1. CHEMISTRY 2013-2014 THE BEHAVIOR OF GASES

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 10 23 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 P eas and V egetables P 1 V 1 = P 2 V 2 You do not need to convert pressure or volume to specific units for these problems because they use ratios.

9. 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. P 1 V 1 = P 2 V 2 P 1 = 87.6 kPa V 1 = 242 mL P 2 = 101.3 kPa V 2 = ? V 2 = 87.6 * 242 = 209 mL 101.3 V 2 = P 1 V 1 P 2

10. 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. V 1 = V 2 T 1 T 2 Temperature must be in Kelvin. K = °C + 273 Remember it by Charles is a V e T

12. 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? V 1 = V 2 T 1 T 2 V 1 = 2.58 LT 1 = 15 + 273 = 288 K V 2 = ? T 2 = 38 + 273 = 311 K V 2 = 2.58 L * 311 K 288 K = 2.8 L V 2 = V 1 * T 2 T 1

13. 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: “ P eas and V egetables on the T able”

14. 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? V 1 = 3.40 L T 1 = 25.0 ºC P 1 = 120.0 kPa T 2 = 273 K P 2 = 101.3 kPa + 273 = 298 K V 2 = ? P 1 V 1 P 2 V 2 T 1 T 2 P 1 V 1 T 2 T 1 P 2 V2V2 = = V2V2 =(120.0 kPa) (3.40 L) (298 K ) (101.3 kPa) (273 K ) V2V2 = 3.69 L