Macroscopic vs. Microscopic Representation

Transcription

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2 Agenda Day 80 Gases- Intro & 11.7 & 12.2 Lesson: PPT Handouts: 1.PPT Handout, 2. ISU - Kinetic Molecular Theory and Real vs. Ideal Gases Sections ( W1)

3 Macroscopic vs. Microscopic Representation

4 States of Matter Solids have greater attractions, that s why they stay together (whereas gases disperse). Liquids have forces that are strong enough to keep their molecules together (thus, they are incompressible with a definite volume). However, these forces are not strong enough to fix molecules in place (shape is not fixed) Gases have forces so weak that, at room temperature, molecules move freely (shape & volume change). The large spaces between molecules makes gases compressible.

5 Kinetic Energy Temperature affects state. Types of Motion: - vibrational (small motion from side to side), - rotational (spinning), - translational (straight line motion). Solids: vibrational, Liquids: all, Gases: mostly translational, also rotational (and vibrational). Kinetic energy is the energy of motion. Temperature = average kinetic energy The average speed will increase with higher temperature.

6 How do solids and liquids differ from gases? Gases molecules/particles in chaotic motion whose average kinetic energy is greater than the attractive forces between molecules Liquids attractive intermolecular forces between molecules is comparable to kinetic energies of molecules: these molecules are held in close proximity, but still move in a rather chaotic motion. Solids intermolecular forces are sufficiently strong relative to kinetic energy. Molecules are virtually locked in place. Often this is very orderly (crystalline structures).

7 IntErmolecular Forces Review Intermolecular Forces (IM) - These are weak forces that exist between molecules and are responsible for the fact that the substance can exist as a liquid (and/or solid). Molecules of gases have little to no intermolecular forces (which is why gases have no definite shape or volume). Dipole-dipole forces, London dispersion, and hydrogen bonding are intermolecular forces.

8 IntrAmolecular Forces Review IntrAmolecular Forces are the bonds or forces within the atoms of a molecule; the two bonds connecting two hydrogen atoms to an atom of oxygen in water are intramolecular forces. The force holding water molecules together so they don t evaporate are intermolecular forces and are About 1/6 to 1/10th as strong as single covalent bonds. Covalent, ionic, and metallic bonding are intramolecular forces. Metallic bonding is the strongest of the three.

9 Types of (intermolecular) forces (aka van derwaals forces): London Dispersion Forces (aka induced dipoleinduced dipole IM forces) - weakest of the IM forces (1) found in pure non-polar liquids (2) found in a solution of 2 different non-polar liquids

10 London dispersion forces are found holding any type of matter together, but these are the only ones found holding nonpolar molecules together!) Dispersion forces result from the constant movement of all particles. When molecules come close to one another, their electron clouds repel each other and a temporary dipole is formed. A dipole is like a magnet; it is any object that has a negative charge on one end and a positive charge on the other end. These dipoles are attracted to each other. Example of I 2 molecules being held together by dispersion forces:

11 Dipole Dipole IM Forces (stronger then dispersion forces) (1) found in pure polar liquids. Ex: between two molecules of methanol (2) found in solutions of 2 different polar liquids. Ex: between water and ammonia

12 Dipole-dipole IM forces occurs when one polar molecule is attracted to another polar molecule. Remember that polar molecules are those that are unequally sharing electrons, because of this all polar molecules are dipoles! Dipole dipole forces occur when two or more polar molecules are together, the molecules orient themselves so that the oppositely charged regions line up with each other. (Dipole-dipole forces are just like London dispersion forces, but much stronger because the dipole is constant and not induced). Example of HCl molecules being held together by dipole-dipole forces:

13 Hydrogen Bonding (strongest of all) (1) occurs in all water solutions along with other forces (2) are NEVER found as the only IM force. Dipole-dipole will always be one of the other IM forces here.

14 Hydrogen bonds are a special case of very strong dipole-dipole interactions. They are not really chemical bonds in the formal case. Strong hydrogen bonding occurs among polar covalent molecules containing H and one of the three small, highly electronegative elements F, O, or N. Hydrogen bonding is exactly the same as dipole-dipole forces except that a hydrogen atom from one molecule (which is in a polar bond) is attracted to a fluorine, oxygen or nitrogen atom (FON) of another molecule. Example of H 2 O molecules being held together by hydrogen bonds:

15 Practice: Draw the dot structure and shape of the following molecules and then determine its polarity and which type of intermolecular force would hold two or more of the following molecules together. 1. Br 2 2. HBr 3. CO 2 4. H 2 S 5. HF

16 Characteristics/Descriptions of Liquids Volatile when 2 liquids are compared, the one that evaporates more readily is said to be volatile. Boiling Point when the vapor pressure of the liquid equals the atmospheric pressure the liquid boils. Water boils at 100. C. Water has a relatively high boiling point. Draw water s dot structure and determine its molar mass: What accounts for the difference in boiling points between water and H 2 S, with a boiling point of 60.0 C?

17 Boiling Point and KE Only the fastest moving molecules are able to overcome the attractive forces of their neighbors and leave the surface of the liquid. The slower molecules are left behind. Thus the average kinetic energy (i.e. temperature) decreases.

