1. Gases & Liquids Ch.12

3. (12-1) Properties of Gases Fluids Low density Compressible Fill a container & exert P equally in all directions Influenced by T

5. Kinetic-Molecular Theory Explains behavior of gases 2 major assumptions: 1. Collisions are elastic 2. V of individual gas molecules is negligible KE is lost KE is maintained

6. Ideal Gas Describes the behavior of gases under most conditions High T & low P gases act more ideal

7. Kinetic Energy T of a gas determines the avg. KE of its particles KE = ½ mv 2 –Where, m = mass, v = speed

8. Pressure Pressure (P) = force (F) area (A) SI units: –F: newtons (N) –P: pascal (Pa) = 1 N/m 2

9. Standard Temp. & Pressure STP: std. conditions for a gas Temp. (T): 0 °C (273 K) Pressure (P): 1 atm (101.325 kPa) –See Table 12-1, p.428 for more P units

10. Greenhouse Effect Inc. in the T of Earth caused by reflected solar radiation that’s trapped in the atmosphere Inc. in greenhouse gases such as CO 2 & CFCs

12. Free Radical Atom or molecule that has 1 or more unpaired e - & is very reactive –UV radiation breaks apart CFCs, making Cl Chain rxn: self-sustaining rxn in which the product from 1 step acts as a reactant for the next step –Cl + O 3  ClO + O 2 + O –ClO + O  Cl + O 2

14. (12-2) The Gas Laws Symbols: –P = pressure –T = temp in K –V = volume –n = # of moles

15. Boyle’s Law At constant T: –Inc. P, dec. V –Dec. P, inc. V P 1 V 1 = P 2 V 2

16. Boyle’s Law Practice If the P exerted on a 300 mL sample of H 2 gas at constant T is inc. from 0.500 atm to 0.750 atm, what will be the final V of the sample? 1.List known V 1 = 300 mLP 1 = 0.500 atm V 2 = ? mLP 2 = 0.750 atm

17. Boyle’s Law Practice 2.Write eq. P 1 V 1 = P 2 V 2  V 2 = P 1 V 1 P 2 3. Substitute & solve V 2 = (0.500 atm)(300 mL) = 200 mL 0.750 atm

18. Dalton’s Law of Partial P’s Total P in a gas mixture is the sum of the partial P’s of the individual components P total = P A + P B + P C… –Where, P total = total P, P A = partial P of A

19. Dalton’s Law Practice A mixture of O 2, N 2, & H 2 gases exerts a total P of 278 kPa. If the partial P’s of O 2 & H 2 are 112 kPa & 101 kPa respectively, what would be the partial P of the N 2 ? 1.List the known P total = 278 kPaP N 2 = ? kPa P O 2 = 112 kPaP H 2 = 101 kPa

20. Dalton’s Law Practice 2.List the eq. & rearrange P total = P O 2 + P H 2 + P N 2 P N 2 = P total - P O 2 - P H 2 3.Substitute & solve P N 2 = 278 kPa – 112 kPa – 101 kPa = 65 kPa

21. Mole Fraction # of moles of 1 component compared w/ the total # of moles in the mixture Mole fraction (X) = ____mol A___ total mols To calc. partial P: P A = P T X A

22. Mole Fraction Example The total P of a mixture of gases is 0.97 atm. The mole fraction for N 2 is 0.78. What’s the partial P of N 2 ? 1.List the known P total = 0.97 atmX N 2 = 0.78

23. Mole Fraction Example 2.Write the eq. P N 2 = P total X N 2 3.Substitute & solve P N 2 = (0.97 atm)(0.78) = 0.76 atm

24. Charles’ Law At constant P: –V inc., T inc. –V dec., T dec. V 1 = V 2 T 1 T 2

25. Charles’ Law Practice Gas in a balloon occupies 2.5 L at 300 K. The balloon is dipped into liquid N 2 at 80 K. What V will the gas in the balloon occupy at this T? 1.List known V 1 = 2.5 LT 1 = 300 K V 2 = ? LT 2 = 80 K

26. Charles’ Law Practice 2. Write eq. V 1 = V 2  V 2 = V 1 T 2 T 1 T 2 T 1 3. Substitute & solve V 2 = (2.5 L)(80 K) = 0.67 L (300 K)

