1. Kinetic Molecular Theory

2. What if… b Left a basketball outside in the cold… would the ball appear to be inflated or deflated? b Which picture box do you think will have more pressure?

3. A. Kinetic Molecular Theory

4. Characteristics of Gases b Gases expand to fill any container. random motion, no attraction b Gases are fluids (like liquids). no attraction b Gases have very low densities. no volume = lots of empty space

5. TemperatureTemperature K = ºC + 273 b Temp: measure of avg. KE (speed) b Always use Kelvin when working with gases. Kelvin Is directly proportional to temp. Always positive

6. PressurePressure b Pressure: Amount of force exerted on a container/object b Barometer Measures atmospheric pressure Mercury Barometer

7. STPSTP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

8. Pressure Demo b Experiencing pressure Variables Observations Explanation:

9. PressurePressure b KEY UNITS AT SEA LEVEL 101.325 kPa = 1 atm = 760 mm Hg =760 torr = 14.7 psi kPa: kilopascal Atm: atmosphere mmHg: millimeters Mercury Psi: pounds per square inch How would you find a pascal?

10. PracticePractice b If the atmospheric pressure in Colorado is 525 mmHg, what is the pressure in kPa?

11. Real Gases b Particles in a REAL gas… have their own volume attract each other b Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

12. II. The Gas Laws BOYLES CHARLES GAY- LUSSAC Ch. 12 - Gases

13. Boyles Law Demo b Experiencing pressure Variables Observations Explanation:

14. A. Boyle’s Law P V P 1 V 1 = P 2 V 2

15. A. Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp P V P 1 V 1 = P 2 V 2

16. B. Charles Law Demo b Experiencing pressure Variables Observations Explanation:

17. V T B. Charles’ Law

18. V T b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

19. P T C. Gay-Lussac’s Law

20. P T b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

21. D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

22. GIVEN: V 1 = 473 cm 3 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 E. Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

23. 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 E. Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

24. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 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) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

25. GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

26. III. Ideal Gas Law (p. 334-335, 340-346) Ch. 10 & 11 - Gases

27. V n A. Avogadro’s Principle

28. V n b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

29. PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R

30. B. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

31. 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 B. Ideal Gas Law b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

32. GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K B. Ideal Gas Law 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 IDEAL GAS LAW