General Chemistry CHM 1032
University of Central Florida
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Introductory Chemistry 3rd Edition Nivaldo Tro Chapter 11 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills MA 2009 Prentice Hall Properties of Gases Expand to completely ll their container Take the shape of their container Low density Much less than solid or liquid state Compressible Mixtures of gases are always homogeneous Fluid Tro s Introductory Chemistry 2 Chapter 11 Kinetic Molecular Theory The particles of the gas either atoms or molecules are constantly moving The attraction and repulsion between particles is negligible When the moving particles hit another particle or the container they do not stick but they bounce off Without and continue moving in another direction There is a lot of empty space between the particles in a gas The average kinetic energy of the particles is directly proportional to the Kelvin temperature Tro s Introductory Chemistry 3 Chapter 11 Gas Particles Pushing Gas molecules are constantly in Gas molecules a motion 0 5 If we could measure the total amount 4 a 9 of force exerted by gas molecules 9 39 hitting the entire surface at any one 9 L 9 g instant we would know the pressure 2 0 359 the gas is exerting m enayrgmozooe Fearsnr Prentice Hall In Pressure force per unit area Tro s Introductory Chemistry 4 Chapter 11 The Effect of Gas Pressure A gas will ow from area of high pressure to low pressure The bigger the difference in pressure the stronger the ow of the gas Tro s Introductory Chemistry Chapter 11 Which Way Would Air Flow Two lled balloons are connected with a long pipe One of the balloons is plunged down into the water Which way will the air ow Will air ow from the lower balloon toward the top balloon or will it ow from the top balloon to the bottom one Tro39s Introductory Chemistry Chapter 1 1 Is This Possible at 21 Depth of 20 m Copyrighl 2006 Peavson Prenll Tro39s Introductory Chemistry Chapter 1 1 Soda Straws and Gas Pressure The pressure of the air inside the straw is the same as the pressure of the air outside the straw so liquid levels are the same on both sides Pressure lowered la b Copyright 2009 Pearson Prentice Hall Inc Tro s Introductory Chemistry Chapter 1 l The pressure of the air inside the straw is lower than the pressure of the air outside the straw so liquid is pushed up the straw by the outside air Gas Properties Explained Gases have taken the shape and volume of their containers because the particles don t stick together allowing them to move and ll the containers they re in In solids and liquids the particles are attracted to each other strongly enough so they stick together Gases are compressible and have low density because of the large amount of unoccupied space between the particles Tro s Introductory Chemistry Chapter 11 Properties Inde nite Shape and Inde nite Volume Because the gas molecules have enough kinetic energy to overcome attractions they keep moving around and spreading out until they ll the container k Copyright 2009 Pearson Prentice Hall Inc 11 A Tro s Introductory Chemistry Chapter As a result gases take the shape and the volume of the container they are in Properties Compressibility Copyright 2009 Pearson Prentice Hall Inc cupyrigm zung Pearson Prentice Hall IN Because there is a lot of unoccupied space in the structure of a gas the gas molecules can be squeezed closer together Tro39s Introductory Chemistry Chapter 11 1 1 Gas Properties Explained Low Density Because there is a lot of o 9 unoccupied space in the structure of a gas gases do 0 not have a lot of mass in a 9 given volume the result is that they have low density 0 Copyright 2009 Pearson Prentice Hall Inc Tro s