Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering the air density, making the balloon buoyant.

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1 1 12 The Gaseous State of Matter Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering the air density, making the balloon buoyant. Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Copyright 2 Chapter Outline 12.1 Properties of Gases A. Measuring the Pressure of a Gas B. Pressure Dependence: Number of Molecules and Temperature 12.2 Boyle s Law 12.3 Charles Law 12.4 Avogadro s Law A. Mole-Mass-Volume Calculations 12.5 Combined Gas Laws 12.6 Ideal Gas Law A. Kinetic-Molecular Theory B. Real Gases 12.7 Dalton s Law of Partial Pressures 12.8 Density of Gases 12.9 Gas Stoichiometry 3 Gases: Properties of Gases i) Have indefinite volume Expand to fill a container ii) Have indefinite shape Assume the shape of a container iii) Have low densities Example d air = 1.2 g/l at 25 C d water = 1.0 g/ml at 25 C iv) Have high velocities and kinetic energies Volume occupied by 1 mol of H 2 O: as a liquid (18 ml) as a gas (22.4 L)

2 4 Measuring Pressure Pressure: Force per unit area force Pressure = area Pressure depends on: 1) The number of gas molecules 2) Gas temperature 3) Volume occupied by the gas Pressure results from gas molecule collisions with the container walls. SI unit of pressure is the pascal (Pa) = 1 newton/meter 2 Unit Conversions: 1 atm = 760 mm Hg = 760 torr = kpa = bar = psi 5 Practicing Pressure Conversions Convert 740. mm Hg to a) atm and b) kpa. a) Use the conversion factor: 1 atm = 760 mm Hg 740. mm Hg 740. mm Hg 1 atm 760 mm Hg kpa 760 mm Hg = atm b) Use the conversion factor: kpa = 760 mm Hg = kpa 6 Atmospheric Pressure Definition: total pressure exerted by gases in the atmosphere Due to the mass of the atmospheric gases pressing downward on the Earth s surface. Major Components of Dry Air

3 7 Use a Barometer Measuring Pressure Measuring Pressure 1) Invert a long tube of Hg over an open dish of Hg. 2) Hg will be supported (pushed up) by the pressure of the atmosphere. 3) Height of Hg column can be used to measure pressure. 8 Pressure Dependence 1) On the Number of Molecules Pressure (P ) is directly proportional to the number of gas molecules present (n ) at constant temperature (T ) and volume (V ). Increasing n creates more frequent collisions with the container walls, increasing the pressure V = 22.4 L T = 25.0 C 0.5 mol H 2 P = 0.5 atm 1 mol H 2 P = 1 atm 2 mol H 2 P = 2 atm 9 Pressure Dependence 2) On Temperature Pressure is directly proportional to temperature when moles (n ) and volume (V ) are held constant. Increasing T causes: a) more frequent and b) higher energy collisions 0.1 mol of gas in a 1L container T = 0 C T = 100 C 2.24 atm 3.06 atm

4 10 Boyle s Law The volume of a fixed quantity of gas is inversely proportional to the pressure exerted by the gas at constant mass and temperature. Graph showing inverse PV relationship PV = constant (k ) or P 1 V P = k V 1 Most common form: P 1 V 1 = P 2 V 2 11 Boyle s Law Problems What volume will 3.5 L of a gas occupy if the pressure is changed from 730. mm Hg to 600. mm Hg? P 1 V 1 = P 2 V 2 V 1 = 3.5 L P 1 = 730. mm Hg P 2 = 600. mm Hg P 1 V 1 Solve For V 2 V 2 = P2 V 2 = 3.5 L 730. mm Hg 600. mm Hg = 4.3 L 12 Boyle s Law Problems A sample of Ne gas occupies 250. ml at 880. torr. the P Ne if the volume is increased to 1.0 L, assuming constant temperature. (Note: Convert ml to L.) P 1 V 1 = P 2 V 2 V 1 = L V 2 = 1.0 L P 1 = 880. mm Hg P 1 V 1 Solving For P 2 P 2 = V L P 2 = 880. torr 1.0 L = 220 mm Hg

