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Chapter 11: Properties of Gases

Chapter 11: Properties of Gases. Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop. Group Problem. In groups of 2-3 brainstorm how to describe a gas . What are some observable properties? What variables would you use to describe a gas?. Group Problem.

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Chapter 11: Properties of Gases

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  1. Chapter 11: Properties of Gases Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

  2. Group Problem • In groups of 2-3 brainstorm how to describe a gas. • What are some observable properties? • What variables would you use to describe a gas?

  3. Group Problem • Describe a gas: • Will expand to fill a volume • Mostly empty space so can be compressed • Can expand & contract with temperature • Particles constantly in motion & constantly colliding • Some gases are heavier then others and sink to the floor rather then rise to the ceiling

  4. Properties of Common Gases • Despite wide differences in chemical properties, all gases more or less obey the same set of physicalproperties. Four Physical Properties of Gases • Inter-related • Pressure (P) • Volume (V ) • Temperature (T ) • Amount = moles (n)

  5. Review: The Mole • Avagadro’s number (NA) allows us to measure the number of particles of a gas as the number of moles: • NA = 6.02214129 × 1023 particles/mole • We can measure the number of moles of a gas by measuring its mass and knowing its Molar Mass • Molar Mass = mass / (# of moles) • M = m/n Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  6. Group Problem How many moles of the CFC pollutant CCl2F2 are in 50.0g?

  7. Group Problem How many moles of the CFC pollutant CCl2F2 are in 50.0g? 50.0 g CCl2F2(1 mol/121 g CCl2F2) = 0.41 mol CCl2F2

  8. Group Problem Calculate the mass of 3 moles of nerve agent VX: CH3CH2 C11H26NO2PS

  9. Group Problem Calculate the mass of 3 moles of nerve agent VX: CH3CH2 3 mol C11H26NO2PS ( 267 g C11H26NO2PS /mol C11H26NO2PS) = 801 g C11H26NO2PS

  10. Review: Temperature • Temperature is measured with a thermometer usually in °C, °F, or Kelvin • Fahrenheit • O is freezing point of salt water/96 is temperature of life • Celsius • O is freezing point of water/100 is boiling point of water • Kelvin • Uses absolute 0 where all motion stops • O°C = 273 K • °C = (°F -32) × (5/9)

  11. Group Problem If room temperature is 25°C, what is room temperature in Kelvin? In °F?

  12. Group Problem If room temperature is 25°C, what is room temperature in Kelvin? RT = 25°C + 273 K = 298 K In °F? 25°C = (°F -32) × (5/9) °F = [(25°C) 9/5] +32 = 77 °F

  13. Pressure: Measurement and Units • Pressure is force per unit area • Earth exerts gravitational force on everything with mass near it • Weight • Measure of gravitational force that earth exerts on objects with mass • What we call weight is gravitational force acting on object (weight ≠ mass)

  14. Force vs. Pressure • Consider someone wearing flat shoes vs. high "spike" heels • Weight of person is the same F = 120 lbs • Pressure on floor differs greatly (F/A) This is why snow shoes have a large footprint

  15. Pressure • Atmospheric Pressure • Resulting force per unit area • When earth's gravity acts on molecules in air • Pressure due to air molecules colliding with object • Barometer • Instrument used to measure atmospheric pressure

  16. Vaccum A vacuum exerts zero pressure on a containers walls

  17. Toricelli Barometer • Simplest barometer • Tube that is 80 cm in length • Sealed at one end • Filled with mercury • In dish filled with mercury

  18. Toricelli Barometer • Air pressure • Pushes down on mercury • Forces mercury up tube • Weight of mercury in tube • Pushes down on mercury in dish • When two forces balance • Mercury level stabilizes • Read atmospheric pressure

  19. Toricelli Barometer • If air pressure is high • Pushes down on mercury in dish • Increase in level in tube • If air pressure is low • Pressure on mercury in dish less than pressure from column • Decrease level in tube Result: • Height of mercury in tube is the atmospheric pressure

  20. Standard Atmospheric Pressure • Typical range of pressure for most places where people live 730 to 760 mm Hg • Top of Mt. Everest Air pressure = 250 mm Hg Standard atmosphere (atm) • Average pressure at sea level • Pressure needed to support column of mercury 760 mm high measured at 0 °C

  21. Units of Pressure • SI unit for pressure • Pascal = Pa = 1 N/m2 • 1 atm = 101,325 Pa = 101 kPa • 100 kPa = 0.9868 atm • Other units of pressure • 1.013 Bar = 1013 mBar = 1 atm • 760 mm Hg = 1 atm • 760 torr = 1 atm • At sea level 1 torr = 1 mm Hg

  22. Group Problem Express Pressure in atm and kPa for a gas at 705 mmHg

  23. Group Problem Express Pressure in atm and kPa for a gas at 705 mmHg. 705 mmHg(1 atm/760 mmHg) = 0.927 atm 0.927 atm (101 kPa/1 atm) = 93.6 kPa

  24. Manometers • Used to measure pressure inside closed reaction vessels • Pressure changes caused by gases produced or used up during chemical reaction • Open-end manometer • U tube partly filled with liquid (usually mercury) • One arm open to atmosphere • One arm exposed to trapped gas in vessel

  25. Open Ended Manometer Pgas = Patm Pgas > Patm Gas pushes mercury up tube Pgas < Patm Atmosphere pushes mercury down tube

  26. Ex. Using Open Ended Manometers A student collected a gas in an apparatus connected to an open-end manometer. The mercury in the column open to the air was 120 mm higher and the atmospheric pressure was measured to be 752 torr. What was the pressure of the gas in the apparatus? This is a case of Pgas > Patm Pgas = 752 torr + 120 torr = 872 torr

  27. Ex. Using Open Ended Manometers In another experiment, it was found that the mercury level in the arm of the manometer attached to the container of gas was 200 mm higher than in the arm open to the air. What was the pressure of the gas? This is a case of Pgas < Patm Pgas = 752 torr – 200 torr = 552 torr

  28. Group Problem CO2 collected in a monometer in a lab with a barometric reading of 97 kPa. What is the Pressure of CO2? 33 mmHg

  29. Group Problem CO2 collected in a monometer in a lab with a barometric reading of 97 kPa. What is the Pressure of CO2? Pgas < Patm 33mm Hg (101 kPa/760 mmHg) =4.4 kPa Pgas = 97 kPa – 4.4 kPa = 92.06 kPa 33 mmHg

  30. Closed-end Manometer • Arm farthest from vessel (gas) sealed • Tube filled with mercury • Then open system to flask and some mercury drains out of sealed arm • Vacuum exists above mercury in sealed arm

  31. Closed-end Manometer • Level of mercury in arm falls, as not enough pressure in the flask to hold up Hg • Patm = 0 • Pgas = PHg • So directly read pressure

  32. Your Turn Gas pressure is measured using a close-ended mercury manometer. The height of fluid in the manometer is 23.7 in. Hg. What is this pressure in atm? • 23.7 atm • 0.792 atm • 602 atm • 1.61 atm 0.792 atm

  33. Group Problem What is the pressure of an unknown gas in atm within this closed monometer? Closed monometer 437 mm 205 mm

  34. Group Problem What is the pressure of an unknown gas in atm within this closed monometer? PHg = 437mm-205mm = 232mm 232mmHg (1 atm/160 mmHg) = 0.31 atm Closed monometer 437 mm 205 mm

  35. Comparison of Hg and H2O • Pressure of 1 mm column of mercury and 13.6 mm column of water are the same • Mercury is 13.6 times more dense than water • Both columns have same weight and diameter, so they exert same pressure d = 13.6 g/mL d = 1.00 g/mL

  36. Using Liquids Other Than Mercury in Manometers and Barometers • Simple relationship exists between two systems. • For example, use water (d = 1.00 g/mL) instead of mercury (d = 13.6 g/mL) in the tube • Use this relationship to convert pressure change in mm H2O to pressure change in mm Hg For converting from mm Hg to mm H2O In general

  37. Ex. Converting mm Acetone to mm Hg - Solution Acetone has a density of 0.791 g/mL. Acetone is used in an open-ended manometer to measure a gas pressure slightly greater than atmospheric pressure, which is 756 mm Hg at the time of the measurement. The liquid level is 20.4 mm higher in the open arm than in the arm nearest the gas sample. What is the gas pressure in torr?

  38. Ex. Converting mm Acetone to mm Hg - Solution First convert mm acetone to mm Hg Then add PHg to Patm to get Ptotal • Pgas = Patm + PHg • = 756.0 torr + 1.19 torr • Pgas = 757.2 torr

  39. Boyle’s Law • Studied relationship between P and V • Work done at constant T as well as constant number of moles (n) • T1= T2 • AsV decreases,P increases

  40. Ideal Gas Law Charles Law If Pressure is constant but freeze a balloon, it decreases in V Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  41. Charles’s Law • Charles worked on relationship of how V changes with T • Kept P and n constant • Demonstrated Vincreases as T increases

  42. Low T, Low P P High T, High P T (K) Gay-Lussac’s Law • Worked on relationship between pressure and temperature • Volume (V ) and number of moles (n) are constant • P increases as T increases • This is why we don’t heat canned foods on a campfire without opening them • Showed that gas pressure is directly proportional to absolute temperature

  43. Group Problem What happens to gas pressure when you raise the temperature?

  44. Group Problem What happens to gas pressure when you raise the temperature? Pressure increases because the faster moving molecules hit the walls of the container with greater force No change in pressure is observed because the area increased.

  45. Combined Gas Law • Boyle’s law: • Charles Law: • Guy-Lussac’s Law: • is equivalent to • For any two conditions: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  46. Combined Gas Law • All T 's must be in K • Value of P and V can be any units as long as they are the same on both sides • Can derive Boyle’s Law, Charle’s Law, and Guy Lussac’s Law from this equation • Gives all relationships needed for fixed amount of gas under two sets of conditions

  47. How Other Laws Fit into Combined Gas Law

  48. Used for calculating effects of changing conditions T in Kelvin P and V any units, as long as units cancel Example: If a sample of air occupies 500. mL at 273.15 K and 1 atm, what is the volume at 85 °C and 560 torr? Combined Gas Law V2 =890. mL

  49. Ex. Using Combined Gas Law • What will be the final pressure of a sample of nitrogen gas with a volume of 950. m3 at 745 torr and 25.0 °C if it is heated to 60.0 °C and given a final volume of 1150 m3? • First, number of moles is constant even though actual number is not given • You are given V, P and T for initial state of system as well as T and V for final state of system and must findPfinal • This is a clear case for combined gas law

  50. Ex. Using Combined Gas Law • List what you know and what you don’t know • Convert all temperatures to Kelvin • Then solve for unknown—here P2 P2 = 688 torr

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