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Gases and gas laws Chapters 13.1 and 14
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Kinetic molecular theory The theory of moving molecules
Basic assumptions describing the behavior of gases: Gas consist of small particles that are very far apart Most of the volume of a gas is empty space Individual particles have essentially zero volume compared to the total volume of the gas Gases particles are in constant, random, straight line motion Attractive and repulsive forces between particles are negligible
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Kinetic molecular theory The theory of moving molecules
Basic assumptions: Random motion of gas particles is constantly interrupted by collisions between particles and between particles and surfaces Such collisions are perfectly elastic Kinetic energy may be transferred between particles but is not lost (conserved) The average kinetic energy of gas particles is directly proportional to absolute temperature (kelvin) The rate of motion of particles is related to temperature
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Kinetic molecular theory The theory of moving molecules
Characteristics of gases: Gases readily change volume Container containing gas is always full Molecules separate and spread out uniformly until fill available space Compressibility: can be compressed or reduced to small fractions of original volume Expandability: can expand to fill virtually any volume Gases typically have low densities (mass to volume ratio)
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Kinetic molecular theory The theory of moving molecules
Characteristics of gases: Gases exert pressure Due to constant random motion, molecules collide with each other and with the walls of their container Collisions exert pressure on surroundings
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Kinetic molecular theory The theory of moving molecules
Characteristics of gases: Gas Temperature and Kinetic Energy Relationship between average kinetic energy of group of gas molecules and the temperature measured in degrees Kelvin is directly proportional As temperature increases, average kinetic energy of molecules ____________ As temperature decreases, average kinetic energy of molecules ____________ Relationship only exists when temperature is expressed in Kelvin scale Molecules have 0 kinetic energy when temp is 0 K Only have kinetic energy at temps above 0 K increases decreases
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Movement of gas molecules
Describes movement of gases through other materials Gas molecules move from areas of high concentration to low concentration Examples: Perfume molecules spreading across a room Aromas of cooking foods waft through the house DIFFUSION Animation by Caitlin Claunch
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Movement of gas molecules
Gas molecules escape through tiny openings Gas molecules move from an area of high pressure to low pressure Examples: Tire flattens due to a puncture Balloon deflates when stuck with a pin EFFUSION Animation by Caitlin Claunch
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Gases Review Gas particles are in constant, ________ motion. Particles travel in ____________. The size of the particles is ___________ compared to the volume of ______ space. Attraction and repulsion between particles is __________. Collisions are ______ because particles transfer energy but the total kinetic energy is ________. ____________ is the measure of average kinetic energy of particles. At a given temperature, ___ gas particles have the same ______________ energy. In diffusion, gas particles move from high ____________ to low ____________, whereas in effusion, particles move from high ________ to low ________. random straight lines insignificant empty negligible elastic constant Temperature all average kinetic concentration concentration pressure pressure
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Variables describing gases
Pressure (P): force per unit area Volume (V): amount of space taken up Temperature (T): measure of average kinetic energy Number of Moles (n): amount of substance
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pressure Results from billions of gas particles in motion and colliding against their container Barometer: instrument used to measure atmospheric pressure Manometer: instrument used to measure gas pressure in a closed container Worksheet: Pressure Unit Conversions
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pressure UNITS: The relationships among the different units of pressure can be used as conversion factors Values expressed in table above describe the same amount of pressure atm mm Hg torr psi kPa Pa 1 760 14.7 101,325 SI UNIT Worksheet: Pressure Unit Conversions
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Dalton’s law of partial pressures
The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture Ptotal = P1 + P2 + P3 + … Pn 101.3 kPa kPa kPa kPa kPa kPa
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Finding the partial pressure
A mixture of oxygen (O2), carbon dioxide (CO2), and nitrogen (N2) has a total pressure of atm. What is the partial pressure of O2, if the partial pressure of CO2 is 0.70 atm and the partial pressure of N2 is 0.12 atm? Known PN2 = 0.12 atm PCO2 = 0.70 atm Ptotal = 0.97 atm Unknown PO2 = ? atm Practice Problems #4, #5, #6 – p. 392 Ptotal = PO2 + PCO2 + PN2 PO2 = Ptotal – PCO2 – PN2 PO2 = 0.97 atm – 0.70 atm – 0.12 atm = 0.15 atm
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U12-3 PRACTICE/EXAMPLES p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13
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U12-3: p. 392, #4 What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 mm Hg? Known PHe = 439 mm Hg Ptotal = 600 mm Hg Unknown PH2 = ? mm Hg U12-3 Practice Examples: #4, #5, #6 – p. 392 Ptotal = PH2 + PHe PH2 = Ptotal – PHe PH2 = 600 mm Hg – 439 mm Hg = 161 mm Hg
