Properties of Gases & Gas Laws

1. Properties of Gases & Gas Laws Unit 10

2. Kinetic Molecular Theory • Gases are composed of a large number of particles that behave like small, solid, spherical objects in a state of constant, random motion. • These particles move in a straight line until they collide with another particle or the walls of the container (they follow Newtonian Physics). • These particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space. • There is no force of attraction between gas particles or between the particles and the walls of the container. • Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container. • The average kinetic energy of a collection of gas particles depends on the temperature of the gas and nothing else.

3. Kinetic Molecular Theory • Kinetic Molecular theory requires: • An ideal gas that perfectly fits all the assumptions of the kinetic-molecular theory. • A real gas is a gas that does not behave completely according to the assumptions of the kinetic-molecular theory. • Ideal gases aren’t real, but they offer a good approximation at normal temperatures and pressures.

4. Properties of Gases • Gases expand to fill their containers • Gases are compressible • Gases have relatively low densities • Gases “flow” like liquids so both liquids and gases are termed “fluid”. • Gases undergo diffusion which is the spontaneous mixing of the particles of two substances due to their random motion.

5. Pressure • Pressure (P) is defined as the force per unit area on a surface. • Pressure = force/area • The unit for force is thenewton (N). • Some common units of pressure include: • mmHg, atm, kPa, mbar, torr • The equality related to these: • 760 mmHg = 1 atm = 101.325 kPa = 1013 mbar = 760 torr

6. Familiar Pressure Measurments • A barometer is a device used to measure atmospheric pressure. • Because atmospheric pressure is often measured with a mercury barometer, a common unit of pressure is millimeters of mercury, mmHg.

7. Standard Temperature/Pressure • The volume of a gas varies based on temperature and pressure. • In order to make valid comparisons of volumes, scientists have agreed on standard conditions for gases. • Standard Temperature and Pressure (STP) is equal to exactly 1 atm pressure and 0º C (273K). • Memorize this!!!

8. Avogadro’s Principle • Equal volumes of gases under the same conditions have equal numbers of molecules. • The Molar Volume (1mole)of any gas at STP is 22.4 L.

9. Volume Stoichiometry • If a gas is at STP, we can use the 4 steps of stoichiometry along with Avogodro’s Principle to calculate the volume of gas produced in a chemical equation. • Remember: the conversion factor is 1 mole = 22.4 L @ STP.

10. Sample Problem #1 • How many liters of oxygen gas are needed to completely react 150L of hydrogen at STP according to this equation: H2 + O2 H2O First, balance the equation and add the given and unknown values. 150 L X L 2 2

11. Sample Problem #1 (cont.) Step 2: 150 L H2 x 1 mole = 6.7 moles H2 22.4 L Step 3: 6.7 moles H2 x 1 mole O2 = 3.4 moles O2 2 mole H2 Step 4: 3.4 moles O2 x 22.4 L = 75L O2 1 mole

12. Sample Problem #1 (Short Way) • Set up a proportion with the given on top and the unknown on the bottom: 2 moles H2 = 150 L H2 1 mole O2 X L O2 Now, cross multiply and solve for X: 2 X = 150 X = 75 L

13. Sample Problem #2 • How many liters of nitrogen are needed to completely react with 10.0 L of hydrogen at STP according to this equation: H2 + N2 NH3 • Solution: 3 H2 + N2 2 NH3 10.0 L X L

14. Sample Problem #2 Solution 3 moles H2 = 10.0 L H2 1 mole N2 X L N2 3 X = 10.0 X = 3.33 L N2 How many grams of nitrogen? 3.33 L N2 x 1 mole = 0.149 mole N2 22.4 L 0.149 mol x 28.0134 g/mol = 4.17 g N2

15. Practice • Stoichiometry Diagram • WS #1 Gas Stoich @ STP • WS #2 Gas Stoich @ STP

16. Boyle’s Law • The volume of a given sample of a gas at constant temperature is inversely proportional to its pressure. • P1 V1 = P2 V2

17. Boyle’s Law

18. Boyle’s Law Problem #1 • The volume of a gas at 1.0 atm is 75.3 mL. If temperature remains constant, what will the new volume be if the pressure is increased to 2.5 atm? V1= P1= V2= P2= 75.3 mL (75.3)(1) = X(2.5) 1.0 atm 75.3 = 2.5 X X mL X = 30. mL 2.5 atm

19. Boyle’s Law Problem #2 • At 780. mmHg, a gas has a volume of 27.8 mL. What pressure is needed to increase the volume to 45.5 mL if the temperature remains constant? V1= P1= V2= P2= 27.8 mL (27.8)(780) = 45.5X 780 mmHg 21684 = 45.5 X 45.5 mL X = 477 mmHg X mmHg

20. Practice • WS #3 Boyle’s Law

21. Kelvin Temperature Scale & Gas Laws • When using temperature for ANY gas law problem, you must always convert Celsius to Kelvins. • To convert to Kelvins: • K = C + 273. • K = Kelvin temperature; C = degrees Celsius. • To convert back to Celsius: • C = K – 273.

