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Properties of Gases

Properties of Gases
Chapter 11 GasesProperties of Gases • Expand to completely fill their container = Compressible • Take the shape of their container. • Low density. Much less than solid or liquid state. • Mixtures of gases are always homogeneous • http://www.youtube.com/watchv=Ehci9vrvfw www.ThesisScientist.comThe Structure of a Gas • Gases are composed of particles that are flying around very fast in their container(s). • They move in straight lines until they encounter either the container wall or another particle, then they bounce off. • If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there. • It’s why balloons look round www.ThesisScientist.comKinetic Molecular Theory • The particles of the gas (either atoms or molecules) are constantly moving The attraction between particles is negligible. • Therefore: 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. www.ThesisScientist.comKinetic Molecular Theory of Gases • 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 www.ThesisScientist.comKinetic Molecular Theory www.ThesisScientist.comWhat does all this Pushing do • 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= pounds per square inch www.ThesisScientist.comThe Effect of Gas Pressure • The pressure exerted by a gas can cause some amazing and startling effects. www.ThesisScientist.comWhich Way Would Air Flow Two filled balloons are connected with a long pipe. One of the balloons is plunged down into the water. Which way will the air flow Will air flow from the lower balloon toward the top balloon; or will it flow from the top balloon to the bottom one Tro' ww s In w tr .T odu hes citor sSc yi e C nh tiesm t.c is otm ry, Chapter 9 11Is This Possible at a Depth of 20 m = 65.6 ft Tro' ww s In w tr .T odu hes citor sSc yi e C nh tiesm t.c is otm ry, Chapter 10 11Soda Straws and Gas Pressure The pressure of The pressure of the the air inside the air inside the straw straw is the same is lower than the as the pressure pressure of the air outside of the air outside the straw—so the straw—so liquid levels are liquid is pushed the same on both up the straw by sides. the outside air. www.ThesisScientist.comAir Pressure • The atmosphere exerts a pressure  At sea level it is 14.7 psi.  The atmosphere goes up about 370 miles, but 80 is in the first 10 miles from Earth’s surface. • This is the same pressure that a column of water would exert if it were about 10.3 m high. • Therefore water can not be pumped higher, in a straw for instance, than 10.3 m/33.8 ft www.ThesisScientist.comMeasuring Air Pressure • Use a barometer. • Column of mercury supported by air gravity pressure. • Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury. www.ThesisScientist.comAtmospheric 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 14.7 psi, but at 29,028 ft it is only 4.9 psi (1/3) • Rapid changes in atmospheric pressure may cause your ears to ―pop‖ due to an imbalance in pressure on either side of your ear drum. www.ThesisScientist.comPressure 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.‖ www.ThesisScientist.comCommon Units of Pressure Unit Average air pressure at sea level Pascal (Pa) 101,325 Kilopascal (kPa) 101.325 Atmosphere (atm) 1 (exactly) Millimeters of mercury (mmHg) 760 (exactly) Inches of mercury (inHg) 29.92 Torr (torr) 760 (exactly) 2 Pounds per square inch (psi, lbs./in ) 14.7 www.ThesisScientist.comExample 11.1—A HighPerformance Bicycle Tire Has a Pressure of 125 psi. What Is the Pressure in mmHg Given: 125 psi Find: mmHg Solution Map: psi atm mmHg 1 atm 760 mmHg 14.7 psi 1 atm Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: 1 atm 760 mmHg 3 125 psi 6.4610 mmHg 14.7 psi 1 atm Check: Since mmHg are smaller than psi, the answer makes sense. www.ThesisScientist.comBoyle’s Law • Pressure of a gas is inversely proportional to its volume. Constant T and amount of gas. Graph P vs. V is curved. Graph P vs. 1/V is in a straight line. • As P increases, V decreases by the same factor. • P x V = constant. • P x V = P x V . 