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U12-3: p. 392, #5 Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa, and 1.20 kPa. Known P1 = 5.00 kPa P2 = 4.56 kPa P3 = 3.02 kPa P4 = 1.20 kPa Unknown Ptotal = ? kPa U12-3 Practice Examples: #4, #5, #6 – p. 392 Ptotal = P1 + P2 + P3 + P4 Ptotal = 5.00 kPa kPa kPa kPa = kPa
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U12-3: p. 392, #6 Find the partial pressure of carbon dioxide in a gas mixture with a total pressure of 30.4 kPa if the partial pressures of the other two gases in the mixture are 16.5 kPa and 3.7 kPa. Known P1 = 16.5 kPa P2 = 3.7 kPa Ptotal = 30.4 kPa Unknown PCO2 = ? kPa U12-3 Practice Examples: #4, #5, #6 – p. 392 Ptotal = P1 + P2 + PCO2 PCO2 = Ptotal – P1 – P2 PCO2 = 30.4 kPa – 16.5 kPa – 3.7 kPa = 10.2 kPa
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p. 415: #68, #69, #70, #71 Chapter 13 homework
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Pressure Unit Conversions Worksheet
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volume Gases expand to fill their container
That is, they do not have a definite volume Gas laws explain how and when gas volume changes UNITS: L mL cm3 quart 1 1000 1.057 SI UNIT
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temperature Measure of average kinetic energy of particles Recall KMT: At any given temperature, all gases have same average kinetic energy UNITS: Measured using the Kelvin scale (absolute temperature scale) developed by Lord Kelvin 0 degrees kelvin is defined as absolute zero Absolute zero is the hypothetical temperature at which all motion stops K = °C + 273
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Stp Standard Temperature and Pressure 0°C or 273 K
1 atm or kPa
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Combined gas law States the relationship among pressure (P), volume (V), and temperature (T) when amount of gas (n) remains constant Pressure and Temperature are _______________ proportional directly Pressure and Volume are _______________ proportional inversely Volume and Temperature are _______________ proportional directly
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Using Combined gas law A gas occupies 7.84 cm3 at 71.8 kPa and 25°C. What volume will it occupy at STP? Rearrange to solve for unknown variable, V2 VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 71.8 kPa 7.84 cm3 25°C + 273 K = 298 K kPa unknown 273 K
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Using Combined gas law V2 = (71.8 kPa)(7.84 cm3)(273K)
A gas occupies 7.84 cm3 at 71.8 kPa and 25°C. What volume will it occupy at STP? V2 = (71.8 kPa)(7.84 cm3)(273K) ( kPa)(298K) V2 = 5.1 cm3 VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 71.8 kPa 7.84 cm3 25°C + 273 K = 298 K kPa unknown 273 K p #19, #20, #21, #22, #23
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p. 449: #92, #93, #94 Chapter 14 homework
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U12-3 PRACTICE/EXAMPLES p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13
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U12-3: p. 430, #19 A helium-filled balloon at sea level has a volume of 2.1 L at atm and 36°C. If it is released and rises to an elevation at which the pressure is atm and the temperature is 28°C, what will be the new volume of the balloon? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 0.998 atm 2.1 L 36°C + 273 K = 303 K 0.900 atm V2 = 2.3 L unknown 28°C + 273 K = 301 K
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U12-3: p. 430, #20 At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 20.0-mL container, what will be the gas pressure inside the container? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 1.00 atm P2 = P1V1T2 V2T1 30.0 mL 0.00°C + 273 K = 273 K unknown P2 = 2.3 L 20.0 mL 30.0°C + 273 K = 303 K
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Boyle’s law Boyle’s Law
States the relationship between pressure (P) and volume (V), when temperature (T) and amount of substance (n) remain constant Boyle’s Law Pressure and Volume are __________ proportional As pressure increases, volume __________ As pressure decreases, volume __________ inversely decreases increases
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Using boyle’s law A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. Rearrange to solve for unknown variable, V2 VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 150. kPa 100. mL remains constant 200. kPa unknown remains constant
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Using boyle’s law A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 150. kPa 100. mL remains constant 200. kPa unknown V2 = (150. kPa)(100. mL) (200. kPa) V2 = mL p #1, #2, #3, #4, #5
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Charles’ law Charles’ Law
States the relationship between volume (V) and temperature (T), when pressure (P) and amount of substance (n) remain constant Charles’ Law Volume and Temperature are __________ proportional As volume increases, temperature __________ As volume decreases, temperature __________ directly increases decreases
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Using charles’ law A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. Rearrange to solve for unknown variable, V2 VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = remains constant 473 cm3 36°C + 273 K = 309 K remains constant unknown 94°C + 273 K = 367 K
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Using charles’ law A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = V2 = (473 cm3)(367 K) (309 K) V2 = 560 cm3 remains constant 473 cm3 p #6, #7, #8 36°C + 273 K = 309 K remains constant unknown 94°C + 273 K = 367 K
