Chapter 10: Properties of Gases: The Air We Breathe

Chapter 10: Properties of Gases: The Air We Breathe Sept, 2006 Sept, 2016 http://ozonewatch.gsfc.nasa.gov 1

Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 2

The Properties of Gases Neither definite shape nor definite volume The Properties of Gases Gases can be compressed. 3

The Properties of Gases All gases are miscible with all other gases. http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch13_s01 Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 4

Kinetic Molecular Theory of Gases 1. Gas particles have tiny volumes compared with their container s volume Kinetic Molecular Theory of Gases 2. They don t interact with other gas molecules, e.g. no intermolecular forces. 5

Kinetic Molecular Theory of Gases 3. They move randomly and constantly Kinetic Molecular Theory of Gases 4. Elastic collisions with walls of container and other gas molecules 6

Kinetic Molecular Theory of Gases 5. Have average kinetic energy that is proportional to absolute Kelvin temperature: KE = ½ mu 2 rms <KE> A = <KE> B since both at the same temperature ½ m A u 2 rms,a = ½ m B u 2 rms,b m A u 2 rms,a = m B u 2 rms,b u 2 rms,b u 2 rms,b m A u 2 rms,a = m B u 2 rms,b m A u 2 rms,a = m B u rms,a u rms,b = m B m A u rms,a u rms,b = MM B MM A MM = molar mass m A u 2 rms,b u 2 rms,a = m B m A Graham s Law of Effusion: We will return to this in Section 10.11 u 2 rms,b m A u rms,a u rms,b = m B m A rate effusion,a rate effusion,b = MM B MM A 7

rms velocity inversely proportional to the Molar Mass = 32 g/mol = 28 = 18 = 4 Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 8

Pressure = force/unit area Molecules collide with the inside surface of the container. The force of the collision is measured as pressure. Pressure at Sea Level Pounds/in 2 (psi) 14.7 psi Atmospheres (atm) 1 atm Pascals (N/m 2 ) 101.325 X 10 3 Pa Torr (mmhg) 760 mmhg Torricelli s Barometer 760 mm Hg vacuum Column of mercury Atmospheric pressure The pressure of the atmosphere on the surface of the mercury in the dish is balanced by the downward pressure exerted by the mercury in the column. 9

Elevation and Atmospheric Pressure 0.35 atm 0.62 atm 0.83 atm Sea level Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 10

yaxis 5/16/2018 State Variables for a Gas P = pressure T = temperature V = volume n = number of moles Some math before continuing - y = mx + b 60 50 40 m 30 20 10 0 b 0 5 10 15 20 25 xaxis 11

Boyle s Law: P and V (n and T held constant) P and V are inversely proportional - P then V If you plot P vs. 1/V the result is a straight line P = mx + b = m(1/v) + b so P = m/v and PV = m P 1 V 1 = m and P 2 V 2 = m, so P 1 V 1 = P 2 V 2 Boyle s Law and Respiration 12

Applying Boyle s Law Example A bubble of oxygen at the bottom of a lake floats up to the surface. The pressure at the bottom of the lake is 4.75 atm and the volume is 5.65 ml. At the surface, the new pressure is 0.985 atm. Assuming that the temperature and number of moles remained constant, what is the final volume of the bubble? P 1 = 4.75 atm V 1 = 5.65 ml P 2 = 0.985 atm V 2 =? ml Explaining Boyle s Law Using Kinetic Molecular Theory 13

