Kinetic Theory of Gases. 6.1 Properties of Gases 6.2 Gas Pressure. Properties That Describe a Gas. Gas Pressure. Learning Check.


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1 Chapter 6 Gases Kinetic Theory of Gases 6.1 Properties of Gases 6.2 Gas Pressure A gas consists of small particles that move rapidly in straight lines. have essentially no attractive (or repulsive) forces. are very far apart. have very small volumes compared to the volume of the container they occupy. have kinetic energies that increase with an increase in temperature. 1 2 Properties That Describe a Gas Gas Pressure Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Gas pressure is a force acting on a specific area. Pressure (P) = force area has units of atm, mmhg, torr, lb/in. 2, and kilopascals(kpa). 1 atm = 760 mm Hg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in. 2 1 atm = Pa 1 atm = kpa 3 4 A. What is 475 mmhg expressed in atm? 1) 475 atm 2) atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mmhg? 1) 2.00 mmhg 2) 1520 mmhg 3) mmhg A. What is 475 mmhg expressed in atm? 2) atm 475 mmhg x 1 atm = atm 760 mmhg B. The pressure in a tire is 2.00 atm. What is this pressure in mmhg? 2) 1520 mmhg 2.00 atm x 760 mmhg = 1520 mmhg 1 atm 5 6
2 Atmospheric Pressure Altitude and Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Atmospheric pressure is about 1 atmosphere at sea level. depends on the altitude and the weather. is lower at higher altitudes, where the density of air is less. is higher on a rainy day than on a sunny day. 7 8 Barometer A barometer measures the pressure exerted by the gases in the atmosphere. indicates atmospheric pressure as the height in mm of the mercury column. A. The downward pressure of the Hg in a barometer is than (as) the pressure of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/ml) because 1) H 2 O is less dense than mercury. 2) H 2 O is heavier than mercury. 3) air is more dense than H 2 O Chapter 6 Gases A.The downward pressure of the Hg in a barometer is 3) the same (as) the pressure of the atmosphere. 6.3 Pressure and Volume (Boyle s Law) B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/ml) because 1) H 2 O is less dense than mercury
3 Boyle s Law PV Constant in Boyle s Law Boyle s law states that the pressure of a gas is inversely related to its volume when T and n are constant. if volume decreases, the pressure increases. In Boyle s law, the product P x V is constant as long as T and n do not change. P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyle s law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) Solving for a Gas Law Factor Boyles Law and Breathing The equation for Boyle s law can be rearranged to solve for any factor. P 1 V 1 = P 2 V 2 Boyle s law To solve for V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 P 2 P 2 V 1 x P 1 = V 2 P 2 During an inhalation, the lungs expand. the pressure in the lungs decreases. air flows towards the lower pressure in the lungs Boyles Law and Breathing Calculations with Boyle s Law During an exhalation, lung volume decreases. pressure within the lungs increases. air flows from the higher pressure in the lungs to the outside
