2. BEHAVIOR OF GASES Gases have weight Gases take up space Gases exert pressure Gases fill their containers Gases are mostly empty space (the molecules in a gas are separate, very small, and very far apart) Gases doing all of these things!

3. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  Gases are mostly empty space  The molecules in a gas are separate, very small and very far apart

4. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  Gas molecules are in constant, chaotic motion  Collisions between gas molecules are elastic (there is no energy gain or loss)

5. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  The average kinetic energy of gas molecules is directly proportional to the absolute temperature  Gas pressure is caused by collisions of molecules with the walls of the container

6. Measurements of Gases  To describe a gas, its volume, amount, temperature, and pressure are measured. Volume: measured in L, mL, cm 3 (1 mL = 1 cm 3 ) Amount: measured in moles (mol), grams (g) Temperature: measured in KELVIN (K) K = ºC + 273 Pressure: measured in mm Hg, torr, atm, etc. P = F / A (force per unit area)

7. Moderate Force (about 100 lbs) Small Area (0.0625 in 2 ) Enormous Pressure (1600 psi) P = F / A

8. Bed of Nails Large Surface Area (lots of nails) Moderate Force Small Pressure P = F / A

9. Units of Pressure  Units of Pressure:  1 atm = 760 mm Hg  1 atm = 760 torr  1 atm = 1.013 x 10 5 Pa  1 atm = 101.3 kPa  1 atm = 1.013 bar

10. Boyle’s Law As P , V  (when T and n are constant) and vice versa…. INVERSE RELATIONSHIP V  1/P P 1 V 1 = P 2 V 2 For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure.

11. Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant)  P 1 V 1 = P 2 V 2  (1.2 atm)(12 L) = (3.6 atm)V 2  V 2 = 4.0 L

12. Charles’ Law Jacques Charles (1746-1828) The volume of a given number of molecules is directly proportional to the Kelvin temperature. As T , V  (when P and n are constant) and vice versa…. DIRECT RELATIONSHIP V  T

13. Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C)  V 1 / T 1 = V 2 / T 2  (117 mL) / (373 K) = (234 mL) / T 2  T 2 = 746 K  T 2 = 473 ºC

14. Combined gas law This is for one gas undergoing changing conditions of temp, pressure, and volume. Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp):

15. Example 1: A sample of neon gas occupies 105 L at 27°C under a pressure of 985 torr. What volume would it occupy at standard conditions?  P 1 = 985 torr  V 1 = 105 L  T 1 = 27 °C = 300. K  P 2 = 1 atm = 760 torr  V 2 = ?  T 2 = 0 °C = 273 K P 1 V 1 T 2 = P 2 V 2 T 1 (985 torr)(105 L)(273K) = (760torr)(V 2 )(300K) V 2 = 124 L

16. Example 2: A sample of gas occupies 10.0 L at 240°C under a pressure of 80.0 kPa. At what temperature would the gas occupy 20.0 L if we increased the pressure to 107 kPa?  P 1 = 80.0 kPa  V 1 = 10.0 L  T 1 = 240 °C = 513 K  P 2 = 107 kPa  V 2 = 20.0 L  T 2 = ? P 1 V 1 T 2 = P 2 V 2 T 1 (80.0kPa)(10.0L)(T 2 ) = (107kPa)(20.0L)(513K) T 2 = 1372K≈ 1370K

17. Example 3: A sample of oxygen gas occupies 23.5 L at 22.2 °C and 1.3 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C?  P 1 = 1.3 atm  V 1 = 23.5 L  T 1 = 22.2 °C = 295.2 K  P 2 = ?  V 2 = 11.6 L  T 2 = 12.5 °C = 285.5 K P 1 V 1 T 2 = P 2 V 2 T 1  P 2 = P 1 V 1 T 2 /V 2 T 1 P 2 =(1.3 atm x (760mm Hg/1atm))(23.5L)(285.5K) (11.6L)(295.2K) = P 2 = 1936 mm Hg ≈ 1900 mmHg

18. Gases: Standard Molar Volume & The Ideal Gas Law  Avogadro’s Law: at the same temperature and pressure, equal volumes of all gases contain the same # of molecules (& moles).  Standard molar volume = 22.4 L @STP  This is true of “ideal” gases at reasonable temperatures and pressures,the behavior of many “real” gases is nearly ideal.