18 Vapor Pressure the pressure exerted by a gas above a liquid in a closed container. Vapor pressure increases with increasing temperature. Because the rate of evaporation increases with increasing temperature, vapor pressures of liquids always increases as temperature increases. Vapor pressure decreases with increased IM forces. Water has low vapor pressure. It is good that water has a low vapor pressure, otherwise what would happen to the rivers & lakes? An easy way to remember vapor pressure is to ask yourself how fast the liquid evaporates, the faster the rate of evaporation, the higher the vapor pressure. Example: water and rubbing alcohol

19 Other facts about boiling points for all liquids: a) normal boiling point is the boiling point of a substance at 1 atm of pressure b) boiling point is unique for each liquid; it does not depend on volume or surface area c) boiling point increases with increased molecular mass if IM forces are the same d) boiling point increases with increased IM forces

20 Heat Capacity (or specific heat) the amount of heat needed to change the temperature of 1.0 g of a substance by 1.0 C. Because water has such a high heat capacity, it is cooler at the beach on a warm day. The water absorbs the heat and lowers the air temperature.

21 SUMMARY WHEN PREDICTING PROPERTIES Remember water has a: (1) high boiling point (2) high surface tension (3) high heat capacity (4) low vapor pressure because of dipole-dipole and hydrogen bonding

22 Next, look at the type of intramolecular forces (metallic, ionic, covalent). If all molecules being compared are covalent, then look at any intermolecular forces present. From the strongest to the weakest are: Metallic Ionic Covalent a. Hydrogen bonding with dipole-dipole b. dipole-dipole c. London dispersion Intramolecular forces Intremolecular forces

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24 Let s look at some of the Nature of Gases: 1. Expansion gases do NOT have a definite shape or volume. 2. Fluidity gas particles glide past one another, called fluid just like a liquid. 3. Compressibility can be compressed because gases take up mostly empty space. 4. Diffusion gases spread out and mix without stirring and without a current. Gases mix completely unless they react with each other.

25 Kinetic Molecular Theory of Gases 1. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

26 Kinetic Molecular Theory of Gases 2. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 3. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 4. Gas molecules exert neither attractive nor repulsive forces on one another. 5. Each gas molecule behaves as if it were alone in container (due to #3 and #4)

27 Demo 1: Diffusion of KMnO 4 Cold water Hot water Observations: KMnO 4 (potassium permanganate) diffuses faster in hot water. Explanation: KMnO 4 dissolves in water. As the KMnO 4 dissolves, it collides with H 2 O molecules and spreads out. Because molecules in hot water are moving faster, the solid dissolves faster and diffuses faster.

28 Demo 2: thermal expansion of a liquid Glass tube Rubber stopper Coloured water in florence flask Hot plate Observations: coloured water moves slowly up the tube Explanation: heat increases the kinetic energy of liquid particles. The particles move faster (greater vibrational, rotational, and translational energy). This greater movement increases the distance between molecules. Thus, the volume expands.

29 Demo 3: Heating Mercury Air Glass tube Beads Mercury Heat source Observations: beads are propelled upwards when mercury is heated. Explanation: heat increases the kinetic energy, so that the particles move faster. The fastest moving mercury molecules (that boil off) transfer their energy to the beads, causing them to jump. Would this work with water?

30 Will a gas diffuse faster or slower in a vacuum? Gas particles travel in a straight line until they hit another particle. A vacuum is devoid of molecules, so particles of a gas placed in a vacuum will not bump into anything until they hit the side of the container. Thus, a gas will diffuse faster in a vacuum.

31 Pressure KMT Viewpoint Origin of Pressure Gas molecules hitting container walls Temp, KE, # collisions, P Volume, # collisions, P

32 Pressure Macroscopic Viewpoint Pressure = Force Area Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmhg = 760 torr 1 atm = 101,325 Pa Temp, KE, Force, P Volume, Area, P

33 Atmospheric Pressure Barometer

34 Ideal vs. Real Gases Ideal gases always obey the kinetic theory. (Closest to ideal would be the noble gases.) Real gases vary from the kinetic theory at various temperatures and pressures. The greatest deviation from ideal gas behavior occurs at: high pressure higher density of gas molecules Molecules are closer together so:» finite volume of gas molecules more important» attraction between molecules more important This can also occur when the temperature is low the gas will change state form gas to liquid

35 Real Gases In the real world, the behavior of gases only conforms to the idealgas equation at relatively high temperature and low pressure. Curves for 1 mol of gas

36 Even the same gas (e.g. nitrogen) will show wildly different behavior under high pressure at different temperatures. Real Gases

37 Definitions Volume refers to the space matter (gas) occupies. Measured in liters (L). Pressure the number of times particles collide with each other and the walls of the container (force exerted on a given area). Measured in atmospheres (atm). 1atm = 760 millimeters Hg ( Barometers use Hg) 1atm = 760 torr (Named after Torricelli for the invention of the barometer) 1atm = kpa kilopascals Memorize Table 1. P. 542 STP: standard temperature and pressure (0 C, kpa or 1atm). SATP: standard ambient temperature and pressure (25 C, 100 kpa).

38 How Much Pressure? Q - A balloon is filled with pure oxygen. What is the pressure of the oxygen in the balloon? A - Atmospheric pressure. If it was not, then the balloon would expand or shrink. Q - A windbag is blown up with exhaled air. What is the pressure of oxygen in the bag? A Around 16-21% of atmospheric pressure (O 2 is 16% of exhaled air, 21% of atmosphere) Q - A solid container is filled with pure oxygen. What is the pressure in the container? A - It could be anything. The container is solid and therefore cannot shrink or expand.

