1. Unit 6 Gases and Gas Laws

2. Gases in the Atmosphere The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity. By volume, dry air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO 2, and small amounts of other gases.

3. Air Pollution Human activity has polluted the air with other gases: – Sulfur Oxides (SO 2 & SO 3 ) – produced from coal burning. Contribute to acid rain. – Nitrogen Oxides (NO & NO 2 ) – produced by burning fossil fuels. Contribute to acid rain. – Carbon Monoxide (CO) – emitted by motor vehicles. – Ground-level Ozone (O 3 ) – produced when products of fossil fuel combustion react in the presence of sunlight.

4. The Ozone Layer O 3 in the troposphere (ground-level ozone) is a pollutant, but O 3 in the stratosphere is a necessary part of our atmosphere. Stratospheric O 3 protects us by absorbing UV light. CFCs destroy stratospheric O 3, and have been banned in the US. Ozone: Good up high, bad nearby.

5. Atmospheric Pressure Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it. Pressure = Force Area

6. Measuring Pressure A common unit of pressure is millimeters of mercury (mm Hg). 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli who invented the barometer (used to measure atmospheric pressure). The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.

7. Measuring Pressure (continued) The pressure of a gas sample in the laboratory is often measured with a manometer. the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere. For this sample, the gas has a larger pressure than the atmosphere.

8. Measuring Pressure (continued) Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m 2. One pascal is very small, so usually kilopascals (kPa) are used instead. One atm is equal to 101.3 kPa. 1 atm = 760 mm Hg (Torr) = 101.3 kPa

9. Units of Pressure

10. Converting Pressure Sample Problem The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in: a. millimeters of mercury (mm Hg) b. kilopascals (kPa) 0.830 atm atm mm Hg 1 760 x = 0.830 atm atm kPa 1 101.3 x = 631 mm Hg 84.1 kPa

11. Dalton’s Law of Partial Pressures Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases. P T = P 1 + P 2 + P 3 …

12. Dalton’s Law of Partial Pressures Sample Problem A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm. a.What is the total pressure of this system? b. What is the total pressure in mm Hg? P T = P 1 + P 2 + P 3 … P T = 0.5 atm + 0.7 atm + 1.2 atm = 2.4 atm atm mm Hg 1 760 x = 1800 mm Hg

13. The Kinetic-Molecular Theory: A Model for Gases Matter is composed of particles which are constantly moving. The average kinetic energy of a particle is proportional to its Kelvin temperature. The size of a particle is negligibly small. Collisions are completely elastic – energy may be exchanged, but not lost (like billiard balls.)

14. Ideal Gases The kinetic-molecular theory assumes: 1)no attractions between gas molecules 2)gas molecules do not take up space An Ideal Gas is a hypothetical gas that perfectly fits the assumptions of the kinetic-molecular theory. Many gases behave nearly ideally if pressure is not very high and temperature is not very low.

15. Properties of Gases: Fluidity Gas particles glide easily past one another. Because liquids and gases flow, they are both referred to as fluids.

16. Properties of Gases: Expansion Since there’s no significant attraction between gas molecules, they keep moving around and spreading out until they fill their container. As a result, gases take the shape and the volume of the container they are in.

17. Properties of Gases: Low Density Gas particles are very far apart. There is a lot of unoccupied space in the structure of a gas. Since gases do not have a lot of mass in a given volume, they have a very low density The density of a gas is about 1/1000 the density of the same substance as a liquid or solid.

18. Properties of Gases: Compressibility Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can easily be squeezed closer together.

19. Diffusion and Effusion Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion. Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

20. Rate of Diffusion Light molecules move faster than heavy ones. The greater the molar mass of a gas, the slower it will diffuse and/or effuse.

21. Graham’s Law of Effusion Graham’s Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas. The ratio of effusion rates of two different gases is given by the following equation:

22. Graham’s Law of Effusion Sample Problem Calculate the molar mass of a gas that effuses at a rate 0.462 times N 2. Solution: (0.462) 2 = 28.0 g/mol MM unknown == 131 g/mol 28.0 g/mol (0.462) 2 = 28.0 g/mol (0.213)

23. Gases and Pressure Gas pressure is caused by collisions of the gas molecules with each other and with the walls of their container. The greater the number of collisions, the higher the pressure will be.

24. Pressure – Volume Relationship When the volume of a gas is decreased, more collisions will occur. Pressure is caused by collisions. Therefore, pressure will increase.

25. Boyle’s Law Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature. P 1 and V 1 represent initial conditions, and P 2 and V 2 represent another set of conditions. P 1 V 1 = P 2 V 2

26. Boyle’s Law Sample Problem A sample of oxygen gas has a volume of 150.0 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? Solution: P 1 V 1 = P 2 V 2 (0.947 atm) (150.0 mL) = (0.987 atm)V2V2 V2V2 = (0.947 atm) (150.0 mL) (0.987 atm) = 144 mL

27. Volume – Temperature Relationship the pressure of gas inside and outside the balloon are the same. at low temperatures, the gas molecules don’t move as much – therefore the volume is small. at high temperatures, the gas molecules move more – causing the volume to become larger.

28. Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. Charles’s Law V1V1 V2V2 = T1T1 T2T2 V 1 and T 1 represent initial conditions, and V 2 and T 2 represent another set of conditions.

29. The Kelvin Temperature Scale Absolute zero – The theoretical lowest possible temperature where all molecular motion stops. The Kelvin temperature scale starts at absolute zero (-273 o C.) This gives the following relationship between the two temperature scales: K = o C + 273

30. Charles’s Law Sample Problem 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? Solution: V1V1 V2V2 = T1T1 T2T2 752 mL V2V2 = 298 K 323 K K = o C + 273 T1T1 =25 + 273= 298 T2T2 =50 + 273= 323 752 mL V2V2 = 298 K 323 K x= 815 mL

31. Pressure – Temperature Relationship Increasing temperature means increasing kinetic energy of the particles. The energy and frequency of collisions depend on the average kinetic energy of the molecules. Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.

32. Gay-Lussac’s Law Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature. P 1 and T 1 represent initial conditions. P 2 and T 2 represent another set of conditions. P1P1 P2P2 = T1T1 T2T2

33. Gay-Lussac’s Law Sample Problem The gas in a container is at a pressure of 3.00 atm at 25°C. What would the gas pressure in the container be at 52°C? Solution: P1P1 P2P2 = T1T1 T2T2 3.00 atm P2P2 = 298 K 325 K K = o C + 273 T1T1 =25 + 273= 298 T2T2 =52 + 273= 325 3.00 atm P2P2 = 298 K 325 K x= 3.27 atm

34. The Combined Gas Law The combined gas law is written as follows: Each of the other simple gas laws can be obtained from the combined gas law when the proper variable is kept constant. P1P1 P2P2 = T1T1 T2T2 V1V1 V2V2

35. The Combined Gas Law Sample Problem A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Solution: (1.08 atm) (0.855 atm) = 298 K 283 K K = o C + 273 T1T1 =25 + 273= 298 T2T2 =10 + 273= 283 (1.08 atm) V2V2 = (298 K) (283 K) = 60.0 L P1P1 P2P2 = T1T1 T2T2 V1V1 V2V2 (50.0 L) V2V2 (0.855 atm)

36. Avogadro’s Law In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.) Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or: V1V1 V2V2 = n1n1 n2n2

37. The Ideal Gas Law All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law: R represents the ideal gas constant which has a value of 0.0821 (L atm)/(mol K). PV = nRT

38. The Ideal Gas Law Sample Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Solution: PV = nRT P (10.0 L)= (0.500 mol)(0.0821 L atm/mol K) P = (0.500 mol) (298 K) (10.0 L) = 1.22 atm (298 K) (0.0821 L atm/mol K)

39. Molar Mass of a Gas The ideal gas law can be used in combination with mass measurements to calculate the molar mass of an unknown gas. Molar mass is calculated by dividing the mass (in grams) by the amount of gas (in moles.) Molar Mass g = mol

40. Molar Mass of a Gas Sample Problem Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg. Solution: PV = nRT PV RT = 886 mmHg T = P = (1.17 atm) (328 K) (0.225 L) = 31.8 g/mol 55 + 273 = 328 K (0.0821 L atm/mol K) n = 760. mmHg 1 atm 1.17 atm = 0.00978 mol grams mole = MM = 0.311 g 0.00978 mol =

41. Standard Molar Volume Standard Temperature and Pressure (STP) is 0 o C (273 K) and 1 atm. The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

42. Molar Volume Conversion Factor Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

43. Molar Volume Conversion Sample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? b. What volume does 0.768 moles of a gas occupy at STP? 5.00 L L mol 22.4 1 x = 0.223 mol 0.768 mol mol L 1 22.4 x = 17.2 L

44. The Mole Map Revisited As you recall, you can convert between number of particles, mass (g), and volume (L) by going through moles.

45. Stoichiometry Revisited Remember that you can use mole ratios and volume ratios (gases only) as conversion factors: 2CO(g) + O 2 (g) → 2CO 2 (g) 2 molecules 1 molecule2 molecules 2 mole 1 mole2 mol 2 volumes 1 volume2 volumes Example: What volume of O 2 is needed to react completely with 0.626 L of CO at the same temperature & pressure conditions to form CO 2 ? 0.626 L CO L CO L O 2 2 1 x = 0.313 L O 2

46. Gas Stoichiometry Sample Problem What volume (in L) of H 2 at 355 K and 738 mmHg is required to synthesize 35.7 g of methanol, given: CO(g) + 2H 2 (g) → CH 3 OH(g) Solution: First, use stoichiometry to solve for moles of H 2 : Then, use the ideal gas law to find the volume of H 2 : 35.7 g CH 3 OH g CH 3 OH mol CH 3 OH 32.0 1 = 2.23 mol H 2 66.9 L H 2 mol H 2 2 mol CH 3 OH 1 nRT P = (0.971 atm) (355 K)(2.23mol) = (0.0821 L atm/mol K) V =