Introductory Chemistry Chapter 12 1 1 The Pressure of a Gas Pressure is the result of the i constant movement of the gas 0 molecules and their collisions 0 o o o with the surfaces around them 0 0 9 The pressure of a gas depends 0 on several factors Number of gas particles in a 9 o a given volume o o Higher Volume of the container 0 of Average speed of the gas 0 0 particles Tro s Introductory Chemistry 13 Chapter 11 Density and Pressure Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them When more molecules are added more molecules hit the container at any one instant resulting in higher pressure Also higher density Tro s Introductory Chemistry Chapter 11 C Lower pressure 000 Higher pressure Copyright 2009 Pearson Prentice Hall Inc 14 Vacuunn Air Pressure The atmosphere exerts a pressure on everything it contacts Vacu m On average 147 psi The atmosphere goes up about 370 W pounds miles but 80 is in the rst 10 miles per from Earth s surface Square inch This is the same pressure that a column of water would exert if it were about 103 m high Copyright 2009 Pearson Prentice Hall Inc Tro s Introductory Chemistry 15 Chapter 11 Measuring Air Pressure Useabarometer Vacuum I Column of mercury supported by air Glass tube pressure gravity Force of the air on the 3399 surface of the mercury Atmospheric balanced by the pull of 1 pressure gravity on the column of mercury 7 Mercury Tro39s Introductory Chemistry Chapter 16 1 1 Atmospheric Pressure and Altitude The higher up in the atmosphere you go the lower the atmospheric pressure is around you At the surface the atmospheric pressure is 147 psi but at 10000 ft is is only 100 psi Rapid changes in atmospheric pressure may cause your ears to pop due to an imbalance in pressure on either side of your ear drum Tro s Introductory Chemistry 17 Chapter 11 Pressure Imbalance in Bar If there is a difference in pressure across the eardrum membrane the membrane will be pushed out What we commonly call a popped eardrum Copyright 2009 Pearson Prentice Hall Inci Tro39s Introductory Chemistry Chapter 3918 1 1 Common Units of Pressure Unit Average air pressure at sea level Pascal Pa 101325 Kilopascal kPa 101325 Atmosphere atm 1 exactly Millimeters of mercury mmHg 760 exactly Inches of mercury ian 2992 Torr torr 760 exactly Pounds per square inch psi 1bsin2 147 Tro s Introductory Chemistry Chapter 19 1 1 Example 111 A HighPerformance Bicycle Tire Has a Pressure of 125 psi What Is the Pressure in mmHg Given 125 psi Find mmHg Solution Map 147 psi latm Relationships 1 atm 147 psi 1 atm 760 mmHg Solution latm 760 mmH 125 pSlX x g 646x103 mmHg 1471731 133311 Check Since mmHg are smaller than psi the answer makes sense Practice Convert 455 psi into kPa Unit Average air pressure at sea level Pascal Pa 101325 Kilopascal kPa 101325 Atmosphere atm 1 exactly Millimeters of mercury mmHg 760 exactly Inches of mercury ian 2992 Torr torr 760 exactly Pounds per square inch psi lbsin2 147 Tro s Introductory Chemistry Chapter 28 1 1 Practice Convert 455 psi into kPa Continued Given Find Concept Plan 147 psi latm Relationships 1 atm 147 psi 1 atm 101325 kPa S m39 m 1 am 101 325 kP a 455 p61gtlt gtlt 314kPa 147 1951 1 am Check Since kPa are smaller than psi the answer makes sense Boyle s Law Pressure of a gas is inversely proportional to its volume Constant T and amount of gas Graph P vs Vis curved Graph P vs lVis in a straight line As P increases V decreases by the same factor P X V constant P1X V12P2X V2 Tro s Introductory Chemistry Chapter 11 ll valve 1 U 1stroke Volume increases Pressure decreases l l l l l Oneway I 1 Downstroke Volume decreases Pressure increases Boyle s Experiment Added Hg to a J tube with inside mi Used length of air column in in as a measure of