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Unit 1 Gases The Nature of Gases Objectives: 1. Describe the assumption of the kinetic theory as it applies to gases. 2. Interpret gas pressure in terms of kinetic theory 3. Define the relationship between Kelvin temperature and average kinetic energy. 4. Explain why gases are easier to compress than solids or liquids. 5. Describe the 3 factors that affect gas pressure. The Kinetic Theory Properties of Gases Expand to fill their container take the shape of their container low density Compressible • Compressibility measures how much the volume of matter decreases under pressure. mixtures of gases are always homogeneous Fluids (flow) Gas pressure Results from collisions of gas particles with an object. In empty space where there are no particles, there is no pressure and is called a vacuum. Atmospheric pressure (air pressure): due to atoms and molecules in air. Barometer: used to measure atmospheric pressure. Units for measuring pressure: Pascal (Pa) Standard atmosphere (atm) Millimeters of mercury (mmHg) 1 atm = 760 mmHg = 101.3 kPa 1kpa = 1000 pa Standard pressure: 1 atm Factors affecting gas pressure Amount of gas Volume Temperature Standard temperature : 0C (273K) Converting between units of pressure A pressure gauge records a pressure of 450 kPa. What is the measurement expressed in atmospheres and millimeters of mercury? For converting to atm: 450 kpa x 1 atm = 4.4 atm 1013.kPa For converting to mmHg: 1. 450kPa x 760 mmHg = 3.4 x 103 mmHg 101.3 kPa 2. What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg? 3. The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm? 2. What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg? 51.3 kPa, 0.507 atm 3. The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm? 33.7 kPa is greater than 0.25 atm Kinetic energy and temperature The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the particles of the substance. Directly proportional means that temperatures increases as the average kinetic energy increases or decreases as the average kinetic energy decreases. K = C + 273 The Atmosphere an “ocean” of gases mixed together Composition nitrogen (N2)………….. ~78% oxygen (O2)…………… ~21% argon (Ar)……………... ~1% Trace amounts of: carbon dioxide (CO2)… water vapor (H2O)……. ~0.04% ~0.1% He, Ne, Rn, SO2, CH4, NxOx, etc. Depletion of the Ozone Layer O3 depletion is caused by chlorofluorocarbons (CFCs). Ozone (O3) in upper atmosphere blocks ultraviolet (UV) light from Sun. UV causes skin cancer and cataracts. Uses for CFCs: refrigerants CFCs aerosol propellants -- banned in U.S. in 1996 O3 is replenished with each strike of lightning. Reaction_to_Air_Pressure_Below_Sea_Level.asf Classwork: Read pages 103-105 Do problems 1,4,5,6 Gas Laws Objectives Describe the relationships among the temperature, pressure, and volume of a gas 2. Use the gas laws to solve problems 1. Boyle’s Law : Pressure and Volume States that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure. If pressure increases, volume decreases; if pressure decreases, volume increases. P1 x V1 = P2 x V2 P: pressure V: volume 1: initial condition 2: final condition Using Boyle’s Law 1. A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)? Using Boyle’s Law A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)? P1=103 kPa V1= 30.0L P2= 25.0kPa V2=? 1. P1V1= P2V2 V2= P1V1 = (103 kPa)(30.0L) = 124 L P2 (25.0 kPa) Since pressure decreases, you expect volume to increase. 2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)? 2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)? P1= 205 kPa V1= 4.00L P2=? V2=12.0L P1V1=P2V2 P2= P1V1 = (205 kPa)( 4.00 L) = 68.3 kPa V2 (12.0L) Classwork: p 121 # 2-6 Charles’s Law: Temperature and Volume States that the temperature of an enclosed gas varies directly with the volume at constant pressure. As temperature increases, volume increases. V1 = V2 T1 T2 V1: initial volume V2: final volume T1: initial temperature T2: final temperature Temperature has to be in Kelvin scale. Volume and Temperature As a gas is heated, it expands. This causes the density of the gas to decrease. 23 Using Charles’s Law 1. A balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ? Using Charles’s Law A balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ? V1 = V2 V1= 4.00L V2= ? T1 T2 T1= 24C +273= 297 K T2= 58C + 273 = 331 K V2= V1T2 = (4.00L)(331K) = 4.46 L T1 (297K) Since temperature increases, you expect the volume to increase. Classwork: p124 # 11, 12 (a-c), 13 1. Combined Gas Law Describes the relationship among the pressure, temperature and volume, when the amount of gas is constant. P1V1 = P2V2 T1 T2 Standard temperature and pressure (STP): 0C, 1 atm Useful conversions: 1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L Using the combined gas law: 1. The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)? Using the combined gas law: 1. The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)? P1= 153 kPa V1= 30.0 L T1= 313K P2= 1 atm=101.3kPa V2= ? T2= 0C= 273K P1V1 = P2V2 T1 T2 V2= P1V1T2 = (153 kPa x 30.0L x 273K)= 39.5 L P2T1 (101.3kPa x 313K) Classwork: p 126 #14(a-c), 15 (a-c) , 16, 17, 19 Ideal Gases Objectives Compute the value of an unknown using the ideal gas law. 2. Compare and contrast real and ideal gases. 