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# Chapter 10.3 : The Gas Laws

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### Chapter 10.3 : The Gas Laws

1. 1. The Gas lawsChapter 10.3<br />Objectives:<br />Use the kinetic-molecular theory to explain the relationships between gas volume, temperature, and pressure.<br />Use Boyle’s law to calculate volume-pressure changes at constant temperature.<br />Use Charles’s law to calculate volume-temperature changes at constant pressure.<br />Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.<br />Use the combined gas law to calculate volume-temperature-pressure changes.<br />Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.<br />
2. 2. The Gas Laws<br />Simple mathematical relationships between<br />Volume<br />Temperature<br />Pressure<br />Amount of gas<br />Gas Law Program: <br />Shows the relationship of all four of the above on gases<br />http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm<br />Constant :<br />Volume and amount of gas<br />Shows:<br />Change in pressure and temperature<br />
3. 3. Boyle’s Law<br />States – volume of a fixed mass of gas varies inversely with the pressure at constant temperature.<br />Constant temperature and amount of gas<br />If you double volume, pressure is cut in half<br />If you cut volume in half, pressure doubles<br />
4. 4. Boyle’s Law<br />Pressure caused by<br /> Moving molecules hitting container walls<br />Speed of particles (force) and number of collisions<br />Both increase pressure<br />Mathematically:<br />Volume-Pressure Data for Gas Sample<br />1<br />Volume Pressure P x V<br /> (mL) (atm) <br />V<br />k<br />=<br />PV<br />or<br />=<br />k<br />P<br />k is constant <br /> 1200 0.5 600 <br />P is pressure <br /> 600 1.0 600 <br />V is volume <br /> 300 2.0 600 <br /> 200 3.0 600 <br /> 150 4.0 600 <br /> 120 5.0 600 <br /> 100 6.0 600 <br />Interactive graph<br />
5. 5. Boyle’s Law<br />Boyle’s Law Equation:<br />P1V1<br />=<br />k<br />P2V2<br />=<br />k<br />So:<br />P1V1<br />=<br />P2V2<br />Sample Problem 1<br />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.<br />P1<br />=<br />0.947 atm<br />P1V1<br />=<br />P2V2<br />V1<br />=<br />150. mL<br />(0.947 atm)<br />(150. mL)<br />P2<br />P2<br />V2<br />=<br />P2<br />=<br />0.987 atm<br />0.987 atm<br />P1V1<br />V2<br />=<br />?<br />=<br />V2<br />=<br />144 mL<br />P2<br />
6. 6. Charles’s Law<br />Volume-temperature Relationship<br /> When using temperature, Must use absolute zero, Kelvin Temperature scale<br />Temperature -273.15oC is absolute zero<br />All molecular movement would stop.<br />K = oC + 273<br />212<br />100<br />373<br />So : How many Kelvin is 10 oC?<br />K = oC + 273<br /> = 10oC + 273<br />32<br />0<br />273<br /> = 283 K<br />Fahrenheit<br />Celsius<br />Kelvin<br />Temperature Scales<br />
7. 7. Charles’s Law<br />Volume-Temperature Data for Gas Sample<br />Temperature Kelvin Volume V/T <br /> (oC) (K) (mL) (mL/K)<br /> 273 546 1092 2 <br /> 100 373 746 2 <br /> 10 283 566 2 <br /> 1 274 548 2 <br /> 0 273 546 2 <br /> -1 272 544 2 <br /> -73 200 400 2 <br /> -173 100 200 2 <br /> -223 50 100 2 <br />