39 Sample Problems Convert 4.40 atm to mmhg. Convert 212.4kPa to mmhg.

40 Definitions Temperature average kinetic energy - as temperate increases gas particles move faster, as temperature decreases gas particles move slower. Measured in Kelvin (K). K = C Number of Moles tells you how much of a certain gas you have 1 mole = number of grams of the compound or element (molar mass)

41 Demo 4: Kill da wabbit Question: what will happen when the air pressure surrounding a balloon is decreased? Why? Explanation: A balloon normally stays the same size because the pressure inside the balloon is equal to the pressure outside the balloon (rate of collisions with the wall of the balloon is equal). Reducing the pressure on the outside eliminates opposing collisions, allowing the balloon to expand.

42 Agenda Day 81 Gas Laws Lesson: PPT, Handouts: 1.PPT Handout 2. ISU- The Gas Laws ( W2-3)

43 Gas Laws How do all of pressure, temperature, volume, and amount of a gas relate to each other? Rules for solving gas law problems: 1 st write down what is given and what is unknown, 2 nd identify the gas law you want to use, and 3 rd rearrange the formula to solve for the unknown and then solve the problem. (If temperature is involved, it MUST be converted to Kelvin! K = C)

44 Boyle s Law - Pressure and Volume (when temperature remains constant) V 1 P 1 = V 2 P 2 V 1 = initial or old volume V 2 = final or new volume P 1 = initial or old pressure P 2 = final or new pressure Inverse Relationship (As pressure increases, volume decreases and as pressure decreases, volume increases.)

45 Effect of Pressure on Volume Boyle s Law 1 atm 2 atm 5 atm

46 Which picture represents what the gas will look like when the pressure is doubled? (Assume constant n, T)

47 Boyle s Law P a 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas 5.3

48 Kinetic Molecular theory of gases and Boyle s Law P a collision rate with wall Collision rate Increases with decreased volume P a 1/V Increase P, decrease volume

49 An increase in pressure decreases the volume of air in the lungs.

50 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 P 2 =? V 1 = 946 ml V 2 = 154 ml P 2 = P 1 x V 1 V mmhg x 946 ml = = 4460 mmhg 154 ml

51 Sample Problems What is the new pressure when 80.0mL of gas at 500.mmHg is moved to a 100.mL container? A gas at 800.torr of pressure has a volume of 5.00L. What volume does this gas occupy at 1.00X10 3 torr of pressure?

52 Charles Law -Volume and Temperature (when pressure is constant) V 1 /T 1 = V 2 /T 2 V 1 = initial or old volume V 2 = final or new volume T 1 = initial or old temperature T 2 = final or new temperature Direct Relationship (As temperature increases, volume increases and as temperature decreases, volume decreases.)

53 V a T V = kt V/T = k V 1 /T 1 = V 2 /T 2 Constant pressure Constant amount of gas As T increases, V Increases

54 The volume of a gas increases with and increase in temperature.

55 How Volume Varies With Temperature If we place a balloon in liquid nitrogen it shrinks: So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon). Let s take a closer look at temperature before we try to find the exact relationship of V vs. T.

56 Which picture represents what the gas will look like when the temperature is increased? (Assume constant n, P)

57 Kinetic theory of gases and Charles Law -Average kinetic energy a T -Increase T, Gas Molecules hit walls with greater Force, this Increases the Pressure BUT since pressure must remain constant, and only volume can change -Volume Increase to reduce Pressure -Increase Temperature, Increase Volume

58 Temperature scales Is 20 C twice as hot as 10 C? No. 68 F (20 C) is not double 50 F (10 C) Is 20 kg twice as heavy as 10 kg? Yes. 44 lb (20 kg) is double 22 lb (10 kg) What s the difference? Weights (kg or lb) have a minimum value of 0. But the smallest temperature is not 0 C. We saw that doubling P yields half the V. Yet, to investigate the effect of doubling temperature, we first have to know what that means. An experiment with a fixed volume of gas in a cylinder will reveal the relationship of V vs. T

59 Temperature vs. Volume Graph Volume (ml) Temperature ( C)

60 Determination of Absolute Zero

61 The Kelvin Temperature Scale If a volume vs. temperature graph is plotted for gases, most lines can be interpolated so that when volume is 0 the temperature is -273 C. Naturally, gases don t really reach a 0 volume, but the spaces between molecules approach 0. At this point all molecular movement stops. 273 C is known as absolute zero (no E K ) Lord Kelvin suggested that a reasonable temperature scale should start at a true zero value. He kept the convenient units of C, but started at absolute zero. Thus, K = C C =? K: K= C+273 = = 335 K Notice that kelvin is represented as K not K.

