1. Chapter 14TEST Thursday May 17th,2012; there will also be a Partner lab identification component (w/ some problems). Show your work, box your answers for credit Some problems require you to know that the following form diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. Do The practice problems on the worksheets and review in your book

2. Labs: - Charles Law Graph -Molar mass of Unknown Gas -Confirm molar volume of H2 gas -Collect data on properties/reactivity of selected gases (2 days to collect data) Assignments: Mixed Gas Law Worksheet - Vapor pressure worksheet

3. Gas laws Charles Law relationship

4. Gas laws – volume and temperature Temperature review Temperature is a measure of the average kinetic energy in a substance Substances at the same temperature have the same average kinetic energy As substances heat up, their particles move faster; particles slow down when cooled

5. Most substances expand when heated and contract when cooled Thermometers contain a liquid which expands and contracts (apx) linearly Temperature scales 1.Fahrenheit: water freezes @ 32 o F, boils @ 212 o F this scale was invented by Gabriel Fahrenheit, who was a chemist and glassblower – he made the best early thermometers (ca 1714) 2.Celsius: water freezes @ 0 o C, boils @ 100 o C 3.Kelvin: water freezes @ 273 o K, boils @ 373 o K :same magnitude as Celsius scale

6. Absolute zero All molecular motion stops -459 o F, -273 o C, 0 o K Gas volume and temperature What happens when a balloon is taken out into the cold? Gases linearly expand when heated and contract when cooled

8. If you take a balloon with a volume of 5570 mL at 21.0 o C outside on a day when the temp is -25.0 o C. Calculate new volume. Convert C to K T 1 = 21 + 273 = 294 K T 2 = -25 + 273 = 248 K V 2 = (V 1 )(T 2 ) T 1 = (5570mL)(248 K) = 4698 mL 294 K

9. Charles Law V 2 = V 1 T 2 T 1 T 2 xV 2 = V 1 x T 2 T 2 T 1 V 2 = V 1 x T 2 T 1

10. An unopened bag of potato chips initially at 19.0 C has a volume of 259 mL. If left in a closed car during the summer and the temp rises to 58.0 C, what would be its new volume? V 2 = (V 1 )(T 2 )/(T 1 ) = (259 mL)(58 + 273)K / (19 + 273)K = 294 mL

11. I make a thermometer which measures temperature by compressing and expanding gas in a piston. At 100 °C the volume of the piston is 20.00 L. If I take the piston outside and the volume is 15.48 L, which activity would be most appropriate? a) Swimming at the beach b) rock climbing c) snow skiing

13. T 2 = V 2 x T 1 V 1 = 15.48 L x 373 K 20.00 L T 2 = 288.7 K 288.7 K - 273 K = 15.7 C 15.7C x 1.8F/C + 32 o F = 60 o F (let’s go rock climbing; too cols for swimming, too hot for snow skiing)

14. Grahams law of Gas diffusion Lighter molecules (at the same temperature) move faster (K.E. = ½ mv 2 ) The faster one is (the square root of the ratio of the heavier divided by the lighter mass) times faster ( page 436 of text)

15. Combined gas law and ideal gas law Boyles,Charles, ‘the other one; aka Gay-Lussac’s”

16. Pressure and temperature As T ↑ (increases), P ↑ (increases) at a constant V Why? Particles are moving faster, hitting the walls of the container with greater force and with greater frequency And as T ↓, P ↓ because particles slow down – hit the wall less often and with less force P 1 = P 2 T 1 T 2.

17. Combined gas law P 1 V 1 = P 2 V 2 T 1 T 2. What would P 2 be equal to? P 2 V 2 = P 1 V 1 T 2 T 1. P 2 V 2 (T 2 ) = P 1 V 1 (T 2 ) T 2 (V 2 ) T 1 (V 2 ) P 2 = P 1 V 1 (T 2 ) T 1 (V 2 )

18. Combined gas law P 1 V 1 = P 2 V 2 T 1 T 2. Boyles law (at constant temperature) P 1 V 1 = P 2 V 2 Charles law (at constant pressure) V 1 = V 2 T 1 T 2 Gay Lusac law ( at constant volume) P 1 = P 2 T 1 T 2

19. A helium balloon is accidentally released. Its initial volume is 3.95 L, the ground temp is 21.0 o C and pressure is 102 kPa. The balloon rises to 9,900 feet where the temp is 5.00 o C and the pressure is 70.4 kPa. Calculate the volume of the balloon. Remember to convert temp to o K: ( o C + 273 = o K)

20. V 2 = P 1 V 1 (T 2 ) T 1 (P 2 ) = 102 kPa x 3.95 L x 278 K 294 K x 70.4 kPa = 5.41 L

21. The deepest spot in Lake Harriet is 82 feet. An air bubble with a volume of 2.35 mL rises from this spot. The pressure down there is 337.4 kPa and the temp is 4.0 o C Calculate its volume at the surface where the pressure is 99.23 kPa and the temp is 21.0 o C.

22. V 2 = P 1 V 1 (T 2 ) T 1 (P 2 ) = 337.4 kPa x 2.35 mL x 294 K 277 K x 99.23 kPa = 8.48 mL

23. Ideal gas law 1 mole of any gas has the same volume at the same P and T In other words, the same number of molecules of any gas take up the same amount of space under the same conditions 1 mole of any gas at STP is 22.4 liters  PV=nRTn is number of moles R is a constant

24. What is the value of “R” P 1 V 1 = P 2 V 2 = nR T 1 T 2. (the number of moles times some constant); select the units of pressure and volume you like and calculate “R” R = PV/nT R= 760 mmHg (22.4 liters)/1 mole(273K) R = 62.36 mmHg L / mol. K

25. What is the value of “R” P 1 V 1 = P 2 V 2 = nR T 1 T 2. (the number of moles times some constant); select the units of pressure and volume you like and calculate “R” R = PV/nT R= 101.3 kPa (22.4 liters)/1 mole(273K) R = 8.31 kPa L / mol. K

26. What is the value of “R” P 1 V 1 = P 2 V 2 = nR T 1 T 2. (the number of moles times some constant); select the units of pressure and volume you like and calculate “R” R = PV/nT R= 1.0 atmosphere (22.4 liters)/1 mole(273K) R = 0.082 atm. L / mol. K

27. What is the value of “R” R = 62.36 mmHg L / mol. K R = 8.31 kPa L / mol. K R = 0.082 atm. L / mol. K Or any other number and unit combination that equals this value