volume 48 00 Mercury 44 28 added 4 O 2 36 101 32 151 28 2 l 2 Gas 24 297 22 350 h l o39s Introductory Chemistry Chapter 31 Hg 11 Pressure ian 1407 120 100 Boyle39s Experiment 10 20 30 40 50 60 Volume of Air in3 Tro s Introductory Chemistly Chapter 32 1 1 Pressure ian 140 r 120 i 100 i 80 60 407 20 Inverse Volume vs Pressure of Air Boyle39s Experiment 001 002 003 004 005 006 007 lnv Volume in393 Tro s Introductory Chemistry Chapter 3 3 11 Boyle s Experiment P X V Pressure Volume P x V 2913 48 1400 3350 42 1400 4163 34 1400 5031 28 1400 6131 23 1400 7413 19 1400 8788 16 1400 11556 12 1400 Tro s Introductory Chemistry Chapter 1 1 When you double the pressure on a gas the volume is cut in half as long as the temperature and amount of gas do not change V05L Copyright 2009 Pearson Prentice Hall Inc Tro39s Introductory Chemistry Chapter 35 1 1 Gas Laws Explained Boyle s Law 0 Boyle s law says that the volume of a gas is inversely proportional to the pressure Decreasing the volume forces the molecules into a smaller space 0 More molecules Will collide with the container at any one instant increasing the pressure Tro39s Introductory Chemistry 36 Chapter 11 Boyle s Law and Diving Since water 1s more dense than air for each 10 rn you dive below the surface the pressure on your lungs increases 1 atm At 20 In the total pressure is 3 atm If your tank contained air at 1 atm of pressure you would not be able to inhale it into your lungs You can only generate enough force to overcome about 106 atm Depth 0 m Scuba tanks have a1atm regulator so 1 air from tl1e ta Copyright 2006 Pearson Prentice Hall Inc Tro s Introductory Chemistry 37 Chapter 11 Boyle s Law and Diving Continued If a diver holds her breath and rises to the surface quickly the outside pressure drops to 1 atm According to Boyle s law What should happen to the volume of air in the lungs Since the pressure is decreasing by a factor of 3 the volume will expand by a factor of 3 causing damage to internal organs Always Exhale When Risingll Tro s Introductory Chemistry 3 8 Chapter 11 Example 112 A Cylinder with a Movable Piston Has a Volume of 60 L at 40 atm What Is the Volume at 10 atm Given V1 60 L P1 40 atm P2 10 atm Find Solution Map Relationships Solution 40 am 60 L 24 L 10 at i Check Since P and V are inversely proportional When the pressure decreases 4X the volume should increase 4X and it does Practice A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0500 atm If the Volume of the Balloon Is Now 2780 mL What Was It Originally 1 atm 760 torr Tro s Introductory Chemistry 47 Chapter 11 Practice A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0500 atm If the Volume of the Balloon Is Now 2780 mL What Was It Originally Continued Given V2 2780 mL P1 762 torr P2 0500 atm Find V1 mL Solution Map Pm Pa 1 a V1 2 2 2 Pl Relationships P1 V1 P2 V2 1 atm 760 torr exactly Solution V P2 0 V2 1 P 1 atm 1 782 X 103 atm M 760 0500 atr o2780 L1350mL 103 atr Check Since P and V are inversely proportional When the pressure decreases 2X the volume should increase 2X and it does Temperature Scales 100 C 373 K 212 F 671 R BP Water 0 C 273 K 32 F 459 R MP Ice 389 C 2341 K 38 F 421 R BP Mercury 183 C 90 K 297 F 162R BP Oxygen 269 C 2730C 4K 0K 452 F 45901 7R BP Helium OR Absolute Celsms Kelvm Fahrenhe1t Rankme Zero Gas Laws and Temperature Gases expand when heated and contract when cooled so there is a relationship between volume and temperature Gas molecules move faster when heated causing them to strike surfaces with more force so there is a relationship between pressure and temperature In order for the relationships to be proportional the temperature must be measured on an absolute scale When doing gas problems always convert your temperatures