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The Kinetic Theory, Pressure & Gas Laws

The Kinetic Theory, Pressure & Gas Laws. GAS The four quantities needed to describe a gas : 1) number of particles 2) temperature 3) pressure 4) volume. The Kinetic Theory. The Kinetic Theory explains the effects of temperature & pressure on matter.

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The Kinetic Theory, Pressure & Gas Laws

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  1. The Kinetic Theory, Pressure & Gas Laws

  2. GAS • The four quantities needed to describe a gas: • 1) number of particles • 2) temperature • 3) pressure • 4) volume

  3. The Kinetic Theory • The Kinetic Theory explains the effects of temperature & pressure on matter. • Idea that all particles of matter are constantly in motion • 4 Assumptions: of gases for IDEAL GASES • 1) All gases are composed of small particles and have no volume • 2) These small particles move in continual, random, and rapid motion with no attraction or repulsion forces • 3) All collisions are perfectly elastic • 4) The average kinetic energy of the particles is directly proportional to its Kelvin temperature

  4. Real vs. Ideal Gases • An ideal gas is one that follows the gas laws at all conditions of pressure and temperature and follows all the assumptions of the KMT • An ideal gas does not really exist!!!!! • Real gases can be liquefied and sometimes solidified • The behavior of real gases under many conditions is similar to ideal gases. Each gas is different. • Real gases behave like ideal gases at most conditions except for extremely high pressures and low temperatures

  5. Kinetic Theory con’t • The physical behavior of a gas depends on its volume, temperature, and pressure. • Temperature: Kelvin • the average kinetic energy of the particles in an object • ie. 25oC, O2 molecules 443 m/s; 1700 km/h; 1057 mi/h • Absolute Zero: • temperature at which all molecular motion should cease; lowest possible temperature • -273.15oC or 0K

  6. Kinetic Theory con’t • Temperature is not the total amount of thermal energy a substance has absorbed. • Question 1: • How is the average kinetic energy of water molecules affected when hot water from a kettle is poured into cups at the same temperature as the water? • IT IS UNAFFECTED

  7. Kinetic Theory con’t • Question 2: • By what factor does the average kinetic energy of the molecules of a gas in an aerosol container increase when the temperature is raised from 300K (27oC) to 900K? • The average kinetic energy triples

  8. Pressure • Pressure: • the force per unit area 1 object exerts on another • SI Unit – Pascal (Pa) • Gas Pressure: • the pressure exerted by a gas • it is created by collisions of gas particles with an object

  9. What happens to the pressure when you add more gas particles without changing its volume or temperature? Why? • What happens to the pressure when you decrease the volume of the container for a fixed mass of a gas at a constant temperature? Why? • What must happen to a gas’s temperature in order for its pressure to decrease? Why?

  10. Pressure • Pascal: (Pa) • the SI unit for pressure • equivalent to 1N/m2 • Millimeter of Mercury: (mmHg) • pressure needed to support a column of mercury 1 mm high

  11. Pressure con’t • Atmosphere: (atm) • the pressure required to support 760 mm of mercury in a mercury barometer • Torr: (torr) • named after Evangelista Torricelli who invented the barometer • Barometer: • a closed-arm manometer used to measure pressure

  12. Pressure con’t • 1 atm = 760 mmHg • 1 atm = 101.325 kPa • 760 mmHg = 101.325 kPa • 1 torr = 1 mmHg • 1 atm = 760 torr • 1 atm = 760 mmHg = 760 torr = 101.325 kPa • Question 3: Convert the following: • a. 4.328 atm to kPa b. 328 kPa to mmHg • c. 3290 Pa to atm

  13. 101.325 kPa 1.0 atm • 4.328 atm x • = 438.5 kPa • 328 kPa x • = 2460 mmHg • 3290 Pa x • = 0.0325 atm 760 mmHg 101.325 kPa 1.0 atm 101325 Pa

  14. manometer: • a devise used to measure pressure of a gas • Two Types of Manometers • 1) closed-arm: • used to measure the actual or “absolute” gas pressure • *known as a barometer • 2) open-arm manometer: • one arm of the manometer is open to the atmospheric air to measure the gas pressure of a confined gas

  15. Avogadro’s Hypothesis • Equal volumes of gases at the same temperature and pressure contain equal numbers of particles

  16. States of Matter • 1)GASES: • are independent of one another moving in straight lines until they collide with something that affects its direction and possibly its speed • have no definite shape or volume • gases assume the shape of the container

  17. 2) Liquids: • have a definite volume but will take the shape of their container • particles are attracted to one another but have enough energy to slide past each other • no bond is formed between the particles

  18. 3) Solids: • particles possess relatively fixed positions and vibrate around that fixed point • attractive forces hold the particles extremely close but the particles of the solid are still traveling in straight paths between colliding with its exceedingly close neighboring particles • the physical state of a substance at STP depends on the attractive forces verses the energy of the particles

  19. Ionic compounds: • tend to be solids with strong electric charges • Molecular compounds • are attracted by van der Waals forces • high molecular mass compounds tend to be solids • nonpolar molecules of low molecular mass tend to be gases • *the greater the mass and polarity, the more likely the compound will be a solid or liquid

  20. 4) Plasma: • matter at temperatures greater than 5000oC causing a state where the matter is composed of electrons and positive ions • magnetohydrodynamics