19. The IDEAL GAS LAW  Shows the relationship among the pressure, volume, temp. and # moles in a sample of gas.  P = pressure (atm)  V = volume (L)  n = # moles  T = temp (K)  R = universal gas constant = 0.0821 The units of R depend on the units used for P, V & T

20. Example 1: What volume would 50.0 g of ethane, C 2 H 6, occupy at 140 ºC under a pressure of 1820 torr?  P = (1820 torr)(1 atm/760 torr) = 2.39 atm  V = ?  n = (50.0 g)(1 mol / 30.08 g) = 1.66 mol  T = 140 °C + 273 = 413 K PV = nRT  V = nRT/P V = (1.66 mol) (0.0821 L·atm/mol·K)(413 K) (2.39 atm) V = 23.6 L

21. Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions.  P = 1.00 atm  V = 8.96 L  n = ?  T = 273 K PV = nRT  n = PV/RT n = (1 atm)(8.96 L)/(0.0821 L·atm/mol·K)(273 K) n = 0.400 mol (a)

22. Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions.  Or the easier way… (a)

23. Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions.  Convert moles to grams… (b)

24. Example 3: Calculate the pressure exerted by 50.0 g ethane, C 2 H 6, in a 25.0 L container at 25 ºC?  P = ?  V = 25.0 L  n = (50.0 g)(1 mol / 30.08 g)  T = 25 °C + 273 = 298 K PV = nRT  P = nRT/V P= (1.66 mol)(0.0821 L·atm/mol·K)(298 K) (25.0 L) P = 1.62 atm

25. zDALTON’S LAW OF PARTIAL PRESSURES

26. Partial Pressures and Mole Fractions  In a mixture of gases each gas exerts the pressure it would exert if it occupied the volume alone.  The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases:  P total = P 1 + P 2 + P 3 + …

27. Example: If 100.0 mL of hydrogen gas, measured at 25  C and 3.00 atm, and 100.0 mL of oxygen, measured at 25  C and 2.00 atm, what sould be the pressure of the mixture of gases?  P total = P 1 + P 2 + P 3 + …  P T = 3.00 atm + 2.00 atm  P T = 5.00 atm Notice the two gases are measured at the same temp. and vol.

28. Vapor Pressure of a Liquid  The pressure exerted by its gaseous molecules in equilibrium with the liquid; increases with temperature

29. Vapor Pressure of a Liquid  P atm = P gas + P H 2 O  or  P gas = P atm - P H 2 O

30. Vapor Pressure of a Liquid (calculated using WATER DISPLACEMENT) Temp. (  C) v.p. of water (mm Hg) Temp. (  C) v.p. of water (mm Hg) 1815.482118.65 1916.482219.83 2017.542321.07

31. Example 1: A sample of hydrogen gas was collected by displacement of water at 25  C (vapor pressure of water at 25  C is 23.76 mm Hg). The atmospheric pressure was 748 mm Hg. What pressure would the dry hydrogen exert in the same conditions?  P H 2 = P atm - P H 2 O  P H 2 =748 mm Hg – 23.76 mm Hg  P H 2 = 724.24 mm Hg  P H 2  724 mm Hg

32. Example 2: A sample of oxygen gas was collected by displacement of water. The oxygen occupied 742 mL at 27  C (the vapor pressure of water at 27  C is 26.74 mm Hg). The atmospheric pressure was 753 mm Hg. What volume would the dry oxygen occupy at STP?  P O2 = P atm - P H2O  P O2 =753 mm Hg – 26.74 mm Hg  P O2 = 726 mm Hg  P 1 V 1 T 2 = P 2 V 2 T 1  V 2 = P 1 V 1 T 2 /P 2 T 1  V2 = (726 mm Hg)(742 mL)(273K)/(760 mm Hg)(300.K)  V 2 = 645 mL

33. Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25  C (the vapor pressure of water at 25  C is 23.76 mm Hg). She collects 152 mL of H 2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected.  P H 2 = P atm - P H 2 O  P H 2 =758 mm Hg – 23.76 mm Hg  P H 2 = 734 mm Hg

34. Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25  C. She collects 152 mL of H 2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected. PV = nRT  n = PV/RT n = (734mmHg x(1 atm/760 mmHg))(0.152 L) (0.0821 L·atm/mol·K)(298 K) n = 0.00600 mol H 2

35. Mole Fraction Moles of Substance Total Moles x Total Pressure= Partial Pressure of that substance

36. Graham’s Law of Diffusion & Effusion  Where,  Rate = rate of diffusion or effusion  MM=molar mass

37. WS: Graham’s Law z1. Under the same conditions of temperature and pressure, how many times faster will hydrogen effuse compared to carbon dioxide?

38. Answer to #1

39. Extra Question zRank the following gases in terms of fastest to slowest effusion: Kr, He, Rn, N 2

40. ANSWER zThe lighter mass a gas has, the faster it will move. zThe heavier mass a gas has, the slower it will move. Fastest = He, N 2, Kr, Rn = Slowest