to kelvins K C273 amp CK273 0F 18 C 32 amp C 0556OF32 Tro s Introductory Chemistry 50 Chapter 11 Standard Conditions Common reference points for comparing Standard pressure 100 atm Standard temperature O 0C V 273 K STP Tro s Introductory Chemistry Chapter 11 51 Volume and Temperature In a rigid container raising the temperature increases the pressure For a cylinder With a piston the pressure outside and inside stay the same To keep the pressure from rising the piston moves out increasing the volume of the cylinder VAs volume increases pressure decreases Tro s Introductory Chemistry 52 Chapter 11 Volume and Temperature Continued Copyright 2009 Pearson Prentice Hall Inc As a gas is heated it expands This causes the density of the gas to decrease Because the hot air in the balloon is less dense than the surrounding air it rises 39fro39s Introductory Chemistry Chapter 53 1 1 Charles s Law Volume is directly proportional to temperature Constant P and amount of gas Graph of Vvs T is a straight line As T increases Valso increases Kelvin T Celsius T 273 V constant X T If T is measured in kelVin Tro s Introductory Chemistry Chapter 1 1 VLE T172 54 Volume L 05 0 J 0 co 0 N 01 0 300 250 Charle39s Law amp Absolute Zero 2 00 150 100 50 Temperature C lVqume L OH 9 02 1500 torr OVolume L OH 9 02 2500 torr A Volume L of05 g 02 1500 torr 0 Volume L of05 g 802 1500 torr 55 Absolute Zero Theoretical temperature at which a gas would have zero volume and no pressure Kelvin calculated by extrapolation O K 27315 0C 459 OF O R Never attainable Though we ve gotten real close All gas law problems use the Kelvin temperature scale Tro s Introductory Chemistry Chapter 11 57 50 4O 30 Volume L N o O 10 Determining Absolute Zero Absolute zero of temperature 273 C0K l O O O O O O O O O O C lt14 00 N O O l 00 lt3 0 l H x 4 I H x 4 N l l Temperature C Copyright 2009 Pearson Prentice Hall Incl Tro s Introductory Chemistry Chapter 1 1 William Thomson the Lord of Kelvin extrapolated the line graphs of volume vs temp erature to determine the theoretical temperature that a gas would have given a volume of O 58 Example 113 A Gas Has a Volume of 257 L at 0 0C What Was the Temperature at 280 L Given V1 280 L V2 257 L t2 0 C Find 1 K and 0C Solution Map E Relationships TK t C 273 11 T2 Solution Tl T21 V1 t1 T1 273 T2 0273 2 1 22916 273 T22273K t 240C 1 28035 Check Since T and V are directly proportional When the volume decreases the temperature should decrease and it does Practice The Temperature Inside a Balloon Is Raised from 250 0C to 2500 0C If the Volume of Cold Air Was 100 L What Is the Volume of Hot Air Tro s Introductory Chemistry 68 Chapter 11 Practice The Temperature Inside a Balloon Is Raised from 250 CC to 2500 C C Ifthe Volume of Cold Air Was 100 L What Is the Volume of Hot Air Continued Given V1 100 L t 250 C L t2 2500 C Find Solution Map 1 V1 V2 Relationships TK t C 27315 1 T2 Solution T 0 V T1 25027315 V2 2 T1 2982 K 1 T22SOO27315 5232Ko100 L 17 SL T2 5232 K 2982K 0 Check Since T and V are directly proportional when the temperature increases the volume should increase and it does The Combined Gas Law Boyle s law shows the relationship between pressure and volume At constant temperature Charles s law shows the relationship between volume and absolute temperature At constant pressure The two laws can be combined together to give a law that predicts what happens to the volume of a sample of gas when both the pressure and temperature change As long as the amount of gas stays constant P1V1P2V2 Ti T2 70 Example ll4 A Sample of Gas Has an Initial Volume of 158 mL at a Pressure of 735 mmHg and a Temperature of 34 CC If the Gas Is Compressed to 108 mL and Heated to 85 CC What Is the Final Pressure Given