62 Kelvin Practice What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Absolute zero is 273 C or 0 K Calculate the missing temperatures 0 C = 273 K 100 C = 373 K 100 K = 173 C 30 C = 243 K 300 K = 27 C 403 K = 130 C 25 C = 298 K 0 K = 273 C

63 A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 /T 1 = V 2 /T 2 V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 =? T 2 = V 2 x T 1 V L x K = = 192 K 3.20 L

64 2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? V 1 = 3.5 L, T 1 = 300K, V 2 =?, T 2 = 200K Using Charles law: V 1 /T 1 = V 2 /T L / 300 K = V 2 / 200 K V 2 = (3.5 L/300 K) x (200 K) = 2.3 L 3. If a 1 L balloon is heated from 22 C to 100 C, what will its new volume be? V 1 = 1 L, T 1 = 22 C = 295 K V 2 =?, T 2 = 100 C = 373 K V 1 /T 1 = V 2 /T 2, 1 L / 295 K = V 2 / 373 K V 2 = (1 L/295 K) x (373 K) = 1.26 L

65 Gay-Lussac s Law - Pressure and Temperature (when volume is constant) P 1 /T 1 = P 2 /T 2 P 1 = initial or old pressure P 2 = final or new pressure T 1 = initial or old temperature T 2 = final or new temperature Direct Relationship - As the temperature of the gas increases the pressure of the gas Increases.

66 Sample Problems The gas in an aerosol can is at 3atm of pressure at 298K. What would the gas pressure in the can be at 325K? At 120. C the pressure of a sample of nitrogen gas is 769torr. What will the pressure be at 205 C?

67 The Combined Gas Law

68 Combining the gas laws So far we have seen three gas laws: Robert Boyle Jacques Charles P 1 V 1 = P 2 V 2 V 1 V = 2 T 1 T 2 These are all subsets of a more encompassing law: the combined gas law Joseph Louis Gay-Lu P 1 P = 2 T 1 T 2 P 1 V 1 P 2 V 2 = T 1 T 2

69 Combined Gas Law Equations P 1 = P 2 T 1 V 2 T 2 V 1 P 2 = P 1 T 2 V 1 T 1 V 2 T 1 = P 1 T 2 V 1 P 2 V 2 T 2 = P 2 T 1 V 2 P 1 V 1 V 1 = P 2 T 1 V 2 T 2 P 1 V 2 = P 1 T 2 V 1 P 2 T 1

70 Sample Problems A helium filled balloon has a volume of 50.0mL at 298K and 1.08atm. What volume will it have at 0.855atm and 203K? Given 700.mL of oxygen at 7.00 C and 7.90atm of pressure, what volume does is occupy at 27.0 C and 4.90atm of pressure? Note: any unit for pressure will work, provided the same units are used throughout. The only unit that MUST be used is K for temperature.

71 V 1 = 50.0 ml, P 1 = 101 kpa V 2 = 12.5 ml, P 2 =? T 1 = T 2 P 1 V 1 T 1 = (101 kpa)(50.0 ml) (T 1 ) (P 2 ) = = (101 kpa)(50.0 ml)(t 2 ) (T 1 )(12.5 ml) P 2 V 2 T 2 (P 2 )(12.5 ml) (T 2 ) = 404 kpa Notice that T cancels out if T 1 = T 2

72 V 1 = 0.10 L, T 1 = 298 K V 2 =?, T 2 = 463 P 1 = P 2 (V 2 ) = (P 1 )(0.10 L) (298 K) P 1 V 1 T 1 = = P 2 V 2 T 2 (P 1 )(0.10 L)(463 K) (P 2 )(298 K) (P 2 )(V 2 ) (463) = 0.16 L Notice that P cancels out if P 1 = P 2

73 P 1 = 150 kpa, T 1 = 308 K P 2 = 250 kpa, T 2 =? V 1 = V 2 (T 2 ) P 1 V 1 T 1 = (150 kpa)(v 1 ) (308 K) = = (250 kpa)(v 2 )(308 K) (150 kpa)(v 1 ) P 2 V 2 T 2 (250 kpa)(v 2 ) (T 2 ) Notice that V cancels out if V 1 = V 2 = 513 K = 240 C

74 (V 2 ) P 1 = 100 kpa, V 1 = 5.00 L, T 1 = 293 K P 2 = 90 kpa, V 2 =?, T 2 = 308 K P 1 V 1 P = 2 V 2 T 1 T 2 (100 kpa)(5.00 L) (293 K) = = (100 kpa)(5.00 L)(308 K) (90 kpa)(293 K) (90 kpa)(v 2 ) (308 K) = 5.84 L

75 P 1 = 800 kpa, V 1 = 1.0 L, T 1 = 303 K P 2 = 100 kpa, V 2 =?, T 2 = 298 K (V 2 ) P 1 V 1 T 1 = (800 kpa)(1.0 L) (303 K) = = (800 kpa)(1.0 L)(298 K) (100 kpa)(303 K) P 2 V 2 T 2 (100 kpa)(v 2 ) (298 K) = 7.9 L

76 P 1 = 6.5 atm, V 1 = 2.0 ml, T 1 = 283 K P 2 = 0.95 atm, V 2 =?, T 2 = 297 K (V 2 ) P 1 V 1 T 1 = (6.5 atm)(2.0 ml) (283 K) = = (6.5 atm)(2.0 ml)(297 K) (0.95 atm)(283 K) P 2 V 2 T 2 (0.95 atm)(v 2 ) (297 K) = 14 ml 33. The amount of gas (i.e. number of moles of gas) does not change.

