Chapter 11 - Gases Properties of Gases Gas Pressure Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 2 of 77 Gases We Breathe The air we breathe Is a gas mixture. Contains mostly N 2 and O 2 and small amounts of other gases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 3 of 77 Kinetic Theory of Gases Gases consist of _____ particles that Move rapidly in ___________ lines. Have essentially no attractive (or repulsive) ___________. Are very ______ ____________. Have very small volumes compared to the volume of the ____________ they occupy. Have kinetic energies that increase with an __________ in temperature.

Chapter 11 Slide 4 of 77 Properties of Gases Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n, number of moles). Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 5 of 77 Units of Pressure Gas pressure Is described as a force acting on a specific area. Pressure (P) = Force Area Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 6 of 77 Atmospheric Pressure The atmospheric pressure Is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Is about 1 atmosphere or a little less at sea level. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 7 of 77 Altitude and Atmospheric Pressure Atmospheric pressure Depends on the altitude and the weather. Is lower at high altitudes where the density of air is less. Is higher on a rainy day than on a sunny day. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 8 of 77 What is 475 mm Hg expressed in atm? 475 mm Hg x 1 atm = atm 760 mm Hg The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm Learning Check

Chapter 11 Slide 9 of 77 Barometer A barometer Measures the pressure exerted by the gases in the atmosphere. Indicates atmospheric pressure as the height in mm of the mercury column. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 10 of 77 Chapter 11Gases Pressure and Volume Boyle’s Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 11 of 77 Boyle’s Law Boyle’s Law states that The pressure of a gas is ___________ related to its volume when T and n are constant. If volume decreases, the pressure ______________. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 12 of 77 In Boyle’s Law The product P x V is constant as long as T and n do not change. P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyle’s Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyle’s Law

Chapter 11 Slide 13 of 77 Solving for a Gas Law Factor The equation for Boyle’s Law can be rearranged to solve for any factor. P 1 V 1 = P 2 V 2 Boyle’s Law To obtain V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 P 2 P 2 P 1 V 1 = V 2 P 2

Chapter 11 Slide 14 of 77 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 15 of 77 Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T? 1. Set up a data table Conditions 1Conditions 2 P 1 = 550 mm HgP 2 = 2200 mm Hg V 1 = 8.0 LV 2 = (predict smaller V 2 ) Calculation with Boyle’s Law ?

Chapter 11 Slide 16 of Because pressure increases, we predict that the volume will decrease. Solve Boyle’s Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 V 2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume Calculation with Boyle’s Law (Continued)

Chapter 11 Slide 17 of 77 Learning Check For a cylinder containing helium indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases 2) Pressure increases Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 18 of 77 Learning Check If the helium in a cylinder has a volume of 120 mL and a pressure of 850 mm Hg, what is the new volume if the pressure is changed to 425 mm Hg inside the cylinder? A) 60 mL B) 120 mLC) 240 mL P 1 = 850 mm Hg P 2 = 425 mm Hg V 1 = 120 mL V 2 = ?? V 2 = P 1 V 1 = 120 mL x 850 mm Hg = 240 mL P mm Hg

Chapter 11 Slide 19 of 77 Learning Check A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T constant), is the new volume represented by A, B, or C? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 20 of 77 If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 LB) 6.4 LC) 12.8 L V 2 = V 1 P 1 P 2 V 2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant.) Learning Check

Chapter 11 Slide 21 of 77 A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg Learning Check Practice At Home

Chapter 11 Slide 22 of 77 Chapter 11Gases Temperature and Volume (Charles’ Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 23 of 77 Charles’ Law In Charles’ Law The Kelvin temperature of a gas is directly related to the volume. P and n are constant. When the temperature of a gas increases, its volume increases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 24 of 77 For two conditions, Charles’ Law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles’ Law to solve for V 2 T 2 xV 1 = V 2 x T 2 T 1 T 2 V 2 = V 1 T 2 T 1 Charles’ Law: V and T

Chapter 11 Slide 25 of 77 Learning Check Solve Charles’ Law expression for T 2. V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 =V 2 T 1 Isolate T2 by dividing through by V1 V 1 T 2 =V 2 T 1 V 1 T 2 =V 2 T 1 V 1 T2T2

Chapter 11 Slide 26 of 77 A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0°C, what is the new volume of the balloon (P constant)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? (decrease) T 1 = 21°C = 294 KT 2 = 0°C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations. Calculations Using Charles’ Law

Chapter 11 Slide 27 of 77 Calculations Using Charles’ Law (continued) 2. Solve Charles’ law for V 2 V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K When temperature decreased, the volume decreased as predicted.

