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, 14th Ed. Morris Hein and Susan Arena Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 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 © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. Properties of Gases Gases: i) Have indefinite volume Expand to fill a container ii) Have indefinite shape Assume the shape of a container iii) Have low densities Volume increase 1200 fold when 1 mole of water goes from a liquid to a gas. Example Volume occupied by 1 mol of H2O: as a liquid (18 mL) as a gas (22.4 L) dair = 1.2 g/L at 25 °C dwater = 1.0 g/mL at 25 °C iv) Have high velocities and kinetic energies © 2014 John Wiley & Sons, Inc. All rights reserved.

gas molecule collisions with the container walls. Measuring Pressure Pressure: Force per unit area Pressure = area force Pressure depends on: 1) The number of gas molecules 2) Gas temperature Pressure results from gas molecule collisions with the container walls. 3) Volume occupied by the gas SI unit of pressure is the pascal (Pa) = 1 newton/meter2 Unit Conversions: 1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 1.013 bar = 14.69 psi © 2014 John Wiley & Sons, Inc. All rights reserved.

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 × 1 atm 760 mm Hg = 0.974 atm b) Use the conversion factor: 101.3 kPa = 760 mm Hg 740. mm Hg × 101.3 kPa 760 mm Hg = 98.63 kPa © 2014 John Wiley & Sons, Inc. All rights reserved.

Atmospheric Pressure Due to the mass of the atmospheric gases pressing 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 © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. Measuring Pressure Measuring Pressure Use a Barometer 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. Barometer invented by E. Torricelli- hence the name Torr 3) Height of Hg column can be used to measure pressure. © 2014 John Wiley & Sons, Inc. All rights reserved.

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 H2 P = 0.5 atm 1 mol H2 P = 1 atm 2 mol H2 P = 2 atm © 2014 John Wiley & Sons, Inc. All rights reserved.

Pressure Dependence 2) On Temperature Pressure is directly proportional to temperature when moles (n ) and volume (V ) are held constant. T = 0 °C T = 100 °C 2.24 atm 3.06 atm Increasing T causes: a) more frequent and b) higher energy collisions 0.1 mol of gas in a 1L container © 2014 John Wiley & Sons, Inc. All rights reserved.

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. P a V 1 PV = constant (k ) or P = k × V 1 Most common form: P1V1 = P2V2 Graph showing inverse PV relationship © 2014 John Wiley & Sons, Inc. All rights reserved.

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? P1V1 = P2V2 Knowns V1 = 3.5 L P1 = 730. mm Hg P2 = 600. mm Hg V2 = P2 P1V1 Solve For V2 730. mm Hg 600. mm Hg × Calculate V2 = 3.5 L = 4.3 L © 2014 John Wiley & Sons, Inc. All rights reserved.

Boyle’s Law Problems A sample of Ne gas occupies 250. mL at 880. torr. Calculate the PNe if the volume is increased to 1.0 L, assuming constant temperature. (Note: Convert mL to L.) P1V1 = P2V2 Knowns V1 = 0.250 L V2 = 1.0 L P1 = 880. mm Hg P2 = V2 P1V1 Solving For P2 0.250 L 1.0 L Calculate P2 = 880. torr × = 220 mm Hg © 2014 John Wiley & Sons, Inc. All rights reserved.

Boyle’s Law Problems 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) 0.458 mm Hg P2 = V2 P1V1 65.0 L 25.0 L = 745 torr × = 1940 mm Hg Sense check: As volume decreases, pressure should increase! © 2014 John Wiley & Sons, Inc. All rights reserved.

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. Cannot achieve 0 volume as a real gas condenses before you can reach 0K. V -T relationship of methane (CH4) with extrapolation (-----) to absolute zero. © 2014 John Wiley & Sons, Inc. All rights reserved.

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 TK = T°C + 273 Kelvin temperature © 2014 John Wiley & Sons, Inc. All rights reserved.

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 a T V T = k Most common form: V1 V2 T1 T2 = © 2014 John Wiley & Sons, Inc. All rights reserved.

Charles’ Law Problems 3.0 L of H2 gas at −15 ºC is allowed to warm to 27 ºC at constant pressure. What is the gas volume at 27 ºC? V1 V2 T1 T2 = Knowns V1 = 3.0 L T1 = −15 ºC = 258 K T2 = 27 ºC = 300. K V2 = T1 V1T2 Solving For V2 V2 = T1 V1T2 300. K Calculate = 3.0 L × = 3.5 L 258 K © 2014 John Wiley & Sons, Inc. All rights reserved.

