kmol
• 53. Activity Problem 1: How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mmHg? Solution: P = 1.00 x 10-6 mmHg T = 0.0° C + 273 = 273 K V = 985 mL R = 0.0821 L·atm/mol·K
• 54. PV = nRT n = PV/RT n = 1.00 x 10-6 mm x 1 atm/760 mm x 985 mL x 1 L/103 mL/ (0.0821 L·atm/mol·K x 273 K) = 5.78 x 10-11 moles N₂ n= 5.78 x 10-11 moles N₂ x 6.02 x 1023 N₂ molecules/1 mol N₂ = 3.48 x 1013 N₂ molecules
• 55. Problem 2: Calculate the mass of 15.0 L of NH3 at 27° C and 900. mm Hg. Solution: P = 900. mm Hg T = 27° C + 273 = 300 K V = 15.0 L R = 0.0821 L·atm/mol·K
• 56. PV = nRT n = PV/RT n = 900. mm x 1 atm/760 mm x 15.0 L/(0.0821 L·atm/mol·K x 300 K) n = 0.721 moles NH3 x 17.04 g NH3/ 1 mol NH3 = 12.3 g NH3
• 57. Assessment I. Choose the best answer 1. At the water’s surface, the pressure on your body due to the mass of air around you is about a. 760 KPa c. 100 mm b. 101.3 KPa d. 760 mm
• 58. 2. If the volume of mole of gas molecules remains constant, lowering the temperature will make the pressure a. increase b. increase then decrease c. decrease d. decrease then increase
• 59. 3. If the volume available to the gas is increased, the pressure exerted by one mole of gas molecules will a. increase b. increase then decrease c. decrease d. decrease then increase
• 60. 4. How many moles of O₂ are present in 44.8 L of O₂ at STP? a. 1.2 moles b. 1.4 moles c. 2.0 moles d. 2.8 moles
• 61. 5. What pressure must be applied to 225 ml of gas at 1 atm to reduce its volume to 100 ml? a. 0.44 atm b. 2.25 atm c. 22500 atm d. 1250 atm
• 62. Answer 1. c 2. a 3. b 4. d 5. c
• 63. II. Problem solving 6. Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. 7. Calculate the mass of 400. mL of carbon dioxide collected over water at 30.° C and 749 mm Hg.
• 64. 8. Find out the volume contained by 6.8 g of ammonia at STP. (N=14, H=1) 9. What volume of hydrogen will react with 22.4 liters of oxygen to form water? 10. Find the volume from the 0.250 moles gas at 200kpa and 300K temperature.
• 65. Solution ⑥ P = 671 mm Hg T = 47° C + 273 = 320. K V = 478 mL R = 0.0821 L·atm/mol·K
• 66. PV = nRT n = m/MM D = m/V = P x MM/R x T D = 671 mm x 1 atm/760 mm x 83.80 g/mol/(0.0821 L·atm/mol·K x 320. K) = 2.82 g/L
• 67. ⑦ PT = Pgas + Pwater = 749 mm Hg R = 0.0821 L·atm/mol·K Pgas = 749 mm Hg – 31.8 mm Hg = 717 mm Hg V = 400.0 L T = 30.° C + 273 = 303 K
• 68. n = PV/RT n = 717 mm x 1 atm/760 mm x 400.0 mL x 1 L/10³ mL/(0.0821 L·atm/mol·K x 303 K) n = 0.0152 mol CO2 n = 0.0152 mol CO2 x 44.01 g CO2/ 1 mol CO2 = 0.669 g CO2
• 69. ⑧ Gram molecular mass of NH3 = [N = 1 x 14)] + [H = (3 + 1)] = 14 + 3 = 17 g Molar volume = 22.4 liters Volume of 6.8 g of ammonia at STP = ?
• 70. The ratio between mass and volume is as follows: 17 g : 22.4 liters 6.8 g : x x = (6.8 x 22.4) / 17 = 8.96 Liters
• 71. ⑨ 2H2 (g) + O2 (g) → 2H2O (l) From the equation, 2 volumes of hydrogen react with 1 of oxygen or 2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen. The volume of hydrogen that will react is 44.8 liters.
• 72. ⑩ P = 200 kPa n = 0.250 mol T = 300K R = 8.314 J K-1 mol-1 Volume(V) = nRT / P = (0.250 x 8.314 x 300) / 200 = 623.55 / 200 V = 3.12 L