5 13 A sample of gaseous nitrogen in a 65.0 L automobile air bag has a pressure of 745 mm Hg. If the sample is transferred to a 25.0 L bag at the same temperature, what is the pressure in the bag? a) 2.18 mm Hg b) 1940 mm Hg c) 287 mm Hg d) mm Hg P 1 V 1 P 2 = = V2 Boyle s Law Problems 65.0 L 745 torr = 1940 mm Hg 25.0 L Sense check: As volume decreases, pressure should increase! 14 Temperature in Gas Law Problems Kelvin Temperature Scale Derived from the relationship between temperature and volume of a gas. As a gas is cooled by 1 ºC increments, the gas volume decreases in increments of 1/273. All gases are expected to have zero volume if cooled to 273 ºC. V T relationship of methane (CH 4 ) with extrapolation ( ) to absolute zero. 15 Temperature in Gas Law Problems This temperature ( 273 ºC) is referred to as absolute zero. Absolute zero is the temperature (0 K) when the volume of an ideal gas becomes zero. All gas law problems use the Kelvin temperature scale! Celsius temperature T K = T C Kelvin temperature

6 16 Charles Law The volume of a fixed quantity of gas is directly proportional to the absolute temperature of the gas at constant pressure. V = k T or V T = k V T Most common form: V 1 V = 2 T 1 T 2 17 Charles Law Problems 3.0 L of H 2 gas at 15 ºC is allowed to warm to 27 ºC at constant pressure. What is the gas volume at 27 ºC? V 1 V = 2 T 1 T 2 V 1 = 3.0 L T 1 = 15 ºC = 258 K T 2 = 27 ºC = 300. K V 1 T 2 Solving For V 2 V 2 = T 1 V 2 = V 1 T 2 = 3.0 L 300. K 258 K = 3.5 L T 1 18 Charles Law Problems A gas has a volume of 3.00 L at 10.0 ºC. What is the temperature of the gas if it expands to 6.00 L, assuming constant pressure? V 1 V = 2 T 1 T 2 V 1 = 3.00 L T 1 = 10.0 ºC = 283 K T 1 V 2 Solving For T 2 T 2 = V1 V 2 = 6.00 L T 2 = T 1 V 2 = 283 K 6.00 L 3.00 L = 566 K V 1

7 19 a) 916 ml b) 109 ml c) 943 ml d) 427 ml Charles Law Problems At 321 K, a gas occupies 635 ml of volume. If the temperature is decreased to 216 K, what is the new gas volume? V 2 = V 1 T 2 = 635 ml T K 321 K = 427 ml Sense Check: As temperature decreases, volume decreases! 20 Avogadro s Law Equal volumes of different gases at constant T and P contain the same number of molecules. 1 volume unit 4 molecules 1 volume unit 4 molecules 2 volume units 8 molecules 21 Given the following gas phase reaction: N H 2 2 NH 3 If 12.0 L of H 2 gas are present, what volume of N 2 gas is required for complete reaction? T and P are held constant. By Avogadro s Law, we can use the reaction stoichiometry to predict the N 2 gas needed. Solving For V N2 V H2 = 12.0 L Avogadro s Law 12.0 L H 2 1 L N 2 3 L H 2 = 4.00 L N 2 required

8 22 At constant T and P, how many liters of O 2 are required to make 45.6 L of H 2 O? a) 11.4 L b) 45.6 L c) 22.8 L d) 91.2 L Avogadro s Law Given the following gas phase reaction: 2 H 2 + O 2 2 H 2 O 45.6 L H 2 O 1 L O 2 2 L H 2 O = 22.8 L O 2 required Sense Check: Less moles of O 2 equal less L of O 2! 23 Molar Volume: volume 1 mol of gas occupies at STP molar volume = 22.4 L/mol at STP Molar volume can be used as a conversion factor if the mass and volume occupied by a gas are known. Example: Mole/Mass/Volume Relationships 1.0 L of O 2 at STP has a mass of g. Show that the molar mass of O 2 is 32.0 g/mol g O L O L O 2 1 mol O 2 = 32.0 g/mol O 2 24 Mole/Mass/Volume Relationships If 3.00 L of a gas measured at STP has a mass of 5.35 g, calculate the molar mass. a) 39.9 g/mol b) 79.6 g/mol c) 12.6 g/mol d) 25.0 g/mol 5.35 g gas 3.00 L gas 22.4 L O 2 1 mol gas = 39.9 g/mol Unit Check: Molar mass has units of g/mol, so use dimensional analysis when setting up the problem!