27. Pressure & Temp. P inc. w/ inc. in T at constant V P 1 = P 2 T 1 T 2

28. P & T Practice Gas in a sealed can has a P of 3.00 atm at 25°C. A warning says not to store the can in a place where the T will exceed 52°C. What would the gas P in the can be at 52°C? 1.List known P 1 = 3.00 atmT 1 = 25°C = 298 K P 2 = ? atmT 2 = 52°C = 325 K

29. P & T Practice 2. Write eq. P 1 = P 2  P 2 = P 1 T 2 T 1 T 2 T 1 3. Substitute & solve P 2 = (3.00 atm)(325 K) = 3.27 atm (298 K)

30. Avogadro’s Law V’s of different gases under the same T & P’s have the same # of molecules V 1 = V 2 n 1 n 2

31. Gay-Lussac’s Law Law of Combining Volumes: at constant T & P, gases react in whole # V proportions H 2 + Cl 2  2 HCl 1 V + 1 V  2 V

32. Effusion Motion of a gas through a small opening Diffusion: gas particles disperse from areas of high to low conc.

33. Graham’s Law of Effusion At the same T & P, 2 gases rates of effusion can be measured by: ½ M A v A 2 = ½ M B v B 2 or v A = M B v B M A Where: –v = speed of effusion (2 gases, A & B) –M = molar mass

34. Graham’s Law Practice O 2 has an avg. speed of 480 m/s at room T. On avg., how fast is SO 3 traveling at the same T? 1.List known v O 2 = 480 m/sM O 2 = 32 g/mol v SO 3 = ? m/sM SO 3 = 80.07 g/mol

35. Graham’s Law Practice 2.Write eq. v SO 3 = M O 2  v SO 3 = v O 2 M O 2 v O 2 M SO 3 M SO 3 3. Substitute & solve v SO 3 = (480 m/s) (32 g/mol) (80.07 g/mol) = 300 m/s

36. More Graham’s Practice Compare the rate of effusion of H 2 O vapor w/ O 2 gas at the same T & P. 1. List known M H 2 O = 18.02 g/mol M O 2 = 32 g/mol

37. More Graham’s Practice 2. Write eq. v H 2 O = M O 2 v O 2 M H 2 O 3. Substitute & solve v H 2 O = 32 g/mol = 1.33 v O 2 18.02g/mol H 2 O effuses 1.33X faster than O 2

38. (12-3) Ideal Gas Law PV = nRT Where: –R = 8.314 LkPa / molKor –R = 0.0821 Latm / molK

39. Ideal Gas Law Practice Calculate the V of 1.00 mol of CO 2 gas at STP. 1.List known P = 1.00 atmV = ? L n = 1.00 mol R = 0.0821 Latm/molK T = 273 K

40. Ideal Gas Law Practice 2.Write eq. PV = nRT  V = nRT P 3.Substitute & solve V = (1.00 mol)(0.0821 Latm/molK)(273 K) (1.00 atm) = 22.4 L

41. Combined Gas Law Moles remain constant, but other conditions change P 1 V 1 = P 2 V 2 T 1 T 2

42. Combined Gas Law Practice A sample of CO 2 gas occupies 45 L at 750 K & 500 kPa. What’s the V of this gas at STP? 1.List known P 1 = 500 kPaP 2 = 101.325 kPa V 1 = 45 LV 2 = ? L T 1 = 750 KT 2 = 273 K

43. Combined Gas Law Practice 2.Write eq. P 1 V 1 = P 2 V 2  V 2 = P 1 V 1 T 2 T 1 T 2 T 1 P 2 3. Substitute & solve V 2 = (500 kPa)(45 L)(273 K) (750 K)(101.325 kPa) = 81 L

44. Gas Stoichiometry Gas V’s can be determined from mole ratios in bal. eqs. 3H 2 + N 2  2NH 3 3 L 1 L2 L 22 L N 2 x 3 L H 2 = 66 L H 2 1 L N 2

45. (12-4) Changes of State Evaporation: l  g Condensation: g  l Sublimation: s  g

46. Vapor Pressure P exerted by a vapor in equilibrium w/ its liquid state at a given T –H 2 O(l) H 2 O(g)

47. Phase Diagrams Shows T’s & P’s at which a substance exists in different phases Phases are at equilibrium along the lines Phase: substance has uniform composition & properties

49. Phase Diagrams (cont.) Normal bp: boiling T at 1 atm Critical point: T & P above which the properties of vapor can’t be distinguished from a liquid –Supercritical fluids Triple point: T & P where 3 phases exist in equil.

50. C.P. N.B.P.N.M.P.