2. Gases Properties of gases and gas laws. Chapter 14

3. Kinetic Molecular Theory: Gases p114 Warm up: describe characteristics of a gas. New: All collisions are perfectly elastic. (no slowing after collision)

4. Characteristics of Gases Pressure: gas particles hitting the sides of the container Pressure units: 1 atmospheres (symbol = atm) 2 millimeters of mercury (symbol = mmHg), also referred to as torr. 3 kiloPascals (symbol = kPa)

5. Converting Between Units of Pressure 1 atm = 101.3 kPa = 760.0 mmHg (torr) EX: Convert 0.875 atm to mmHg. 0.875atm ( 760.0mmHg ) = 665mmHg ( 1 atm )

6. EX: Convert 253.4 kPa to mmHg 253.4 kPa ( 760.0 mmHg ) = 1901 mmHg ( 101.3 kPa )

7. Standard Temperature and Pressure of Gasses (STP) Standard atmospheric pressure = 1 atm Standard temperature = 0°C

8. Temperature Units – EVERY TEMPERATURE MUST BE IN KELVIN. K = °C + 273 EX: Convert 25°C to K K = 25 + 273 K = 298

9. Conversion practice p. 113 1. Convert 1.2 atm to kpa 2. Convert 450 mmHg to atm 3. Convert 123 kpa to atm 4. Convert -34 o C to K 5. Convert 230 o C to K

10. Gas LawsGas Laws: Boyle’s Law p116 Warm up: Define inverse proportion and direct proportion. Robert Boyle, a British chemist, examined the relationship between volume and pressures of gases.

11. Boyle’s Law p116 At constant temp, volume goes up as pressure goes down. (inverse proportion)

12. Boyle’s Law Equation p.116 P 1 V 1 = P 2 V 2 P1=initial pressure, p2=final pressure PRESSURES HAVE TO BE IN SAME UNITS! V1=initial volume, v2=final volume

13. Example: p116 EX: A gas fills 2.85 L at 245 kPa. What’s the volume at standard pressure?.

14. More Boyle’s Problems p115 5.00 L of O 2 gas has a pressure of 1110 mmHg. What’s the volume at 1435 mmHg? If a 3.50 L bottle of N 2 has a pressure of 450. kPa, what’s the new pressure if the volume was 2.25 L? 10.0 L of Helium gas has a pressure of 1500.0 torr, what is the pressure at 7.50L?

15. Charles’s Law p118 Warm up: what happens to the volume of a gas as it is heated? (claim, reasons) Jacques Charles, a French physicist, noted the relationship between the volume of a gas and temperature. His work led to the Kelvin scale.

16. Charles’s Law p118 At constant pressure, volume increases with temperature. (direct proportion)

17. p.118 Charles’s Law equation: V1=initial volume, v2= final volume T1=initial temp, t2=final temp TEMPS MUST BE IN KELVIN! V 1 T 2 = V 2 T 1

18. Example p118 EX: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature?

19. More Charles’s Problems p117 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C? 5.00 L of a gas is collected at 100. K and then expands to 20.0 L. What’s the new temp? Calculate the new temperature when 2.00 L of gas at 20.0 °C is compressed to 1.00 L.

20. Combined Gas Law p120 Warm up: what does it mean if a problem states “ the pressure is constant”? We can combine Boyle’s and Charles’s Laws to derive: P 1 V 1 T 2 = P 2 V 2 T 1 We use this equation when pressure, volume, and temp are all changing.

21. Example: p120 2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? Step 1: Make a table for known values: P 1 = 745.0 mmHg P 2 = 760.0 mmHg V 1 = 2.00 LV 2 = X T 2 = 273 KT 1 = 298 K

22. Combined Gas Law Problems p119 A gas has a volume of 800.0 mL at - 23.00 °C and 300.0 torr. What’s the volume at 227.0 °C and 600.0 torr? 500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. When the tank cools to 20.0 °C, the pressure is 30.0 atm. What is the volume? The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume is increased from 85.0 mL to 350.0 mL. What’s the final temp if the original temp was 90.0 °C?

24. The Ideal Gas Equation p122 Warm up: convert 25.0 g NaCl to moles. Convert 2.5 moles NaCl to grams. Convert 450 ml to L. Used to show how amount of gas particles affects the other variables: PV = nRT *P = pressure, V = volume (Liters), n = moles, T = temperature, and R =gas constant

25. Gas Constants p122 R=0.0821(when pressure in atm) R=8.31 (kpa) R=62.4 (mmHg)

26. Solving Problems Using The Ideal Gas Equation p122 EX: Calculate the volume of 15.5 g of CO 2 gas at STP. Ex: How many grams of hydrogen would be contained in a 2.5 L flask at 98 C and 456 kpa?

28. More Ideal Gas Problems p121 1. A sample carbon dioxide is collected in a 0.250 L flask. The gas has a pressure of 1.3 atm at a temp of 31°C. How many moles of CO 2 were generated? Grams? 2. A flashbulb contains 2.4 moles of O 2 gas at a pressure of 2.01 kPa and a temperature of 19°C. What is the volume?