77 The Ideal Gas Law PV = nrt

78 Agenda Day 82 Ideal Gas Law Read: Section Make your own notes on this section Lesson: PPT, Handouts: 1.PPT Handout 2. ISU- The Ideal Gas Law ( W4)

79 Ideal Gas Equation Boyle s law: V a 1 (at constant n and T) P Charles law: V a T (at constant n and P) Avogadro s law: V a n (at constant P and T) V a nt P V = constant x nt P = R nt P PV = nrt R is the gas constant R = L atm / (mol K)

80 Ideal Gases An ideal gas exhibits certain theoretical properties. Specifically, an ideal gas Obeys all of the gas laws under all conditions. Does not condense into a liquid when cooled. Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph. In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations. We have done calculations using several gas laws (Boyle s Law, Charles s Law, Combined Gas Law). There is one more to know

81 Sample problems How many moles of H 2 is in a 3.1 L sample of H 2 measured at 300 kpa and 20 C? PV = nrt P = 300 kpa, V = 3.10 L, T = 293 K (300 kpa)(3.1 L) = n (8.31 kpa L/K mol)(293 K) (300 kpa)(3.1 L) = n = 0.38 mol (8.31 kpa L/K mol)(293 K) How many grams of O 2 are in a 315 ml container that has a pressure of 12.0 atm at 25 C? PV = nrt P= kpa, V= L, T= 298 K ( kpa)(0.315 L) = n = mol (8.31 kpa L/K mol)(298 K) mol x g/mol = 4.95 g

82 Argon is an inert gas used in lightbulbs to retard the 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 nr V P 1 T 1 = P T = constant = P 2 T 2 P 2 = P 1 x T 2 T 1 n, V and R are constant P 1 = 1.20 atm T 1 = 291 K = 1.20 atm x 358 K 291 K = 1.48 atm P 2 =? T 2 = 358 K

83 Solving for Density and /or Molar Mass of a gas using the Ideal Gas Law Density (units are g/l) Use the Ideal Gas Law to find moles (n), convert n to grams OR use the Ideal Gas Law to find the volume. Divide m (in grams) by the volume. Molar Mass (units are g/mol) If density is given, use the density of the gas to determine the molar mass (use 1 L at the volume and solve for n). If a mass is given, use the Ideal Gas Law to solve for n and then find the molar mass.

84 Ideal Gas Law Questions 1. How many moles of CO 2 (g) is in a 5.6 L sample of CO 2 measured at STP? 2. a) Calculate the volume of 4.50 mol of SO 2 (g) measured at STP. b) What volume would this occupy at 25 C and 150 kpa? (solve this 2 ways) 3. How many grams of Cl 2 (g) can be stored in a 10.0 L container at 1000 kpa and 30 C? 4. At 150 C and 100 kpa, 1.00 L of a compound has a mass of g. Calculate its molar mass ml of an unknown gas weighs g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?

85 Sample Problems What is the density of a sample of ammonia gas, NH 3, if the pressure is atm and the temperature is 63.0 C? What is the density of argon gas at a pressure of 551 torr and a temperature of 25.0 C?

86 Sample Problems The density of a gas was found to be 2.00g/L at 1.50atm and 27.0 C. What is the molar mass of the gas? What is the molar mass of a gas if 0.427g of the gas occupies a volume of 125mL at 20.0 C and 0.980atm?

87 Types of Problems PV 1 1 PV 2 2 Make Substitution into PV = nrt n T 1 1 n 2 T 2 moles( n) mass, g MolarMass, g / mole Given initial conditions, determine final conditions; Cancel out what is constant Density mass Volume

88 Density (d) Calculations d = m V = PM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance M = drt P d is the density of the gas in g/l

89 1. Moles of CO 2 is in a 5.6 L at STP? P= kpa, V=5.6 L, T=273 K PV = nrt (101.3 kpa)(5.6 L) = n (8.31 kpa L/K mol)(273 K) ( kpa)(5.6 L) (8.31 kpa L/K mol)(273 K) = n = 0.25 mol 2. a) Volume of 4.50 mol of SO 2 at STP. P= kpa, n= 4.50 mol, T= 273 K PV=nRT (101.3 kpa)(v)=(4.5 mol)(8.31 kpa L/K mol)(273 K) (4.50 mol)(8.31 kpa L/K mol)(273 K) V = = L (101.3 kpa)

90 2. b) Volume at 25 C and 150 kpa (two ways)? Given: (4.50 P mol)(8.31 = 150 kpa, kpa L/K mol)(298 n = 4.50 mol, T = K) 298 K V = = 74.3 L (150 kpa) From a): P = kpa, V = L, T = 273 K Now P = 150 kpa, V =?, T = 298 K P 1 V 1 P = 2 V 2 T 1 T 2 (101.3 kpa)(100 L) (150 kpa)(v = 2 ) (273 K) (298 K) (101.3 kpa)(100.8 L)(298 K) (V 2 ) = = 74.3 L (273 K)(150 kpa)

91 3. How many grams of Cl 2 (g) can be stored in a 10.0 L container at 1000 kpa and 30 C? PV = nrt P= 1000 kpa, V= 10.0 L, T= 303 K (1000 kpa)(10.0 L) = n = 3.97 mol (8.31 kpa L/K mol)(303 K) 3.97 mol x 70.9 g/mol = 282 g 4. At 150 C and 100 kpa, 1.00 L of a compound has a mass of g. Calculate PV molar = nrt mass. P= 100 kpa, V= 1.00 L, T= 423 K (100 kpa)(1.00 L) = n = mol (8.31 kpa L/K mol)(423 K) g/mol = g / mol = 88.1 g/mol

