1. Gases Chapter 13

2. Kinetic-Molecular Theory of Matter Model for gases Explains why gases behave the way that they do Based on the idea that particles of matter are always in motion.

3. Kinetic-Molecular Theory of Gases Provides a model for the ideal gas An “ideal gas” is an imaginary gas that perfectly fits all the assumptions of the kinetic-molecular theory.

4. 5 Assumptions of the KMT of gases 1.Gases consist of large numbers of tiny particles that are far apart relative to their size Explains the lower density of gases compared with solids and liquids Explains why gases are easily compressed

5. 5 Assumptions of the KMT of gases cont… 2.Collisions between gas particles and container walls are elastic collisions Elastic collision- collision in which there is no net loss of kinetic energy

6. 5 Assumptions of the KMT of gases cont… 3.Gas particles are in continuous, rapid, random motion. They posses kinetic energy. 4. There are no forces of attraction or repulsion between gas particles. 5. The average kinetic energy of gas particles is directly proportional to the Kelvin temperature of the gas.

7. Kinetic-Molecular Theory and the Nature of Gases Real gases behave like ideal gases at low pressure and high temperature Expansion: Gases do not have a definite shape or volume – Completely fill any container in which enclosed

8. Kinetic-Molecular Theory and the Nature of Gases Characteristics of Gases – Fluidity: Gas particles easily glide past one another – Low Density: density of a substance in the gaseous state = about 1/1000 the density of the same substance in the solid/liquid state – Compressibility: gas particles can be easily compressed under pressure to increase the amount of gas stored in a container

9. KMT and the Nature of Gases Diffusion and Effusion Diffusion: Spontaneous mixing of the particles of two substances caused by random motion Example- removing the cap from a closed container of ammonia will cause the ammonia gas to mix with the air and spread throughout the room Rate of diffusion depends on three properties of the gas particles involved: – Speed – Diameter – Attractive forces present Effusion: process by which gas particles pass through a tiny opening

10. Deviations of Real Gases from Ideal Behavior KMT applies only to ideal gases Real Gas: a gas that does not behave completely according to the assumptions of the kinetic-molecular theory. **No gas does!!

11. 13.1 Pressure Pressure and Force – Pressure (P): defined as the force per unit area on a surface. P = force / area SI Unit for force: Newton (N) – Depends on the volume, temperature and number of molecules present – Example: Gas in a small container (low volume) at a very high temperature… will the pressure be high or low? – High temperatures make the gas molecules move very quickly and because the gas is in a small container there will be many collisions between the gas molecules and container walls creating high pressure

12. Pressure Measuring Pressure – Barometer: device used to measure atmospheric pressure Introduced by Torricelli in the early 1600’s Placed a glass rod in a tub of Mercury (Hg). When placed in the mercury the glass rod would fill to 760 mm above the dish. This measurement would become the standard for atmospheric pressure -- 760 mm Hg – Manometer: device used to measure the pressure of an enclosed gas sample

13. Pressure Units of pressure 1.Millimeter of Mercury (mm Hg) - 1 mm Hg is now called 1 torr, honoring Torricelli - A.P at 0°C = 760 mm Hg or torr 2.Atmosphere of Pressure (atm) - exactly equal to 760 mm Hg - 1 atm = 760 mm Hg - A.P. at 0°C = 1 atm 3.Pascal (Pa) - the pressure exerted by a force of one newton (1N) acting on an area of one square meter - A.P. at 0°C = 1.01325 x 10 5 Pa or 101.3 KPa

14. Pressure Conversion Factors: 1.00 atm =760. mm Hg 760. torr 1.01325 x 10 5 Pa 101.3 KPa **Hint: write these down on a separate sheet of paper entitled CONVERSIONS, you will be permitted to use this on quizzes and tests

15. Standard Temperature and Pressure For purposes of comparison, scientists have agreed on standard conditions of… – Pressure = 1.00 atm – Temperature = 0°C = 273K Abbreviated STP

16. Example Problems The average atmospheric pressure in Denver is 0.830 atm. Express the pressure in mm Hg and kPa. **Hint: look at your conversion factors – atm → mm Hg: Use times sign fraction bar! Start with given: 0.830 atm x 760 mm Hg 1 atm = 631 mm Hg – atm → kPa: Use times sign fraction bar! Start with given: 0.830 atm x 101.325 kPa 1 atm = 84.1 kPa

17. You Try!! Complete the following problems on your outline or on a separate sheet of paper using your pressure conversion factors: 1.Convert a pressure of 748 torr to mmHg 748 mmHg 2. Convert a pressure of 500. torr to atm and kPa 0.658 atm66.6 kPa 3.Convert a pressure of 1.87 atm to torr 1420 torr 4.Convert a pressure of 1.75 atm to kPa and mmHg 177 kPa1330 mmHg

18. 13.2 Boyles Law Robert Boyle discovered that gas pressure and volume are related mathematically Gas laws deal with: volume, temperature, pressure and amount of a gas

