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Chapter 16 Section 3

Chapter 16 Section 3. Behavior of Gases. Pressure. Gas particles are constantly moving and colliding, which results in pressure (Force/area) Containers (balloons, tires) remain inflated because particles collide with the walls of the container

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Chapter 16 Section 3

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  1. Chapter 16 Section 3 Behavior of Gases

  2. Pressure • Gas particles are constantly moving and colliding, which results in pressure (Force/area) • Containers (balloons, tires) remain inflated because particles collide with the walls of the container • If more particles of gas are pumped into the container, there will be more collisions and the walls will be pushed further outward

  3. Pascal • Pressure is measured in pascals (Pa) • 1 Pa = 1 N/1m2 • 1000 Pa = 1 kPa • At Earth’s surface, the atmosperic pressure = 101.3 kPa • 101,300 N per square meter

  4. Boyle’s Law • What happens to gas pressure if you decrease the size of the container? • Particles will strike each other and the walls more often, increasing pressure • If you give the gas particles more space, they will hit the walls less often, pressure will decrease • Weather balloons

  5. Boyle’s Law in Action (Volume-Pressure Equation) • A balloon has a volume of 10.0 L at a pressure of 101 kPa. What will be the new volume when the pressure drops to 43.0 kPa? • P1V1 = P2V2 • 101 x 10.0 = 43.0 x V2 • 1010 = 43.0 x V2 43.0 43.0 • 23.5 L = V2

  6. The Pressure-Temperature Relationship • Why do you need to keep pressurized spray canisters away from heat? • Hotter temp. = faster moving particles (more collisions with the walls) • Volume can’t be increased (rigid canister) • Pressure increases • Canister will explode

  7. Charles’s Law • Gases expand when they are heated (hot air balloons) • Hot air is less dense than cool air • The volume of gas increases with increasing temp. (also, the volume of gas decreases with decreasing temp) • Gas is heated  particles move faster  particles strike the walls of their container more often and with more force  Larger volume

  8. Using Charles’s Law • Temperature must be in Kelvin • V1 = V2 T1 T2 • What would be the resulting volume of a 2.0 L balloon at 25 C that was placed in a container with ice water at 3 C ? • T1 = 25 C + 273 = 298 K • T2 = 3 C + 273 = 276 K • 2 = V2 298 276 • 2 x 276 = 298 x V2 • 552 = 298 x V2 • 1.9 L = V2

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