1. I. Behavior of Gases (Read p. 66-69) Topic 4- Gases S.Panzarella

2. Review of Characteristics of Gases b Gases expand to fill any container. random motion, no attraction b Gases have very low densities. no volume = lots of empty space S.Panzarella

3. b Gases can be compressed. no volume = lots of empty space b Gases undergo diffusion & effusion. random motion; entropy S.Panzarella

4. Ever see a fast food commercial. Don't those burgers look great. Then you get one, and you are disappointed because it is not the perfect one on the TV. So you lie to yourself. No matter how many burgers you order in your life, no one will be as perfect at the one on TV. But if you think it is perfect, who really cares. -VS- Ideal gas are like the commercials and real gases are what you get. Nothing is ever perfect, but close is good enough. S.Panzarella

5. I. Kinetic Molecular Theory – I. Kinetic Molecular Theory – (under ideal circumstances) - explains how an “ideal” gas acts A. Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight-line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature (KE = ½ MV 2 ) S.Panzarella

6. B. Gas behavior is most ideal…  at low pressures  at high temperatures  in nonpolar atoms/molecules of low molecular mass ie. H 2 and He  think about the conditions for a beach vacation (high temps and low pressure) C. Real Gases Particles…. b Real gases deviate from an ideal gas at low temperatures and high pressures. Have their own volume (due to increased pressure) Attract each other (due to increased temps S.Panzarella

7. HomeworkHomework b Read page 66, 68-69 b pg 4-5 of guide, #’s 1-11 S.Panzarella

8. CHEM DO review – You try 5. Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas? (1)He(g) (2) NH3(g) (3) Cl2(g) (4) CO2(g) 6. Which gas has properties that are most similar to those of an ideal gas? (1) N2 (2) O2 (3) He (4) Xe 7. One reason that a real gas deviates from an ideal gas is that the molecules of the real gas have (1) a straight-line motion (2) no net loss of energy on collision (3) a negligible volume (4) forces of attraction for each other 1. Which gas will most closely resemble an ideal gas at STP? 1) SO2 2) NH3 3) Cl2 4) H2 2. At STP, which gas would most likely behave as an ideal gas? (1) H2 (2) CO2(3) Cl2(4) SO2 3. Which gas has properties that are most similar to those of an ideal gas? (1) O2 (2) H2(3) NH3(4) HCl 4. Under which conditions does a real gas behave most like an ideal gas? (1) at high temperatures and low pressures (2) at high temperatures and high pressures (3) at low temperatures and low pressures (4) at low temperatures and high pressures

9. Chem Review Do Answers 1. 4 2. 1 3. 2 4. 1 5. 1 6. 3 7. 4 S.Panzarella

10. III. The Gas Laws b Shows the relationships between TEMPERATURE, PRESSURE, VOLUME and number of moles of a gas b Used to determine what effect changing one of those variables will have on any of the others. S.Panzarella

11. Normal conditions…… Standard Temperature & Pressure - STP Standard Temperature & Pressure - *STP 273 K (or 0◦C) and 101.3 kPa or 1 ATM Temperature MUST be (KELVIN) when working with gases. K = ºC + 273 S.Panzarella * Found in Reference Table A 1 Atm = 760 torr =760 mmHg = 101.3 kPa = 14.7 lb/in 2

12. Bill Nye Video S.Panzarella

13. 1. Combined Gas law b Combines 3 gas laws b # moles are held constant b P, T and V change b Formula: found on Table T P1V1T1P1V1T1 = P2V2T2P2V2T2 S.Panzarella

14. Combined Gas Law – Table T P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1 S.Panzarella

15. GIVEN: V 1 = 2.00 L P 1 = 80.0 kPa T 1 = 300 K V 2 = 1.00 L P 2 = 240. kPa T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (80.0 kPa)(2.00 L)(T 2 ) = (240.kPa)(1.00 L)(300 K) T 2 = 450 K COMBINED Gas Law Problem #1 b 2.00 L sample of gas at 300. K and a pressure of 80.0 kPa is placed into a 1.00 L container at a pressure of 240. kPa. What is the new temperature of the gas? S.Panzarella

16. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V 2 = ? P 2 = 101.3 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) = (101.3 kPa) V 2 (298 K) V 2 = 5.09 cm 3 COMBINED Gas Law Problem #2 b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. S.Panzarella

17. 2. Boyle’s Law b Video: Let’s watch (1:01): http://www.youtube.com/watch?feature=player _embedded&v=DcnuQoEy6wA Video: Let’s watch (1:01): http://www.youtube.com/watch?feature=player _embedded&v=DcnuQoEy6wA b Pressure vs. Volume (Constant Temperature): (think: squeezing a balloon) b As pressure is INCREASED, volume is DECREASED S.Panzarella

18. Boyle’s Law con’t.  INVERSE relationship:  PV = K  formula P 1 V 1 = P 2 V 2 P V S.Panzarella

19. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 BOYLE’S Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL S.Panzarella

20. 3. Charles’ Law Let’s Watch Video (33 sec) : http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded b Volume vs. Temperature (Constant Pressure) (think: hot air balloon) b As temperature is INCREASED, volume is INCREASED As the temperature of the water increases, the volume of the balloon increases. S.Panzarella

21. Charles’ Law con’t. b DIRECT relationship: b V/T = K b Formula V 1 T 2 = V 2 T 1 V T S.Panzarella

22. GIVEN: V 1 = 5.00 L T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 CHARLES’ Law Problems CHARLES’ Law Problems b A gas occupies 5.00L at 36°C. Find its volume at 94°C. TT VV (5.00 L)(367 K)=V 2 (309 K) V 2 = 5.94 L S.Panzarella