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? V1 V2 T1 T2 = Knowns V1 = 3.00 L V2 = 6.00 L T1 = 10.0 ºC = 283 K T2 = V1 T1V2 Solving For T2 T2 = V1 T1V2 6.00 L Calculate = 283 K × = 566 K 3.00 L © 2014 John Wiley & Sons, Inc. All rights reserved.

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? a) 916 mL b) 109 mL c) 943 mL d) 427 mL V2 = T1 V1T2 216 K = 635 mL × = 427 mL 321 K Sense Check: As temperature decreases, volume decreases! © 2014 John Wiley & Sons, Inc. All rights reserved.

Avogadro’s Law Equal volumes of different gases at constant T and P contain the same number of molecules. Avogadro’s Law provided proof for the concept of diatomic molecules for hydrogen and chlorine. 1 volume unit 4 molecules 1 volume unit 4 molecules 2 volume units 8 molecules © 2014 John Wiley & Sons, Inc. All rights reserved.

Avogadro’s Law Given the following gas phase reaction: N2 + 3 H2 2 NH3 If 12.0 L of H2 gas are present, what volume of N2 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 N2 gas needed. Knowns VH2 = 12.0 L Solving For VN2 Calculate 12.0 L H2 × 1 L N2 = 4.00 L N2 required 3 L H2 © 2014 John Wiley & Sons, Inc. All rights reserved.

Avogadro’s Law Given the following gas phase reaction: 2 H2 + O2 2 H2O At constant T and P, how many liters of O2 are required to make 45.6 L of H2O? a) 11.4 L b) 45.6 L c) 22.8 L d) 91.2 L 45.6 L H2O × 1 L O2 = 22.8 L O2 required 2 L H2O Sense Check: Less moles of O2 equal less L of O2! © 2014 John Wiley & Sons, Inc. All rights reserved.

Mole/Mass/Volume Relationships 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: 1.0 L of O2 at STP has a mass of 1.429 g. Show that the molar mass of O2 is 32.0 g/mol. 1.429 g O2 22.4 L O2 = 32.0 g/mol O2 × 1.0 L O2 1 mol O2 © 2014 John Wiley & Sons, Inc. All rights reserved.

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 22.4 L O2 = 39.9 g/mol × 3.00 L gas 1 mol gas Unit Check: Molar mass has units of g/mol, so use dimensional analysis when setting up the problem! © 2014 John Wiley & Sons, Inc. All rights reserved.

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. P1V1 P2V2 T1 T2 = 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. P and V units, as long as consistent, are ok. T must always be in K. Standard Temperature and Pressure (STP): 0.00 °C (273.15 K) and 1 atm (760 torr) © 2014 John Wiley & Sons, Inc. All rights reserved.

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? P1V1 P2V2 T1 T2 = Knowns V1 = 0.125 L P1 = 760 torr T1 = 273 K P2 = 320 torr T2 = 65 ºC = 338 K V2 = T1P2 P1V1T2 Solving For V2 Calculate P2 V2 = V1 × P1 ×T2 T1 0.125 L = × 760. torr × 338 K = 0.368 L 320. torr 273 K © 2014 John Wiley & Sons, Inc. All rights reserved.

Combined Gas Law Problems What is the volume at STP for a gas that occupies 1.62 L at 616 torr and 42 °C? P1V1 P2V2 T1 T2 = Knowns V1 = 1.62 L P1 = 616 torr T1 = 42 °C = 315 K P2 = 760. torr T2 = 273 K V2 = T1P2 P1V1T2 Solving For V2 Calculate V2 = V1 ×P1 × T2 P2 1.62 L = T1 × 616 torr × 273 K = 1.14 L 760. torr 315 K © 2014 John Wiley & Sons, Inc. All rights reserved.

Combined Gas Law Problems A balloon is filled with 266 L of He gas, measured at 38 °C and 0.995 atm. What will its volume be when the temperature is lowered to −76 ° C and the pressure is 0.561 atm? a) 299 L b) 95.0 L c) 745 L d) 237 L V2 = V1 × P1 × T2 P2 266 L = T1 × 0.995 atm × 197 K = 299 L 0.561 atm 311 K © 2014 John Wiley & Sons, Inc. All rights reserved.

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 a 1/P Boyle’s Law V a T Charles’ Law V a n Avogadro’s Law © 2014 John Wiley & Sons, Inc. All rights reserved.

Ideal Gas Constant R is derived from conditions at STP. Calculate R. PV = nRT Knowns P = 1.00 atm V = 22.4 L T = 273 K n = 1.00 mol R PV nT = Solving For R Calculate R = P × V = 1.00 atm × 22.4 L = 0.0821 L . atm n × T 1.00 mol × 273 K mol . K Units are critical in ideal gas problems! © 2014 John Wiley & Sons, Inc. All rights reserved.

