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Chapter 11 Gases 2006, Prentice Hall CHAPTER OUTLINE Properties of Gases Pressure & Its Measurement The Gas Laws Vapor Pressure & Boiling Point Combined Gas Law Avogadro’s Law STP & Molar Volume Ideal Gas Law Partial Pressures 2 PROPERTIES OF GASES Gases occupy are the much least dense greater and space mostthan mobile the of same the three amount phases of liquid of matter. or solid. This Particles is because of matter the gas in the particles gas phase are are spaced spaced apart far apart fromfrom one one another another and and are therefore move rapidly compressible. and collide with each other often. Solid or liquid particles are spaced much closer and cannot be compressed. 3 PROPERTIES OF GASES Gases are characterized by four properties. These are: 1. Pressure (P) 2. Volume (V) 3. Temperature (T) 4. Amount (n) 4 KINETIC-MOLECULAR THEORY The Scientists KMT use consists the kinetic-molecular of several postulates: theory (KMT) to describe the behavior of gases. 1. Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities. 2. Gas particles have little attraction for one another. Therefore attractive forces between gas molecules can be ignored. 3. The distance between the particles is large compared to their size. Therefore the volume occupied by gas molecules is small compared to the volume of the gas. 5 KINETIC-MOLECULAR THEORY 4. Gas particles move in straight lines and collide with each other and the container frequently. The force of collision of the gas particles with the walls of the container causes pressure. 5. The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin). Animation 6 Kinetic Molecular Theory Tro's Introductory Chemistry, Chapter 7 PRESSURE & ITS MEASUREMENT Pressure is the result of collision of gas particles with the sides of the container. Pressure is defined as the force per unit area. Pressure is measured in units of atmosphere (atm) or mmHg or torr. The SI unit of pressure is pascal (Pa) or kilopascal (kPa). 1 atm = 760 mmHg = 101.325 kPa 1 mmHg = 1 torr 8 Common 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) inches of mercury (inHg) torr (torr) 760 (exactly) 29.92 760 (exactly) pounds per square inch (psi, lbs./in2) Tro's Introductory Chemistry, Chapter 14.7 9 PRESSURE & ITS MEASUREMENT Atmospheric pressure can be measured with the use of a barometer. Mercury is used in a barometer due to its high density. At sea level, the mercury stands at 760 mm above its base. 10 Atmospheric Pressure & 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 10,000 ft is 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's Introductory Chemistry, Chapter 11 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's Introductory Chemistry, Chapter 12 Example 1: The atmospheric pressure at Walnut, CA is 740. mmHg. Calculate this pressure in torr and atm. 1 mmHg = 1 torr 740. mmHg = 740. torr 740. mmHg x 1 atm = 0.974 atm 760 mmHg 13 Example 2: The barometer at a location reads 1.12 atm. Calculate the pressure in mmHg and torr. 1.12 atm x 760 mmHg = 851 mmHg atm 1 1 mmHg = 1 torr 851 mmHg = 851 torr 14 PRESSURE & MOLES OF A GAS The pressure of a gas is directly proportional to the number of particles (moles) present. The greater the moles of the gas, the greater the pressure 15 BOYLE’S LAW At Theconstant relationship temperature, of pressure the and volume volume of a in fixed amount gases is called of gas Boyle’s is inversely Law.proportional to its pressure. As pressure Pressure and increases, volume of athe volume gas areof the gas decreases inversely proportional 16 BOYLE’S LAW Boyle’s Law can be mathematically expressed as P1 x V1 = P2 x V2 P and V under initial condition P and V under final condition 17 Example 1: A sample of gas has a volume of 12-L and a pressure of 4500 mmHg. What is the volume of the gas when the Pressure pressure is reduced to 750 mmHg? Volume decreases P1= 4500 mmHg P2= 750 mmHg V1 = 12 L V2 = ??? increases P1 x V1 = P2 x V2 P1V1 (4500 mmHg)(12 L) V2 = = P2 (750 mmHg) V2 = 72 L 18 Example 2: A sample of hydrogen gas occupies 4.0 L at 650 mmHg. What volume would it occupy at 2.0 atm? Pressure P1= 650 mmHg P2= 2.0 atm V1 = 4.0 L V2 = ??? Volume Pincreases = 2.0 atm = 1520 mmHg 2 decreases P1 x V1 = P2 x V2 P1V1 (650 mmHg)(4.0 L) V2 = = P2 (1520 mmHg) Pressures must be in the sameV2 = 1.7 L unit 19 CHARLES’S LAW At Theconstant relationship pressure, of temperature the volume and of a volume fixed in amount gases is called of gas Charles’ is directlyLaw. proportional to its absolute temperature. Temperature and As temperature volume of a gas increases, the are directly volume of the proportional gas increases 20 CHARLES’S LAW Charles’ Law can be