1 1 2 2 www.ThesisScientist.comWhen 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). www.ThesisScientist.comGas 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. www.ThesisScientist.comBoyle’s Law and Diving • Since water is more dense Scuba tanks have a than air, for each 10 m you regulator so that the dive below the surface, the pressure on your lungs air from the tank is increases 1 atm. delivered at the same  At 20 m the total pressure is pressure as the water 3 atm. surrounding you. • If your tank contained air at 1 atm of pressure, you This allows you to would not be able to inhale take in air even when it into your lungs. the outside pressure is  You can only generate large. enough force to overcome about 1.06 atm. www.ThesisScientist.comBoyle’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 Rising www.ThesisScientist.comExample 11.2—A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm Given: V =6.0 L, P = 4.0 atm, P = 1.0 atm 1 1 2 Find: V , L 2 Solution Map: V , P , P V 1 1 2 2 PV 1 1 V 2 P 2 Relationships: P ∙ V = P ∙ V 1 1 2 2 Solution: PV 1 1 V 2 P 2 4.0 atm6.0 L  24 L 1.0 atm Check: Since P and V are inversely proportional, when the pressure decreases 4x, the volume should increase 4x, and it does. www.ThesisScientist.comPractice—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 (1 atm = 760 torr) www.ThesisScientist.comPractice—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, Continued Given: V =2780 mL, P = 762 torr, P = 0.500 atm 2 1 2 Find: V , mL 1 Solution Map: V , P , P V 2 1 2 1 PV 2 2 V 1 P 1 Relationships: P ∙ V = P ∙ V , 1 atm = 760 torr (exactly) 1 1 2 2 Solution: PV 2 2 V 1 P 1 atm 1 782 torr1.03 atm 0.500 atm2780 mL 760 torr 1350 mL 1.03 atm Check: Since P and V are inversely proportional, when the pressure decreases 2x, the volume should increase 2x, and it does. www.ThesisScientist.comGas 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 = °C + 273 °C = K 273 °F = 1.8 °C + 32 °C = 0.556(°F32) www.ThesisScientist.comStandard Conditions • Common reference points for comparing. • Standard pressure = 1.00 atm. • Standard temperature = 0 °C. 273 K. • STP. www.ThesisScientist.comVolume 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. As volume increases, pressure decreases. www.ThesisScientist.comVolume and Temperature, Continued 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. www.ThesisScientist.comCharles’s Law • Volume is directly proportional to temperature. V V Constant P and amount of gas. 1 2  Graph of V vs. T is a straight line. T T • As T increases, V also increases. 1 2 • Kelvin T = Celsius T + 273. • V = constant x T. If T is measured in kelvin. www.ThesisScientist.comWe’re losing altitude. Quick, Professor, give your lecture on Charles’s law www.ThesisScientist.comAbsolute Zero • Theoretical temperature at which a gas would have zero volume and no pressure. Kelvin calculated by extrapolation. • 0 K = 273.15 °C = 459 °F • Never attainable. Though we’ve gotten real close • All gas law problems use the Kelvin temperature scale. www.ThesisScientist.comDetermining Absolute Zero 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 0. www.ThesisScientist.comExample 11.3—A Gas Has a Volume of 2.57 L at 0 °C. What Was the Temperature at 2.80 L Given: V =2.80 L, V = 2.57 L, t = 0°C 1 2 2 Find: t , K and °C 1 Solution Map: V , V , T T 1 2 2 1 V 1 T T 1 2 V V V 1 2 2  Relationships: T(K) = t(°C) + 273, T T 1 2 Solution: TV 2 1 t T 273 T 1 1 1 V 2 T 0 273 2 t 297.6 273 1 273 K2.80 L T 273 K  297.6 K 2 t 24 C 2.57 L 1 Check: Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does. www.ThesisScientist.comThe 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. PVPV 1 1 2 2   T T 1 2 www.ThesisScientist.comExample 11.4—A Sample of Gas Has an Initial Volume of 158 mL at a Pressure of 735 mmHg and a Temperature of 34 °C. If the Gas Is Compressed to 108 mL and Heated to 85 °C, What Is the Final Pressure Given: V = 158 mL, t = 34 °C , P = 735 mmHg 1 1 1 V = 108 mL, t = 85 °C 2 2 Find: P , mmHg 2 Solution Map: P ,V , V , T , T P 1 1 2 1 2 2 PVT 1 1 2 P 2 PV PV VT 1 1 2 2 2 1  Relationships: T(K) = t(°C) + 273, T T 1 2 Solution: PVT 1 1 2 P 2 T 34 273 1 VT 2 1 T 307 K 1 735 mmHg158 mL358 K P 2 T 85 273 108 mL307 K 2 3 T 358 K P1.2510 mmHg 2 2 Check: Since T increases and V decreases we expect the pressure should increase, and it does. www.ThesisScientist.comPractice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions www.ThesisScientist.comAvogadro’s Law • Volume is directly proportional to the number of gas molecules. V = constant xn. Constant P and T. V V 1 2 More gas molecules = larger  volume. n n 1 2 • Count number of gas molecules by moles, n. • Equal volumes of gases contain equal numbers of molecules. The gas doesn’t matter. www.ThesisScientist.comAvogadro’s Law, Continued www.ThesisScientist.comExample 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L Given: V =4.8 L, V = 6.4 L, n = 0.22 mol 1 2 1 Find: n , and added moles 2 Solution Map: V , V , n n 1 2 1 2 V 2 n n 1 2 V V V 1 1 2  Relationships: mol added = n – n , n n 2 1 1 2 Solution: nV moles added 0.29 0.22 1 2 n 2 V moles added 0.07 mol 1  0.22 mol 6.4 L  0.29 mol 4.8 L Check: Since n and V are directly proportional, when the volume increases, the moles should increase, and it does. www.ThesisScientist.comPractice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy www.ThesisScientist.comPractice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy, Continued Given: V =22.4 L, n = 1.00 mol, n = 0.750 mol 1 1 2 Find: V 2 Solution Map: V , n , n V 1 1 2 2 n 2 VV 1 2 n V V 1 1 2  Relationships: n n 1 2 Solution: V n 1 2 V 2 n 1 22.4 L0.750 mol 16.8 L 1.00 mol Check: Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does. www.ThesisScientist.comIdeal 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. atm L  We will use 0.0821 and convert P to atm and V to L. mol K • 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. PV  R or PV nRT nT www.ThesisScientist.comExample 11.7—How Many Moles of Gas Are in a Basketball with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C Given: V = 3.2 L, P = 24.2 psi, t = 25 °C, Find: n, mol Solution Map: P, V, T, R n PV n  RT Relationships: atmL 1 atm = 14.7 psi PV nRT, R  0.0821 molK T(K) = t(°C) + 273 Solution: PV n 1 atm 24.2 psi1.6462 atm RT 14.7 psi 1.6462 atm3.2 L T (K)  25C 273  0.22 mol atmL  0.0821 298 K molK T 298 K Check: 1 mole at STP occupies 22.4 L at STP; since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas. www.ThesisScientist.comMolar 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 www.ThesisScientist.comExample 11.8—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 m= =0.31 0.31 1g 1g , , VV =0.225 L = 0.225 L , P, =P 886 mm = 1.1658 a Hg, tm t=55 , T°C = , 328 K Find: Mola Molar r Ma Mass, ss, g g/mol /mol Solution Map: P, V, T, R n n, m MM PV m n  MM  RT n atmL 1 atm = 760 mmHg, PV nRT, R  0.0821 m molK Relationships: MM  T(K) = t(°C) + 273 n Solution: m 0.311 g PV MM 1 atm 3 n n 9.745410 mol 886 mmHg1.1658 atm RT 760 mmHg  31.9 g/mol T (K)  55C 273  1.1658 atm 0.225 L 3  9.745410 mol atmL T 328 K 0.0821328 K molK Check: The value 31.9 g/mol is reasonable. www.ThesisScientist.comPractice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C Tro's Iww ntrodu w.T ch tor esy is S Cch ie em ntiis stt. rc yo , m Chapter 121 11Mixtures 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.  All gases in the mixture have the same volume and temperature. 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, N 78 Argon, Ar 0.9 2 Oxygen, O 21 Carbon dioxide, CO 0.03 2 2 www.ThesisScientist.comPartial 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. Dalton’s law of partial pressures. P = P + P + P +... total gas A gas B gas C P  P P  P  0.78 atm  0.21 atm  0.01 atm  1.00 atm air N O Ar 2 2 www.ThesisScientist.comExample 11.9—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: P = 341 mmHg, P = 112 mmHg, P = 558 mmHg He Ne tot Find: P , mmHg Ar Solution Map: P , P , P P tot He Ne Ar P = P – (P + P ) Ar tot He Ne P = P + P + etc. Relationships: tot a b Solution: P 558341112 mmHg Ar 105 mmHg Check: The units are correct, the value is reasonable. www.ThesisScientist.comMountain Climbing and Partial Pressure • Our bodies are adapted to breathe O 2 at a partial pressure of 0.21 atm.  Sherpa, people native to the Himalaya mountains, have adapted to the much lower partial pressure of oxygen in their air. • Partial pressures of O —lower than 2 0.1 atm—leads to hypoxia.  Unconsciousness or death. • Climbers of Mt. Everest must carry O in cylinders to prevent hypoxia. 