19. Boyles Law Boyles Law: Pressure-Volume Relationship Boyles Law: the volume of a fixed mass of a gas is inversely related to the pressure at constant temperature Mathematically: P 1 V 1 = P 2 V 2 – P 1 V 1 : initial conditions – P 2 V 2 : new conditions Used: To compare the changing conditions of a gas

20. Boyles Law Use KMT to understand why this occurs:

21. Using Boyles Law Example 1: Suppose that 1.0 L of gas is initially at 1.0 atm of pressure (V 1 = 1.0 L P 1 = 1.0 atm). The gas is allowed to expand fivefold at constant temperature to 5.0 L (V 2 = 5.0 L), calculate the new pressure. P 1 V 1 = P 2 V 2 → (1.0 atm x 1.0 L) = P 2 x 5.0L 1.0 atm x 1.0 L = P 2 5.0 L P 2 = 0.20 atm

22. You Try!! Complete the following problems on your outline or on a separate sheet of paper using Boyle’s Law: 1.A sample of oxygen gas has a volume of 150 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? 140 mL 2.A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature? 3.18 atm 3.A sample of gas collected in a 350 cm 3 container exerts a pressure of 103 kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.) 240 cm 3

23. Charles Law Charles Law: Volume- Temperature Relationship Discovered: At constant pressure, when the temperature of gas molecules increase, the volume of the gas increases as well

24. Charles Law Charles Law: the volume of a fixed mass of a gas at constant pressure varies directly with the Kelvin temperature – Kelvin (K) = 273.15 + °C – Kelvin scale gets rid of (-) temperatures Mathematically: V 1 / T 1 = V 2 / T 2 – V 1 and T 1 = initial conditions – V 2 and T 2 = new conditions

25. Charles Law Example 1: A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? V 1 / T 1 = V 2 / T 2 __________________________________________________________________________________________________________________________________________________________ First- Identify each variable: given and unknown V 1 = 752 mL V 2 = ? T 1 = 25°C + 273K = 298K T 2 = 50°C + 273K = 323K __________________________________________________________________________________________________________________________________________________________ 752 mL / 298K = V 2 / 323K (752 mL / 298K) 323K = V 2 V 2 = 815 mL Ne

26. You Try!! Complete the following problems on your outline or on a separate sheet of paper using Charles Law: 1. A helium filled balloon has a volume of 2.75 L at 20.0°C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? In °C? 262K-11°C 2.A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? (**look at your notes on STP) 2.61L 3.4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C? 4.06L

27. Gay-Lussac’s Law Pressure-Temperature Relationship Gay Lussac’s Law: the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature Mathematically: P 1 / T 1 = P 2 / T 2

28. Gay-Lussac’s Law Example 1: The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure on the can be at 52°C? P 1 / T 1 = P 2 / T 2 ___________________________________________________ First- Identify each variable: given and unknown P 1 = 3.00 atm P 2 = ? T 1 = 25°C + 273K = 298K T 2 = 52°C + 273K = 325K ___________________________________________________ _ 3.00 atm / 298K = P 2 / 325K (3.00atm / 298K) 325K = P 2 P 2 = 3.27 atm

29. You Try!! Complete the following problems on your outline or on a separate sheet of paper using Gay-Lussac’s Law: 1.A sample of helium gas has a pressure of 1.20 atm at 22.0°C. At what Celsius temperature will the helium reach a pressure of 2.00 atm? 219°C 2.Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. 1.03 atm… pressure change = 1.03 – 1.00 = 0.03 atm 3.A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard pressure? 340.K

30. Avogadro’s Law Volume-Mole Relationship Avogadro’s Law: for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas. Mathematically: V 1 / n 1 = V 2 / n 2

31. Avogadro’s Law Example 1: Suppose we have a 12.2L sample containing 0.50 mol of O 2 gas at a pressure of 1atm and a temperature of 25 ° C. If all of this O 2 is converted to ozone, O 3, at the same temperature and pressure, what will be the volume of the ozone be if 0.33 mol is formed? Equation: 3O 2 (g)  2O 3 (g) V 1 / n 1 = V 2 / n 2 ___________________________________________________ First- Identify each variable: given and unknown V 1 = 12.2 L V 2 = ? n 1 = 0.50 mol n 2 = 0.33 mol ___________________________________________________

32. Avogadro’s Law Example 1 cont. Determine n 2 : 0.50mol O 2 x (2 mol O 3 / 3 mol O 2 ) = 0.33 mol O 3 Plug into equation: 12.2 L / 0.50 mol = V 2 / 0.33 mol Solve for V 2 : V 2 = 8.1 L

33. You Try! Consider two samples of N 2 (g). Sample 1 contains 1.5 mol N 2 and has a volume of 36.7 L at 25 ° C and 1 atm. Sample 2 has a volume of 16.5 L at 25 ° C and 1 atm. Calculate the number of moles of N 2 in sample 2.