4 Calculation with Boyle s Law Freon12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mmhg after its pressure is changed to 2200 mmhg at constant T and n? 1. Set up a data table: Conditions 1 Conditions 2 P 1 = 550 mmhg P 2 = 2200 mmhg V 1 = 8.0 L V 2 =? Calculation with Boyle s Law (continued) 2. When pressure increases, volume decreases. Solve Boyle s law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 x P 1 P 2 V 2 = 8.0 L x 550 mmhg = 2.0 L 2200 mmhg pressure ratio decreases volume For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant). 1) pressure decreases 2) pressure increases For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant). 1) pressure decreases B 2) pressure increases A If a sample of helium gas has a volume of 120 ml and a pressure of 850 mmhg, what is the new volume if the pressure is changed to 425 mmhg? 1) 60 ml 2) 120 ml 3) 240 ml 3) 240 ml P 1 = 850 mmhg P 2 = 425 mmhg V 1 = 120 ml V 2 =?? V 2 = V 1 x P 1 = 120 ml x 850 mmhg = 240 ml P mmhg Pressure ratio increases volume 23 24
5 A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is constant), is the new volume represented by A, B, or C? A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon A If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L B) 6.4 L C) 12.8 L If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L V 2 = V 1 x P 1 P 2 V 2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (temperature is constant.) A sample of oxygen gas has a volume of 12.0 L at 600. mmhg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mmhg 2) 400. mmhg 3) 1200 mmhg 1) 200. mmhg Data Table Conditions 1 Conditions 2 P 1 = 600. mmhg P 2 =??? V 1 = 12.0 L V 2 = 36.0 L P 2 = P 1 x V 1 V mmhg x 12.0 L = 200. mmhg 36.0 L 29 30
6 If the sample of nitrogen (gas) has a volume of 360 ml at a pressure of 720 mmhg, what is the new volume when the pressure is increased to 1.20 atm (T constant)? A) 284 ml B) 456 ml C) 2160 ml We need to make the units for initial and final pressure the same: 1.20 atm x 760 mmhg = 912 mmhg 1 atm V 2 = V 1 x P 1 P 2 V 2 = 360 ml x 720 mmhg = 284 ml (A) 912 mmhg Volume decreases when there is an increase in the pressure (temperature is constant.) Chapter 6 Gases Charles s Law 6.4 Temperature and Volume (Charles s Law) In Charles s Law, the Kelvin temperature of a gas is directly related to the volume. P and n are constant. when the temperature of a gas increases, its volume increases Charles s Law: V and T For two conditions, Charles s law is written V 1 = V 2 (P and n constant) Rearranging Charles s law to solve for V 2 : T 2 x V 1 = V 2 x T 1 T 1 T 1 Solve Charles s law expression for T 2. V 1 = V 2 V 2 = V 1 x T 2 T
7 Calculations Using Charles s Law V 1 = V 2 Crossmultiply to give: V 1 T 2 = V 2 T 1 Isolate T 2 by dividing through by V 1 : V 1 T 2 = V 2 T 1 V 1 V 1 T 2 = T 1 x V 2 V 1 A balloon has a volume of 785 ml at 21 C. If the temperature drops to 0 C, what is the new volume of the balloon (P constant)? 1. Set up data table: Conditions 1 Conditions 2 V 1 = 785 ml V 2 =? T 1 = 21 C = 294 K T 2 = 0 C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations Calculations Using Charles s Law (continued) 2. Solve Charles s law for V 2 : V 1 = V 2 V 2 = V 1 x T 2 T 1 V 2 = 785 ml x 273 K = 729 ml 294 K A sample of oxygen gas has a volume of 420 ml at a temperature of 18 C. At what temperature (in C) will the volume of the oxygen be 640 ml (P and n constant)? 1) 443 C 2) 170 C 3)  82 C ) 170 C T 2 = T 1 x V 2 V 1 T 2 = 291 K x 640 ml = 443 K 420 ml = 443 K 273 = 170 C Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure when V decreases. B. When T decreases, V. C. Pressure when V changes from 12 L to 24 L. D. Volume when T changes from 15 C to 45 C