22. Charles’ Law • The volume of a given sample of a gas is directly proportional to its temperature at constant pressure. • V1 = V2 T1 T2

23. Charles’ Law Problem #1 • A sample of gas at 45.0ºC has a volume of 25.3 mL. At constant pressure, what would the new volume be at 25.0ºC? V1= T1= V2= T2= 25.3 mL 25.3 = X . 318298 45.0ºC + 273 318 X = 7539.4 X mL 25.0ºC + 273 X = 23.7 mL

24. Charles’ Law Problem #2 • At 37.0ºC a gas has a volume of 48.8 mL.If the pressure is constant, what Celsius temperature is necessary to reduce the volume to 25.0 mL? V1= T1= V2= T2= 48.8 mL 48.8 = 25.0 310 X 37.0ºC + 273 48.8 X = 7750 X = 158.81 K 25.0 mL X 158.81 – 273 = -114 ºC

25. Practice • WS #4 Charles’ Law

26. Combined Gas Law • Combined gas law formula is used when temperature or pressure are not constant. • The formula combines Boyle’s and Charles’ laws: P1V1 = P2V2 T1 T2

27. Combined Gas Law Problem #1 • At 37.0ºC and 790.0 mmHg, a gas has a volume of 355.0 mL. What would the new pressure be if the temperature was reduced to 25.0ºC and the volume to 250.0 mL? V1= P1= T1= V2= P2= T2= 355.0 mL (790)(355) = 250X . 310298 790.0 mmHg 77500 X = 83574100 37.0ºC + 273 250.0 mL X = 1078 mmHg X mmHg 25.0ºC + 273

28. Combined Gas Law Problem #2 • At 28.0ºC and 980.5 kPa, a gas has a volume of 285.0 mL. What would the new volume be at STP? V1= P1= T1= V2= P2= T2= 285.0 mL (980.5)(285) = 101.325X . 301273 980.5 kPa 30498.825 X = 76287802.5 28.0ºC + 273 X mL X = 2501 mL = 2.50 x 103 mL 101.325 kPa 0ºC + 273

29. Practice • WS #5 Combined Gas Law

30. Dalton’s Law of Partial Pressures • Dalton’s law states that the pressure of a system is equal to the sum of the pressures of the gases in the system. • Ptotal = Pgas1 + Pgas2 + … • The pressures of gases in our lungs at 37ºC at sea level would bePtotal = PN2 + PO2 + PCO2 + PH2O • 573 mmHg + 100 mmHg + 40 mmHg + 47 mmHg = • 760 mmHg (atmospheric pressure)

32. Gases Collected by Water Displacement • When a gas is collected over water, water vapor molecules are mixed with the gas. • Just as the gas exerts pressure, so does the water vapor. • To find the pressure of just the collected gas, use Dalton’s Law and subtract the pressure exerted by the water vapor. • 1mL=1cm3

33. Gases Collected by Water Displacement

34. Dalton’s Law Problem #1 • Oxygen gas was collected over water when the barometer read 97.5 kPa, and 20.0ºC. What is the partial pressure of the oxygen collected? • At 20.0ºC, the pressure of water is 2.3 kPa. • Ptotal – PH2O = PO2 • 97.5 – 2.3 = 95.2 kPa.

35. Dalton’s Law Problem #2 • A chemist collects 96.0 mL of gas over water at 27.0ºC when the pressure is 1220 kPa. What volume would the dry gas occupy at 70.0 ºC and 1270 kPa? V1= P1= T1= V2= P2= T2= 96.0 mL (1216.4)(96) = 1270X . 300343 1220 kPa - 3.6 381000 X = 40053619.2 27.0ºC + 273 X mL X = 105 mL 1270 kPa 70.0ºC + 273

36. Practice • WS #6 Dalton’s Law—Gas Collection

37. Ideal Gas Law • The Ideal Gas Lawis a mathematical relationship between pressure, volume, temperature and number of moles of gas. • The formula is: • PV = nRT • P = Pressure (in atm, mmHg or kPa) • V = Volume (must be in Liters) • n = number of moles • R = Proportionality constant depending on units of pressure: (8.314 L/kPa, 0.0821 L/atm or 62.4 L/mmHg) • T = temperature (in Kelvins)