34. The Combined Gas Law The combined gas law: P 1 V 1 / T 1 = P 2 V 2 / T 2 –F–From this equation, any value can be calculated as long as the other five values are known Preferred Units: –P–P = atm –V–V = L or mL –T–T = K MEMORIZE THIS EQUATION!!!

35. Example 1: A helium filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will this balloon have at 0.855 atm and 10°C? P 1 V 1 / T 1 = P 2 V 2 / T 2 __________________________________________________________________________________________________________________________________________________________________________________ First- Identify each variable: given and unknown P 1 = 1.08 atm P 2 = 0.855 atm T 1 = 25°C + 273K = 298K T 2 = 10°C + 273K = 283K V 1 = 50.0 L V 2 = ? _________________________________________________________________________________________ (1.08 atm x 50.0 L) / 298 K = (0.855 atm x V 2 ) / 283 K [(1.08 atm x 50.0 L) / 298 K] x 283 K = 0.855 atm x V 2 { [(1.08 atm x 50.0 L) / 298 K] x 283 K } / 0.855 atm = V 2 P 2 = 60.0 L He

36. You Try!! Complete the following problems on your outline or on a separate sheet of paper using the Combined Gas Law: 1.A 350. mL air sample collected at 35.0°C has a pressure of 550. torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57.0°C? 485 torr 2.A gas has a volume of 800. mL at -23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? 800. mL 3.A gas sample occupies 3.25 liters at 24.5 °C and 1825 mm Hg. Determine the temperature at which the gas will occupy 4250 mL at 1.50 atm. 243 K

37. The Ideal Gas Law Combination of Boyle’s, Charles, and Gay-Lussac’s Laws Uses universal gas constant: R – Always has the value 0.08206 L atm / mol K Written: PV = nRT –P: pressure –V: volume –n: number of moles –R: constant –T: temperature

38. Ideal Gas Law Calculations A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0 ° C and a pressure of 1.5 atm. Calculate the number of moles of H 2 present in the gas sample. Use: PV = nRT – P = 1.5 atmn = ? – V = 8.56 LR = 0.08206 L atm / mol K – T = 0 ° C + 273 = 273 K Rearrange equation and solve n = PV / RT n = 0.57 mol

39. YOU TRY! DO: Self-Check Exercise 13.5 on page 403 in book DO: Example 13.9 on page 403 in book DO: Self-Check Exercise 13.5 on page 403 in book

40. Dalton’s Law of Partial Pressures Dalton’s Law of Partial Pressures: the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases Partial pressure: the pressure of each gas in a mixture Daltons Law expressed: P T = P 1 + P 2 + P 3 + P 4 … – P T = total pressure – P 1 + P 2 + P 3 + P 4 … = the partial pressures of component gases 1, 2, 3, 4, and so on…

41. Calculating the total pressure of a gas and water vapor inside a collection bottle is done using Dalton’s Law: – The collection bottle is raised until the water levels inside and outside of the bottle are the same – At this point, the total pressure inside the bottle = atmospheric pressure (P atm ) – According to Dalton’s Law: P atm = P gas + P H 2 O – The pressure of water at the temperature it was collected would have to be found using your table Dalton’s Law of Partial Pressures

42. Water Displacement Gases produced in the laboratory are often collected over water The gas produced by the reaction displaces the water, which is more dense A gas collected by water displacement is not pure but is always mixed with water vapor Water vapor, like other gases, exerts pressure called water- vapor pressure The collection bottle is raised so that the water level inside and outside of the container are equal and at this point the total pressure inside the collection bottle must equal P atm If we know the atmospheric pressure and the water temperature, we can figure out the pressure of the gas inside the container.

43. Dalton’s Law of Partial Pressures Example 1: Oxygen gas from the decomposition of KClO 3 was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. What was the partial pressure of the oxygen collected? P atm = P gas + P H 2 O First identify each variable: P T = P atm = 731.0 torr P H 2 O = 17.5 torr (vapor pressure of water at 20°C) P gas = P O 2 = ? Plug into equation: P atm = P O 2 + P H 2 O → P O 2 = P atm - P H 2 O P O 2 = 731.0 torr – 17.5 torr = 713.5 torr

44. Example 2: Some hydrogen gas is collected over water at 20.0°C. The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is 742.5 torr. What is the barometric pressure at the time the gas is collected? P atm = P gas + P H 2 O First identify each variable: P T = P atm = ? P H 2 O = 17.5 torr (partial pressure of water at 20.0°C P gas = P H 2 = 742.5 torr Plug into equation: P atm = P H 2 + P H 2 O P H 2 O = 742.5 torr + 17.5 torr = 760.0 torr Dalton’s Law of Partial Pressures

45. Example 3: Helium gas is collected over water at 30°C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mmHg? P atm = P gas + P H 2 O First identify each variable: P T = P atm = 750.0 mmHg P H 2 O = 31.8 mmHg (partial pressure of water at 30.0°C) P gas = P He = ? Plug into equation: P atm = P He + P H 2 O → P He = P atm - P H 2 O P He = 750.0 mmHg – 31.8 torr = 718.2 mmHg