mathematically expressed as V1 V2 = T1 T2 V and T under T must be in initial units of K condition V and T under final condition 21 Example 1: A 2.0-L sample of a gas is cooled from 298K to 278 K, at constant pressure. What is the new volume of the gas? Temperature Volume decreasesV1 V2 decreases = V1 = 2.0 L V2 = ??? T1 = 298 K T2 = 278 K T1 T2 T2 V2 = V1 x = 2.0 L x 278 K T1 298 K V2 = 1.9 L 22 Example 2: If 20.0 L of oxygen is cooled from 100ºC to 0ºC, what is the new volume? Temperature s must be in + 273 = 373 Volume Temperature T = 100ºC K V1 = 20.0 L 1 K decreases decreases V2 = ??? T1 = 100ºC T2 = 0ºC T2= 0ºC + 273 = 273 K T2 273 K V2 = V1 x = 20.0 L x T1 373 K V2 = 14.6 L 23 GAY-LUSSAC’S LAW The At constant relationship volume, of temperature the pressureand of apressure fixed in amount gases is ofcalled gas is Gay-Lussac’s directly proportional Law. to its absolute temperature. As temperature Temperature and increases, the pressure of a gas pressure of the are directly gas increases proportional 24 GAY-LUSSAC’S LAW Gay-Lussac’s Law can be mathematically expressed as P1 P2 = T1 T2 P and T under initial T must be in condition units of K P and T under final condition 25 Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire Temperatures and reaches temperature of 400C ? must be in K P1 = 4.0 atm P2 = ??? T1= 25ºC + 273 = 298 K T2= 400ºC + 273 = 673 K T1 = 25ºC T2 = 400ºC 26 Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire Temperature and reaches temperature of 400C ? Pressure increases increases P1 P2 = P1 = 4.0 atm T1 T2 P2 = ??? T1 = 298 K T2 673 K P2 = P1 x = 4.0 L x T1 298 K T2 = 673 K P2 = 9.0 atm 27 Example 2: A cylinder of gas with a volume of 15.0-L and a pressure of 965 mmHg is stored at a temperature of 55C. To what temperature must the cylinder be cooled to reach a Pressure Temperature pressure of 850 decreases mmHg? decreases P1 = 965 mmHg T1= 55ºC + 273 = 328 K P2 = 850 mmHg 850 mmHg P2 T2 = T1 x = 328 K x 965 mmHg P1 T1 = 55ºC T2 = ??? TT22==289 16ºC K 28 VAPOR PRESSURE & BOILING POINT In an open container, liquid molecules at the surface that possess sufficient energy, can break away from the surface and become gas particles or vapor. In a closed container, these gas particles can accumulate and create pressure called vapor pressure. Vapor pressure is defined as the pressure above a liquid at a given temperature. Vapor pressure varies with each liquid and increases with temperature. 29 VAPOR PRESSURE & BOILING POINT Listed below is the vapor pressure of water at various temperatures. 30 VAPOR PRESSURE & BOILING POINT A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure). For example, at sea level, water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature. 31 VAPOR PRESSURE & BOILING POINT At higher altitudes, where atmospheric pressure is lower, water reaches boiling point at temperatures lower than 100C. For example, in Denver, where atmospheric pressure is 630 mmHg, water boils at 95C, since its vapor pressure is 630 mmHg at this temperature. 32 COMBINED GAS LAW This All pressure-volume-temperature law is useful for studying the effect relationships of changes can be combined in two variables. into a single relationship called the combined gas law. P1V1 P2 V2 = T1 T2 Initial condition Final condition 33 COMBINED GAS LAW The individual gas laws studied previously are embodied in the combined gas law. P1 V1 P2 V2 = T1 T2 Charles’s Law Boyle’s Gay-Lussac’s Law Law 34 Example 1: A 25.0-mL sample of gas has a pressure of 4.00 atm at a temperature of 10C. What is the volume of the gas at a pressure of 1.00 atm and a temperature of 18C? P1= 4.00 atm V1= 25.0 mL T1= 283 K P2= 1.00 atm V2= ??? T2 = 291 K P1V1 P2 VBoth pressureP1 T2 2 V2 = V1 x x = T1 T2 and temp. P2 T1 change 4.00 atm 291 K V2 = 25.0 mL x x 1.00 atm 283 K Must use V2 =combined 103 mL gas law 35 AVOGADRO’S LAW At Theconstant relationship temperature of molesand andpressure, volume inthe gases volume is calledofAvogadro’s a fixed amount Law.of gas is directly proportional to the number of moles. As number of moles increases, Number of moles thevolume volumeof ofa and the directly gas gas are increases proportional 36 Avogadro’s Law Tro's Introductory Chemistry, Chapter 37 AVOGADRO’S LAW As a result of Avogadro’s Law, equal volumes of different gases at the same temp. and pressure contain equal number of moles (molecules). 2 Tanks of gas of equal volume at the same T & P contain the same number of molecules 38 AVOGADRO’S LAW Avogadro’s Law also allows chemists to relate volumes and moles of a gas in a chemical reaction. This relationship is For example: only valid for gaseous substances (g) 2H O 2 H2 (g) + 1 O2 2 (g) 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 2 Liters 1 Liter 2 Liters 39 Example 1: A sample of helium gas with a mass of 18.0 g occupies 1.6 Liters at a particular temperature and pressure. Same Temp. L at the same What mass of oxygen would occupy1.6 temperature and pressure?