  21. Vaporization • Vaporization: • the conversion of a liquid to a gas or vapor below its boiling point (bp) • Evaporation: • the vaporization of an uncontained liquid • Vapor Pressure: • pressure created by a vapor in equilibrium with its liquid

  22. Vaporization con’t • Boiling Point: (bp) • temperature at which the vapor pressure of the liquid is equal to the external pressure • Normal Boiling Point: • boiling point for a substance at 1 atm • * Boiling is a cooling process

  23. Vaporization con’t • evaporation • Liquid Vapor (gas) condensation

  24. Heat of Fusion: • the additional amount of energy required to cause a phase change from a solid to liquid once a substance reaches its melting/freezing point • Heat of Vaporization: • the additional amount of energy required to cause a phase change from a gas to liquid once a substance reaches its boiling/condensation point

  25. Phase Changes

  26. Phase Diagram

  27. Triple Point: • temperature and pressure where all three phases are at equilibrium • Critical Point • at the critical temperature and critical pressure of the substance, beyond this point the liquid and gas phases become indistinguishable

  28. JOHN DALTON

  29. DALTON’S LAW OF PARTIAL PRESSURES • At a constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases • PARTIAL PRESSURE: • the pressure(contribution) each gas in a mixture makes to the total pressure of the mixture • Ptotal = P1 + P2 + P3 …

  30. Example: • Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other elements. What is the partial pressure of O2 at 1.00 atm if PN2 = 593.4 mmHg, PCO2 = 0.3 mmHg and Ptrace = 7.1 mmHg? • Ptotal = PO2 + PCO2 + PN2 + Ptrace • PO2 = Ptotal - (PCO2 + PN2 + Ptr) • = 760mmHg - (593.4 mmHg + 0.3 mmHg + 7.1 mmHg) • = 159 mmHg

  31. Your Turn: • Determine the total pressure of a mixture of gases if the partial pressures of the gases are PO2 = 242.5 mmHg, PHe= 27.3 kPa, and PNe = 0.021 atm.

  32. PT = PO2 + PHe + PNe • Look for unit agreement!!!! • We will use mmHg as the common unit for our answer • = 27.3 kPa X 760 mmHg = 205 mmHg • 101.325 kPa • = 0.021 atm X 760 mmHg = 16 mmHg • 1 atm • PT = 242.5 mmHg + 205 mmHg + 16 mmHg • 464 mmHg

  33. Robert Boyle

  34. Boyle's Law • When the number of particles and temperature are constant, pressure and volume are inversely proportional. (opposite) • P1 V1 = P2 V2

  35. P1 = • initial pressure • V1 = • initial volume • P2 = • final pressure • V2 = • final volume

  36. Example: • A balloon is filled with 30.0 L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 185.3 mmHg? Assume temperature remains constant.

  37. 30.0 L = V1 • 1.00 atm = P1 • 185.3 mmHg = P2 • ? L = V2 • Look for unit agreement • 1.00 atm = 760 mmHg • P1V1 = P2V2 • Divide both sides by P2

  38. P1V1 P2 • V2 = • V2 = (760 mmHg)(30.0 L) = 185.3 mmHg • = 123 L

  39. Jacques Charles

  40. Charles’ Law • states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant • V1 = V2 T1 T2 • *temperature must always be in Kelvin

  41. A 225 cm3 volume of gas is collected at 58.0oC. What volume, in liters, would this sample of gas occupy at standard temperature? • V1 = 225 cm3 • V2 = ? • T1 = 58.0oC • T2 = 0.0oC • Charles’ Law : 273.15 + 58.0oC = 331.2K : 273.15 + 0.0oC = 273.2K

  42. V2 = V1T2 • T1 • = (225 cm3)(273.2 K) • 331.2 K • = 186 cm3 · 1 L = • 1000 cm3 • = 0.186 L

  43. Your Turn: • A balloon, inflated in an air-conditioned room at 27.0oC, has a volume of 4.0 L. It is heated to a temperature of 57.3oC. What is the new volume of the balloon if the pressure remains constant? • T1 = 27.0oC • V1= 4.0 L • T2 = 57.3oC • V2 = ???? • Convert all temperatures to Kelvin

  44. V1=V2 T1 T2 • V2 = V1 T2 T1 • V2 =(4.0 L)(330.5K) 300.2 K • = 4.4 L

  45. Joseph Louis Gay-Lussac

  46. Gay-Lussac’s Law • the pressure of a given mass of gas is directly proportional to the Kelvin temperature if the volume is held constant P1= P2 T1 T2

  47. An acetylene gas cylinder has a pressure of 24350 mmHg at a temperature of 19.49oC. What would the internal pressure be if the temperature was increased to 100.0oC? Assume no change the volume of the cylinder. • P1 = 24350 mmHg • T1 = 19.49oC + 273.15 = 292.64 K • P2 = X • T2 = 100.00oC + 273.15 = 373.15K

  48. P1= P2 • T1 T2 • P2= P2T1 • T2 • X = (24350 mmHg)(373.15 K) • 294.64 K • 24350 mmHg = X 292.64 K 373.15K • X = 31050 mmHg

  49. Your Turn: • A gas has a pressure of 50.0 atm at 540 K. What will the temperature, in Celsius, be if the pressure is increased to 8330 kPa? • P1 = 50.0 atm • T1 = 540 K • P2 = 8330 kPa • T2 = X

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