92 5. 98 ml of an unknown gas weighs g at SATP. Calculate the molar mass. PV = nrt P= 100 kpa, V= L, T= 298 K (100 kpa)(0.098 L) = n = mol (8.31 kpa L/K mol)(298 K) g/mol = g / mol = g/mol It s probably neon (neon has a molar mass of g/mol)

93 Agenda Day 83 Avogadro s Law of Combining Volumes and Partial Pressures Read Section 12.1 & Make your own study notes on this section. Lesson: PPT, Handouts: 1.PPT Handout 2. ISU- ( W4)

94 Avogadro s Law V a number of moles (n) V = constant x n Constant temperat Constant pressure V 1 /n 1 = V 2 /n 2

95 Kinetic theory of gases and Avogadro s law states that equal volumes of gases at the same temperature and pressure contain equal number of particles Avogadro s Law More moles of gas, more collisions with walls of container More collisions, higher pressure BUT since pressure must remain constant and only volume can change Volume increases to decrease pressure to original value

96 Which picture represents what the gas will look like when the moles of gas is doubled? (Assume constant P, T)

97 Avogadro s Law

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99 Daltons Law of Partial Pressures The pressure of each gas in a mixture is called the partial pressure of that gas. Daltons Law of Partial Pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. So far: pure gases Many gases are actually mixtures of two or more gases: air: O 2, N 2, H 2 O, etc How do mixtures of gases behave? Gas Mixtures--Partial Pressure

100 Gas Mixtures--Partial Pressure P= 6 kpa P= 8 kpa P= 9 kpa O 2 (g) N 2 (g) CO 2 (g)

101 Gas Mixtures--Partial Pressure P total = P O2 + P N2 + P CO2 So for this example: P total = 6 kpa + 8 kpa + 9 kpa = 23 kpa

102 Dalton s Law of Partial Pressures V and T are constant P 1 P 2 P total = P 1 + P 2

103 Kinetic theory of gases and Dalton s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total = SP i In other words, at constant T and V, P total depends only on the total number of moles of gas present P total is independent of the type (or types) of gases present. This works according to the KMT because at the same temperature molecules of different gases have the same Kinetic Energy.

104 Partial Pressure-Mole Fraction When describing a mixture of gases, it is useful to know the relative amount of each type of gas. Mole fraction (X): a dimensionless number that expresses the ratio of the number of moles of one component compared to the total number of moles in a mixture.

105 Mole Fraction If a gas mixture contains 5.0 mol O 2(g), 3.0 mol H 2 O (g), and 12.0 mol N 2(g), what is the mole fraction of oxygen? X O = n O2 n t = 5.0 mol 20.0 mol = 0.25 On the exam, you must be able to calculate the mole fraction of each component of a gas mixture.

106 Partial Pressure The partial pressure of a gas in a mixture can be found: P A = X A P total where P A = partial pressure of gas A X A = mole fraction of gas A P total = total pressure of mixture

107 Partial Pressure Calculation A mixture of gases contains 0.51 mol N 2, 0.28 mol H 2, and 0.52 mol NH 3. If the total pressure of the mixture is 2.35 atm, what is the partial pressure of H 2? P H2 = X H2 P total X H2 = 0.28 mol 0.28 mol mol mol = 0.21 P H2 = 0.21 x 2.35 = 0.50 atm

108 Sample Problems A mixture of gases has the following partial pressure for the component gases at 20.0 C in a volume of 2.00L: oxygen 180.torr, nitrogen 320.torr, and hydrogen 246torr. Calculate the pressure of the mixture. What is the final pressure of a 1.50L mixture of gases produced from 1.50L of neon at atm, 800.mL of nitrogen at 150.mmHg and 1.2oL of oxygen at 25.3kPa? Assume constant temperature. (Hint use Boyle s law.)

109 In the lab Chemical reaction producing gas eg: NH 4 NO 2 (s) N 2 (g) + H 2 O (l) Determine number of moles (amount) of gas collected?

110 When a gas is collected over water; you always have a mixture of that gas and water.

111 Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. P total = P gas + P H2 O To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure. Table1 P. 596 shows water vapor pressure (T dep)

112

113 Vapour Pressure Defined Vapour pressure is the pressure exerted by a vapour. E.g. the H 2 O(g) in a sealed container. Eventually the air above the water is filled with vapour pushing down. As temperature, more molecules fill the air, and vapour pressure. Yet, molecules both leave and join the surface, so vapour pressure also pushes molecules up. To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the pressure it exerts at different C.

114 Measuring Vapour Pressure Vapour pressure Temperature Vapour pressure for H 2 O C kpa C kpa See pg. 464 for more When the vapour pressure is equal to the atmospheric pressure (P atm ), the push out is enough to overcome P atm and boiling occurs. Thus, water will boil at a temperature below 100 C if the atmospheric pressure is reduced.

115 Collecting gases over water Many times gases are collected over H 2 O Often we want to know the volume of dry gas at STP (useful for stoichiometry). For this we must make 3 corrections: 1. The level of water inside and outside the tube must be level (so pressure inside is equal to the pressure outside). 2. The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas). 3. Finally, values are converted to STP using the combined gas law.