VI 158 mL t1 34 0C L P 735 mmHg V2 108 mL t2 85 OC Find Solution Map Relationships TK t C 273 T1 T2 Solution P P1 0 V1 0 T2 T1 34 273 2 V2 T1 T1 307K P 735mmHgo158irrl o358 l T2 85 273 2 108 mt 307x T22358K P2l25gtltlO3mmHg Check Since Tincreases and Vdecreases we expect the pressure should increase and it does Practice A Gas Oeeupies 100 L When Its Pressure Is 300 atm and Temperature Is 27 C What Volume Will the Gas Oceupy Under Standard Conditions Tro39s Introductory Chemistry Chapter 79 1 1 Practice A Gas Occupies 100 L When Its Pressure Is 300 Atm and Temperature Is 27 0C What Volume Will the Gas Occupy Under Standard Conditions Continued Given V1 1oo Lt1 27 0C L P 300 atm t2 o 0C P2 100 atm Find V2 L Solution Map V09 Phl a l gt a V1 T2 2 P2 T1 Relationships T t C 273 Solution V P1 0 V1 0 T2 T1 27 273 2 p2 T1 T1300K V 300amf100L273K T2 o 273 2 100 atn o 300K Check When T decreases Vshould decrease whenP decreases Vshould increase opposite trends make it hard to evaluate our answer Avogadro s Law Volume is directly proportional to the number of gas molecules V constant X n Constant P and T V More gas molecules larger volume Count number of gas molecules by moles n Equal volumes of gases contain equal numbers of molecules The gas doesn t matter Tro s Introductory Chemistry Chapter 11 81 Avogadro s Law Continued I III I I 0 02 04 06 08 1 12 14 Number of moles n Copyrlghl 2009 Pearson Prentice Hail Inc Copyright 2009 Pearson Prentice Hall Inc Tro39s Introductory Chemistry Chapter 82 1 1 Example 115 A 022 M01 Sample of He Has a Volume of 48 L How Many Moles Must Be Added to Give 64 L Given V1 48 L V2 64 L 721 022 mol Find n2 and added moles Solution Map V2 Relationships mol added 712 n1 n1 n2 Sohgimlz n1 0 V2 moles added 2 029 022 2 V1 moles added 2 007 mol 022 mol 0 6415 2 029 mol 48m Check Since n and V are directly proportional when the volume increases the moles should increase and it does Practice If 100 Mole of a Gas Oooupies 224 L at STP What Volume Would 0750 Moles Oooupy Tro39s Introductory Chemistry Chapter 92 1 1 Practice If 100 Mole of a Gas Occupies 224 L at STP What Volume Would 0750 Moles Occupy Continued Given V1 224 L n1 100 mol n2 0750 mol Find Solution Map E Relationships n2 Solution 224 L 0 0750 61 168 L 100 mol Check Since 11 and V are directly proportional When the moles decreases the volume should decrease and it does Ideal Gas Law By combining the gas laws we can write a general equation R is called the Gas Constant The value of R depends on the units of P and V We will use 00821 Use the ideal gas law when you have a gas at one condition use the combined gas law when you have a gas whose condition is changing and convert P to atm and Vto L PW 0 nT R PV RT Tro s Introductory Chemistry 94 Chapter 11 Example 117 How Many Moles of Gas Are in a Basketball with Total Pressure 242 Psi Volume of32 L at 25 OC Given V 32 L9 P 242 psi 1 25 0C Find Solution Map Relationships 1 am 2 147 psi PV nRT R 00821 21ml TK t C 273 m0 39 Solution P 0 V n 242p x 1231 216 62 atm R o T 1 39 l 4 2 0 2 TK 25 C 273 6 6 3 022 mol T 2 298 K omn 298 K Check 1 mole at STP occupies 224 L at STP since there is a much smaller volume than 224 L we expect less than 1 mole of gas Practice Calculate the Volume Occupied by 637 g of SO2 MM 6407 at 608 X 103 mmHg and 23 OC Tro39s Introductory Chemistry Chapter 104 1 1 Practice Calculate the Volume Occupied by 637 g of 802 MM 6407 at 608 X 103 mmHg and 23 0C Continued Given mSO2 637 g P 608 X 103 mmHg f 23 0C Find Solution Map I 6407 g V P Relationships 1 atm 760 mmHg PV nRT R 00821 atmL TK t C 273 1 mol so 6407 g mOI39K Solution lmol SO 3 Mi 637 29942 ISO 608gtlt10 Wing x 760 g 800atm 302gtlt 6407 m0 2 n0R0T TK 23 C 273 P T 259K 9942 mar 