Charles s Law: V and T (n and P held constant) T and V are directly proportional - T then V If you plot V vs. T the result is a straight line V = m T + b = m T so V = m T and V/T = m V 1 /T 1 = m and V 2 /T 2 = m V1 T 1 V T 2 2 Volume of a gas extrapolates to zero at absolute zero (0 K = 273 C) so you MUST use the Kelvin temperature. Jacques Alexandre Charles (1796-1823) The French chemist Charles was most famous in his lifetime for his experiments in ballooning. The first such flights were made by the Montgollier brothers in June 1783, using a large spherical balloon made of linen and paper and filled with hot air. In August 1783, however, a different group. supervised by Jacques Charles, tried a different approach. Exploiting his recent discoveries in the study of gases, Charles decided to inflate the balloon with hydrogen gas. Because hydrogen would escape easily from a paper bag, Charles made a bag of silk coaled with a rubber solution. Inflating the bag to its final diameter took several days and required nearly 500 pounds of acid and 1000 pounds of iron to generate the hydrogen gas. A huge crowd watched the ascent on August 27, 1783. The balloon stayed aloft for almost 45 minutes and travelled about 15 miles. When it landed in a village, however, the people were so terrified they tore if to shreds. 14

Sample Exercise 10.4: Applying Charles Law Several students at a northern New England campus are hosting a party celebrating the mid-january start of spring semester classes. They decide to decorate the front door of their apartment building with party balloons. The air in the inflated balloons is initially 70 o F. After an hour outside, the temperature of the balloons is -12 o F. Assuming no air leaks from the balloons and the pressure in them does not change significantly, how much does their volume change? Express your answer as a percentage of the initial volume. Explaining Charles Law Using Kinetic Molecular Theory 15

Avogadro s Law: V and n (T and P held constant) V and n are directly proportional n then V If you plot V vs. n the result is a straight line V = m n + b = m n so V = m n and V/n = m V 1 /n 1 = m and V 2 /n 2 = m V1 n 1 V n 2 2 Avogadro s Law Problem (none given in text, p. 427) V 1 = 3.5 L n 1 = 0.140 mol V 2 = 10.1 L n 2 =? mol 16

Chapter Outline Note: we are skipping ahead to Sec. 10.6, then back to Sec. 10.5 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 Ideal Gas Equation Boyle s law: V a (at constant n and T) 1 P Charles law: V a T (at constant n and P) Avogadro s law: V a n (at constant P and T) 17

Standard Temperature and Pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nrt Sample Exercise 10.7: Applying the Ideal Gas Law Bottles of compressed O 2 carried by climbers ascending Mt. Everest are designed to hold one kilogram of the gas. What volume of O 2 can one bottle deliver to a climber at an altitude where the temperature is -38 o C and the atmospheric pressure is 0.35 atm? Assume that each bottle contains 1.00 kg of O 2. 18

Chapter Outline Note: we are now back to Sec. 10.5 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 Deriving the Combined Gas Law P, V, T, and n 19

Sample Exercise 10.6: Applying the Combined Gas Law The pressure inside a weather balloon as it is released is 798 mmhg. If the volume and temperature of the balloon are 131 L and 20 o C, what is the volume of the balloon when it reaches an altitude where its internal pressure is 235 mmhg and T = -52 o C? Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 20

Densities and Molecular Weights of Gases Using PV = nrt Sample Exercise 10.8: Calculating the Density of a Gas According to the U.S National Weather Service, the air temperature in Phoenix, AZ reached 25.56 o C on January 1, 2012, when the atmospheric pressure was 1.0106 atm. What is the density of the air? Assume the average molar mass of air is 28.8 g/mol, which is the weighted average of the molar masses of the various gases in dry air. 21

Example: Calculating the Molecular Weight from PV = nrt 1.018 g of Freon-113 gas is trapped in a 145 ml container at 760 mmhg and 50.0 C. What is the molar mass of Freon-113? Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 22

Gas Laws & Stoichiometry Example: Combining Stoichiometry and the Ideal Gas Law Chlorine gas can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid: MnO 2 (s) + 4 HCl(aq) MnCl 2 (aq) + 2 H 2 O(l) + Cl 2 (g) How many grams of MnO 2 should be added to excess HCl to obtain 275 ml of chlorine gas at 5.0 C and 650 mmhg? 23