1. Avogadro’s Principle and Molar Volume Avogadro’s principle states that equal volume of all gases, measured under the same conditions of pressure and temperature, contain the same number of particles. At STP, the volume of one mole of gas is 22.4 L(this is called the molar volume) From last year, to convert between grams and moles of a substance we used its molar mass. Avogadro’s Law 31 Converting between moles and grams: How many moles are 98.32g CO2? Calculate molar mass CO2 (use periodic table) Molar mass= 12 + (2 x 16) =44.0 g/1 mol To convert grams to moles: 98.32g x 1 mol = 2.23 mol CO2 44g Using Avogadro’s Principle How many moles of carbon monoxide, CO, will occupy a volume of 250.0 mL at STP? V= 250.0mL =0.250 L At STP, molar volume is : 22.4L/1 mol 1. 0.25 L x 1 mol = 0.0112 mol CO 22.4L Conversions: 1L =1dm3 ; 1 m3 = 1000 L ; 1 cm3 =1mL 2. How many grams of carbon dioxide, CO2, will occupy a volume of 500.0 mL at STP? V= 500.0mL =0.5 L Molar mass CO2= 44g/mol At STP, molar volume is : 22.4L/1 mol 0.5 L x 1 mol x 44 g = 0.982g CO2 22.4L 1 mol Classwork: p 132 # 1, 2, 3 The ideal gas law Considers that amount of gas varies. New variable n: number of moles of gas (mol) Ideal gas constant (R) R= 8.314 L kPa (when pressure is measured in kPa) mol K R= 0.0821 L atm (when pressure is measured in atm) mol K PV= nRT (T must be in Kelvin) Using the ideal gas law 1. A deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many moles of CH4 does the cavern contain? Using the ideal gas law 1. A deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many moles of CH4 does the cavern contain? P= 1500kPa V= 2.24x106 L R= 8.314 L kPa n= ? T= 315 K mol K PV=nRT n= PV = (1500 kPa)x (2.24x106 L )x (mol K) RT (8.314 L KPa)x(315K) =1.28x106 moles Using the ideal gas law 2. How many moles of oxygen will occupy a volume of 2.5 L at 1.2 atm and 25C? (0.12 moles) Using the ideal gas law 2. How many moles of oxygen will occupy a volume of 2.5 L at 1.2 atm and 25C? (0.12 moles) P= 1.2 atm V=2.5 L T=25C= 298K R= 0.0821 L atm mol K Using the ideal gas law 2. How many moles of oxygen will occupy a volume of 2.5 L at 1.2 atm and 25C? (0.123 moles) P= 1.2 atm V=2.5 L T=25C= 298K R= 0.0821 L atm mol K PV=nRT n=PV =(1.2atm)x(2.5L)x (mol K) RT (0.0821 L atm) x(298K) = 0.123 moles Classwork: p 133 # 7,8 and p141 # 3,4 Ideal gas variations For calculating molar mass M= mRT PV M: molar mass (g/mol) m: mass (g) For calculating density D= MP D: density (g/L) RT Sample problem 1. What is the molar mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP? Sample problem What is the molar mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP? M= ? V= 300mL=0.3 L T=273K, P= 101.3kPa m=0.855g 1. M= mRT = (0.855g )x(8.314 L kPa)x(273K) PV (mol K ) (101.3kPa)(0.3L) = 63.8g/mol 2. At what temperature will 5.00g of Cl2 exert a pressure of 115 kPa at a volume of 750 mL? M= mRT PV 3. If the density of a gas is 1.2 g/L at 0.920 atm and 20C, what is its molar mass? D= PM RT At what temperature will 5.00g of Cl2 exert a pressure of 115 kPa at a volume of 750 mL? M= mRT M= (2x35.5)=71.0g/mol V=750mL=0.750L PV T=MPV = 71.0g 115kPa 0.750 L mol K = 147 K mR 5.00 g 8.314 L Kpa 3. If the density of a gas is 1.2 g/L at 0.920 atm and 20C, what is its molar mass? D= PM RT M= DRT = 1.2g 0.0821 L atm 293 K =31.4g/mol P L mol K 0.920 atm 2. Classwork: p 133 #5,6, 9 and p141 #2,5 Gases: Mixtures and Movements Objectives: 1. Relate the total pressure of a mixture of gases to the partial pressures of the component gases. 2. Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses Dalton’s Law Partial pressure: the contribution each gas in a mixture makes to the total pressure. Dalton’s law of partial pressure: at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component of the gases. P total= P1 + P2 + P3 + … Graham’s Law Diffusion: the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout. Effusion: a gas escapes through a tiny hole in its container. Particles of lower molar mass diffuse and effuse faster than gases of higher molar mass. Using Dalton’s and Graham’s Laws: Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10kPa, 0.040kPa, and 0.94kPa respectively? Ptotal = 101.3 kPa PN2= 79.10kPa PCO2= 0.040kPa Pothers= 0.94kPa PO2= ? Ptotal = PN2 + PCO2 + PO2 + Pothers 1. PO2 = Ptotal- (PN2+ PCO2+ POthers) = 101.3kPa-(79.1kpa+0.040kPa+0.94kPa) = 21.22 kPa 2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature. 2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature. rateHe /rateN2= (molar mass N2/molar massHe) rateHe = rateN2 (molar mass N2/molar massHe) = 505m/s (28/4) = 1336 m/s Classwork: dalton’s and graham’s law handout.