116 Sample calculation A gas was collected over 21 C H 2 O. After equalizing water levels, the volume was 325 ml. Give the volume of dry gas at STP (P atm =102.9 kpa). Step 1: Determine vapour pressure (pg. 596) At 21 C vapour pressure is 2.49 kpa Step 2: Calculate the pressure of dry gas P gas = P atm - P H2O = = kpa Step 3: List all of the data T 1 = 294 K, V 1 = 325 ml, P 1 = kpa Step 4: Convert to STP = 299 ml (P 1 )(V 1 )(T 2 ) (100.4 kpa)(325 ml)(273 K) V 2 = = (P 2 )(T 1 ) ( kpa)(294 K)

117 Assignment ml of O 2 is collected by the downward displacement of water at 24 C and an atmospheric pressure of kpa. What is the volume of dry oxygen measured at STP? ml of H 2 is collected over water at 22 C and at an atmospheric pressure of 99.8 kpa. What is the volume of dry H 2 at STP? 3. If H 2 is collected over water at 22 C and an atmospheric pressure of kpa, what is the partial pressure of the H 2 when the water level inside the gas bottle is equal to the water level outside the bottle?

118 1) Vapor pressure at 24 C = 2.98 kpa Pgas = Patm - Pvapor = kpa kpa = kpa = P 1 V 1 = 37.8 ml, P 1 = kpa, T 1 = 297 K V 2 =?, P 2 = kpa, T 2 = 273 K P 1 V 1 = T 1 (99.42 kpa)(37.8 ml) = (297 K) P 2 V 2 T 2 (101.3 kpa)(v 2 ) (273 K) (99.42 kpa)(37.8 ml)(273 K) (V 2 ) = = 34.1 ml (297 K)(101.3 kpa)

119 2) Vapor pressure at 22 C = 2.64 kpa Pgas = Patm - Pvapor = 99.8 kpa kpa = kpa = P 1 V 1 = 236 ml, P 1 = kpa, T 1 = 295 K V 2 =?, P 2 = kpa, T 2 = 273 K P 1 V 1 T 1 = (97.16 kpa)(236 ml) (295 K) P 2 V 2 T 2 (101.3 kpa)(v 2 ) = (273 K) (97.16 kpa)(236 ml)(273 K) (V 2 ) = = 209 ml (295 K)(101.3 kpa)

120 Answers 3 - Total pressure = PH 2 + PH 2 O kpa = PH kpa kpa kpa = PH 2 = kpa

121 Gas Stoichiometry

122 Agenda Day 84 Stoichiometry and Gases Look over Tutorial 1 on P. 598 and Tutorial 2 on P Lesson: PPT, Handouts: 1.PPT Handout 2. ISU- Stoichiometry and gases ( W4)

123 Molar Volume of Gases Recall that 1 mole of a compound contains X molecules of that compound it doesn t matter what the compound is. One mole of any gas, at STP, will occupy the same volume as one mole of any other gas at the same temperature and pressure, despite any mass differences. The volume occupied by one mole of a gas at STP is known as the standard molar volume of a gas. It has been found to be 22.4liters. We can use this as a new conversion factor 1mol of gas/22.4l of same gas. (Avogadro s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules).

124 Sample Problems What volume, in L, is occupied by 32.0 grams of oxygen gas at STP?

125 If all the reactants and products are gases in a chemical reaction and the pressure and temperature remain constant throughout the problem: In stoichiometry problems, the coefficients of a balanced chemical equation will also represent the volumes of gases found at a fixed temperature and pressure When solving such problems gas volumes are equivalent to their molar ratios

126 Gas Stoichiometry We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations. We can use stoichiometry for gas reactions. As before, we need to consider mole ratios when examining reactions quantitatively. molar mass of x mole ratio from balanced equation molar mass of y grams (x) moles (x) moles (y) grams (y) PV = nrt P, V, T (x) P, V, T (y) At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts.

127 Sample problem 1 CH 4 burns in O 2, producing CO 2 and H 2 O(g). A 1.22 L CH 4 cylinder, at 15 C, registers a pressure of 328 kpa. a) What volume of O 2 at SATP will be required to react completely with all of the CH 4? First: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) PV = nrt P = 328 kpa, V = 1.22 L, T = 288 K (328 kpa)(1.22 L) = n = mol (8.31 kpa L/K mol)(288 K) # mol O 2 = mol CH 2 mol O 2 4 x = mol 1 mol CH 4 PV = nrt P= 100 kpa, n= mol, T= 298 K (0.334 mol)(8.31 kpa L/K mol)(298 K) =V = 8.28 L (100 kpa) or # L = mol x 24.8 L/mol = 8.28 L

128 Sample problem 1 continued CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) b) How many grams of H 2 O(g) are produced? # g H 2 O= mol CH 2 mol H 4 2O g H x 2 O x = 6.02 g 1 mol CH 4 1 mol H 2 O H 2 O c) What volume of CO 2 (at STP) is produced if only 2.15 g of the CH 4 was burned? # mol CO 2 = 2.15 g CH 1 mol CH 4 1 mol CO 4 x x 2 = g CH 4 1 mol CH 4 mol CO 2 PV = nrt P = kpa, n = mol, T = 273 K (0.134 mol)(8.31 kpa L/K mol)(273 K) = V = 3.00 L CO 2 (101.3 KPa) or # L = mol x 22.4 L/mol = 3.00 L