Mole Fraction & Partial Pressure Mole Fraction: Ratio of the # of moles of a given component in a mixture to the total # of moles in a mixture: x 1 n1 n1 n n n n... total 1 2 3 Mole Fraction in Terms of Pressure: When V and T are constant, P n Mole Fraction & Partial Pressure Since P n And: x x 1 1 n n 1 1 total P1 P total P P total Then P x P 1 1 total 25

Sample Exercise 10.11: Calculating Mole Fractions and Partial Pressures Scuba divers who dive to depths below 50 meters may breathe a gas mixture called Trimix during the deepest parts of their dives. One formulation of the mixture, called Trimix 10/70, is 10% oxygen, 70% helium, and 20% nitrogen by volume. What is the mole fraction of each gas in this mixture, and what is the partial pressure of oxygen in the lungs of a diver at a depth of 60 meters (where the ambient pressure is 7.0 atm)? Collecting a Gas over Water 2 KClO 3 (s) 2 KCl(s) + 3 O 2 (g) Gases collected: O 2 (g) and H 2 O(g) P P P total O H O 2 2 26

Sample Exercise 10.12: Calculating the Quantity of a Gas Collected by Water Displacement During the decomposition of KClO 3, 92.0 ml of gas is collected by the displacement of water at 25.0 o C. If atmospheric pressure is 756 mmhg, what mass of O 2 is collected? Chapter Outline 10.1 The Properties of Gases 10.2 The Kinetic Molecular Theory of Gases* 10.3 Atmospheric Pressure 10.4 Relating P,T, and V: The Gas Laws 10.5 The Combined Gas Law (we will add moles to the equation) 10.6 Ideal Gases and the Ideal Gas Law 10.7 Densities of Gases 10.8 Gases in Chemical Reactions 10.9 Mixtures of Gases 10.10 Solubilities of Gases and Henry s Law 10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly 10.12 Real Gases *Effusion has been added to Section 10.11 27

Solubility of Gases Solubility of gases depends on T and P Solubility as Pressure Solubility as Temperature Henry s Law: The higher the partial pressure of the gas above a liquid, the more soluble C gas P gas Henry s Law C gas k H P gas 28

Sample Exercise 10.13: Calculating Gas Solubility Using Henry s Law Lake Titicaca is located high in the Andes Mountains between Peru and Bolivia. Its surface is 3811 m above sea level, where the average atmospheric pressure is 0.636 atm. During the summer, the average temperature of the water s surface rarely exceeds 15 o C. What is the solubility of oxygen in Lake Titicaca at that temperature? Express your answer in molarity and in mg/l. 29

Graham s Law of Effusion u rms,a u rms,b = MM B MM A MM = molar mass rate effusion,a rate effusion,b = MM B MM A Sample Exercise 10.1: Calculating Relative Rates of Effusion An odorous gas emitted by a hot spring was found to effuse at 0.342 times the rate at which helium effuses. What is the molar mass of the emitted gas? rate effusion,a rate effusion,b = MM B MM A A = X (MM =?) B = He (MM = 4.003 g/mol) rate effusion X rate effusion,he = MW He MW X 0.342 = 4.003 g/mol MM X (0.342) 2 = 4.003 g/mol MM X and so MM X = 34.2 g/mol or H 2 S 31

Diffusion - Same Formula as Graham s Law of Effusion: U rms,a effusion rate A diffusion rate A = = U rms,b effusion rate B diffusion rate B = M B M A Problem 10.121 A flask of ammonia is connected to a flask of an unknown acid HX by a 1.00 m glass tube. As the two gases diffuse down the tube, a white ring of NH 4 X forms 68.5 cm from the ammonia flask. Identify element X. Gas Diffusion: Molecules Moving Rapidly Maxwell s Equation u rms = 3RT M R = 8.314 J/mol K 32

Practice Exercise 10:14 Calculating Root-Mean-Square Speeds Calculate the root-mean-squared speed of nitrogen molecules at 25 o C. 33