23. 4. Gay-Lussac’s Law b Temperature vs. Pressure (Constant Volume)- (think: car tires or pressure cooker) b As temperature is INCREASED, pressure is INCREASED S.Panzarella

24. Gay-Lussac’s Law con’t b DIRECT relationship: b P/T = K b Formula: P 1 T 2 = P 2 T 1 P T S.Panzarella

25. GIVEN: P 1 = 1.00 atm T 1 = 200 K P 2 = ? T 2 = 800 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 GAY-LUSSAC’S Law Problem b A 10.0 L sample of gas in a rigid container at 1.00 atm and 200. K is heated to 800. K. Assuming that the volume remains constant, what is the new pressure of the gas? PP TT (1.00 atm) (800 K) = P 2 (200 K) P 2 = 4.00 atm S.Panzarella

26. HomeworkHomework b See guide pg 5-6 Part A (use your RB) Part B #12-21 Quiz on Thursday S.Panzarella

27. 2 more Gas laws (TEXT BOOK p. 350-353) Read these pages first! S.Panzarella

28. 5. Graham’s Law b States that the rate of effusion (diffusion) of a gas is inversely proportional to the square root of the gas’s molar mass. b Helium effuses (and diffuses) nearly three times faster than nitrogen at the same temperature. S.BarryS.Panzarella

29. Graham’s Law cont. b Gases of SMALL MASS (or DENSITY) diffuse faster than gases of higher molar mass. b ex. At STP, which gas will diffuse more rapidly? Use Table S a) He b) Ar c) Kr d) Xe S.Barry

30. 6. Dalton’s Law b At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. b Formula b Three gases are combined in container T P total = P 1 + P 2 + P 3... S.Panzarella

31. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa Dalton’s Law example #1 b Hydrogen gas is collected over water at 22.5°C and 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. S.Panzarella

32. GIVEN: P O2 = ? P N2 = 79 kPa P CO2 = 0.034 kPa P others = 0.95 kPa P total = 101.3 kPa WORK: P total = P O2 + P N2 + P CO2 + P oth 101.3 kPa = P O2 + 79 kPa + 0.034 kPa + 0.95 kPa 101.3 kPa = P O2 + 79.984 kPa 21.316 kPa = P O2 b Ex. Air contains oxygen, nitrogen, and carbon dioxide and other gases. What is the partial pressure of O 2 at 101.3 kPa of pressure if the P N2 = 79 kPa, the P CO2 = 0.034 kPa and P others = 0.95 kPa? Dalton’s Law example #2 S.Panzarella

33. III. Avogadro’s Hypothesis b Definition: Equal volume of gases at the same temperature and pressure contain equal numbers of particles (molecules) regardless of their mass Let’s watch (2:05) http://www.youtube.com/watch?v=fexEvn0ZOpo&feature=player_embedded S.Panzarella

34. 1) Which rigid cylinder contains the same number of gas molecules at STP as a 2.0-liter rigid cylinder containing H 2(g) at STP? (a)1.0-L cylinder of O 2(g) (b)2.0-L cylinder of CH 4(g) (c)1.5-L cylinder of NH 3(g) (d)4.0-L cylinder of He (g) 2) Which two samples of gas at STP contain the same total number of molecules? (a) 1 L of CO (g) and 0.5 L of N 2(g) (a) 1 L of CO (g) and 0.5 L of N 2(g) (b) 2 L of CO (g) and 0.5 L of NH 3(g) (b) 2 L of CO (g) and 0.5 L of NH 3(g) (c) 1 L of H 2(g) and 2 L of Cl 2(g) (c) 1 L of H 2(g) and 2 L of Cl 2(g) (d) 2 L of H 2(g) and 2 L of Cl 2(g) (d) 2 L of H 2(g) and 2 L of Cl 2(g) 3) At the same temperature and pressure, which sample contains the same number of moles of particles as 1 liter of O 2 (g)? (a)1 L Ne(g) (c) 0.5 L SO 2 (g) (b)(b) 2 L N 2( g) (d) 1 L H 2 O( ℓ ) You Try

35. B. Molar volume: b The number of molecules in 22.4 L of any gas at STP has been chosen as a standard unit called 1 mole b 1 mole = 22.4 L of any gas at STP contains 6.02x10 23 particles b 22.4 L of any gas at STP is = to its molecular mass S.Panzarella

36. Molar volume cont. b Density = molecular mass of gas volume (22.4 L) b ex. What is the density of 1 mole of oxygen gas? b ex. Which gas has a density of 1.70 g/L at STP? a) F2 b) He c) N2 d) SO2 YouTry S.Panzarella

37. IV. Ideal Gas Law PV=nRT Recall - Avogadro’s law, which is derived from this basic idea, says that the volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas Video: http://education-portal.com/academy/lesson/the-ideal-gas-law-and-the- gas-constant.html#lessonhttp://education-portal.com/academy/lesson/the-ideal-gas-law-and-the- gas-constant.html#lesson S.Panzarella

38. PV T VnVn PV nT Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lesson http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lessonhttp://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lesson Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lesson http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lessonhttp://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lesson = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R You don’t need to memorize these values! Merge the Combined Gas Law with Avogadro’s Principle: n is the number of moles of the gas S.Panzarella

39. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law Problem #1 b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. S.Panzarella

40. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K Ideal Gas Law Problem #2 b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3 S.Panzarella