1. Kinetic Molecular Theory High kinetic energy. Move rapidly in straight lines. Are very far apart. Have essentially no attractive (or repulsive) forces. Have very small volumes compared to the volume of the container they occupy. Can be easily compressed. 1

2. Atmospheres mmHgmmHg psi Grams molesmoles mL L cc ˚C K ˚F Temperature Volume PressureAmount Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). 2

3. Atmospheric pressure Is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Barometer Barometer, for measuring atmospheric pressure. Evangelista Torricelli Italian physicist 1 atm = 760 mm Hg (exact) 1 atm = 760 torr 1 atm = 14.7 lb./in. 2 1 atm = 101.325 kPa

4. Pressure and Volume Compressibility = ability to be squeezed into a smaller volume by the application of pressure. Liquids and solids are relatively incompressible. Robert Boyle-1661 Robert Boyle-1661= quantitative study of gas compressibility. Used mercury and a J shaped tube. 4

5. Pressure (atm) Volume (L) 0.2500 2.801 0.5000 1.400 0.7500 0.9333 1.000 0.6998 2.000 0.3495 3.000 0.2328 P increases V decreases Data: Conclusion: volume of a gas is inversely proportional to the applied pressure. Math follows: P 1 V 1 = P 2 V 2 final conditions P 2 V 2 = constant If there is a change: initial conditions 5 P 1 V 1 = constant

6. Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.c Boyle’s Law  6

7. Boyle’s Law To inhale: The rib cage expands and the diaphragm lowers. This increases the volume of the lungs. Increasing the volume causes the pressure to decrease. Air is drawn into the lungs to equalize the pressure. To exhale: The rib cage contracts and the diaphragm is raised. This decreases the volume of the lungs. Decreasing the volume causes the pressure to increase. Air is expelled out of the lungs to equalize the pressure. 7

8. Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of an 8.0 L sample of Freon gas after its pressure is changed from 550 mmHg to 2200 mmHg at constant T? STEP 1 Set up a data table: Conditions 1Conditions 2 P 1 = 550 mmHgP 2 = 2200 mmHg V 1 = 8.0 LV 2 = ? STEP 2 Solve Boyle’s law for V 2. P 1 V 1 = P 2 V 2 V 2 = V 1 x P 1 P 2 STEP 3 Set up problem V 2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg Example: Boyle’s Law 8

9. Quantitative observations of gases at different temperatures. Jacques Alexandre Charles (1787) French physicist 9

10. y = m x + b V = m ˚C + b Data shows that @ T = -273.15 ˚C, the V is 0. 0 = m ( -273.15 ) + b Therefore: b = m (273.15) Substituting: V = m ˚C + m (273.15 ) Rearrange: V = m(˚C + 273.15) Gives: V = m (K) 10 T = -273.15 ˚C

11. Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.c  Charles’ Law 11

12. 12

13. = Combine 13

14. Ideal Gas Equation R = 0.0821 L atm mole K R = 62.4 L mmHg mole K 14

15. P 2 V 2 = n 2 RT 2 P 1 V 1 = n 1 RT 1 STEP 2 PV = nRT Summary STEP 1 Organize the data in a table of initial and final conditions. STEP 2 Use PV = nRT to find the needed gas law. STEP 3 Substitute values into the gas law eq. and solve for the variable. A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). STEP 1 1 2 P 1 = 600. mm Hg P 2 = ? V 1 = 12.0 L V 2 = 36.0 L P 2 V 2 = P 1 V 1 P 2 = P 1 x V 1 V 2 STEP 3 P 2 = 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L 15 P 2 V 2 = 1 P 1 V 1 P 2 V 2 = n 2 RT 2 P 1 V 1 = n 1 RT 1

16. STP (Standard Temperature and Pressure) Standard pressure 1 atm or 760 mm Hg Standard temperature 0°C or 273 K A convenient way of looking at gases in chemical reactions. The Molar Volume of gas @ STP = 22 4 Lvolume occupied by 1 mole of any gas 1 mole= 22.4 L What is the volume, in liters, of 64.0 g of O 2 gas at STP? 16

17. The partial pressure of a gas Is the pressure of each gas in a mixture. Is the pressure that gas would exert if it were by itself in the container. Dalton’s Law of Partial Pressures Pressure depends on the total number of gas particles, not on the types of particles. The total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases. P T = P 1 + P 2 + P 3 +..... John Dalton 17 Collecting a gas over water P t = P atm = P gas + P wv P gas = P atm – P wv The pressure of the water vapor depends on the T of the water and is tabulated. Partial Pressure