9 25 Combined Gas Laws A combination of Boyle s and Charles Laws. Used in problems involving changes in P, T, and V with a constant amount of gas. P 1 V 1 P 2 V = 2 T 1 T 2 The volume of a fixed quantity of gas depends on the temperature and pressure. It is not possible to state the volume of gas without stating the temperature and pressure. Standard Temperature and Pressure (STP): 0.00 C ( K) and 1 atm (760 torr) 26 Combined Gas Law Problems A sample of gas occupies 125 ml at STP. What is the volume of the gas at 65 ºC and 320. torr? V 1 = L P 1 = 760 torr T 1 = 273 K P 2 = 320 torr T 2 = 65 ºC = 338 K Solving For V 2 V 2 = P 1 V 1 T 2 T1 P 2 V 2 = V 1 P 1 T 2 = L 760. torr 338 K = L 320. torr 273 K P 2 T 1 P 1 V 1 P 2 V = 2 T 1 T 2 27 What is the volume at STP for a gas that occupies 1.62 L at 616 torr and 42 C? V 1 = 1.62 L Solving For V 2 V 2 = T1 P 2 V 2 = V 1 P 1 T 2 = Combined Gas Law Problems P 2 T L P 1 V 1 P 2 V = 2 T 1 T 2 P 1 = 616 torr T 1 = 42 C = 315 K P 2 = 760. torr T 2 = 273 K P 1 V 1 T torr 760. torr 273 K 315 K = 1.14 L

10 28 A balloon is filled with 266 L of He gas, measured at 38 C and atm. What will its volume be when the temperature is lowered to 76 C and the pressure is atm? a) 299 L b) 95.0 L c) 745 L d) 237 L Combined Gas Law Problems V 2 = V 1 P 1 = 266 L T 2 P 2 T atm atm 197 K 311 K = 299 L 29 Ideal Gas Law A single equation relating all properties of a gas. PV = nrt where R is the universal gas constant Constant n and T Constant n and P Constant P and T V 1/P Boyle s Law V T Charles Law V n Avogadro s Law 30 R is derived from conditions at STP. R. Solving For R Ideal Gas Constant P = 1.00 atm V = 22.4 L T = 273 K n = 1.00 mol R = PV nt PV = nrt R = P V= 1.00 atm 22.4 L = L. atm n 1.00 mol 273 K mol. K T Units are critical in ideal gas problems!

11 31 How many moles of He are contained in a L container at 30. ºC and atm? Solving For n P = atm V = L T = 30. ºC = 303 K n = P V= R T Ideal Gas Law Practice n = PV RT PV = nrt atm L L. atm 303 K mol. K = mol 32 What volume will be occupied by mol of N 2 at atm and 24 C? Solving For V V = nrt P Ideal Gas Law Practice P = atm n = mol T = 24 ºC = 297 K V = nrt P PV = nrt = mol L. atm 297 K = 9.87 L mol. K atm 33 Ideal Gas Law Practice The ideal gas law can also be written in terms of molar mass of a gas. PV = nrt n = mass in grams (g) molar mass ( ) PV = grt

12 34 A g gas sample has a pressure of 432 torr in a 333 ml container at 23 ºC. What is the molar mass of the gas? Solving For Ideal Gas Law Practice PV = grt P = 432 torr = atm V = L T =296 K mass = g = grt = g L atm/mol K 296 K= 27.0 g/mol PV atm John L Wiley & Sons, Inc. All rights reserved. 35 the molar mass ( ) of an unknown gas if g occupies a volume of 754 ml at 30. ºC and 342 torr. a) 35.4 g/mol b) 21.9 g/mol c) 87.3 g/mol d) 55.0 g/mol = grt PV Ideal Gas Law Practice = g L atm/mol K 303 K= 56.3 g/mol atm L 36 Kinetic Molecular Theory A general theory developed to explain the behavior and theory of gases, based on the motion of particles. Assumptions of Kinetic Molecular Theory (KMT): 1) Gases consist of tiny particles. 2) The distance between particles is large when compared to particle size. The volume occupied by a gas is mostly empty space. 3) Gas particles have no attraction for one another. 4) Gas particles move linearly in all directions, frequently colliding with the container walls or other particles.

