Report

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Air Pressure & Shallow Wells • water for many homes is • • • supplied by a well less than 30 ft. deep with a pump at the surface the pump removes air from the pipe, decreasing the air pressure in the pipe the outside air pressure then pushes the water up the pipe the maximum height the water will rise is related to the amount of pressure the air exerts Tro, Chemistry: A Molecular Approach 2 Atmospheric Pressure • pressure is the force • exerted over an area on average, the air exerts the same pressure that a column of water 10.3 m high would exert 14.7 lbs./in2 so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Tro, Chemistry: A Molecular Approach Force Pressure Area 3 Gases Pushing • gas molecules are constantly in motion • as they move and strike a surface, they push on that surface push = force • if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting pressure = force per unit area Tro, Chemistry: A Molecular Approach 4 The Effect of Gas Pressure • the pressure exerted by a gas can cause some amazing and startling effects • whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure the bigger the difference in pressure, the stronger the flow of the gas • if there is something in the gas’s path, the gas will try to push it along as the gas flows Tro, Chemistry: A Molecular Approach 5 Atmospheric Pressure Effects • differences in air pressure result in weather and wind patterns • the higher up in the atmosphere you climb, the lower the atmospheric pressure is around you at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 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, Chemistry: A Molecular Approach 6 Pressure Imbalance in Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum.” Tro, Chemistry: A Molecular Approach 7 The Pressure of a Gas • result of the constant movement of the gas molecules and their collisions with the surfaces around them • the pressure of a gas depends on several factors number of gas particles in a given volume volume of the container average speed of the gas particles Tro, Chemistry: A Molecular Approach 8 Measuring Air Pressure • use a barometer • column of mercury • supported by air pressure force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury Tro, Chemistry: A Molecular Approach gravity 9 Common Units of Pressure Unit pascal (Pa), 1 Pa 1 Average Air Pressure at Sea Level N m 2 101,325 kilopascal (kPa) 101.325 atmosphere (atm) 1 (exactly) millimeters of mercury (mmHg) inches of mercury (inHg) torr (torr) pounds per square inch (psi, lbs./in2) Tro, Chemistry: A Molecular Approach 760 (exactly) 29.92 760 (exactly) 14.7 10 Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? Given: 132 psi Find: mmHg Concept Plan: psi atm 1 atm 14.7 psi Relationships: mmHg 760 mmHg 1 atm 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: 1 atm 760 mmHg 132 psi 6.82103 mmHg 14.7psi 1 atm Check: since mmHg are smaller than psi, the answer makes sense Manometers • the pressure of a gas trapped in a container can be • • • measured with an instrument called a manometer manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other a competition is established between the pressure of the atmosphere and the gas the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere Tro, Chemistry: A Molecular Approach 12 Manometer for this sample, the gas has a larger pressure than the atmosphere, so Pressure gas Pressure atmosphere Pressure h Pressure gas (mmHg) Pressure atmosphere(mmHg) difference in Hg levels (mm) Tro, Chemistry: A Molecular Approach 13 Boyle’s Law • pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line • as P increases, V decreases by the same factor • P x V = constant • P1 x V1 = P2 x V2 Tro, Chemistry: A Molecular Approach 14 Boyle’s Experiment • added Hg to a J-tube with • air trapped inside used length of air column as a measure of volume Tro, Chemistry: A Molecular Approach Length of Air in Column (in) 48 44 40 36 32 28 24 22 Difference in Hg Levels (in) 0.0 2.8 6.2 10.1 15.1 21.2 29.7 35.0 15 Boyle's Expt. 140 120 Pressure, inHg 100 80 60 40 20 0 0 10 20 30 40 50 60 3 Volume of Air, in Tro, Chemistry: A Molecular Approach 16 Inverse Volume vs Pressure of Air, Boyle's Expt. 140 120 Pressure, inHg 100 80 60 40 20 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Inv. Volume, in-3 Tro, Chemistry: A Molecular Approach 17 Boyle’s Experiment, P x V Pressure Volume P x V 29.13 48 1400 33.50 42 1400 41.63 34 1400 50.31 28 1400 61.31 23 1400 74.13 19 1400 87.88 16 1400 115.56 12 1400 Tro, Chemistry: A Molecular Approach 18 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) Tro, Chemistry: A Molecular Approach 19 Boyle’s Law and Diving • since water is denser than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs at 20 m the total pressure is 3 atm Tro, Chemistry: A Molecular Approach 20 Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? Given: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm Find: V2, L Concept Plan: V1, P1, P2 V2 Relationships: P1 Solution: P1 V1 P2 V2 ∙ V1 = P2 ∙ V2 P1 V1 V2 P2 4.52atm 7.25L 27.1 L 1.21atm 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 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Tro, Chemistry: A Molecular Approach 22 A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Given: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm Find: V1, mL Concept Plan: V1, P1, P2 V1 Relationships: P1 P2 V2 P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) Solution: 782 torr 1 atm 1.03 atm 760 torr Check: V2 P2 V2 V1 P1 0.500atm 2780L 1350mL 1.03atm since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does Charles’ Law • volume is directly proportional to temperature constant P and amount of gas graph of V vs T is straight line • as T increases, V also increases • Kelvin T = Celsius T + 273 • V = constant x T V1 V2 T1 T2 if T measured in Kelvin Tro, Chemistry: A Molecular Approach 24 Charles’ Law – A Molecular View • the pressure of gas inside and outside the balloon are the same • at high low temperatures, temperatures,the the gas molecules are moving not movingsoasthey faster, fast,hit sothe they sides don’t of the hit balloon the sides harder of the – balloon as causing thehard volume – to thereforelarger become the volume is small Tro, Chemistry: A Molecular Approach 25 Charles' Law & Absolute Zero 0.6 Volume (L) of 1 g O2 @ 1500 torr Volume (L) of 1 g O2 @ 2500 torr 0.5 Volume (L) of 0.5 g O2 @ 1500 torr Volume (L) of 0.5 g SO2 @ 1500 torr 0.4 Volume, L The data fall on a straight line. If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero 0.3 0.2 0.1 0 -300 -250 -200 -150 -100 -50 Temperature, °C 0 50 100 150 26 Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L? Given: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C Find: t1, K and °C Concept Plan: V1, V2, T2 T1 T2 V1 V2 Relationships: T(K) = t(°C) + 273.15, T1 V1 V2 T1 T2 T2 V1 T1 V2 T2 0.00 273.15 T2 273.15K 273.15K 2.57 L 297.6 K 2.80L Solution: Check: t1 T1 273.15 t1 297.6 273.15 t1 24 C 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 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Tro, Chemistry: A Molecular Approach 28 The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Given: V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C Find: V2, L Concept Plan: V1, T1, T2 V2 V1 V2 T2 T1 Relationships: T(K) = t(°C) + 273.15, Solution: T1 25.0 273.15 T1 298.2K T2 250.0 273.15 V2 V1 V2 T1 T2 T2 V1 T1 523.2K 10.0L 17.5 L 298.2K T2 523.2K Check: since T and V are directly proportional, when the temperature increases, the volume should increase, and it does Avogadro’s Law • volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger volume V 1 V2 n1 n 2 • count number of gas molecules by moles • equal volumes of gases contain equal numbers of molecules the gas doesn’t matter Tro, Chemistry: A Molecular Approach 30 Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Given: V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol Find: n2, and added moles Concept Plan: V1, V2, n1 V2 n1 n2 V1 n2 V1 V2 n1 n2 Relationships: mol added = n2 – n1, Solution: n V n2 1 2 V1 molesadded 0.314 0.225 molesadded 0.089 mol 0.225mol 6.48L 0.314 mol 4.65L Check: since n and V are directly proportional, when the volume increases, the moles should increase, and it does Ideal Gas Law • By combing 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 atm L • we will use 0.08206 mol Kand convert P to atm and V to L • the other gas laws are found in the ideal gas law if two variables are kept constant • allows us to find one of the variables if we know the other 3 P V R n T Tro, Chemistry: A Molecular Approach or PV nRT 32 Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? Given: V = 3.24 L, P = 24.3 psi, t = 25 °C, Find: n, mol Concept Plan: P, V, T, R Relationships: 1 atm = 14.7 psi PV n RT n PV nRT, R 0.08206 T(K) = t(°C) + 273.15 atmL molK PV n 1 atm 24.3 psi 1.6531 atm R T 14.7 psi 1.6531atm 3.24 L 0.219 mol T(K) 25C 273.15 atm L 0.08206molK 298 K Solution: T 298 K Check: 1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas Standard Conditions • since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions STP • standard pressure = 1 atm • standard temperature = 273 K 0°C Tro, Chemistry: A Molecular Approach 34 Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Tro, Chemistry: A Molecular Approach 35 A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Given: V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C Find: V2, L Concept Plan: P1, V1, T1, R Relationships: 1 atm = 14.7 psi PV n RT n n R T1 atm V psi 44.1 3.00 atm P 14.7 psi atm L 273 K 1.219 mol 0.08206mol K T(K) 27C 273.15 1.00 at m 127 L .K T .300 nRT V P PV nRT, R 0.08206 T(K) = t(°C) + 273.15 Solution: P2, n, T2, R V2 atmL molK PV R T 3.00atm 10.0 L 1.219 mol atm L 0.08206mol 300.K K n Check: 1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas Molar Volume • solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L 6.022 x 1023 molecules of gas notice: the gas is immaterial • we call the volume of 1 mole of gas at STP the molar volume it is important to recognize that one mole of different gases have different masses, even though they have the same volume Tro, Chemistry: A Molecular Approach 37 Molar Volume Tro, Chemistry: A Molecular Approach 38 Density at Standard Conditions • • • • density is the ratio of mass-to-volume density of a gas is generally given in g/L the mass of 1 mole = molar mass the volume of 1 mole at STP = 22.4 L Molar Mass, g Density 22.4 L Tro, Chemistry: A Molecular Approach 39 Gas Density 1 mol mass mass moles moles molar mass molar mass mass in grams density volume in liters PV nR T mass PV RT molar mass mass P (molar mass) density V RT • density is directly proportional to molar mass Tro, Chemistry: A Molecular Approach 40 Example 5.7 – Calculate the density of N2 at 125°C and 755 mmHg Given: Find: Concept Plan: P = 755 mmHg, t = 125 °C, dN2, g/L P, MM, T, R d P MM R T 1 atm = 760 mmHg, MM = 28.01 g d Relationships: T(K) = t(°C) + 273.15 Solution: 755 mmHg 1 atm 0.99342 atm 760 mmHg T(K) 125C 273.15 T 398 K Check: P MM R T atm L R 0.08206mol K d P MM R T 0.99342 atm 28.01molg d 0.08206 398 K atm L mol K 0.852g/L since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is 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 in grams Molar Mass moles Tro, Chemistry: A Molecular Approach 42 Example 5.8 – Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg Given: m=0.311g, V=0.225 L, P=1.1658 P=886 mmHg, atm, T=328 t=55°C,K, Find: molar mass, g/mol P, V, T, R Concept Plan: n PV R T n n, m MM m n MM atmL PV nRT, R 0.08206mol 1 atm = 760 mmHg, K m MM Relationships: T(K) = t(°C) + 273.15 n Solution: m 0.311g 1 atm MM PV 1.1658 atm n886 mmHg n 9.745410-3 mol 760 mmHg R T 31.9g/mol T(K) 55 C273.15 1.1658 atm 0.225L T 328 KatmL 9.7454103 mol 0.08206molK 328 K Check: the value 31.9 g/mol is reasonable Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Tro, Chemistry: A Molecular Approach 44 Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Given: m=9.988g, n=0.250 mol, P=1.0197 P=775 mmHg, atm, T=300. t=27°C,K Find: density, g/L Concept Plan: P, n, T, R V nR T P V V, m m d V d atmL PV nRT, R 0.08206mol 1 atm = 760 mmHg, K m d Relationships: T(K) = t(°C) + 273.15 V Solution: nR T 1 atm 1.0197 atm 775 torr V 760 torr P T(K)mol 27C0.08206 273.15 atm L 300. K 0.250 mol K T 300. K 6.0355 L 1.0197atm Check: m 9.988g V 6.0355 L 1.65g/L d the value 1.65 g/L is reasonable Mixtures of Gases • when gases are mixed together, their molecules behave independent of each other all the gases in the mixture have the same volume all completely fill the container each gas’s volume = the volume of the container all gases in the mixture are at the same temperature therefore they have the same average kinetic energy • therefore, in certain applications, the mixture can be thought of as one gas even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules Tro, Chemistry: A Molecular Approach 46 Partial Pressure • the pressure of a single gas in a mixture of gases is • called its partial pressure we can calculate the partial pressure of a gas if we know what fraction of the mixture it composes and the total pressure or, we know the number of moles of the gas in a container of known volume and temperature • the sum of the partial pressures of all the gases in the mixture equals the total pressure Dalton’s Law of Partial Pressures because the gases behave independently Tro, Chemistry: A Molecular Approach 47 Composition of Dry Air Tro, Chemistry: A Molecular Approach 48 The partial pressure of each gas in a mixture can be calculated using the ideal gas law for two gases, A and B, mixed together nA x R x T nB x R x T PA PB V V the temperatu re and volume of everything in the mixture are the same n total n A n B n total x R x T Ptotal PA PB V Tro, Chemistry: A Molecular Approach 49 Example 5.9 – Determine the mass of Ar in the mixture = 0.275 mmHg, atm, V PNe = =112 1.00 L, mmHg, T=298Ptot K = 662 mmHg, Given: PAr He=341 V = 1.00 L, T=298 K massAr ,g Find: mass Ar, g PAr Concept Plan: Ptot, PHe, PNe PAr = Ptot – (PHe + PNe) PAr, V, T PV n R T n Ar mAr m n MM atmL Ptot = Pa + Pb + etc., PV nRT, R 0.08206mol m K 1 atm = 760 mmHg, MM = 39.95 g/mol MM Relationships: Ar n Solution: PV PAr 662 341 112 mmHg n R T 209 mmHg 1.125 10 2 39.95 g mol 1 mol 0.449 g Ar 0.275atm 1.00 L 1 atm 1.125102 mol 209 mmHg K 0.275 atm 0.08206 298 760 mmHg atm L mol K Check: the units are correct, the value is reasonable Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe. Tro, Chemistry: A Molecular Approach 51 Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe Given: Find: Concept Plan: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm nXe, V, T, R PXe Relationships: n Xe R T V n Xe R T V atm L 0.17 mol 0.08206mol K 598 K 8.7 L 0.9589 at m PXe PNe Ptotal PXe 3.9 atm 0.9589 atm Check: Ptot, PXe PNe PNe Ptotal PXe atmL PV nRT, R 0.08206mol , Ptotal PNe PXe K Solution: 2.9 atm PXe the unit is correct, the value is reasonable Mole Fraction the fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c PA nA Ptotal n total nA cA n total for gases, = volume % / 100% the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure Tro, Chemistry: A Molecular Approach PA cA Ptotal 53 Mountain Climbing & Partial Pressure • our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air • partial pressures of O2 lower than 0.1 atm will lead to hypoxia unconsciousness or death • climbers of Mt Everest carry O2 in cylinders to prevent hypoxia on top of Mt Everest, Pair = 0.311 atm, so PO2 = 0.065 atm Tro, Chemistry: A Molecular Approach 54 Deep Sea Divers & Partial Pressure • its also possible to have too much O2, a condition called oxygen toxicity PO2 > 1.4 atm oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions • its also possible to have too much N2, a condition called nitrogen narcosis also known as Rapture of the Deep • when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases at a depth of 55 m the partial pressure of O2 is 1.4 atm divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air Tro, Chemistry: A Molecular Approach 55 Partial Pressure & Diving Tro, Chemistry: A Molecular Approach 56 Ex 5.10 – Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K Given: Find: Concept Plan: nmHe 24.2mol, g, mO2 nO2==43.2 0.135 g Vmol = 12.5 V = L, 12.5 T =L,298 T =K298 K He==6.05 cHe=0.97817, , cO2, PHe, atm, cO2=0.021827, PO2, atm, PPtotal atm, PO2, atm, Ptotal, atm He, atm mgas cgas ngas c gas A cgas, Ptotal Relationships: n gas A ntot, V, T, R Ptotal n total n total R T V Pgas PA c A Ptotal atmL PV nRT, R 0.08206mol , PA c A Ptotal K Solution: Ptot c He MMHe = 4.00 g/mol MMO2 = 32.00 g/mol PHe cmol Ptotal 6.05 He He 0.97817 6.05mol 0.135 0He .97 817 12mol .099Oatm 2 n total R T 1 mol He P total 24.2 g He V g 6.05 mol He 4.00 11.8atm atmL 1 mol O 6.185 mol 20.08206 molOK 2 298 K 4.32 PO0.135 c Omol POtotal 2 g O2 32.00 g 0.135 mol c 0.021827 O 12.5 L 6.05mol He 827 0.135 0.021 12mol .099Oatm 2 12.099 atm 0.264atm 2 2 2 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 find out the partial pressure of the water vapor in the gas you collect • if you collect a gas sample with a total pressure of 758.2 mmHg* at 25°C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg Table 5.4* Tro, Chemistry: A Molecular Approach 58 Vapor Pressure of Water Tro, Chemistry: A Molecular Approach 59 Collecting Gas by Water Displacement Tro, Chemistry: A Molecular Approach 60 Ex 5.11 – 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2. Given: V=1.02 L, P P=755.2 mmHg, mmHg, T=293 T=293 K K O2=737.65 Find: mass O2, g Concept Plan: Ptot, PH2O PO2 PO2 Ptotal PH2O @ 20C PO2,V,T PV n R T nO2 gO2 atmL PV nRT, R 0.08206mol 1 atm = 760 mmHg, K Relationships: Ptotal = PA + PB, O2 = 32.00 g/mol Solution: PO2 P755.2 V 17.55 (Table 5.4) n PO2 R 737 .65 mmHg T 0.97059 atm 1atm 1.02 L 737.65 mmHg atm L 0.97059 atm 0.08206mol K760 mmHg 293. K 4.117510 2 mol 4.117510 2 mol 1.32g 32.00g 1 mol Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Tro, Chemistry: A Molecular Approach 62 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Given: V=10.0 L, nH2=0.12 mol, T=323 K Find: Ptotal, atm Concept Plan: nH2,V,T P nR T V PH2 PH2, PH2O Ptotal Ptotal PH2 PH2O @ 50C 1 atm = 760 mmHg Relationships: Ptotal = PA + PB, PV nRT, R 0.08206atmL molK Solution: 760 mmHg 0.3181 atm 241.8 mmHg 1 atm nR T V atm L 0.12mol 0.08206mol K 323 K 10.0L 0.3181at m PH 2 Ptotal 241.8 92.6 (Table 5.4) Ptotal 330 mmHg Reactions Involving Gases • the principles of reaction stoichiometry from • Chapter 4 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, 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 coefficients in the equation as a mole ratio when gases are at STP, use 1 mol = 22.4 L P, V, T of Gas A Tro, Chemistry: A Molecular Approach mole A mole B P, V, T of Gas B 64 Ex 5.12 – What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g) Given: n mH2 = 2.2284 = 37.5g, mol,P=738 P=0.97105 mmHg, atm, T=355 T=355 KK CH3OH Find: VH2, L Concept Plan: g CH3OH mol CH3OH 1 mol CH 3OH 32.04g mol H2 P, n, T, R 2 mol H 2 1 mol CH 3OH V V nR T P Relationships: 1 atm = 760 mmHg, CH3OH = 32.04 g/mol atmL 1 mol CH3OH : 2 mol H2 PV nRT, R 0.08206mol K Solution: R HT2 1 mol CH3OH n 2 mol 37.5 g CH3OH V 32.04 g 1 molPCH3OH 2.2284 mol H 2 738 mmHg 2.2284 mol 0.08206 355 K 0.97105 atm 1 atm 0.97105 atm 760 mmHg 66.9 L atm L mol K Ex 5.13 – How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g) Given: VH2 = 1.24 L, P=1.00 atm, T=273 K Find: massH2O, g Concept Plan: L H2 1 mol H 2 22.4 L mol H2 mol H2O 2 mol H 2 2 mol H 2O g H2 O 18.02 g 1 mol H 2O Relationships: H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP 2 mol H2O : 2 mol H2 Solution: 1 mol H 2 2 mol H 2O 18.02 g H 2O 1.24 L H 2 22.4 L H 2 2 mol H 2 1 mol H 2O 0.998 g H 2O Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) 2 Hg(l) + O2(g) (MMHgO = 216.59 g/mol) Tro, Chemistry: A Molecular Approach 67 What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) 2 Hg(l) + O2(g) Given: n mO2 == 0.023085 10.0g, P=0.750 mol, P=0.750 atm, T=313 atm, T=313 K K HgO Find: VO2, L Concept Plan: g HgO mol HgO 1 molHgO 216.59g mol O2 P, n, T, R 1 molO 2 2 mol HgO V V nR T P Relationships: 1 atm = 760 mmHg, HgO = 216.59 g/mol atmL 2 mol HgO : 1 mol O2 PV nRT, R 0.08206mol K Solution: nR T 1 mol HgO 1 mol V O2 10.0 g HgO 216.59 g 2 mol HgO P 0.023085 mol 0.023085 mol O 2 0.791L 0.08206 0.750at m atm L mol K 313K Properties of Gases • expand to completely fill 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, Chemistry: A Molecular Approach 69 Kinetic Molecular Theory • the particles of the gas (either atoms or molecules) are constantly moving • the attraction between particles is negligible • when the moving particles hit another particle or the container, they do not stick; but they bounce off and continue moving in another direction like billiard balls Tro, Chemistry: A Molecular Approach 70 Kinetic Molecular Theory • there is a lot of empty space between the particles compared to the size of the particles • the average kinetic energy of the particles is directly proportional to the Kelvin temperature as you raise the temperature of the gas, the average speed of the particles increases but don’t be fooled into thinking all the particles are moving at the same speed!! Tro, Chemistry: A Molecular Approach 71 Gas Properties Explained – Indefinite Shape and Indefinite Volume Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container. Tro, Chemistry: A Molecular Approach As a result, gases take the shape and the volume of the container they are in. 72 Gas Properties Explained Compressibility Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together Tro, Chemistry: A Molecular Approach 73 Gas Properties Explained – Low Density Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is they have low density Tro, Chemistry: A Molecular Approach 74 Density & Pressure • 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, Chemistry: A Molecular Approach 75 Gas Laws Explained Boyle’s Law • 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 more molecules will collide with the container at any one instant, increasing the pressure Tro, Chemistry: A Molecular Approach 76 Gas Laws Explained Charles’s Law • Charles’s Law says that the volume of • a gas is directly proportional to the absolute temperature increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently on average • in order to keep the pressure constant, the volume must then increase Tro, Chemistry: A Molecular Approach 77 Gas Laws Explained Avogadro’s Law • Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules • increasing the number of gas molecules causes more of them to hit the wall at the same time • in order to keep the pressure constant, the volume must then increase Tro, Chemistry: A Molecular Approach 78 Gas Laws Explained – Dalton’s Law of Partial Pressures • Dalton’s Law says that the total pressure of a mixture of • • • gases is the sum of the partial pressures kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact therefore the molecules behave independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy since the average kinetic energy is the same, the total pressure of the collisions is the same Tro, Chemistry: A Molecular Approach 79 Dalton’s Law & Pressure • since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side Tro, Chemistry: A Molecular Approach 80 Deriving the Ideal Gas Law from Kinetic-Molecular Theory • pressure = Forcetotal/Area • Ftotal = F1 collision x number of collisions in a particular time interval F1 collision = mass x 2(velocity)/time interval no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall Ftotal α mass∙velocity2 x Area x no. molecules/Volume • Pressure α mv2 x n/V • Temperature α mv2 • P α T∙n/V, PV=nRT Tro, Chemistry: A Molecular Approach 81 Calculating Gas Pressure Tro, Chemistry: A Molecular Approach 82 Molecular Velocities • all the gas molecules in a sample can travel at different • • • speeds however, the distribution of speeds follows a pattern called a Boltzman distribution we talk about the “average velocity” of the molecules, but there are different ways to take this kind of average the method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities v2 urms u2 n Tro, Chemistry: A Molecular Approach 83 Boltzman Distribution Distribution Function Fraction of Molecules O2 @ 300 K Molecular Speed Tro, Chemistry: A Molecular Approach 84 Kinetic Energy and Molecular Velocities • average kinetic energy of the gas molecules depends on the average mass and velocity KE = ½mv2 • gases in the same container have the same temperature, • the same average kinetic energy if they have different masses, the only way for them to have the same kinetic energy is to have different average velocities lighter particles will have a faster average velocity than more massive particles Tro, Chemistry: A Molecular Approach 85 Molecular Speed vs. Molar Mass • in order to have the same average kinetic energy, heavier molecules must have a slower average speed Tro, Chemistry: A Molecular Approach 86 Temperature _and Molecular Velocities • KEavg = ½NAmu2 NA is Avogadro’s number • KEavg = 1.5RT R is the gas constant in energy units, 8.314 J/mol∙K 1 J = 1 kg∙m2/s2 • equating and solving we get: NA∙mass = molar mass in kg/mol urms 3RT 3RT NA m MM • as temperature increases, the average velocity increases Tro, Chemistry: A Molecular Approach 87 Temperature vs. Molecular Speed • as the absolute temperature increases, the average velocity increases the distribution function “spreads out,” resulting in more molecules with faster speeds Tro, Chemistry: A Molecular Approach 88 Ex 5.14 – Calculate the rms velocity of O2 at 25°C Given: O2, t = 25°C Find: urms Concept Plan: MM, T urms urm s 3RT MM urm s 3RT MM Relationships: T(K) = t(°C) + 273.15, O2 = 32.00 g/mol Solution: T (K) t (C) 273.15 T 25 273.15 T 298 K 3RT urms MM kg m 2 2 s 3 8.314 298K molK 482 m/s -3 kg 32.0010 mol Mean Free Path • molecules in a gas travel in • • straight lines until they collide with another molecule or the container the average distance a molecule travels between collisions is called the mean free path mean free path decreases as the pressure increases Tro, Chemistry: A Molecular Approach 90 Diffusion and Effusion • the process of a collection of molecules spreading out • • • from high concentration to low concentration is called diffusion the process by which a collection of molecules escapes through a small hole into a vacuum is called effusion both the rates of diffusion and effusion of a gas are related to its rms average velocity for gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of the molar mass rate Tro, Chemistry: A Molecular Approach 1 MM 91 Effusion Tro, Chemistry: A Molecular Approach 92 Graham’s Law of Effusion • for two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation: rategas A rategas B Tro, Chemistry: A Molecular Approach Molar Massgas B Molar Massgas A 93 Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2 Given: rateunknown gas 0.462 rateN 2 Find: MM, g/mol Concept Plan: rateA/rateB, MMN2 MMunknown Molar Massunknown Relationships: N2 = 28.01 g/mol Molar MassN2 rateunknown rateN 2 2 rat egas A rat egas B Molar Massgas B Molar Massgas A Solution: Molar Massunknown Molar MassN 2 rateunknown rateN 2 2 28.01 g m ol 131g 0.462 2 m ol • • • Ideal vs. Real Gases Real gases often do not behave like ideal gases at high pressure or low temperature Ideal gas laws assume 1) no attractions between gas molecules 2) gas molecules do not take up space based on the kinetic-molecular theory at low temperatures and high pressures these assumptions are not valid 95 The Effect of Molecular Volume • at high pressure, the amount of space occupied by the molecules is a significant amount of the total volume • the molecular volume makes the real volume larger than the ideal gas law would predict • van der Waals modified the ideal gas equation to account for the molecular volume b is called a van der Waals constant and is different for every gas because their molecules are different sizes nRT V Tro, Chemistry: A Molecular Approach P nb 96 Real Gas Behavior • because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures Tro, Chemistry: A Molecular Approach 97 The Effect of Intermolecular Attractions • at low temperature, the attractions between the • • molecules is significant the intermolecular attractions makes the real pressure less than the ideal gas law would predict van der Waals modified the ideal gas equation to account for the intermolecular attractions a is called a van der Waals constant