2. 1 © 2006 Brooks/Cole - Thomson General Chemistry Gases and Their Properties

3. KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

4. 3 © 2006 Brooks/Cole - Thomson Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 5.2 Barometer Pressure = Force Area (force = mass x acceleration)

5. 4 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases — and so does speed.

6. 5 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases — and so does speed.

7. 6 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory where u is the speed and M is the molar mass. speed INCREASES with Tspeed INCREASES with T speed DECREASES with Mspeed DECREASES with M Maxwell’s equation

8. 7 © 2006 Brooks/Cole - Thomson Pressure according to Kinetic Molecular Theory Pressure is caused by the constant bombardment of gas molecules against the wall of container. The energy of gas molecules (Temperature)The energy of gas molecules (Temperature) The frequency of gas molecules collideThe frequency of gas molecules collide The frequency depends on Speed (Temperature dependent) and Concentration.The frequency depends on Speed (Temperature dependent) and Concentration. Analogy: Speed bag Analogy: Speed bag The energy of gas molecules (Temperature)The energy of gas molecules (Temperature) The frequency of gas molecules collideThe frequency of gas molecules collide The frequency depends on Speed (Temperature dependent) and Concentration.The frequency depends on Speed (Temperature dependent) and Concentration. Analogy: Speed bag Analogy: Speed bag

9. 8 © 2006 Brooks/Cole - Thomson 5.3 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

10. 9 © 2006 Brooks/Cole - Thomson P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 5.3 Boyle’s Law Constant temperature Constant amount of gas

11. 10 © 2006 Brooks/Cole - Thomson 10 Puffing MushmellowPuffing Mushmellow & Ballon Soda bottle SubmarineBallon Pressure surge when bottle being squeezed  Volume of the air inside the dropper decreases  Water filling up the dropper  Increased density of dropper (Mass dropper /Volume  )  Dropper sinking!

12. 11 © 2006 Brooks/Cole - Thomson P 1 x V 1 = P 2 x V 2 5.3 Boyle’s Law Explained in KMT Pressure depends on speed of gas molecule and frequency of collision At constant temperature, the speed of gas molecule remains the same As the volume decrease, same amount of molecules bombard smaller area of wall, thus higher frequency of collision Eventually, the pressure increases.

13. 12 © 2006 Brooks/Cole - Thomson 12 Information Given:P 1 = 4.0 atmV 1 = 6.0 L P 2 = 1.0 atm Find:V 2 = ? L Example: A cylinder equipped with a moveable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm? Answer: 24 L

14. 13 © 2006 Brooks/Cole - Thomson 5.3 Charles’ Law Explained in KMT Pressure depends on speed of gas molecule and frequency of collision To maintain constant pressure, the speed of gas molecule must increase to compensate the loss of frequency As the volume increases, same amount of molecules bombard larger area of wall, thus lower frequency of collision Eventually, the temperature increases.

15. 14 © 2006 Brooks/Cole - Thomson 5.3 V  TV  TV  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 Charles’ Law at Different Temperatures Temperature must be in Kelvin

16. 15 © 2006 Brooks/Cole - Thomson 15 Information Given:V 1 = 2.80 L V 2 = 2.57 Lt 2 = 0°C Find: temp 1 in K and °C Eq’n: SM: V 1, V 2 T 2 → T 1 Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius? T1 = 297 K = 24°C

17. 16 © 2006 Brooks/Cole - Thomson 16

18. 17 5.3 Gay-Lussac’s Law Explained in KMT Pressure depends on speed of gas molecule and frequency of collision As the temperature increases, same amount of molecules bombard the same area of wall with higher speed, also higher frequency of collision Eventually, the pressure increases. Example: The pressure of tire increases at Temperature increases.

19. 18 © 2006 Brooks/Cole - Thomson 5.3 Avogadro’s Law Explained in KMT Pressure depends on speed of gas molecule and frequency of collision At constant pressure and temperature, as the number of gas particle increases, the frequency of collision increases Eventually, the volume increases. To maintain the pressure constant, the area of collision must increases.