How many moles of He are contained in a 0.900 L Ideal Gas Law Practice How many moles of He are contained in a 0.900 L container at 30. ºC and 0.800 atm? PV = nRT Knowns P = 0.800 atm V = 0.900 L T = 30. ºC = 303 K n PV RT = Solving For n Calculate n = P × V = 0.800 atm × 0.900 L = 0.0289 mol R × T 0.0821 L . atm mol . K × 303 K © 2014 John Wiley & Sons, Inc. All rights reserved.

What volume will be occupied by 0.393 mol of N2 Ideal Gas Law Practice What volume will be occupied by 0.393 mol of N2 at 0.971 atm and 24 °C? PV = nRT Knowns P = 0.971 atm n = 0.393 mol T = 24 ºC = 297 K V nRT P = Solving For V Calculate V = nRT = 0.393 mol × 0.0821 L . atm mol . K × 297 K = 9.87 L P 0.971 atm © 2014 John Wiley & Sons, Inc. All rights reserved.

The ideal gas law can also be written in terms of 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 (M ) PV = M gRT © 2014 John Wiley & Sons, Inc. All rights reserved.

Ideal Gas Law Practice A 0.210 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? PV = M gRT Knowns P = 432 torr = 0.568 atm V = 0.333 L T =296 K mass = 0.210 g Solving For M Calculate M = gRT = 0.210 g × 0.0821 L atm/mol K × 296 K = 27.0 g/mol PV 0.568 atm × 0.333 L © 2014 John Wiley & Sons, Inc. All rights reserved.

Ideal Gas Law Practice Calculate the molar mass (M ) of an unknown gas if 0.768 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 M = gRT = 0.768 g × 0.0821 L atm/mol K × 303 K = 56.3 g/mol PV 0.450 atm × 0.754 L © 2014 John Wiley & Sons, Inc. All rights reserved.

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. © 2014 John Wiley & Sons, Inc. All rights reserved.

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 = 1/2mv 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. © 2014 John Wiley & Sons, Inc. All rights reserved.

Real Gases 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 Particles experience intermolecular interactions. © 2014 John Wiley & Sons, Inc. All rights reserved.

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. Ptotal = P1 + P2 + P3… Each gas behaves independently in the mixture. Application of Dalton’s Law Gases collected over H2O contain both the gas and H2O vapor. Vapor pressure of H2O is constant at a given T. Pbottle is equalized so that Pbottle = Patm thus Collecting a gas over water Patm = Pgas + PH2O © 2014 John Wiley & Sons, Inc. All rights reserved.

Partial Pressures Problems A sample of O2 gas is collected over water at 22 ºC and 662 torr. What is the partial pressure of O2 gas? The vapor pressure of water is 19.8 torr at 22 ºC. Knowns Patm = 662 torr PH2O = 19.8 torr Solving For PO2 PO2 = Patm – PH2O Calculate PO2 = 662 torr – 19.8 torr = 642 torr © 2014 John Wiley & Sons, Inc. All rights reserved.

Partial Pressures Problems A 250. mL sample of O2 was collected over water at 23 ºC and 760 torr. What volume will the O2 occupy at 23 ºC when PO2 is 760. torr? The vapor pressure of water at 23 ºC is 21.2 torr. Knowns VO2 + H2O = 250 mL Patm = PO2 + PH2O = 760. torr PH2O= 21.2 torr Solving For VO2 Solve for PO2 using Dalton’s Law Solve for VO2 using Boyle’s Law Calculate PO2 = Ptotal – PH2O = 760 torr – 21.2 torr = 739 torr © 2014 John Wiley & Sons, Inc. All rights reserved.

Partial Pressures Problems (continued) A 250. mL sample of O2 was collected over water at 23 ºC and 760 torr. What volume will the O2 occupy at 23 ºC when PO2 is 760 torr? The vapor pressure of water at 23 ºC is 21.2 torr. Calculate Solve for VO2 with Boyle’s Law V2 = P2 P1V1 P1V1 = P2V2 V2 = P2 P1V1 739 mm Hg 760 mm Hg 0.250 L = × = 0.243 L O2 © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Density Density of a liquid or solid is expressed in g/mL, but gas density is very low, so the standard units are g/L. mass L volume density (d ) = = g The density of a gas at STP can also be related to the compound’s molar mass. ( ) ) g ( 1 mol 22.4 L g dstp = molar mass × = mol L Note: gas densities must be cited at a specific temperature as volume changes as a function of temperature (Charles’ Law). © 2014 John Wiley & Sons, Inc. All rights reserved.

