1. Chapter 09Slide 1 Gases: Their Properties & Behavior 9

2. Chapter 09Slide 2 Gas Pressure01

3. Chapter 09Slide 3 Gas Pressure02 Units of pressure: atmosphere (atm) Pa (N/m 2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in 2 (14.696 lb/in 2 = 1 atm) in Hg (29.921 in Hg = 1 atm)

4. Chapter 09Slide 4 Pressure–Volume Law (Boyle’s Law): Boyle’s Law01

5. Chapter 09Slide 5 Pressure–Volume Law (Boyle’s Law): Boyle’s Law01

6. Chapter 09Slide 6 Boyle’s Law02 Pressure–Volume Law (Boyle’s Law): The volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure. P 1 x V 1 = P 2 x V 2 P 1 x V 1 = K 1 P 2 x V 2 = K 1

7. Chapter 09Slide 7 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3

8. Chapter 09Slide 8 As T increasesV increases

9. Chapter 09Slide 9 Charles’ Law01 Temperature–Volume Law (Charles’ Law): T(K)=Temperature in Kelvin T (K) = t (0C) + 273.15

10. Chapter 09Slide 10 Charles’ Law01 Temperature–Volume Law (Charles’ Law): The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin temperature of the gas.  V  T V 1 T 1 =k 1

11. Chapter 09Slide 11 Variation of gas volume with temperature at constant pressure. V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 Temperature must be in Kelvin P 1 <P 2 <P 3 <P 4 Third Law of Thermodynamics states that it's impossible to reach absolute zero.

12. Chapter 09Slide 12 A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2

13. Chapter 09Slide 13 Avogadro’s Law01 The Volume–Amount Law (Avogadro’s Law):

14. Chapter 09Slide 14 Avogadro’s Law01 The Volume–Amount Law (Avogadro’s Law): At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present.

15. Chapter 09Slide 15 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO

16. Chapter 09Slide 16 Ideal Gas Equation Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

17. Chapter 09Slide 17 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

18. Chapter 09Slide 18 The Ideal Gas Law01 Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. R=The gas constant R = 0.08206 L·atm·K –1 ·mol –1

19. Chapter 09Slide 19 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L

20. Chapter 09Slide 20 Argon is an inert gas used in lightbulbs to retard the oxidation and vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

21. Chapter 09Slide 21 Calculation of Molar Mass 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? Molar Mass = Mass / Number of moles 1. Calc. moles of air. V = 1.00 L, m = 1.23 g, P = 1.00 atm, V = 1.00 L, m = 1.23 g, P = 1.00 atm, T = (273.15+15 )= 288 K n = PV/RT = 0.0423 mol, m = 1.23 g n = PV/RT = 0.0423 mol, m = 1.23 g 2. Calc. molar mass mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

22. Chapter 09Slide 22 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2

23. Chapter 09Slide 23 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B

24. Chapter 09Slide 24 P T = P A + P B n A RT V n B RT V + = (n A + n B ) RT V P T = P A = X A P T P A = n A RT V P B = X B P T P B = n B RT V P A / P T = nAnA n A + n B X A = nAnA n A + n B P A / P T = XAXA X B = nBnB n A + n B P i = X i P T

25. Chapter 09Slide 25 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm

26. Chapter 09Slide 26 Gas Stoichiometry01 In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. Assuming no change in temperature and pressure, calculate the volume of O 2 (in liters) required for the complete combustion of 14.9 L of butane (C 4 H 10 ): 2 C 4 H 10 (g) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(l)

27. Chapter 09Slide 27 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L

28. Chapter 09Slide 28 Gases: P, V, T & n

29. Chapter 09Slide 29 Kinetic Molecular Theory This theory presents physical properties of gases in terms of the motion of individual molecules. Average Kinetic Energy  Kelvin Temperature Gas molecules are points separated by a great distance Particle volume is negligible compared to gas volume Gas molecules are in rapid random motion Gas collisions are perfectly elastic ( No Change of Total Kinetic Energy) Gas molecules experience no attraction or repulsion

30. Chapter 09Slide 30 Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

31. Chapter 09 Slide 31 Average Kinetic Energy (KE) is given by: Kinetic Molecular Theory  U: The Root–Mean–Square Speed

32. Chapter 09Slide 32 Average Kinetic Energy (KE) is given by: Kinetic Molecular Theory03  U: The Root–Mean–Square Speed (1000 miles/hr) rms speed at 25 °C

33. Chapter 09Slide 33 Kinetic Molecular Theory05 Maxwell speed distribution curves.

34. Chapter 09Slide 34 Kinetic Molecular Theory04 The Root–Mean–Square Speed: is a measure of the average molecular speed. R=8.314 J/K.mol 1J = 1Kg.m 2 /s 2 Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in miles/hr at 25°C. Taking square root of both sides gives the equation He : miles/hr

35. Diffusion of Gases Diffusion: The mixing of different gases by molecular motion with frequent molecular collisions.

36. Chapter 09Slide 36 Graham’s Law02 Effusion is when gas molecules escape, through a tiny hole into a vacuum.

37. Chapter 09Slide 37 Graham’s Law03 Graham’s Law: Rate of effusion is proportional to its rms speed, u rms. For two gases at same temperature and pressure:

38. Chapter 09Slide 38 A Problem to Consider How much faster would H 2 gas effuse through an opening than methane, CH 4 ? So hydrogen effuses 2.8 times faster than CH 4

39. Chapter 09Slide 39 Behavior of Real Gases01 Deviations result from assumptions about ideal gases. 1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. 2. Volume of the molecules is negligibly small compared with that of the container.

40. Chapter 09Slide 40 Behavior of Real Gases02 At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures. As a result, the pressure of real gases will be smaller than the ideal value

41. Chapter 09Slide 41 Behavior of Real Gases03 The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

42. Chapter 09Slide 42 Behavior of Real Gases05 Corrections for non-ideality require van der Waals equation. Intermolecular Attractions Excluded Volume

43. Chapter 09Slide 43 A Problem to Consider If sulfur dioxide were an “ideal” gas, the pressure at 0 o C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure. Use the following values for SO 2 a = 6.865 L 2. atm/mol 2 b = 0.05679 L/mol

44. Chapter 09Slide 44 What is the Real pressure of one mole of a gas, with a Volume of 22.4 L at STP? R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L 2. atm/mol 2 b = 0.05679 L/mol First, let’s rearrange the van der Waals equation to solve for pressure.

45. Chapter 09Slide 45 A Problem to Consider The “real” pressure exerted by 1.00 mol of SO 2 at STP is slightly less than the “ideal” pressure.