2  On top of Mt. Everest: P = 0.311 atm, so P = 0.065 atm. air O2 www.ThesisScientist.comDeep Sea Divers and Partial Pressure • It is also possible to have too much O , a condition called 2 oxygen toxicity.  P 1.4 atm. O2  Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions. • It is also possible to have too much N , a condition called 2 nitrogen narcosis.  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.  At a depth of 55 m, the partial pressure of O is 1.4 atm. 2  Divers that go below 50 m use a mixture of He and O called 2 www.ThesisScientist.com heliox that contains a lower percentage of O than air. 2Partial Pressure vs. Total Pressure At a depth of 30 m, the total pressure of air in the divers lungs, and the partial pressure of all the gases in the air, are quadrupled www.ThesisScientist.comReactions 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 coefficients in the equation as a mole ratio. P, V, T of Gas A mole A mole B P, V, T of Gas B www.ThesisScientist.comExample 11.11—How Many Liters of O Are Made from 294 g of 2 KClO at 755 mmHg and 305 K 3 2 KClO (s) → 2 KCl(s) + 3 O (g) 3 2 Given: m n = 3.= 60 2 9 m 4o g l, , P P=7 = 0 5.5 9 9 m 3m 42 H atm g, T,=3 T 0 = 53 K 05 K O2 KClO3 Find: V V ,, L L O2 O2 g KClO mol KClO mol O P, n, T, R V 3 3 2 Solution Map: n RT 1 mol KClO 3 mol O 3 2 V P 122.5 g 2 mol KClO 3 1 atm = 760 mmHg, KClO = 122.5 g/mol Relationships: 3 atmL 2 mol KClO : 3 mol O PV nRT, R  0.0821 3 2 molK Solution: n RT 1 mol KClO 3 mol O 3 2 V 294 g KClO 3 122.5 g 2 mol KClO P 3 atmL  3.60 mol O 3.60 mol0.0821305 K 2 molK  1 atm 0.99342 atm 755 mmHg 0.99342 atm 760 mmHg  90.7 L www.ThesisScientist.comPractice—What Volume of O at 0.750 atm and 313 K is 2 Generated by the Thermolysis of 10.0 g of HgO 2 HgO(s)  2 Hg(l) + O (g) 2 (MM = 216.59 g/mol) HgO Tro's Iww ntrodu w.T ch tor esy is S Cch ie em ntiis stt. rc yo , m Chapter 149 11Practice—What Volume of O at 0.750 atm and 313 K is Generated by 2 the Thermolysis of 10.0 g of HgO 2 HgO(s)  2 Hg(l) + O (g), Continued 2 Given: n m = = 0.1 00 2.3 00 g8 , 5 P m =0 o.l, 75 P 0= atm 0.7 , 5 T0 =3 atm 13, K T = 313 K O2 HgO Find: V V ,, L L O2 O2 g HgO mol HgO mol O P, n, T, R V 2 Solution Map: 1 mol HgO 1 mol O n RT 2 V 216.59 g 2 mol HgO P Relationships: 1 atm = 760 mmHg, HgO = 216.59 g/mol atmL 2 mol HgO : 1 mol O PV nRT, R  0.0821 2 molK Solution: n RT V 1 mol HgO 1 mol O 2 P 10.0 g HgO 216.59 g 2 mol HgO atmL 0.023085 mol0.08206313K molK   0.023085 mol O 2 0.750 atm  0.791 L www.ThesisScientist.comCalculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP. P x V = n x R x T L∙atm (1.00 atm) x V = (1.00 moles)(0.0821 )(273 K) mol∙K V = 22.4 L • 1 mole of any gas at STP will occupy 22.4 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  22.4 L www.ThesisScientist.comMolar Volume There is so much empty space between molecules in the gas state that the volume of the gas is not effected by the size of the molecules (under ideal conditions). www.ThesisScientist.comExample 11.12—How Many Grams of H O Form When 1.24 L H 2 2 Reacts Completely with O at STP 2 O (g) + 2 H (g) → 2 H O(g) 2 2 2 Given: V = 1.24 L, P = 1.00 atm, T = 273 K H2 Find: mass , g H2O L H mol H mol H O g H O 2 2 2 2 Solution Map: 18.02 g 1 mol H 2 mol H 2 2 1 mol H O 2 mol H O 22.4 L 2 2 Relationships: H O = 18.02 g/mol, 1 mol = 22.4 L STP 2 2 mol H O : 2 mol H 2 2 Solution: 1 mol H 2 mol H O 18.02 g H O 2 2 2 1.24 L H 2 22.4 L H 2 mol H 1 mol H O 2 2 2  0.998 g H O 2 Tro's Iww ntrodu w.T ch tor esy is S Cch ie em ntiis stt. rc yo , m Chapter 153 11Practice—What Volume of O at STP is Generated by the 2 Thermolysis of 10.0 g of HgO 2 HgO(s)  2 Hg(l) + O (g) 2 (MM = 216.59 g/mol) HgO Tro's Iww ntrodu w.T ch tor esy is S Cch ie em ntiis stt. rc yo , m Chapter 161 11Practice—What Volume of O at STP is Generated by the 2 Thermolysis of 10.0 g of HgO 2 HgO(s)  2 Hg(l) + O (g), Continued 2 Given: m = 10.0 g, P = 1.00 atm, T = 273 K HgO Find: V , L O2 g HgO mol HgO mol O L O 2 2 Solution Map: 1 mol O 1 mol HgO 22.4 L 2 2 mol HgO 216.59 g 1 mol O 2 Relationships: HgO = 216.59 g/mol, 1 mol = 22.4 L at STP 2 mol HgO : 1 mol O 2 Solution: 1 mol HgO 1 mol O 22.4 L O 2 2 10.0 g HgO 216.59 g 2 mol HgO 1 mol O 2  0.517 L O 2 Tro's Iww ntrodu w.T ch tor esy is S Cch ie em ntiis stt. rc yo , m Chapter 162 11
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