13 37 Kinetic Molecular Theory Assumptions of KMT (continued): 5) Collisions are perfectly elastic. No energy is lost during collisions. 6) The average kinetic energy for particles is the same for all gases (regardless of molar mass) at the same temperature. KE = 1 / 2 mv 2 where m is the mass and v is the velocity of the particle The average kinetic energy is directly proportional to temperature (in K). Gases which behave under these assumptions are know as ideal gases. 38 Real gases typically behave like ideal gases over a fairly wide range of temperatures and pressures. Conditions where real gases deviate from ideal gases: 1) At high pressure (small volumes) Distance between particles is small and the particles do not behave independently. 2) At low temperature Real Gases Particles experience intermolecular interactions. 39 Dalton s Law of Partial Pressures The total pressure of a mixture is the sum of the partial pressures of the different gases in the mixture. P total = P 1 + P 2 + P 3 Each gas behaves independently in the mixture. Application of Dalton s Law Gases collected over H 2 O contain both the gas and H 2 O vapor. Vapor pressure of H 2 O is constant at a given T. Collecting a gas over water P bottle is equalized so that P bottle = P atm thus P atm = P gas + P H2O

14 40 A sample of O 2 gas is collected over water at 22 ºC and 662 torr. What is the partial pressure of O 2 gas? The vapor pressure of water is 19.8 torr at 22 ºC. Solving For P O2 Partial Pressures Problems P atm = 662 torr P O2 = P atm P H2O P H2O = 19.8 torr P O2 = 662 torr 19.8 torr = 642 torr 41 A 250. ml sample of O 2 was collected over water at 23 ºC and 760 torr. What volume will the O 2 occupy at 23 ºC when P O2 is 760. torr? The vapor pressure of water at 23 ºC is 21.2 torr. Solving For V O2 Partial Pressures Problems V O2 + H 2O = 250 ml P H2O= 21.2 torr P O2 = P total P H2O= 760 torr 21.2 torr P atm = P O2 + P H2O = 760. torr 1) Solve for P O2 using Dalton s Law 2) Solve for V O2 using Boyle s Law = 739 torr 42 Partial Pressures Problems (continued) A 250. ml sample of O 2 was collected over water at 23 ºC and 760 torr. What volume will the O 2 occupy at 23 ºC when P O2 is 760 torr? The vapor pressure of water at 23 ºC is 21.2 torr. V 2 = Solve for V O2 with Boyle s Law P 1 V 1 = P 2 V 2 P2 P 1 V 1 V 2 = P2 P 1 V mm Hg = L = L O 760 mm Hg 2

15 43 Density of a liquid or solid is expressed in g/ml, but gas density is very low, so the standard units are g/l. density (d ) = mass volume = g L The density of a gas at STP can also be related to the compound s molar mass. ( ) ( ) d stp = molar mass Gas Density g mol 1 mol 22.4 L = g L Note: gas densities must be cited at a specific temperature as volume changes as a function of temperature (Charles Law). 44 Gas Density Practice the density of Cl 2 at STP. ( ) ( ) g d = molar mass 1 mol = g mol 22.4 L L molar mass Cl 2 = 70.9 g/mol d = 70.9 g Cl 2 1 mol Cl 2 1 mol Cl L Cl 2 = 3.17 g/l Sense Check: Gas densities are expected to be low. 45 Gas Stoichiometry At STP: the molar volume can be used as a conversion factor to convert between moles and volume. Non STP Conditions: use the ideal gas law to convert between moles and volume.