& Pressure 1 mol mol of He = 18.0 g x = 4.50 mol 4.00 g mol of O2 = mol of He = 4.50 mol 32.0 g mass of O2 = 4.50 mol x = 144 g 1 mol 40 Example 2: How many Liters of NH3 can be produced from reaction of 1.8 L of H2 with excess N2, as shown below? N2 (g) + 3 H2 (g) 2 NH3 (g) 1.8 L H2 x 2 L NH 3 = 1.2 L NH3 3 L H2 41 STP & MOLAR VOLUME To better understand the factors that affect gas behavior, a set of standard conditions have been chosen for use, and are referred to as Standard Temperature and Pressure (STP). STP = 0ºC (273 K) 1 atm (760 mmHg) 42 STP & MOLAR VOLUME At STP conditions, one mole of any gas is observed to occupy a volume of 22.4 L. V = 22.4 L Molar Volume at STP 43 Example 1: If 2.00 L of a gas at STP has a mass of 3.23 g, what is the molar mass of the gas? 1 mol mol of gas = 2.00 L x 22.4 L = 0.0893 mol g 3.23 g = Molar Molar mass = = 36.2 g/mol mol 0.0893 mol volume at STP 44 Example 2: A sample of gas has a volume of 2.50 L at 730 mmHg and 20C. What is the volume of this gas at STP? P1= 730 mmHg V1= 2.50 L T1= 293 K P2= 760 mmHg V2= ??? T2 = 273 K P1 T2 V2 = V1 x x P2 T1 730 mmHg 273 K V2 = 2.50 L x x 760 mmHg 293 K V2 = 2.24 L 45 IDEAL GAS LAW Combining all the laws that describe the behavior of gases, one can obtain a useful relationship that relates the volume of a gas to the temperature, pressure and number of moles. n R T V= P Universal gas constant R = 0.0821 L atm/mol K 46 IDEAL GAS LAW This relationship is called the Ideal Gas Law, and commonly written as: P V = n R T Temp. in K Pressure in atm Volume in Liters Number of moles 47 Example 1: A sample of H2 gas has a volume of 8.56 L at a temp. of 0C and pressure of 1.5 atm. Calculate the moles of gas present. PV = nRT P = 1.5 atm V = 8.56 L T = 273 K n = ??? R = 0.0821 (1.5 atm)(8.56 L) PV = n= RT (0.0821 L atm )(273 K) mol K = 0.57 mol 48 Example 2: What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? Moles must 1 mol be calculated mol of N2 = 40.0 g x = 1.43 mol P =750 mmHg from mass 28.0 g V = ??? T = 283 K n = not given m = 40.0 g R = 0.0821 1 atm x P = 750 mmHgPressure 760bemmHg must converted to P = 0.987 atm atm 49 Example 2: What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? nRT V= P P =0.987 atm V = ??? T = 283 K n = 1.43 R = 0.0821 L atm (1.43 mol)(0.0821 )(283 K) mol K V= 0.987 atm V = 33.7 L 50 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 Tro's Introductory Chemistry, Chapter 51 Ideal vs. Real Gases Tro's Introductory Chemistry, Chapter 52 PARTIAL PRESSURES Many gas samples are mixture of gases. For example, the air we breathe is a mixture of mostly oxygen and nitrogen gases. Since gas particles have no attractions towards one another, each gas in a mixture behaves as if it is present by itself, and is not affected by the other gases present in the mixture. In a mixture, each gas exerts a pressure as if it was the only gas present in the container. This pressure is called partial pressure of the gas. 53 DALTON’S LAW In a mixture, the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture. Ptotal = P1 + P2 + P3 + ··· Total pressure of = Sum of the partial pressures a gas mixture of the gases in the mixture This is called Dalton’s law of partial pressures. 54 PARTIAL PRESSURES + PHe = 2.0 atm PAr = 4.0 atm PTotal = ??? PHe + PAr = 6.0 atm Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture 55 PARTIAL PRESSURES The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture. For example, in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas, the partial pressure of each gas is one-half of the total pressure in the container. 56 PARTIAL PRESSURES Twice as many moles of Ar compared to He + PHe = 2.0 atm PPAr = 2 PHe Ar = 4.0 atm PTotal = 6.0 atm The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture 57 Example 1: A scuba tank contains a mixture of oxygen and helium gases with total pressure of 7.00 atm. If the partial pressure of oxygen in the tank is 1140 mmHg, what is the partial pressure of helium in the tank? Ptotal Poxygen = Poxygen + Phelium = 1140 mmHg x 1 atm = 1.50 atm 760 mmHg PTotalatm - P–oxygen 1.50 atm = 5.50 atm Phelium ==7.00 58 Example 2: A mixture of gases contains 2.0 mol of O2 gas and 4.0 mol of N2 gas with total pressure of 3.0 atm. What Twice as many is the partial pressure of each gas in the mixture? moles of N2 compared to O2 Ptotal = Poxygen + Pnitrogen Pnitrogen = 2 Poxygen Pnitrogen = 2/3 (Ptotal ) = 2.0 atm Poxygen = 1/3 (Ptotal ) = 1.0 atm 59 THE END 60