20. 19 © 2006 Brooks/Cole - Thomson 19 Combined Gas Law Boyle’s Law : Pressure and Volume –at constant temperature Charles’ Law : Volume and absolute Temperature –at constant pressure  Volume of a sample of gas when both the Pressure and Temperature change

21. 20 © 2006 Brooks/Cole - Thomson 20 Information Given:V 1 = 158 mL, P 1 = 755 mmHg, t 1 = 34°C V 2 = 108 mL, t 2 = 85°C Find:P 2, mmHg Example: A sample of gas has a volume of 158 mL at a pressure of 755 mmHg and a temperature of 34°C. The gas is compressed to a volume of 108 mL and heated to 85°C, what is the final pressure in mmHg? P 2 = 1.29 x 10 3 mmHg

22. 21 © 2006 Brooks/Cole - Thomson 21 Ideal Gas Law Combined Gas Law + Avgadro’s Law  Ideal Gas Law R is called the Gas Constant the value of R depends on the units of P and V –R = 0.0821 atm · L/K · mol –convert P to atm and V to L

23. 22 © 2006 Brooks/Cole - Thomson 22 Information Given:V = 3.2 L, P = 1.6462 atm, T = 298 K Find:n, mol Eq’n:PV = nRT Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C 0.22 mol

24. 23 © 2006 Brooks/Cole - Thomson 23 Applying Ideal Gas Law: Molar Mass of a GasIdeal Gas Law Determination of the molar mass of an unknown substance: Heat a weighed sample into gaseous state Measure temperature T, pressure P and volume V of the gas Use the Ideal Gas Law: #mole n = PV/RT

25. 24 © 2006 Brooks/Cole - Thomson 24 Information Given:V = 0.225 L, P = 1.1658 atm, t = 328 K, m = 0.311 g Find:molar mass, (g/mol) Eq’n: PV = nRT; MM = mass/moles SM: P,V,T,R → n & mass → mol. mass Example: A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass. n = 0.00974 mol Molar mass = 31.9 g/mol

26. 25 © 2006 Brooks/Cole - Thomson Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P(H 2 O) + P(O 2 ) = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

27. 26 © 2006 Brooks/Cole - Thomson Dalton’s Law John Dalton 1766-1844

28. 27 © 2006 Brooks/Cole - Thomson 27 Zn metal reacts with HCl(aq) to produce H 2 (g). Because water evaporates, some water vapor gets mixed in with the H 2. The gas flows through the tube and bubbles into the jar, where it displaces the water in the jar. Collecting gas over water P gas = P H 2 O + P H 2

29. 28 © 2006 Brooks/Cole - Thomson 28 Reactions Involving Gases Reaction stoichiometry (Chapter 8) can be combined with the Gas Laws for reactions involving gases in reactions of gases, the amount of a gas is often given as a Volume –instead of moles –as we’ve seen, must state pressure and temperature Ideal Gas Law: from Volume of the gas + P & T  Moles; then we can use the coefficients in the equation as a mole ratio

30. 29 © 2006 Brooks/Cole - Thomson 29 Information Given:294 g KClO 3 P O2 = 755 mmHg, T O2 = 308 K Find: V O2, L Eq’n: PV=nRT CF: 1 mole KClO 3 = 122.5 g SM: g → mol KClO 3 → mol O 2 → L O 2 Example: How many liters of O 2 (g) form when 294 g of KClO 3 completely reacts? Assume the O 2 (g) is collected at P = 755 mmHg and T = 308 K 3.60 mole O 2 V = 90.7 L O 2

31. 30 © 2006 Brooks/Cole - Thomson T(ºC) 0 5 10 12 14 16 18 20 22 24 26 28 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0 4.6 6.5 9.2 10.5 12.0 13.6 15.5 17.5 19.8 22.4 25.2 28.3 31.8 42.2 Table 5.2 Vapor Pressure of Water (P ) at Different T H2OH2O P (torr) H2OH2O H2OH2O

32. 31 © 2006 Brooks/Cole - Thomson Collecting a water-insoluble gaseous product and determining its pressure: P total = P gas + P H2O

33. 32 © 2006 Brooks/Cole - Thomson Calculating the Amount of Gas Collected over Water PROBLEM:Acetylene (C 2 H 2 ) is produced in the laboratory when calcium carbide (CaC 2 ) reacts with water: CaC 2 (s) + 2H 2 O (l) → C 2 H 2 (g) + Ca(OH) 2 (aq) A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23ºC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P total = P gas + P H2O P C2H2C2H2 = 717 torr V = 0.523 L n = 0.0203 mol 0.529 g C 2 H 2

34. 33 © 2006 Brooks/Cole - Thomson 33 Standard Conditions (STP) Common reference points for comparing Standard Temperature & Pressure Standard Pressure = 1.00 atm Standard Temperature = 0°C = 273 K

35. 34 © 2006 Brooks/Cole - Thomson 34 Molar Volume of a Gas at STP Definition: The volume of 1 (exact) mole gas at STP Use the Ideal Gas Law: PV = nRT 1 mole of any gas at STP will occupy 22.4 L1 mole of any gas at STP will occupy 22.4 L ==> Molar volume can be used as a conversion factorcan be used as a conversion factor as long as you work at STP as long as you work at STP 1 mol  22.4 L

36. 35 © 2006 Brooks/Cole - Thomson 35 Molar Volume So much empty space between molecules in the gas state,  the volume of the gas is not effected by the size of the molecules, (under ideal conditions).