Chapter 11 Slide 28 of 77 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? A) 443°C B) 170°C C) - 82°C T 2 = T 1 V 2 V 1 T 2 = 291 K x 640 mL = 443 K 420 mL = 443 K K = 170°C Learning Check 18 o C = 291 K

Chapter 11 Slide 29 of 77 Chapter 11Gases Temperature and Pressure (Gay-Lussac’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 30 of 77 Gay-Lussac’s Law: P and T In Gay-Lussac’s Law The pressure exerted by a gas is directly related to the Kelvin temperature. V and n are constant. P 1 = P 2 T 1 T 2 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 31 of 77 A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) 1. Set up a data table. Conditions 1Conditions 2 P 1 = 2.0 atmP 2 = (increase) T 1 = 18°C T 2 = 62°C = 291 K = 335 K Calculation with Gay- Lussac’s Law ?

Chapter 11 Slide 32 of 77 Calculation with Gay- Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P 2 P 1 = P 2 T 1 T 2 P 2 = P 1 T 2 T 1 P 2 = 2.0 atm x 335 K = 2.3 atm 291 K

Chapter 11 Slide 33 of 77 Chapter 11 Gases The Combined Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 34 of 77 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2 V 2 T 1 T 2 Combined Gas Law

Chapter 11 Slide 35 of 77 A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1Conditions 2 P 1 = atm P 2 = 3.20 atm V 1 = L (180 mL) V 2 = 90.0 mL T 1 = 29°C = 302 KT 2 = ?? Combined Gas Law Calculation

Chapter 11 Slide 36 of Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 2 = T 1 P 2 V 2 P 1 V 1 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm mL T 2 = 604 K = 331 °C Combined Gas Law Calculation (continued)

Chapter 11 Slide 37 of 77 A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at -95°C and a pressure of 802 mm Hg (n constant)? Learning Check

Chapter 11 Slide 38 of Data Table Conditions 1Conditions 2 T 1 = 308 KT 2 = -95°C = 178K V 1 = 675 mL V 2 = ??? P 1 = 646 mm Hg P 2 = 802 mm Hg 2. Solve for V 2 V 2 =V 1 P 1 T 2 P 2 T 1 V 2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume (mL) of the gas at -95°C and a pressure of 802 mm Hg (n constant)? P 1 V 1 = P 2 V 2 T 1 T 2

Chapter 11 Slide 39 of 77 Chapter 11 Gases Volume and Moles (Avogadro’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 40 of 77 Avogadro's Law: Volume and Moles In Avogadro’s Law The volume of a gas is directly related to the number of _____ of gas. T and P are constant. V 1 = V 2 n 1 n 2 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 41 of 77 Learning Check If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 42 of 77 If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L Conditions 1Conditions 2 V 1 = 1.5 LV 2 = ??? n 1 = 0.75 mol Hen 2 = 1.2 mol He V 2 = V 1 n 2 n 1 V 2 = 1.5 L x 1.2 mol He = 2.4 L 0.75 mol He V 1 = V 2 n 1 n 2

Chapter 11 Slide 43 of 77 The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP, Standard Temperature and Pressure). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg) STP

Chapter 11 Slide 44 of 77 Molar Volume At standard temperature and pressure (STP). 1 mol of a gas occupies a volume of 22.4 L, which is called its molar volume. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 45 of 77 The molar volume At STP Can be used to form conversion factors L and 1 mol 1 mol 22.4 L Molar Volume as a Conversion Factor

Chapter 11 Slide 46 of 77 Guide to Using Molar Volume Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 47 of What is the volume at STP of 4.00 g CH 4 ? A) 5.59 LB) 11.2 LC) 44.8 L 4.00 g CH 4 x 1 mol CH 4 x 22.4 L (STP) = g CH 4 1 mol CH 4 Learning Check 5.59 L

Chapter 11 Slide 48 of How many grams of He are present in 8.00 L at STP? A) 25.6 gB) gC) 1.43 g 8.00 L x 1 mol He x g He = 22.4 L 1 mol He Learning Check 1.43 g He

Chapter 11 Slide 49 of 77 AT STP, the density of gas can be calculated using the mass of the gas and its volume. Density = Molar mass Molar volume Density of a Gas at STP

Chapter 11 Slide 50 of 77 Density of a Gas Calculate the density in g/L of O 2 gas at STP. At STP, we know that 1 mol O 2 (32.00 g) occupies a volume of 22.4 L. Density O 2 at STP = g O 2 = 1.43 g/L 22.4 L (STP)

Chapter 11 Slide 51 of 77 Chapter 11 Gases The Ideal Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 52 of 77 The ideal gas law Provides a relationship between the four properties (P, V, n, and T) of gases that can be written equal to a constant R. PV = R nT Rearranges these properties to give the ideal gas law expression. PV = nRT Ideal Gas Law

Chapter 11 Slide 53 of 77 The universal gas constant, R Can be calculated using the molar volume of a gas at STP. Calculated at STP uses 273 K,1.00 atm, 1 mol of a gas, and a molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mol) (273K) n T = L atm mol K Universal Gas Constant, R