( ) ( ) Gas Density Practice Calculate the density of Cl2 at STP. L 1 mol 22.4 L d = molar mass = g × mol ) ( ) molar mass Cl2 = 70.9 g/mol d = 70.9 g Cl2 × 1 mol Cl2 = 3.17 g/L 1 mol Cl2 22.4 L Cl2 Sense Check: Gas densities are expected to be low. © 2014 John Wiley & Sons, Inc. All rights reserved.

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. © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Practice at STP For the following reaction: Calculate the number of moles of phosphorus needed to react with 4.0 L of H2 gas at 273 K and 1.0 atm. P4 (s) + 6 H2 (g) 4 PH3 (g) Knowns V =4.0 L T = 273 K P = 1.0 atm Solution Map L H2 mol H2 mol P4 mol P4 = Calculate 4.0 L H2 × 22.4 L H2 1 mol H2 1 mol P4 6 mol H2 × = 0.0030 mol P4 © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Volume Practice Calculate the volume of N2 necessary to react with 9.0 L of H2 gas at 450 K and 5.00 atm. N2 (g) + 3 H2 (g) 2 NH3 (g) a) 9.0 L b) 3.0 L c) 27.0 L d) 1.0 L 3 L H2 9.0 L H2 1 L N2 × = 3.0 L N2 At constant T and P, the volume ratio can be used in place of the mole ratio! © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Practice With the Ideal Gas Law Given the following reaction: 2 NaN3 (s) 2 Na (s) + 3 N2 (g) If an air bag should be filled with a pressure of 1.09 atm at 22 ºC, what amount of solid NaN3 is needed to fill a bag with a volume of 45.5 L? Knowns P = 1.09 atm V = 45. 5L T = 295K Solving for n of N2 then find the mass of NaN3 needed. Calculate n = PV RT = 0.0821 L atm/mol K × 295 K 1.09 atm × 45.5 L = 2.05 mol N2 © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Practice With the Ideal Gas Law (continued) Given the following reaction: 2 NaN3 (s) 2 Na (s) + 3 N2 (g) If an air bag should be filled with a pressure of 1.09 atm at 22.0 ºC, what amount of solid NaN3 is needed to fill a bag with a volume of 45.5 L? Calculate Use the reaction stoichiometry! 2.05 mol N2 × 2 mol NaN3 3 mol N2 × 64.99 g NaN3 1 mol NaN3 = 88.8 g NaN3 © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Practice What volume of O2 at 760. torr and 25 ºC is needed to react fully with 3.2 g of C2H6 (propane)? 2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l) Knowns m = 3.2 g T = 298 K P = 1.00 atm Solution Map m C2H6 mol C2H6 mol O2 volume O2 Calculate 3.2 g C2H6 × 30.08 g C2H6 1 mol C2H6 2 mol C2H6 7 mol O2 × = 0.37 mol O2 = 0.37 mol × × 298 K 0.0821 L . atm mol . K 1.00 atm V = nRT P = 9.1 L © 2014 John Wiley & Sons, Inc. All rights reserved.

Gas Stoichiometry Practice What volume of H2 at 739 torr and 21 ºC is liberated by 42.7 g of Zn when it reacts with HCl? Zn (s) + 2 HCl (g) ZnCl2 (s) + H2 (g) a) 7.6 L b) 16.2 L c) 3.2 L d) 1.8 L m Zn mol Zn mol H2 volume H2 42.7 g Zn × 1 mol Zn 65.38 g Zn 1 mol Zn × 1 mol H2 = 0.653 mol H2 V = nRT P 0.972 atm 0.653 mol × 0.0821 L atm/mol K × 294 K = = 16.2 L H2 © 2014 John Wiley & Sons, Inc. All rights reserved.

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: http://www.scs.illinois.edu/suslick/smell_seeing.html © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 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. 12.2 Boyle’s Law 3) Use Boyle’s Law to calculate changes in pressure or volume of a gas at constant temperature. 12.3 Charles’ Law 4) Use Charles’ Law to calculate changes in temperature or volume of a gas at constant pressure. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. Learning Objectives 12.4 Avogadro’s Law 5) Solve problems using the relationships between moles, mass, and volume of gases. 12.5 Combined Gas Law 6) Use the combined gas law to calculate changes in pressure, volume, or temperature of a gas sample. 12.6 Ideal Gas Law 7) Use the ideal gas law to solve problems involving pressure, volume, temperature, and moles of a gas. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 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. 12.8 Density of Gases 9) Calculate 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!) © 2014 John Wiley & Sons, Inc. All rights reserved.