29. Calculating Density and Molar Mass We can use the ideal gas equation to calculate gas density. d (density) = PM RT M (molar mass) = dRT P

30. Density and Molar Mass Problems What is the density of carbon tetrachloride vapor at 714 torr and 125°C? What is the density of Sodium hydroxide at 450.0mmHg and 75°C? What is the molar mass of a substance that has a density of 2.55 g/L at a pressure of 785 torr and temperature of 45°C?

31. Deviations from Ideal Gas Behavior When using the ideal gas equation, two assumptions were made. 1. Gases have no volume. 2. Gases have no attractive forces between them. We will now examine how real gases can deviate from these assumptions.

32. Gases occupy no volume At low pressures, both ideal and real gases are far apart separated by empty space. As pressure is applied, real gases begin to take up more of the empty space. Ideal gases are still far apart. An ideal gas can have zero volume, but a real gas will become a liquid under increased pressure.

33. Gases have no attractive forces If gases are made up of polar molecules such as water, the attractive forces are large and the behavior of this real gas is markedly different from an ideal gas. There are even weak attractive forces (dispersion forces) between noble gases. For most gases, the ideal gas laws are accurate to about 1%.

34. Dalton’s Law of Partial Pressure Our calculations so far have been for pure gases. John Dalton formed a hypothesis about pressure exerted by a mixture of gases. Dalton’s Law of Partial Pressure: The total pressure in a container is the sum of the partial pressures of all the gases in the container. P total = P 1 + P 2 + P 3 …

35. Gases in a single container are all the same temperature and have the same volume, therefore, the difference in their partial pressures is due only to the difference in the numbers of molecules present.

36. Partial Pressure of Air Air is an example of a mixture of gases. -nitrogen is 78.084 % -oxygen is 20.948 % -argon is 0.934 % -carbon dioxide is 0.0315 -neon, helium, krypton, and xenon are among the other trace gases.

38. The total pressure of the atmosphere at STP is 101.3 kPa. If 78% of air is nitrogen, then 78% of pressure is due to nitrogen molecules. 0.78 x 101.3 = 79 kPa 21% of air is oxygen so 21% of pressure is due to oxygen molecules. 0.21 x 101.3 = 22 kPa

39. Collecting Gases by Water Displacement One method to collect gases is by water displacement. Gases must be insoluble in water. When collection is complete, water vapor is present in the collection container and must be accounted for in the partial pressures of gases.

41. Volume of a dry gas: P gas = P total – P water EX: A quantity of gas is collected over water at 8°C in a 0.353 L vessel at 84.5 kPa.What volume would the dry gas occupy at standard atmospheric pressure and 8°C?

42. Find the pressure of the dry gas: P gas = P total – P water Obtain P water from table (p.983) P = 84.5 kPa – 1.1 kPa 83.4 kPa The remainder of this problem is a pressure and volume comparison. Boyle’s equation will now be used.

43. P 1 V 1 = P 2 V 2 P 2 P 2 83.4 kPa x 0.353 L = 291 L 101.3 kPa

44. Partial Pressure Practice Problems A gas is collected over water and occupies a volume of 596 cm 3 at 43°C. The atmospheric pressure is 101.1 kPa. What volume will the dry gas occupy at 43°C and standard atmospheric pressure? The vapor pressure of water at 43°C is 8.6 kPa. The following gas volumes were collected over water under the indicated conditions. Correct each given volume to the volume that the dry gas would occupy at standard atmospheric pressure and the indicated temp(constant). a. 888 cm 3 at 14°C and 93.3 kPa. b. 30.0 cm 3 at 16°C and 77.5 kPa. c. 34.0 cm 3 at 18°C and 82.4 kPa.

45. Diffusion and Graham’s Law Kinetic theory states that molecules travel in straight lines. Molecules often collide with other molecules which alters its path an sends it on another straight path. This is the basics of diffusion. As gas molecules diffuse, they become more and more evenly distributed throughout their container.

46. Graham’s Law The relative rates at which two gases under identical conditions of temp and pressure will diffuse vary inversely as the square roots of the molecular masses of the gases.

47. Molecules of small mass diffuse faster than molecules of large mass because they travel faster. Smaller molecules can also pass through more substances with greater ease.

48. Using Graham’s Law, we can derive: v 1 = _m 2 _ v 2 m 1

49. Using Graham’s Law EX: Find the relative rate of diffusion for the gases of krypton and bromine. v Kr = m Br2 v Br2 m Kr v Kr = 159.8 v Br2 83.8

50. The square root of 1.907 = 1.38. Since krypton is the lighter gas, it will diffuse 1.38 times as fast as bromine. EX: Compute the relative rate of diffusion of helium to argon. EX: Compute the relative rate of diffusion of argon to radon.

51. Avogadro’s Law at constant temp, Volume and pressure increase with number of moles of the gas.

52. Avogadro’s equation: V 1 n 2 = V 2 n 1

53. Using Avogadro’s Law EX: A balloon is filled with 2.0 moles of Helium gas, occupying 44.8 L at constant temp. How many liters will be occupied if the number of moles is reduced to 1.5?

54. Isolate unknown: V 1 n 2 = V 2 n 1 n 1 n 1 44.8 L x 1.5 mol = 2.0 mol 33.6 L