Chapter 11 Slide 54 of 77 Another value for the universal gas constant is obtained using mm Hg for the STP pressure. The units of R are dependent on the units of the parameters that are used to calculate R. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? Learning Check

Chapter 11 Slide 55 of 77 What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? R = PV = (760 mm Hg) (22.4 L) nT (1 mol) (273 K) = 62.4 L mm Hg mol K Solution

Chapter 11 Slide 56 of 77 Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.86 mol N 2 O at 23°C, what is the pressure (mm Hg) in the tank? Learning Check Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 57 of Adjust the units of the given properties to match the units of R. V = 20.0 L20.0 L T = 23°C K n = 2.86 mol2.86 mol P = ? ? Solution

Chapter 11 Slide 58 of Rearrange the ideal gas law for P. P = nRT V 3. Substitute quantities and solve. P = (2.86 mol)(62.4 L mm Hg)(296 K) (20.0 L) (mol K) = 2.64 x 10 3 mm Hg Solution

Chapter 11 Slide 59 of 77 Learning Check A cylinder contains 5.0 L O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 60 of Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n ( or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mol K) = 0.18 mol O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = mol O 2 x 32.0 g O 2 = 5.8 g O 2 1 mol O 2 Solution

Chapter 11 Slide 61 of 77 What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) molK = mol RT ( L atm)(303K) 2. Set up the molar mass relationship. Molar mass = g = g mol mol = 35.6 g/mol Molar Mass of a Gas

Chapter 11 Slide 62 of 77 Chapter 11 Gases Gas Laws and Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 63 of 77 Gases in Equations The volume or amount of a gas in a chemical reaction can be calculated from STP conditions or the ideal gas law. Mole factors from the balanced equation.

Chapter 11 Slide 64 of 77 STP and Gas Equations What volume (L) of O 2 gas is needed to completely react with 15.0 g of aluminum at STP? 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Plan: g Al mol Al mol O 2 L O 2 (STP) 15.0 g Al x 1 mol Al x 3 mol O 2 x 22.4 L (STP) g Al 4 mol Al 1 mol O 2 = 9.34 L O 2 at STP

Chapter 11 Slide 65 of 77 Ideal Gas Equation and Reactions What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) Plan? g Al moles Al moles Cl 2 L Cl 2

Chapter 11 Slide 66 of 77 Ideal Gas Equation and Reactions 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mol Al x 3 mol Cl 2 = mol Cl g Al 2 mol Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT =(0.083 mol Cl 2 )( L atm)(300 K) P 1.2 atm mol K = 1.7 L Cl 2

Chapter 11 Slide 67 of 77 What volume (L) of O 2 at 24°C and atm is needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

Chapter 11 Slide 68 of 77 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) 1. Calculate the moles of O 2 needed g NH 3 x 1 mol NH 3 x 5 mol O g NH 3 4 mol NH 3 = 2.06 mol O 2 2. Place the moles of O 2 in the ideal gas equation. V = nRT =(2.06 mol)( L atm/mol K)(297 K) P atm = 52.9 L O 2 Solution

Chapter 11 Slide 69 of 77 Chapter 11 Gases Partial Pressure (Dalton’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 70 of 77 The partial pressure of a gas Is the pressure of each gas in a mixture. Is the pressure that gas would exert if it were by itself in the container. Partial Pressure

Chapter 11 Slide 71 of 77 Dalton’s Law of Partial Pressures states that the total pressure Depends on the total number of gas particles, not on the types of particles. Exerted by a gas mixture is the sum of the partial pressures of those gases. P T = P 1 + P Dalton’s Law of Partial Pressures

Chapter 11 Slide 72 of 77 Illustrating Partial Pressures Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 73 of 77 At STP, One mole of a pure gas in a volume of 22.4 L will exert the same pressure as one mole of a gas mixture in 22.4 L. V STP = 22.4 L Gas mixtures Total Pressure 0.5 mol O mol He 0.2 mol Ar 1.0 mol 1.0 mol N mol O mol He 1.0 mol 1.0 atm

Chapter 11 Slide 74 of 77 Learning Check A scuba tank contains O 2 with a pressure of atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 11 Slide 75 of Convert the pressure in atm to mm Hg atm x 760 mm Hg = 342 mm Hg = P O 2 1 atm 2. Calculate the sum of the partial pressures. P total = PO 2 + P He P total = 342 mm Hg mm Hg = 1197 mm Hg Solution

Chapter 11 Slide 76 of 77 For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium? A) 520 mm Hg B) 2040 mm Hg C) 4800 mm Hg Learning Check

Chapter 11 Slide 77 of 77 A) 520 mm Hg B) 2040 mm Hg C) 4800 mm Hg P Total = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm P Total = P O + P He 2 P He = P Total - P O 2 P He = 6080 mm Hg mm Hg = 4800 mm Hg Solution