37. 36 © 2006 Brooks/Cole - Thomson 36 Density of Gas at STP Since every exactly one mole of any gas has a volume of 22.4 L, whereas the mass of such gas would be as the molar mass in grams Density of Gas = Molar mass / Molar Volume Example: Find the Density of Oxygen gas at STP 1.43 g/L

38. 37 © 2006 Brooks/Cole - Thomson 37 Density of Common Gases At STP, the density of common gases (in g/L) as: H 2 0.0900 He 0.179 CH 4 0.716 N 2 1.25 Air 1.29 O 2 1.43 CO 2 1.96 Cl 2 3.17 Which one, hydrogen gas or helium gas, is better in blimps in providing lift? Why carbon dioxide is used in putting out fire? What if its density is less than the air?

39. 38 © 2006 Brooks/Cole - Thomson GAS DENSITY Screen 12.5 Higher Density air Low density helium

40. 39 © 2006 Brooks/Cole - Thomson GAS DENSITY Screen 12.5 PV = nRT d and M proportional and density (d) = m/V

41. 40 © 2006 Brooks/Cole - Thomson USING GAS DENSITY The density of air at 15 o C and 1.00 atm is 1.23 g/L. What is the molar mass of air? For exactly 1 L air, mol = 0.0423 mol Molar mass = 29.1 g/mol

42. 41 © 2006 Brooks/Cole - Thomson Distribution of Gas Molecule Speeds Boltzmann plotsBoltzmann plots Named for Ludwig Boltzmann doubted the existence of atoms.Named for Ludwig Boltzmann doubted the existence of atoms. This played a role in his suicide in 1906. This played a role in his suicide in 1906.

43. 42 © 2006 Brooks/Cole - Thomson Average Velocity of Gas Molecules vs. Temperature Average velocity of gas molecules increase as temperature increases

44. 43 © 2006 Brooks/Cole - Thomson Average Velocity of Gas Molecules vs. Molecular Weight Average velocity decreases with increasing mass.

45. 44 © 2006 Brooks/Cole - Thomson Gas Diffusion DIFFUSION is the gradual mixing of molecules of different gases.

46. 45 © 2006 Brooks/Cole - Thomson Gas Effusion Figure 12.19 EFFUSION is the movement of molecules through a small hole into an empty container.

47. 46 © 2006 Brooks/Cole - Thomson Gas Diffusion and Effusion depends on Velocity of Gas molecule Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He

48. Gas Diffusion/Effusion: Graham’s law governs effusion and diffusion of gas molecules. Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.

49. 48 © 2006 Brooks/Cole - Thomson Graham’s Law in Action https://www.youtube.com/watch?v=M9pHnzf5cdk HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl The location where they meet depends on the relative velocity of these two gases, closer to the starting point of slower-moving gases. Which molecule, HCl or NH 3, moves faster? https://www.youtube.com/watch?v=M9pHnzf5cdk HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl The location where they meet depends on the relative velocity of these two gases, closer to the starting point of slower-moving gases. Which molecule, HCl or NH 3, moves faster? NH 3 (17.0 g/mol) is lighter than HCl (36.5 g/mol) and moves faster Therefore, NH 4 Cl forms closer to HCl end of tube. NH 3 (17.0 g/mol) is lighter than HCl (36.5 g/mol) and moves faster Therefore, NH 4 Cl forms closer to HCl end of tube.

50. 49 © 2006 Brooks/Cole - Thomson Practice: Graham’s Law Calculate the ratio of the effusion rates of H 2 and UF 6. Molar mass: H 2 = 2.016 g/mol, UF 6 = 352.02 g/mol Rate H2/rate UF6 = 13.2

51. 50 © 2006 Brooks/Cole - Thomson Practice: Graham’s Law If the effusion of gas A is 1.83 times faster than oxygen gas, calculate the molar mass of A. Molar mass: O 2 = 32.00 Rate A/O2 = 2.83 Molar mass A = 4.00 g/mol

52. 51 © 2006 Brooks/Cole - Thomson Using KMT to Understand Gas Laws Recall that KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

53. 52 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Real molecules have volume.Real molecules have volume. There are intermolecular forces.There are intermolecular forces. –Otherwise a gas could not become a liquid.

54. 53 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with VAN DER WAALS’s EQUATION. Measured V = V(ideal) Measured P intermol. forces vol. correction J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. nRT V - nb V 2 n 2 a P + ----- )(

55. 54 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Cl 2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm