Kinetic-Molecular Theory of Matter

Gases Properties of Gases Gas Pressure Gases What gases are important for each of the following: O 2, CO 2 and/or He? A. B. C. D. 1 2 Gases What gases are important for each of the following: O 2, CO 2 and/or He? A. CO 2 B. O 2 /CO 2 C. O 2 D. He Kinetic-Molecular Theory of Matter Particles of Matter are always in motion Ideal Gases-an an imaginary gas that fits all the assumptions of the theory Kinetic Energy (KE) formula Physical properties of gases Real gases- gases in our daily lives 3 4 Ideal Gas-imaginary gas that fits all the following assumptions. Particles in gases: Are very far apart Have collisions that are elastic (no KE loss) Move rapidly Have no attraction (or repulsion) Have energy increases at higher temperatures Kinetic Energy of Gases At the same temperature, all gas particles have the same amount of energy Lighter particles move faster than heavier particles KE = mv 2 KE =kinetic energy 2 v = velocity (speed) m = mass 5 6 1

Physical Properties of Gases Gases are compressible Why can you put more air in a tire, but can t add more water to a glass full of water? Gases have low densities Dsolid or liquid = 2 g/ml Dgas 2 g/l Physical Properties of Gases 1. Why does a round balloon become spherical when filled with air? 2. Suppose we filled this room halfway with water. Where would pressure be exerted? 7 8 9 Physical Properties of Gases Gases fill a container completely and uniformly Gases exert a uniform pressure on all inner surfaces of their containers Fluidity: gas particles can slide past one another (gases and liquids are fluids ) Diffusion: gases move from high concentration to low concentration 10 Real Gases Close to ideal gas at standard conditions Have volume Attraction between particles Deviation from ideal gas is greater when Particles are close together» Low temperatures» High pressures Gas is a compound rather than an element oxygen O 2 argon Ar Some Gases in Our Lives Air: nitrogen N 2 carbon dioxide CO 2 Noble gases: ozone O 3 water H 2 O helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F 2 chlorine Cl 2 methane CH 4 nitrogen dioxide NO 11 2 ammonia NH 3 carbon monoxide CO sulfur dioxide SO 2 12 Pressure Force per unit area P = force area Gas pressure is a result of collisions of gas particles. Depends on: Number of gas particles Temperature Volume 2

Instruments used to measure pressure Barometers Barometer measures atmospheric pressure Manometer measures gas pressure of a container 760 mmhg atm pressure 13 14 Manometer Unit of Pressure 15 One atmosphere (1 atm) Is the average pressure of the atmosphere at sea level Is the standard of pressure P = Force Area 1.00atm=760mmHg=760torr=101.3kPa=14.7psi 16 Pressure Types of Pressure Units 760 mm Hg or 760 torr Used in Chemistry 14.7 lb/in. 2 U.S. pressure gauges 29.9 in. Hg U.S. weather reports 101.3 kpa (kilopascals) Weather in all countries except U.S. 1.013 bars Physics and astronomy 17 18 Conversions 760.mmHg=760.torr 760.torr=1.00atm=101.3kPa=14.7psi 14.7psi What is 2.00 atm expressed in torr? 3

19 Conversions 760.mmHg=760.torr 760.torr=1.00atm=101.3kPa=14.7psi 14.7psi The pressure of a tire is measured as 32.0 psi. What is this pressure in kpa? Learning Check G1 A.The downward pressure of the Hg in a barometer is than (as) the weight of the atmosphere. 1) greater 2) less 3) the same B.A water barometer has to be 13.6 times taller than Hg barometer (D( Hg = 13.6 g/ml) because 1) H 2 O is less dense 3) 20 air is more dense than H 2 O 2) Hg is heavier 21 Solution G1 A.The downward pressure of the Hg in a barometer is 3) the same (as) the weight of the atmosphere. B.A water barometer has to be 13.6 times taller than Hg barometer (D( Hg = 13.6 g/ml) because 1) H 2 O is less dense Learning Check G2 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 2) the liquid is at a lower level 3) there is empty space in the straw Could you drink a soda this way on the moon? 1) yes 2) no 3) maybe Why or why not? 22 Solution G2 Learning Check G3 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 3) there is empty space in the straw Could you drink a soda this way on the moon? 2) no Why or why not? Low atmospheric pressure 23 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 10 5 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3) 22,300 mm Hg 24 4

