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# Applied Chapter 12.2 : The Gas Laws

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### Transcript of "Applied Chapter 12.2 : The Gas Laws"

1. 1. Chapter 12.2<br />The Gas Laws<br />
2. 2. State Boyle’s law, and use it to solve problems involving pressure and volume.<br />State Charles’s law, and use it to solve problems involving volume and temperature.<br />State Gay-Lussac’s law, and use it to solve problems involving pressure and temperature.<br />State Avogadro’s law, and explain its importance in determining the formulas of chemical compounds.<br />Objectives<br />
3. 3. 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 />
4. 4. 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 />
5. 5. 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 />
6. 6. 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 />
7. 7. 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 />
8. 8. 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 />
9. 9. 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 />
10. 10. 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 />=<br />815 mL<br />**Always convert to Kelvin!!<br />
11. 11. 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 />
12. 12. 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 />
13. 13. 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 />
14. 14. Avogadro’s Law<br />Equal Volumes of gases at the same temperature and pressure contain equal numbers of molecules.<br />How does it work!<br />Hydrogen Gas + Chlorine gas Hydrochloric Acid<br /> H2Cl2 2 HCl<br /> 1 Volume 1 Volume 2 Volumes <br /> 1 Molecule 1 Molecule 2 Molecules <br /> 1 Mol 1 Mol 2 Mol <br />V = kn Direct Relationship<br />n = amount of gas<br />
15. 15. Molar Volume of Gas<br />Recall<br />1 mol = 6.022 x 10 23 molecules<br />1 mol O2 = 6.022 x 10 23 molecules O2 = 32.00 g O2<br />1 mol H2 = 6.022 x 10 23 molecules H2 = 2.02 g H2<br />Standard Molar Volume of gases<br />Volume occupied by 1 mol of a gas at STP<br />At STP, 1 mol of any gas = 22. 4 L<br />
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