00821 at o 250K 1 X 2255L 800 atm Practice Calculate the Density of a Gas at 775 torr and 27 0C if 0250 moles Weighs 9988 g Tro39s Introductory Chemistry Chapter 106 1 1 Practice Calculate the Density of a Gas at 775 torr and 27 0C if 0250 moles Weighs 9988 2 Continued 1 Z 9 l ioi97 atfn Givem VIP 99 n20an P37 175 l nel lg 390 K Find densitya gL Solution Map g y no R P V t 0L 1 atm 760 mmHg PV my R 00821 11K Relationships TK t C 273 d mo 39 Solution V mm 10197atm d 939988g p 760 V 60355L T 2 3mL K 2 O Q39 OO m 309 2 6055 L 165 gL Check The value 165 gL is reasonable Molar Mass of a Gas One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas measure the temperature pressure and volume and use the ideal gas law mass 1n grams Molar Mass 2 moles Tro s Introductory Chemistry 108 Chapter 11 Example 118 Calculate the Molar Mass of a Gas with Mass 0311 g that Has a Volume of 0225 L at 55 oC and 886 mmHg GiVem m0331ggI ZBZZS 135886 krit ath 306328 K Find Solution Map RoT I 1 atm 760 mmHg PV 71sz R 00821 I Relationships T l C 273 MM 2 mo 71 solution MM 0311g Vlng x7601a tmzll 58atm n 97454 gtltlO393 mol 0 g 319 mol 2 I39Twa i39maw 97454 x 10 3 mol g om23 328K Check The value 319 gmol is reasonable Practice What Is the Molar Mass of a Gas if 120 g Oooupies 197 L at 380 torr and 127 OC Tro39s Introductory Chemistry Chapter 118 1 1 Practice What Is the Molar Mass of a Gas if 120 g Ooeupies l97 L at 380 torr and 127 OC Continued Given mlllg 71551991LP1980 ohatt lZ794D0 K Find molar mass gmol Solution Map R o T 1atm760 torr PV anT R Relationships TK t C 273 MM n SolutrionzlI t m 12 0 g 0 a m x 3amp9 1276091 0 50 m n 30mol UK t6sli iamp1373197z 40gmol 30 mol T E mag 409K Cheek The value 319 gmol is reasonable Mixtures of Gases According to the kinetic molecular theory the particles in a gas behave independently Air is a mixture yet we can treat it as a single gas Also we can think of each gas in the mixture as independent of the other gases A11 gases in the mixture have the same volume and temperature gt All gases completely occupy the container so all gases in the mixture have the volume of the container in Air in Air Gas Gas by volume by volume Nitrogen N2 78 Argon Ar 09 Oxygen O2 21 Carbon dioxide C02 003 Partial Pressure Each gas in the mixture exerts a pressure independent of the other gases in the mixture The pressure of a component gas in a mixture is called a partial pressure The sum of the partial pressures of all the gases in a mixture equals the total pressure VDalton s law of partial pressures JPlotaZ PgasA PgasB PgasC Pair 2 PN2 PO2 PAr 078 atm 021 atm 001 atm 100 atm Tro s Introductory Chemistry 121 Chapter 11 Example 119 A Mixture of He Ne and Ar Has a Total Pressure of 558 MmHg If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg Determine the Partial Pressure of Ar in the Mixture Given PH 341 mmHg PNe 112 mmHg PM 558 mmHg Find Solution Map PArZPI0tPHePNe Relationships Plot 2 Pa Pb etc Solution PM 558 341 112mmHg 105 mmHg Cheek The units are correct the value is reasonable Finding Partial Pressure To nd the partial pressure of a gas multiply the total pressure of the mixture by the fractional composition of the gas For example in a gas mixture that is 800 He and 200 Ne that has a total pressure of 10 atm the partial pressure of He would be PHe 080010 atm 080 atm Fractional composition percentage divided by 100 Tro s Introductory Chemistry Chapter 11 Gas mixture 80 He 9 20 Ne O Ptot 10 atm PHe 080 atm PNe 020 atm Copyright 2009 Pearson Prentice Hall Inc 123 The Partial Pressure of Each Gas in a Mixture or the Total Pressure of a Mixture Can Be Calculated Using the Ideal Gas Law for gases Aand Bin amixture nAxRxT P anRxT V P A V B the temperature and volume of everything