Solution G3 A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm 760. mm Hg (B) B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 760. mmhg = 1.52 x 10 3 mmhg 14.7 psi (B) Gas Laws Boyle s s Law Charles Law Gay-Lussac Lussac s s Law Combined Gas Law Ideal Gas Law Dalton s s Partial Pressure Graham s s Law of Effusion 25 26 Boyle s s Law Reducing the volume by one-half doubles the pressure The volume of a fixed mass of gas varies inversely with the pressure at constant temperature P 2 V 2 As the pressure increases Volume decreases 27 28 29 Pressure and Volume ExperimentPressure Volume P x V (atm)) (L) (atm( x L) 1 8.0 2.0 16 2 4.0 4.0 3 2.0 8.0 4 1.0 16 Boyle's Law P x V = k (constant) when T remains constant 30 Volume How does Pressure and Volume of gases relate graphically? Pressure PV = k Temperature, # of particles remain constant 5

31 Boyle s s Mathematical Law: What if we had a change in conditions? since PV = k = P 2 V 2 Eg: : A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? 1) determine which variables you have: = 2 atm = 3.0 L P 2 = 4 atm V 2 =? 2)determine which law is being represented: 32 P and V = Boyle s s Law 3) Rearrange the equation for the variable you don t t know P 2 4) Plug in the variables and chug it on a calculator: (2.0 atm)(3.0l) = V 2 33 (4atm) = V 2 V = 1.5L 2 34 Learning Check GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm?? (T constant) Explain. 1) 3.2 L 2) 6.4 L 3) 12.8 L 35 Solution GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm?? (T constant) 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume must decrease to cause an increase in the pressure 36 Learning Check GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmhg 2) 400. mmhg 3) 1200 mmhg 6

Solution GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 600. mm Hg x 12.0 L = 200. mmhg (1) 36.0 L Pressure decrease when volume increases. Volume of balloon at room temperature Volume of balloon at 5 C 5 37 38 Charles Law Charles Law: V and T V = 125 ml V = 250 ml T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature? 39 At constant pressure, the volume of a gas is directly related to its absolute (K) temperature 40 = V 2 or K = C C + 273 = V 2 How does Temperature and Volume of gases relate graphically? V/T = k Volume 41 Temp Pressure, # of particles remain constant 1) determine which variables you have: = 127 C C + 273 = 400K = 3.0 L = 227 C C + 273 = 5ooK V 2 =? 2)determine which law is being represented: 42 T and V = Charles s s Law 7

4) Plug in the variables: 43 3.0L V = 2 400K 500K 5) Cross multiply and chug (500K)(3.0L) = V 2 (400K) V 2 = 3.8L 44 Learning Check GL3 Use Charles Law to complete the statements below: 1. If final T is higher than initial T, final V is (greater,( or less) ) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) ) than the initial T. Solution GL3 V and T Problem 45 = V 2 OR = V 2 1. If final T is higher than initial T, final V is (greater( greater) ) than the initial V. 2. If final V is less than initial V, final T is (lower( lower) than the initial T. 46 A balloon has a volume of 785 ml on a Fall day when the temperature is 21 C. In the winter, the gas cools to 0 C. 0 What is the new volume of the balloon? VT Calculation Complete the following setup: Initial conditions Final conditions = 785 ml V 2 =? = 21 C C = 294 K = 0 C 0 C = 273 K V 2 = ml x K = ml K Learning Check GL4 A sample of oxygen gas has a volume of 420 ml at a temperature of 18 C. What temperature (in C) is needed to change the volume to 640 ml? 1) 443 C 2) 170 C C 3) - 82 C 47 Check your answer: If temperature decreases, V should decrease. 48 8

Solution GL4 A sample of oxygen gas has a volume of 420 ml at a temperature of 18 C. What temperature (in C) is needed to change the volume to 640 ml? 2) 170 C = 291 K x 640 ml = 443 K 420 ml Gay-Lussac Lussac s s Law: P and T Doubling the Kelvin temperature doubles the pressure The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature = P 2 or = P 2 = 443 K - 273 K = 170 C 49 50 Think of a tire... Think of a tire... Car before a trip Let s s get on the road Dude! Pressure Gauge Car after a long trip WHEW! Pressure Gauge 51 52 How does Pressure and Temperature of gases relate graphically? Lussac s Mathematical Law: What if we had a change in conditions? since P/T = k Pressure 53 Temp P/T = k Volume, # of particles remain constant 54 P = 2 Eg: : A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C? 9