38. Ideal Gas Law Problem #1 • A 500. g block of dry ice (solid CO2) vaporizes to a gas at room temperature. Calculate the volume of gas produced at 25.0ºC and 975 kPa. P = V = n = R = T = 975 kPa 975 X = (11.4)(8.314)(298) X L 975 X = 28244.3208 500 g/44.0098 = 11.4 mol X = 28.97 = 29.0 L 8.314 L/kPa 25ºC + 273

39. Ideal Gas Law Problem #2 • A sample of CO2 with a mass of 0.250 g was placed in a 350. mL container at 127ºC. What is the pressure exerted by the gas? (Solve for kPa). .350 X = (.00568)(8.314)(400) P = V = n = R = T = X kPa .350 X = 18.889048 .350 L .250 g/44.0098 = .00568 mol X = 54.0 kPa 8.314 L/kPa 127ºC + 273

40. Practice • WS #7 Ideal Gas Law

41. Ideal Gas Law & Stoichiometry • If areaction occurs at something other than STP, Avogadro’s Principle cannot be used. (22.4L is the molar volume at STP only.) • To solve these problems, you must combine stoichiometry with the Ideal Gas Law. • Key to all problems: Always find moles of what you know and convert to moles of what you want to find. • This may require stoichiometry first or Ideal Gas Law first.

42. Ideal Gas Law & Stoichiometry • How many liters of hydrogen gas are needed to produce 73.3 g of ammonia at 45.0ºC and 3.00 atm? • First, find moles of what you know and then convert to moles of what you want. • N2 + 3H2 2 NH3 • 73.3 g NH3 = 4.30 mol NH3 x 3 mol H2 = 17.0304 g 2 mol NH3 • 6.45 mol H2

43. Ideal Gas Law & Stoichiometry • Now use the Ideal Gas Law to find volume. • How many liters of hydrogen gas are needed to produce 73.3 g of ammonia at 45.0ºC and 3.00 atm? (6.45 mol H2) P = V = n = R = T = 3.00 Atm 3 X = (6.45)(.0821)(318) X L 3 X = 168.39531 6.45 mol X = 56.1 L 0.0821 L/Atm 45.0ºC + 273

44. Ideal Gas Law & Stoichiometry • How many liters of CO2 are produced when 55.6 g of sulfuric acid reacts with sodium bicarbonate at 23.0ºC and 758 mmHg? • First, find moles of what you know and then convert to moles of what you want. • H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2 • 55.6 g H2SO4 = 0.567 mol H2SO4 x 2 mol CO2 = 98.0794 g 1 mol H2SO4 • 1.13 mol CO2

45. Ideal Gas Law & Stoichiometry • How many liters of CO2 are produced when 55.6 g of sulfuric acid reacts with sodium bicarbonate at 23.0ºC and 758 mmHg? (1.13 mol CO2) P = V = n = R = T = 758 mmHg 758 X = (1.13)(62.4)(296) X L 758 X = 20871.552 1.13 mol X = 27.5 L 62.4 L/mmHg 23.0ºC + 273

46. Ideal Gas Law & Stoichiometry

47. Practice • WS #8 Ideal Gas Law Stoich

48. Limiting Reactant • How many grams of CO2 are formed if 10.0 g of carbon are burned in 20.0 dm3 of oxygen gas at STP? (what does dm3 represent?) • Write a balanced equation: C + O2 CO2 • Change both quantities to moles: 10.0 g C = .833 mol C x1 mol CO2 = .833 mol CO2 12.011g/mol 1 mol C

49. Limiting Reactant (Cont.) 2. Convert to moles to find the Limiting Reactant 20.0 dm3 O2 = .893 mol O2 x 1 mol CO2 = .893 22.4 L1 mol O2 mol CO2 3. Carbon is the limiting reactant 4. Complete the problem on the basis of the limiting reactant. .833 mol C x 1 mol CO2 = .833 mol CO2 x 44.0098 = 1 mol C36.7 g CO2

50. Limiting Reactant (Cont.) • How many liters of excess reactant remains? • You have .893 mol O2 • You need .833 mol O2 • Excess: .893 - .833 = .060 mol O2 • .060 mol O2 x 22.4 L = 1.3 L O2 • OR: 0.893 x 22.4L = 20.0 L • 0.833 x 22.4 L = 18.7 L • 20.0 L – 18.7 L =1.3 L O2