in the mixture are the same nlotal I 771 713 P nme x R x T total 124 Practice Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 39 atm Volume 87 L Temperature 598 K and 017 moles Xe Tro39s Introductory Chemistry Chapter 125 1 1 Practice Find the Partial Pressure of Neon in a Mixture of Ne and Xe With Total Pressure 39 Atm Volume 87 L Temperature 598 K and 0 l7 Moles Xe Continued Given PM 39 atm V 87 L T 598 K Xe 017 mol Find Solution Map PNe total PXe Relationships P V nRT R 00821 Ptotal PNe PXe P ZillitllglizT P Ne Brotal P Xe Xe V 39 atm 09 89 atm atm 017m0100821mw5 ZO9 89 atm 29atm 87415 Check The unit is correct the value is reasonable Mountain Climbing and Partial Pressure Our bodies are adapted to breathe 02 at a partial pressure of 021 atm V Sherpa people native to the Himalaya Hypof 00 mountains have adapted to the much 021 Ajr at surface 05 lower partial pressure of oxygen in Safe 11 l0100 oxygen at surface I v 14 Maximum exertion Partlal pressures of Oz lower than 1395 1Mmumret 0 Ox 39 en 01 atm leads to hypOXIa toxicity 20 l 2009 Parson menu Hm lm V Unconsciousness or death Climbers of Mt Everest must carry 02 in cylinders to prevent hypoxia V On top of Mt Everest Pair 0311 atm so P02 0065 atm Tro s Introductory Chemistry 127 Chapter 11 Deep Sea Divers and Partial Pressure It is also possible to have too much 02 a condition called oxygen toxicity VP02 gt 14 atm V Oxygen toxicity can lead to muscle spasms tunnel vision and convulsions It is also possible to have too much N2 a condition called nitrogen narcosis V Also known as rapture of the deep When diving deep the pressure of the air that divers breathe increases so the partial pressure of the oxygen increases V At a depth of 55 m the partial pressure of O2 is 14 atm V Divers that go below 50 m use a mixture of He and 02 called heliox that contains a lower percentage of 02 than air Partial Pressure vs Total Pressure Surface 30 111 PM 1 atm Pm 4 atm PMZ 078 atm PNZ 312 atm P02 2 021 atm P02 2 084 ahn Copyright 2009 Pearson Prentice Hall Inc At a depth of 30 In the total pressure of air in the divers lungs and the partial pressure of all the gases in the air are quadrupled Tro39s Introductory Chemistry Chapter 129 l l Collecting Gases Gases are often collected by having them displace water from a container The problem is that since water evaporates there is also water vapor in the collected gas The partial pressure of the water vapor called the vapor pressure depends only on the temperature So you can use a table to nd out the partial pressure of the water vapor in the gas you collect If you collect a gas sample with a total pressure of 758 mmHg at 25 0C the partial pressure of the water vapor will be 238 mmHg so the partial pressure of the dry gas will be 734 mmHg Tro s Introductory Chemistry 13 0 Chapter 11 Vapor Pressure of Water Temp C Pressure mmHg 400 A 350 39 10 92 m 300 1 20 175 g 250 25 238 V 200 g 150 30 318 100 1 40 553 I 50 50 925 0 1 1 1 1 1 1 1 10 20 25 30 40 50 60 70 80 60 1494 0C 70 233 7 80 3551 Tro s Introductory Chemistry Chapter 131 1 1 If the temperature of the water is 300 the vapor pressure of the water is 318 mmHg Hydrogen plus water vapor the C K an 1 u l a v l w r Kr I 39 l i L 73 kg Liei I F a w w r l Copyright 2009 Pearson Prentice Hall Inc If the total pressure is 760 mmHg the partial pressure of the H2 is 760 318 mmHg 728 mmHg Tro39s Introductory Chemistry Chapter 1 1 Zn metal reacts with HClaq to produce H2g The gas ows through the tube and bubbles into the jar where it displaces the water in the jar Because water evaporates some water vapor gets mixed in with the H2 132 Practice 012 moles of H2 Is Collected Over Water in a 100 L Container at 32339 K Find the Total Pressure