1) determine which variables you have: = 127 C C + 273 = 400K = 3.0 atm = 227 C C + 273 = 500K P 2 =? 2)determine which law is being represented: 55 T and P = Gay-Lussac Lussac s s Law 4) Plug in the variables: 56 3.0atm P = 2 400K 500K 5) Cross multiply and chug (500K)(3.0atm) = P (400K) 2 P 2 = 3.8atm Learning Check GL5 LAW RELAT-IONSHIP LAW CON-STANT Use Gay-Lussac Lussac s s law to complete the statements below: Boyle s Charles P V V T = P 2 V 2 / = V 2 / T, n P, n 1. When temperature decreases, the pressure of a gas (decreases( or increases). 2. When temperature increases, the pressure of a gas (decreases or increases). Gay-Lussac Lussac s 57 P T / = P 2 / V, n 58 Solution GL5 1. When temperature decreases, the pressure of a gas (decreases( decreases). 2. When temperature increases, the pressure of a gas (increases( increases). PT Problem A gas has a pressure at 2.0 atm at 18 C. What will be the new pressure if the temperature rises to 62 C? (V constant) T = 18 C C T = 62 C 59 60 10

61 PT Calculation = 2.0 atm = 18 C C + 273 = 291 K P 2 =?? = 62 C C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P 2 = x P 2 = 2.0 atm x K = atm K Complete with Learning Check GL6 1) Increases 2) Decreases 3) Does not change A. Pressure, when V decreases B. When T decreases, V. C. Pressure when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume when T changes from 15.0 C C to 45.0 C C (constant P and n) 62 Solution GL6 A. Pressure 1) Increases,, when V decreases B. When T decreases, V 2) Decreases C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L (constant n and T) The Combined Gas Law Volume and Moles (Avogadro s s Law) Partial Pressures D. Volume 1) Increases when T changes from 15.0 C C to 45.0 C C (constant P and n) 63 64 Combined Gas Law = P 2 V 2 or = P 2 V 2 Rearrange the combined gas law to solve for V 2 = P 2 V 2 Combined Gas Law = P 2 V 2 or = P 2 V 2 Isolate V 2 = P 2 V 2 65 V 2 = P 2 66 V 2 = P 2 11

Learning Check C1 Solve the combined gas laws for. Solution C1 Solve the combined gas law for. (Hint: cross-multiply first.) = P 2 V 2 = P 2 V 2 = P 2 V 2 67 68 Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29 C. What is the new temperature( C) of the gas at a volume of 90.0 ml and a pressure of 3.20 atm? Set up Data Table = 0.800 atm Data Table = 0.180 L = 302 K P 2 = 3.20 atm V 2 = 90.0 ml =?? 69 70 Solution Calculation Solve for Enter data = 302 K x atm x atm ml = K ml Solve for = 302 K x 3.20 atm x 90.0 ml = 604 K 0.800 atm 180.0 ml = K - 273 = C = 604 K - 273 = 331 C 71 72 12

Learning Check C2 A gas has a volume of 675 ml at 35 C C and 0.850 atm pressure. What is the temperature in C C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? Solution G9 = 308 K =? = 675 ml V 2 = 0.315 L = 315 ml = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg 73 74 = 308 K x 802 mm Hg x 315 ml 646 mm Hg 675 ml P inc, T inc V dec,, T dec = 178 K - 273 = - 95 C 75 Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume? True (1) or False(2) 76 Learning Check C3 1. The P exerted by a gas at constant V is not affected by the T of the gas. 2. At constant P, the V of a gas is directly proportional to the absolute T 3. At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V. True (1) or False(2) 77 Solution C3 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V. 78 Avogadro s s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas = V 2 n 1 n 2 initial final 13

STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0 C 0 C or 273 K Standard pressure 1 atm (760 mm Hg) Learning Check C4 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and 25 C? = = = K P 2 = V 2 =?? = K 79 V 2 = 15 L x atm x K = 6.8 L atm K 80 Solution C4 Ideal Gas Law = 1.0 atm = 15 L = 273 K P 2 = 2.0 atm V 2 =?? = 248 K V 2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K The equality for the four variables involved in Boyle s s Law, Charles Law, Gay-Lussac Lussac s s Law and Avogadro s s law can be written PV = nrt R = ideal gas constant 81 82 PV = nrt Learning Check G15 83 R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nt (1mol) (273K) n T = 0.0821 L-atm mol-k 84 What is the value of R when the STP value for P is 760 mmhg? 14

Solution G15 What is the value of R when the STP value for P is 760 mmhg? R = PV = (760 mm Hg) (22.4 L) nt (1mol) (273K) Learning Check G16 Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23 C, what is the pressure (mmhg) in the tank in the dentist office? 85 = 62.4 L-mm L Hg mol-k 86 87 Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L T = 23 C C + 273 20.0 L 296 K n = 2.86 mol 2.86 mol P =?? 88 Rearrange ideal gas law for unknown P P = nrt V Substitute values of n, R, T and V and solve for P P = (2.86( mol)(62.4l-mmhg mmhg)(296 K) (20.0 L) (K-mol mol) = 2.64 x 10 3 mm Hg Learning Check G17 Solution G17 A 5.0 L cylinder contains oxygen gas at 20.0 C C and 735 mm Hg. How many grams of oxygen are in the cylinder? Solve ideal gas equation for n (moles) n = PV RT = (735 mmhg)(5.0 L)(mol L K) K (62.4 mmhg L)(293 L K) 89 90 = 0. 20 mol O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2 15

