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The Behavior of Gases

The Behavior of Gases. Chemistry 2013-2014. Variables that Describe a Gas. Compressibility : a measure of how much the volume of matter decreases under pressure.

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The Behavior of Gases

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  1. The Behavior of Gases Chemistry 2013-2014

  2. Variables that Describe a Gas • Compressibility: a measure of how much the volume of matter decreases under pressure. • Pressure: a physical force pushing on or against an object; abbreviated P, measured in atmospheres (atm), torr, mmHg, Pascals (Pa), or kilopascals (kPa). • Standard pressure (STP) is 1 atm = 760 torr = 760 mmHg = 101,300 Pa = 101.3 kPa

  3. Volume: the amount of space an object occupies; abbreviated V, measured in liters, milliliters, cubic meters, or cubic centimeters.

  4. Temperature: a measurement of the average kinetic energy in an object; abbreviated T, measured in Celsius or Kelvin (use Kelvin for math problems). • Standard temperature (STP) = 0°C = 273 K

  5. Mole: a measurement of the number of particles in an object; abbreviated n, measured in moles. One mole is equal to 6.02 x 1023 particles (atoms, molecules, or formula units).

  6. The Gas Laws

  7. Boyle’s Law for Pressure-Volume Changes • For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. • Remember it by we BOYLE Peas and Vegetables P1V1= P2V2 • You do not need to convert pressure or volume to specific units for these problems because they use ratios.

  8. P1V1 = P2V2 P1 = 87.6 kPa V1 = 242 mL P2 = 101.3 kPa V2= ? V2= 87.6 * 242 = 209 mL 101.3 V2= P1V1 P2 A gas is collected in a 242 mL container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at 101.3 kPa? Assume the temperature is constant.

  9. Charles’ Law for Temperature-Volume Changes • 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 be in Kelvin. K = °C + 273 • Remember it by Charles is a VeT

  10. V1 = 2.58 L T1 = 15 + 273 = 288 K V2 = ? T2= 38 + 273 = 311 K V1= V2 T1 T2 V2 = V1 * T2 T1 V2 = 2.58 L * 311 K 288 K = 2.8 L A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm?

  11. The Combined Gas Law • This law combines pressure, volume, and temperature. • Temperature must be in Kelvin. K = °C + 273 • By canceling out any constant terms, we can derive Boyle’s, Charles’, and Gay-Lussac’s law from the combined gas law. • Easy way to remember: “Peas and Vegetables on the Table”

  12. If a helium-filled balloon has a volume of 3.40 L at 25.0ºC and 120.0 kPa, what is its volume at STP? V1 = 3.40 L P1V1 P2V2 T1 T2 = + 273 = 298 K T1 = 25.0 ºC P1 = 120.0 kPa P1V1 T2 T1P2 V2 = V2 = ? T2 = 273 K P2 = 101.3 kPa (120.0 kPa) V2 = (273 K ) (3.40 L) (298 K ) (101.3 kPa) V2 3.69 L =

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