16 46 For the following reaction: the number of moles of phosphorus needed to react with 4.0 L of H 2 gas at 273 K and 1.0 atm. Gas Stoichiometry Practice at STP P 4 (s) + 6 H 2 (g) 4 PH 3 (g) V =4.0 L T = 273 K P = 1.0 atm Solution Map L H 2 mol H 2 mol P 4 mol P 4 = 4.0 L H 2 1 mol H 2 1 mol P L H2 6 mol H 2 = mol P 4 47 Gas Stoichiometry Volume Practice the volume of N 2 necessary to react with 9.0 L of H 2 gas at 450 K and 5.00 atm. a) 9.0 L b) 3.0 L c) 27.0 L d) 1.0 L N 2 (g) + 3 H 2 (g) 1 L N L H 2 3 L H 2 2 NH 3 (g) = 3.0 L N 2 At constant T and P, the volume ratio can be used in place of the mole ratio! 48 Given the following reaction: 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) If an air bag should be filled with a pressure of 1.09 atm at 22 ºC, what amount of solid NaN 3 is needed to fill a bag with a volume of 45.5 L? P = 1.09 atm V = 45. 5L T = 295K Solving for n of N 2 then find the mass of NaN 3 needed. n = PV RT Gas Stoichiometry Practice With the Ideal Gas Law = 1.09 atm 45.5 L L atm/mol K 295 K = 2.05 mol N 2

17 49 Gas Stoichiometry Practice With the Ideal Gas Law (continued) Given the following reaction: 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) If an air bag should be filled with a pressure of 1.09 atm at 22.0 ºC, what amount of solid NaN 3 is needed to fill a bag with a volume of 45.5 L? Use the reaction stoichiometry! 2.05 mol N 2 2 mol g NaN 3 NaN 3 mol 3 N 2 1 mol NaN 3 = 88.8 g NaN 3 50 What volume of O 2 at 760. torr and 25 ºC is needed to react fully with 3.2 g of C 2 H 6 (propane)? 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O (l) m = 3.2 g T = 298 K P = 1.00 atm Solution Map m C 2 H 6 mol C 2 H 6 mol O 2 volume O 2 1 mol C 2 H 6 7 mol O g C 2 H 6 = 0.37 mol O g C2 H 6 2 mol C 2 H 6 V = nrt P Gas Stoichiometry Practice = 0.37 mol L. atm 298 K mol. K 1.00 atm = 9.1 L 51 Gas Stoichiometry Practice What volume of H 2 at 739 torr and 21 ºC is liberated by 42.7 g of Zn when it reacts with HCl? Zn (s) + 2 HCl (g) ZnCl 2 (s) + H 2 (g) a) 7.6 L b) 16.2 L c) 3.2 L d) 1.8 L m Zn mol Zn mol H 2 volume H g Zn 1 mol Zn 1 mol H 2 = mol H g Zn 1 mol Zn V = nrt = mol L atm/mol K 294 K= 16.2 L H 2 P atm

18 52 Chemistry in Action What the Nose Knows Dogs use smell to detect many drugs, explosives, etc. based on trace amounts of chemical compounds in the air. Sensing low concentrations of chemicals is useful! Better Coffee Better Science Artificial noses could sniff out cancer or explosives! For more information, see: 53 Learning Objectives 12.1 Properties of Gases 1) Explain atmospheric pressure and how it is measured. 2) Be able to convert between the various units of pressure Boyle s Law 3) Use Boyle s Law to calculate changes in pressure or volume of a gas at constant temperature Charles Law 4) Use Charles Law to calculate changes in temperature or volume of a gas at constant pressure. 54 Learning Objectives 12.4 Avogadro s Law 5) Solve problems using the relationships between moles, mass, and volume of gases Combined Gas Law 6) Use the combined gas law to calculate changes in pressure, volume, or temperature of a gas sample Ideal Gas Law 7) Use the ideal gas law to solve problems involving pressure, volume, temperature, and moles of a gas.

19 55 Learning Objectives 12.7 Dalton s Law of Partial Pressures 8) Use Dalton s Law of Partial Pressures to calculate the total pressure for a mixture of gases or the pressure of a single gas in a mixture of gases Density of Gases 9) the density of a gas. (Pay attention to units!) 12.9 Gas Stoichiometry 10) Solve stoichiometry problems involving gases. (Pay attention to the states of matter and use gas laws only for gases!)