8 Chapter 6 Gases Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure 1) increases when V decreases. B. When T decreases, V 2) decreases. C. Pressure 2) decreases when V changes from 12 L to 24 L. D. Volume 1) increases when T changes from 15 C to 45 C. 6.5 Temperature and Pressure (GayLussac s Law) GayLussac s Law: P and T In GayLussac s law the pressure exerted by a gas is directly related to the Kelvin temperature. Solve GayLussac s law for P 2. P 1 = P 2 V and n are constant. P 1 = P Solve GayLussac s law for P 2. P 1 = P 2 Multiply both sides by T 2 and cancel: P 1 x T 2 = P 2 x T 1 T 1 T 1 P 2 = P 1 x T 2 T 1 Calculation with GayLussac s Law A gas has a pressure at 2.0 atm at 18 C. What is the new pressure when the temperature is 62 C? (V and n constant) 1. Set up a data table: Conditions 1 Conditions 2 P 1 = 2.0 atm P 2 =? T 1 = 18 C T 2 = 62 C = 291 K = 335 K 47 48
9 Calculation with GayLussac s Law (continued) 2. Solve GayLussac s law for P 2 : P 1 = P 2 A gas has a pressure of 645 torr at 128 C. What is the temperature in Celsius if the pressure increases to 824 torr? (n and V remain constant) P 2 = P 1 x T 2 T 1 P 2 = 2.0 atm x 335 K = 2.3 atm 291 K temperature ratio increases pressure A gas has a pressure of 645 torr at 128 C. What is the temperature in Celsius if the pressure increases to 1.50 atm? (n and V remain constant) 1. Set up a data table: Conditions 1 Conditions 2 P 1 = 645 torr P 2 = 1.50 atm x 760 torr = 1140 torr 1 atm T 1 = 128 C T 2 = K 273 =? C = 401 K 2. Solve GayLussac s law for T 2 : P 1 = P 2 T 2 = T 1 x P 2 P 1 T 2 = 401 K x 1140 torr = 709 K = 436 C 645 torr pressure ratio increases temperature Chapter 6 Gases Combined Gas Law 6.6 The Combined Gas Law The combined gas law uses Boyle s law, Charles s law, and GayLussac s law (n is constant). P 1 V 1 = P 2 V
10 A gas has a volume of 675 ml at 35 C and atm pressure. What is the volume (ml) of the gas at 95 C and a pressure of 802 mmhg? (n constant) Data Table Conditions 1 Conditions 2 T 1 = 308 K T 2 = 95 C = 178 K V 1 = 675 ml V 2 =??? P 1 = 646 mmhg P 2 = 802 mmhg Solve for V 2 : V 2 = V 1 x P 1 x T 2 P 2 T 1 V 2 = 675 ml x 646 mmhg x 178 K = 314 ml 802 mmhg x 308 K Combined Gas Law Calculation A sample of helium gas has a volume of L, a pressure of atm, and a temperature of 29 C. At what temperature ( C) will the helium have a volume of 90.0 ml and a pressure of 3.20 atm? (n is constant) 1. Set up data table. Conditions 1 Conditions 2 P 1 = atm P 2 = 3.20 atm V 1 = L (180 ml) V 2 = 90.0 ml T 1 = 29 C = 302 K T 2 =?? Combined Gas Law Calculation (continued) 2. Solve for T 2 : P 1 V 1 = P 2 V 2 T 2 = T 1 x P 2 x V 2 P 1 V 1 T 2 = 302 K x 3.20 atm x 90.0 ml = 604 K atm ml T 2 = 604 K = 331 C Chapter 6 Gases 6.7 Volume and Moles (Avogadro s Law) Avogadro's Law: Volume and Moles In Avogadro s law the volume of a gas is directly related to the number of moles (n) of gas. T and P are constant. V 1 = V 2 n 1 n
11 If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L 3) 2.4 L STEP 1: Conditions 1 Conditions 2 V 1 = 1.5 L V 2 =??? n 1 = 0.75 mole of He n 2 = 1.2 moles of He STEP 2: Solve for unknown V 2. V 2 = V 1 x n 2 n 1 STEP 3: Substitute values and solve for V 2. V 2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He STP The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have the same temperature. standard temperature (T) 0 C or 273 K the same pressure. standard pressure (P) 1 atm (760 mmhg) Molar Volume At standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume Molar Volume as a Conversion Factor The molar volume at STP can be used to write conversion factors L and 1 mole 1 mole 22.4 L Using Molar Volume What is the volume occupied by 2.75 moles of N 2 gas at STP? The molar volume is used to convert moles to liters moles N 2 x 22.4 L = 61.6 L 1 mole 65 66