At STP Molar Volume 4.0 g He 16.0 g CH 4 44.0 g CO 2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L 91 92 Learning Check C5 Solution C5 93 A.What is the volume at STP of 4.00 g of CH 4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g 2) 0.357 g3) g 1.43 g A.What is the volume at STP of 4.00 g of CH 4? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 94 16.0 g CH 4 1 mole CH 4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He 95 Daltons Law of Partial Pressures Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = + P 2 + P 3 +... 96 The % of gases in air P AIR Gases in the Air Partial pressure (STP) 78.08% N 2 593.4 mmhg 20.95% O 2 159.2 mmhg 0.94% Ar 7.1 mmhg 0.03% CO 2 0.2 mmhg = P N + P O + P Ar + P CO = 760 mmhg 2 2 2 16

Learning Check C6 Solution C6 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 97 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.14 3) 0.109 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 98 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 99 Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm P = 1.00 atm 1 mole H 2 0.5 mole O 2 + 0.3 mole He + 0.2 mole Ar 100 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents. Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O 2 and 0.418 mole He at 25 C. What is the partial pressure of each gas, and what is the total pressure in the tank? P Solution C7 = nrt P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atmL x 298 K 5.00 L (K mol) = 7.19 atm 101 102 17

Gases Collected by Displacement of Water Atmospheric pressure equals the pressure of the gas plus the pressure of the water vapor P atmosphere = P gas + P H2O Water vapor pressure table Table of Vapor Pressures for Water 103 104 Practice Find the pressure of oxygen gas collected over water at 25.0 C C if the barometer reading is 752mmHg. Gases and the Mole 105 106 107 Gases Many of the chemicals we deal with are gases. They are difficult to weigh, so we ll measure volume Need to know how many moles of gas we have. Two things affect the volume of a gas Temperature and pressure Compare at the same temp. and pressure. 108 Standard Temperature and Pressure Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. 0ºC C and 1 atmosphere pressure Abbreviated atm 273 K and 101.3 kpa kpa is kilopascal 18

At Standard Temperature and Pressure abbreviated STP At ST mole of gas occupies 22.4 L Called the molar volume Used for conversion factors Moles to Liter and L to mol Conversion factors Used to change units. Three questions What can you change the given into? What unit do you want to get rid of? Where does it go to cancel out? 109 110 111 Examples What is the volume of 4.59 mole of CO 2 gas at STP? 112 Density of a gas D = m /V for a gas the units will be g / L We can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole than the mass is the molar mass (PT) At STP the volume is 22.4 L. Examples Find the density of CO 2 at STP. Quizdom Find the density of CH 4 at STP. 113 114 19

The other way Given the density, we can find the molar mass of the gas. Again, pretend you have a mole at STP, so V = 22.4 L. m = D x V m is the mass of 1 mole, since you have 22.4 L of the stuff. What is the molar mass of a gas with a density of 1.964 g/l? 115 116 Gases and Stoichiometry 2 H 2 O 2 (l) ---> > 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O 2 22.4 L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 117 = 0.36 L O 2 at STP 118 Gas Stoichiometry: Practice! A. What is the volume at STP of 4.00 g of CH 4? B. How many grams of He are present in 8.0 L of gas at STP? What if it s s NOT at STP? 1. Do the problem like it was at STP. ( ) 2. Convert from STP (,, ) to the stated conditions (P 2, ) GAS DIFFUSION & EFFUSION diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container. 119 120 20

GAS DIFFUSION & EFFUSION Graham s s law governs effusion and diffusion of gas molecules. 121 Rate for A Rate for B M of B M of A Rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. 1869. Professor in Glasgow and London. GAS DIFFUSION & EFFUSION 122 Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to T inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse 3 from opposite ends of of tube. Gases meet to to form NH 4 Cl 4 Cl HCl heavier than NH 33 Therefore, NH 4 Cl 4 Cl forms closer to to HCl end of of tube. 123 If neon travels at 400. m/s, estimate the average speed of butane (C 4 H 10 ) at the same temperature. 235 m/s 124 Chlorine has a velocity of 0.0380 m/s. What is the average velocity of sulfur dioxide under the same conditions? 0.0400 m/s 125 21