129 Sample problem 2 Ammonia (NH 3 ) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kpa and 80 C can be obtained from the complete reaction of 7.5 kg hydrogen? First we need a balanced equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) # mol NH 3 = 7500 g H 1 mol H 2 2 mol NH 2 x x 3 = 2475 mol 2.02 g H 2 3 mol H 2 PV = nrt P = 450 kpa, n = 2475 mol, T = 353 K (2475 mol)(8.31)(353 K) = V = L NH 3 (450 KPa)

130 Sample problem 3 Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H 2 at STP? First we need a balanced equation: 2Na(s) + 2H 2 O(l) H 2 (g) + 2NaOH(aq) PV = nrt P= kpa, V= 20.0 L, T= 273 K (101.3 kpa)(20.0 L) = n = mol H (8.31 kpa L/K mol)(273 K) 2 or # mol = 20.0 L x 1 mol / 22.4 L = mol # g Na= mol H 2 mol Na g Na 2 x x = 41.1 g Na 1 mol H 2 1 mol Na

131 Assignment 1. What volume of oxygen at STP is needed to completely burn 15 g of methanol (CH 3 OH) in a fondue burner? (CO 2 + H 2 O are products) 2. When sodium chloride is heated to 800 C it can be electrolytically decomposed into Na metal & chlorine (Cl 2 ) gas. What volume of chlorine gas is produced (at 800 C and 100 kpa) if 105 g of Na is also produced? 3. What mass of propane (C 3 H 8 ) can be burned using 100 L of air at SATP? Note: 1) air is 20% O 2, so 100 L of air holds 20 L O 2, 2) CO 2 and H 2 O are the products of this reaction.

132 4. A 5.0 L tank holds 13 atm of propane (C 3 H 8 ) at 10 C. What volume of O 2 at 10 C & 103 kpa will be required to react with all of the propane? 5. Nitroglycerin explodes according to: 4 C 3 H 5 (NO 3 ) 3 (l) 12 CO 2 (g) + 6 N 2 (g) + 10 H 2 O(g) + O 2 (g) a) Calculate the volume, at STP, of each product formed by the reaction of 100 g of C 3 H 5 (NO 3 ) 3. b) 200 g of C 3 H 5 (NO 3 ) 3 is ignited (and completely decomposes) in an otherwise empty 50 L gas cylinder. What will the pressure in the cylinder be if the temperature stabilizes at 220 C?

133 Answers 1. 3O 2 (g) + 2CH 3 OH(l) 2CO 2 (g) + 4H 2 O(g) # L O 2 = 15 g CH 3 OH 1 mol CH 3 OH 3 mol O x x g CH 3 OH 2 mol CH 3 OH x 22.4 L O 2 1 mol O NaCl(l) 2Na(l) + Cl 2 (g) # mol Cl 2 = 105 g Na 1 mol Na 1 mol Cl x x g Na 2 mol Na = 15.7 L O 2 = mol Cl 2 PV = nrt P = 100 kpa, n = mol, T = 1073 K (2.284 mol)(8.31)(1073 K) = V = 204 L Cl 2 (100 KPa)

134 3. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) # g C 3 H 8 = 20 L O 1 mol O 2 2 x 24.8 L O 2 1 mol C 3 H 8 5 mol O 2 = 7.1 g C 3 H 8 4. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) PV = nrt x g C 3 H x 8 1 mol C 3 H 8 n = (1317 kpa)(5.0 L) = 2.8 mol C 3 H 8 (8.31)(283 K) # mol O 2 = 2.8 mol C 3 H 5 mol O 2 8 x = 14 mol O 2 1 mol C 3 H 8 PV = nrt P = 103 kpa, n = 14 mol, T = 283 K (14 mol)(8.31)(283 K) = V = 320 L O 2 (103 KPa)

135 5. # mol C 3 H 5 (NO 3 ) 3 = 100 g C 3 H 5 (NO 3 ) 1 mol C 3 H 5 (NO 3 ) 3 3 x g C 3 H 5 (NO 3 ) 3 # L CO 2 = = mol mol 12 mol CO L x x = 29.6 C 3 H 5 (NO 3 ) 3 4 mol C 3 H 5 (NO 3 ) 3 1 mol L CO 2 # L N 2 = mol 6 mol N L x x = 14.8 C 3 H 5 (NO 3 ) 3 4 mol C 3 H 5 (NO 3 ) 3 1 mol L N 2 # L H 2 O= mol C 3 H 5 (NO 3 ) 3 10 mol H 2 O 22.4 L x x = mol C 3 H 5 (NO 3 ) 3 1 mol L H 2 O # L O 2 = mol 1 mol O L x x = 2.47 C 3 H 5 (NO 3 ) 3 4 mol C 3 H 5 (NO 3 ) 3 1 mol L O 2

136 5. # mol C 3 H 5 (NO 3 ) 3 = 200 g C 3 H 5 (NO 3 ) 1 mol C 3 H 5 (NO 3 ) 3 3 x g C 3 H 5 (NO 3 ) 3 # mol all gases= mol C 3 H 5 (NO 3 ) 29 mol gases 3 x 4 mol C 3 H 5 (NO 3 ) 3 = mol = mol all gases PV = nrt V = 50 L, n = mol, T = 493 K (6.385 mol)(8.31)(493 K) = P = 523 kpa (50 L)