12 Guide to Using Molar Volume 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 g of He are present in 8.00 L of gas at STP? 1) 25.6 g 2) g 3) 1.43 g Gases in Equations A. 1) 5.60 L 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He The volume or amount of a gas at STP in a chemical reaction can be calculated from STP conditions. mole factors from the balanced equation Guide to Using Molar Volume for Reactions STP and Gas Equations What volume (L) of O 2 gas at STP is needed to completely react with 15.0 g of aluminum? 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Plan: g Al mole Al mole O 2 L O 2 (STP) 15.0 g Al x 1 mole Al x 3 moles O 2 x 22.4 L (STP) 27.0 g Al 4 moles Al 1 mole O 2 = 9.33 L of O 2 at STP 71 72
13 What mass of Fe will react with 5.50 L of O 2 at STP? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s)? 5.50 L at STP 5.50 L O 2 x 1 mole x 4 moles Fe x 55.9 g Fe = 18.3 g of Fe 22.4 L 3 moles O 2 1 mole Fe Chapter 6 Gases 6.8 Partial Pressures (Dalton s Law) Partial Pressure The partial pressure of a gas is the pressure of each gas in a mixture. is the pressure that gas would exert if it were by itself in the container Dalton s Law of Partial Pressures Dalton s Law of Partial Pressures Dalton s law of partial pressures indicates that pressure depends on the total number of gas particles, not on the types of particles. the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases. P T = P 1 + P 2 + P
14 Total Pressure Scuba Diving For example, at STP, 1 mole of a pure gas in a volume of 22.4 L will exert the same pressure as 1 mole of a gas mixture in 22.4 L. V = 22.4 L Gas mixtures 1.0 mole N mole O mole He 1.0 mole 0.5 mole O mole He 0.2 mole Ar 1.0 mole 1.0 atm 1.0 atm 1.0 atm When a scuba diver dives, the increased pressure causes N 2 (g)to dissolve in the blood. If a diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends." Helium, which does not dissolve in the blood, is mixed with O 2 to prepare breathing mixtures for deep descents A scuba tank contains O 2 with a pressure of atm and He at 855 mmhg. What is the total pressure in mmhg in the tank? 1. Convert the pressure in atm to mmhg atm x 760 mmhg = 342 mmhg = PO 2 1 atm 2. Calculate the sum of the partial pressures. P total = PO 2 + P He P total = 342 mmhg mmhg = 1197 mmhg For a deep dive, a scuba diver uses a mixture of helium and oxygen with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mmhg, what is the partial pressure of the helium? 1) 520 mmhg 2) 2040 mmhg 3) 4800 mmhg 3) 4800 mmhg P Total = 8.00 atm x 760 mmhg = 6080 mmhg 1 atm P Total = P O + P He P He = P Total  PO 2 P He 2 = 6080 mmhg mmhg = 4800 mmhg 83 84
15 Gases We Breathe The air we breathe is a gas mixture. contains mostly N 2 and O 2, and small amounts of other gases. A. If the atmospheric pressure today is 745 mmhg, what is the partial pressure (mmhg) of O 2 in the air? 1) ) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mmhg) N 2 in the air? 1) 557 2) ) Blood Gases A. If the atmospheric pressure today is 745 mmhg, what is the partial pressure (mmhg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mmhg) N 2 in the air? 1) 557 In the lungs, O 2 enters the blood, while CO 2 from the blood is released. In the tissues, O 2 enters the cells, which releases CO 2 into the blood Blood Gases Gas Exchange During Breathing In the body, O 2 flows into the tissues because the partial pressure of O 2 is higher in blood, and lower in the tissues. CO 2 flows out of the tissues because the partial pressure of CO 2 is higher in the tissues, and lower in the blood. Partial Pressures in Blood and Tissue Oxygenated Deoxygenated Gas Blood Blood Tissues O or less CO or greater 89 90
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