8. 8. Charles’s Law<br />States that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.<br /> Constant Pressure and amount of gas<br />If you double temperature, the volume will double<br />If you cut the temperature in half, the volume will be half as much.<br />V<br />V<br />k<br />=<br />or<br />kT<br />=<br />T<br />k is constant <br />T is temperature <br />V is volume <br />Charles’s Law Equation:<br />V1<br />V2<br />=<br />k<br />=<br />k<br />T1<br />T2<br />So:<br />V2<br />V1<br />=<br />T2<br />T1<br />
9. 9. Charles’s Law<br />Sample Problem 2<br />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?<br />V2<br />V1<br />V1<br />=<br />752 mL<br />T2<br />=<br />T2<br />x<br />x<br />T2<br />T1<br />T1<br />=<br />25oC <br />+ 273 =<br />298 K <br />V2<br />=<br />?<br />V1<br />T2<br />(752 mL)<br />(323 K)<br />T2<br />=<br />50oC <br />+ 273 =<br />323 K <br />V2<br />=<br />=<br />T1<br />298 K<br />**Always convert to Kelvin!!<br />=<br />815 mL<br />
10. 10. Gay-Lussac’s Law<br />Pressure-temperature Relationship<br /> Increasing temperature, increases the speed of the gas particles<br />Thus, more collisions with the container walls<br />Causing an increase in pressure<br />
11. 11. Gay-Lussac’s Law<br />States that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.<br /> Constant volume and amount of gas<br />If you double temperature, pressure doubles<br />If you cut temperature in half, pressure is also cut in half<br />P<br />P<br />k<br />=<br />or<br />kT<br />=<br />T<br />k is constant <br />T is temperature <br />P is pressure <br />P1<br />P2<br />=<br />k<br />=<br />k<br />T1<br />T2<br />So:<br />P2<br />P1<br />=<br />T1<br />T2<br />
12. 12. Gay-Lussac’s Law<br />Sample Problem 3<br />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?<br />P1<br />P2<br />P1<br />=<br />3.00 atm<br />T2<br />=<br />T2<br />x<br />x<br />T2<br />T1<br />T1<br />=<br />25oC <br />+ 273 =<br />298 K <br />P2<br />=<br />?<br />P1<br />T2<br />(3.00 atm)<br />(325 K)<br />T2<br />=<br />52oC <br />+ 273 =<br />325 K <br />P2<br />=<br />=<br />T1<br />298 K<br />**Always convert to Kelvin!!<br />=<br />3.27 atm<br />
13. 13. The Combined Gas Law<br />Sample Problem 4<br />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?<br />P1V1<br />P2V2<br />P1<br />1.08 atm<br />=<br />T2<br />=<br />x<br />T2<br />x<br />T2<br />T1<br />T1<br />=<br />25oC <br />+ 273 =<br />298 K <br />V1<br />=<br />50.0 L<br />P1V1<br />T2<br />P2V2<br />=<br />P2<br />=<br />0.855 atm<br />T1<br />P2<br />P2<br />T2<br />=<br />10.oC <br />+ 273 =<br />283 K <br />P1V1<br />T2<br />V2<br />=<br />?<br />V2<br />=<br />T1<br />P2<br />(1.08 atm)<br />(283 K)<br />(50.0 L)<br />=<br />(298 K)<br />(0.855 atm)<br />60.0 L<br />=<br />
14. 14. Dalton’s Law of Partial Pressures<br />States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.<br />Partial pressure: pressure of each gas in a mixture<br />PT = p1 + p2 + p3 + ……<br />PT= Total Pressure<br />p1 + p2 + p3 = partial pressures<br />
15. 15. Dalton’s Law of Partial Pressures<br />Gas collected by water displacement.<br />Must include the pressure exerted by water vapor<br />PT = p1 + p2 + p3 + ……<br />So: Patm = pgas + pH2O<br />
16. 16. Dalton’s Law of Partial Pressures<br />Sample Problem 5<br />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?<br />Patm = pO2 + pH2O<br />Patm = 731.0 torr<br />PO2 = ?<br />PH2O = 17.5 torr (from appendix in table A-8, pg. 899)<br />pO2 = Patm - pH2O<br />pO2 = 731.0 torr – 17.5 torr <br />= 713.5 torr<br />