Vapor Pressure of Water at 50 0C 926 mmHg Tro39s Introductory Chemistry Chapter 13 3 1 1 Practice 0 12 moles of H2 Is Collected Over Water in a 100 L Container at 323 K Find the Total Pressure Vapor Pr ssure of Water at 50 CC 926 mmHg Continued Given V100 L mm 012 mol T 323 K Solution Map Ptotal PH2 PH2050 C 1 atm 760 mmHg UmL Relatlonshlps Pmm PA PR PV nRTj R 00821 DIK Solution M P noRoT 031813tnrxm g22 l8mmHg H2 V 13131 I 012 ml o 00821 323 1 100415 Ptotal 2amp18 926 03181atm Ptotal 330 mmHg Reactions Involving Gases The principles of reaction involving stoichiometry from Chapter 8 can be combined with the gas laws for reactions involving gases In reactions of gases the amount of a gas is often given as a volume Instead of moles As we ve seen you must state pressure and temperature The ideal gas law allows us to convert from the volume of the gas to moles then we can use the coef cients in the equation as a mole ratio Tro s Introductory Chemistry 13 5 Chapter 11 Example llll H0W Many Liters of 02 Are Made from 294 g of KClO3 at 755 mmHg and 305 K 2 KClO3s gt 2 KCls 3 02g Given namfo 6l29flcg P7699 gtni39531 5 K Find V02L Solution Map 239 u 1 quot j M V n R 1225 g 2molKClO3 P Relationships 1 atm 760 mmHg KClO3 1225 gmol 2 mol KClO3 3 mol O2 pV nRT R 00821 Solution 1 H E I E BlmdiOZT 294gKClO3 x 3 l225g 2melEC39IO3 armL 360m0102 360 m olo 00821m305 K 755 mpH g x la tm 099 42 atm atm 7601mm 907 L Practice What Volume of 02 at 0750 atm and 313 K is Generated by the Thermolysis of 100 g of HgO 2 HgOS gt 2 HgU 02g MMHgO 21659 gmol Tro39s Introductory Chemistry Chapter 146 1 1 Practice What Volume of 02 at 0750 atm and 313 K is Generated by the Thermolysis of 100 g of HgO 2 HgOs gt 2 Hgl 02g Continued Given M50003TH 12305110117503amp13K 313 K Find V02 L Solution Map 1mol HgO 21659g 2mol HgO RelatiOIIShipS 1 atm 760 mmHg HgO 21659 gmol 2 mol HgO 1 mol 02 PV nRT R 00821 313 Solution n R T 1nadi ge iplToro2 100gHgO gtlt X A 216591002 E W 008206 ML 313K 2 0023085 mol of 0 750 am 0791 L Calculate the Volume Occupied by 100 Mole of an Ideal Gas at STP PXVnXRXT Latm 100 atm X V 100 moles0082l m273 K V 224 L 1 mole of any gas at STP will occupy 224 L This volume is called the molar volume and can be used as a conversion factor As long as you work at STP 1 mol 2 224 L Tro s Introductory Chemistry 148 Chapter 11 Molar Volume There is so much o a 0 0 empty space 9 between molecules 0 Q 0 1n the gas state that 0 9 o o the volume of the o 0 gas is not effected a o o 0 by the s12e of the molecules under 1 mol helium at STP 1 mol xenon at STT Volume 224 L Volume 224 L Mass 400 g Mass 2 1313 g Copyright 2009 Pearson Prentice Hall lnc Tro s Introductory Chemistry Chapter 149 1 1 Example lll2 How Many Grams of H20 Form When 124 L H2 Reacts Completely with 02 at STP 02g 2 H2g gt 2 H20g Given VH2 124 L P 100 atrn T 273 K Flnd massHZO g 7 7 p I 2 mol H2 1802g 224 L 2 mol H20 lmol H20 Solution Map RelatiOIlShipS H20 1802 gmol 1 mol 224 L STP 2 mol H20 2 mol H2 Solution 124161112 X lmol H X 2m91 1 130 X1802ng 224 L Hz 2M2 lmol39HjO 0998 g H20 Tro s Introductory Chemistry Chapter 150 ll Practice What Volume of 02 at STP is Generated by the Thermolysis of 100 g of HgO 2 HgOS gt 2 Hgm 02g MMHgO 21659 gmol Tro39s Introductory Chemistly Chapter 158 1 1 Practice What Volume of 02 at STP is Generated by the Thermolysis of 100 g of HgO 2 HgOS gt 2 Hgm 072 Continued Given mHgO 100 g P 100 atm T 273 K Find V02L molt 321 J Solution Map 21659 g 2 mong0 lm0102 RelatiOIlShipS HgO 21659 gmol 1 mol 224 L at STP 2 mol HgO 1 mol O2 Solution 10O g HgOX111aol I lg0gtlt 113191 92 X224L02 21659g 2mcr1 Hgo 111394 93 0517Lo2 Tro s Introductory Chemistiy Chapter 159 l l
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