1. The Gas lawsChapter 10.3 Objectives: Use the kinetic-molecular theory to explain the relationships between gas volume, temperature, and pressure. Use Boyle’s law to calculate volume-pressure changes at constant temperature. Use Charles’s law to calculate volume-temperature changes at constant pressure. Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume. Use the combined gas law to calculate volume-temperature-pressure changes. Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.
2. The Gas Laws Simple mathematical relationships between Volume Temperature Pressure Amount of gas Gas Law Program: Shows the relationship of all four of the above on gases http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm Constant : Volume and amount of gas Shows: Change in pressure and temperature
3. Boyle’s Law States – volume of a fixed mass of gas varies inversely with the pressure at constant temperature. Constant temperature and amount of gas If you double volume, pressure is cut in half If you cut volume in half, pressure doubles
4. Boyle’s Law Pressure caused by Moving molecules hitting container walls Speed of particles (force) and number of collisions Both increase pressure Mathematically: Volume-Pressure Data for Gas Sample 1 Volume Pressure P x V (mL) (atm) V k = PV or = k P k is constant 1200 0.5 600 P is pressure 600 1.0 600 V is volume 300 2.0 600 200 3.0 600 150 4.0 600 120 5.0 600 100 6.0 600 Interactive graph
5. Boyle’s Law Boyle’s Law Equation: P1V1 = k P2V2 = k So: P1V1 = P2V2 Sample Problem 1 A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant. P1 = 0.947 atm P1V1 = P2V2 V1 = 150. mL (0.947 atm) (150. mL) P2 P2 V2 = P2 = 0.987 atm 0.987 atm P1V1 V2 = ? = V2 = 144 mL P2
6. Charles’s Law Volume-temperature Relationship When using temperature, Must use absolute zero, Kelvin Temperature scale Temperature -273.15oC is absolute zero All molecular movement would stop. K = oC + 273 212 100 373 So : How many Kelvin is 10 oC? K = oC + 273 = 10oC + 273 32 0 273 = 283 K Fahrenheit Celsius Kelvin Temperature Scales
8. Charles’s Law States that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. Constant Pressure and amount of gas If you double temperature, the volume will double If you cut the temperature in half, the volume will be half as much. V V k = or kT = T k is constant T is temperature V is volume Charles’s Law Equation: V1 V2 = k = k T1 T2 So: V2 V1 = T2 T1
9. Charles’s Law Sample Problem 2 A sample of neon gas occupies a volume of 752 mL at 25oC. What volume will the gas occupy at 50oC if the pressure remains constant? V2 V1 V1 = 752 mL T2 = T2 x x T2 T1 T1 = 25oC + 273 = 298 K V2 = ? V1 T2 (752 mL) (323 K) T2 = 50oC + 273 = 323 K V2 = = T1 298 K **Always convert to Kelvin!! = 815 mL
10. Gay-Lussac’s Law Pressure-temperature Relationship Increasing temperature, increases the speed of the gas particles Thus, more collisions with the container walls Causing an increase in pressure
11. Gay-Lussac’s Law States that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature. Constant volume and amount of gas If you double temperature, pressure doubles If you cut temperature in half, pressure is also cut in half P P k = or kT = T k is constant T is temperature P is pressure P1 P2 = k = k T1 T2 So: P2 P1 = T1 T2
12. Gay-Lussac’s Law Sample Problem 3 The gas in an aerosol can is at a pressure of 3.00 atm at 25oC. Directions on the can warn the user not to keep the can in a place where the temperature exceed 52oC. What would the gas pressure in the can be at 52oC? P1 P2 P1 = 3.00 atm T2 = T2 x x T2 T1 T1 = 25oC + 273 = 298 K P2 = ? P1 T2 (3.00 atm) (325 K) T2 = 52oC + 273 = 325 K P2 = = T1 298 K **Always convert to Kelvin!! = 3.27 atm
13. The Combined Gas Law Sample Problem 4 A helium-filled balloon has a volume of 50.0 L at 25oC and 1.08 atm. What volume will it have at 0.855 atm and 10.oC? P1V1 P2V2 P1 1.08 atm = T2 = x T2 x T2 T1 T1 = 25oC + 273 = 298 K V1 = 50.0 L P1V1 T2 P2V2 = P2 = 0.855 atm T1 P2 P2 T2 = 10.oC + 273 = 283 K P1V1 T2 V2 = ? V2 = T1 P2 (1.08 atm) (283 K) (50.0 L) = (298 K) (0.855 atm) 60.0 L =
14. Dalton’s Law of Partial Pressures States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. Partial pressure: pressure of each gas in a mixture PT = p1 + p2 + p3 + …… PT= Total Pressure p1 + p2 + p3 = partial pressures
15. Dalton’s Law of Partial Pressures Gas collected by water displacement. Must include the pressure exerted by water vapor PT = p1 + p2 + p3 + …… So: Patm = pgas + pH2O
16. Dalton’s Law of Partial Pressures Sample Problem 5 Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0oC, respectively. What was the partial pressure of the oxygen collected? Patm = pO2 + pH2O Patm = 731.0 torr PO2 = ? PH2O = 17.5 torr (from appendix in table A-8, pg. 899) pO2 = Patm - pH2O pO2 = 731.0 torr – 17.5 torr = 713.5 torr