and is different for every gas because their molecules are different sizes nRT n P a V V Tro, Chemistry: A Molecular Approach 2 98 Real Gas Behavior • because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures Tro, Chemistry: A Molecular Approach 99 Van der Waals’ Equation • combining the equations to account for molecular volume and intermolecular attractions we get the following equation used for real gases a and b are called van der Waal constants and are different for each gas 2 P a n V - nb nRT V Tro, Chemistry: A Molecular Approach 100 Real Gases • a plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases • it reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideality for “low” pressures – meaning the most important factor is the intermolecular attractions • it reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideality for “high” pressures – meaning the most important factor is the molecular volume Tro, Chemistry: A Molecular Approach 101 PV/RT Plots Tro, Chemistry: A Molecular Approach 102 Structure of the Atmosphere • the atmosphere shows several • layers, each with its own characteristics the troposphere is the layer closest to the earth’s surface circular mixing due to thermal currents – weather • the stratosphere is the next layer up less air mixing • the boundary between the • troposphere and stratosphere is called the tropopause the ozone layer is located in the stratosphere Tro, Chemistry: A Molecular Approach 103 Air Pollution • air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, man’s activities though many of the “pollutant” gases have natural sources as well • pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it and the air mixing in the troposphere means that we all get a smell of it! • pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out Tro, Chemistry: A Molecular Approach 104 Pollutant Gases, SOx • SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining as well as volcanoes • lung and eye irritants • major contributor to acid rain 2 SO2 + O2 + 2 H2O 2 H2SO4 SO3 + H2O H2SO4 Tro, Chemistry: A Molecular Approach 105 Pollutant Gases, NOx • NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants as well as lightning storms • • • • NO2 causes the brown haze seen in some cities lung and eye irritants strong oxidizers major contributor to acid rain 4 NO + 3 O2 + 2 H2O 4 HNO3 4 NO2 + O2 + 2 H2O 4 HNO3 Tro, Chemistry: A Molecular Approach 106 Pollutant Gases, CO • CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants • adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2 • at high levels can cause sensory impairment, stupor, unconsciousness, or death Tro, Chemistry: A Molecular Approach 107 Pollutant Gases, O3 • ozone pollution comes from other pollutant gases reacting in the presence of sunlight as well as lightning storms known as photochemical smog and ground-level ozone • O3 is present in the brown haze seen in some cities • lung and eye irritants • strong oxidizer Tro, Chemistry: A Molecular Approach 108 Major Pollutant Levels • government regulation has resulted in a decrease in the emission levels for most major pollutants Tro, Chemistry: A Molecular Approach 109 Stratospheric Ozone • ozone occurs naturally in the stratosphere • stratospheric ozone protects the surface of the earth from over-exposure to UV light from the sun O3(g) + UV light O2(g) + O(g) • normally the reverse reaction occurs quickly, but the energy is not UV light O2(g) + O(g) O3(g) Tro, Chemistry: A Molecular Approach 110 Ozone Depletion • chlorofluorocarbons became popular as aerosol • • • propellants and refrigerants in the 1960s CFCs pass through the tropopause into the stratosphere there CFCs can be decomposed by UV light, releasing Cl atoms CF2Cl2 + UV light CF2Cl + Cl Cl atoms catalyze O3 decomposition and removes O atoms so that O3 cannot be regenerated NO2 also catalyzes O3 destruction Cl + O3 ClO + O2 O3 + UV light O2 + O ClO + O O2 + Cl Tro, Chemistry: A Molecular Approach 111 Ozone Holes • satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions Tro, Chemistry: A Molecular Approach 112