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Gay-lussac’s law Gay-Lussac’s Law
States the relationship between pressure (P) and temperature (T), when volume (V) and amount of substance (n) remain constant Gay-Lussac’s Law Pressure and Temperature are __________ proportional As pressure increases, temperature __________ As temperature decreases, pressure __________ directly increases decreases
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Using gay-lussac’s law
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? Rearrange to solve for unknown variable, P2 VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 3.20 atm remains constant 22.0°C + 273 K = 295 K unknown remains constant 60.0°C + 273 K = 333 K
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Using gay-lussac’s law
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = 3.20 atm remains constant unknown 22.0°C + 273 K = 295 K 60.0°C + 273 K = 333 K P2 = (3.20 atm)(333.0K) (295.0K) P2 = atm p #9, #10, #11, #12, #13
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p. 448: #88, #89, #90, #91 U12-5: Ch.14 HOMEWORK
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U12-3 PRACTICE/EXAMPLES p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13
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U12-3: p. 430, #21 A sample of air in a syringe exerts a pressure of atm at a temperature of 22.0°C. The syringe is placed in a boiling water bath at 100.0°C. The pressure of the air is increased to 1.23 atm by pushing the plunger in, which reduces the volume to mL. What was the original volume of air? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 422, #1 The volume of a gas at 99.0 kPa is mL. If the pressure is increased to 188 kPa, what will be the new volume? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 425, #6 A gas at 89°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 427, #9 A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0°C. If the pressure in the container is increased to 201 kPa, what is the new temperature? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 422, #2 The pressure of a sample of helium in a 1.00-L container is atm. What is the new pressure if the sample is placed in a 2.00-L container? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 430, #22 An unopened, cold 2.00-L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0°C. If the bottle is dropped into a lake and sinks to a depth at which the pressure is 2.85 atm and the temperature is 2.09°C, what will be the volume of gas in the bottle? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 422, #3 Air trapped in a cylinder fitted with a piston occupies mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 430, #23 A sample of gas of unknown pressure occupies L at a temperature of 298 K. The same sample of gas is then tested under known conditions and has a pressure of 32.6 kPa and occupies L at 303 K. What was the original pressure of the gas? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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U12-3: p. 427, #12 If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa? VARIABLES P1 = V1 = T1 = P2 = V2 = T2 =
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Avogadro’s Law Conversion factor 22.4 L of any gas = 1 mol
Addresses variable n (number of moles) States that equal volumes of gases at the same temperature and pressure contain the same numbers of particles Molar volume of gases: 22.4 L/mol at standard temperature and pressure (STP) Standard temperature: 273 K or 0°C Standard pressure: 1.00 atm, kPa, 760 torr, 760 mm Hg Conversion factor 22.4 L of any gas = 1 mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Is amount of substance (n) involved? Is gas at STP? Use relationship between volume and number of moles (Avogadro’s Law) Yes Yes
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol Unit for volume? Liter (L) L = mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol L = mol mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol L = mol mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol L = mol L mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol L = mol 22.4 L 1 mol
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Using avogadro’s law Calculate the volume that mol of gas at STP will occupy. L = mol 22.4 L 1 mol L = 19.7 L
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Using avogadro’s law Calculate the volume that 2.0 kg of methane gas (CH4) will occupy at STP. L = 2.0 kg L = 2800 L 1000 g 1 mol 22.4 L 1 kg 16.0 g 1 mol P #29, #30, #31, #32, #33
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U12-2: Ch. 13 homework U12-5: Ch. 14 homework p. 449: #95, #96
Express answers in atm, kPa, and torr. U12-2: Ch. 13 homework p. 449: #95, #96 U12-5: Ch. 14 homework
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U12-3 PRACTICE/EXAMPLES p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33
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U12-3: p. 432, #24 Determine the volume of a container that holds 2.4 mol of gas at STP. U12-3: p. 432, #25 What size container do you need to hold mol N2 gas at STP?
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U12-3: p. 432, #26 What volume will 1.02 mol of carbon monoxide gas occupy at STP? U12-3: p. 432, #27 How many moles of nitrogen gas will be contained in a 2.00-L flask at STP?
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U12-3: p. 432, #28 If a balloon will rise off the ground when it contains mole of helium in a volume of L, how many moles of helium are needed to make the balloon rise when its volume is L? Assume that temperature and pressure stay constant.
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U12-3: p. 433, #29 How many grams of carbon dioxide gas are in a 1.0-L balloon at STP? U12-3: p. 433, #30 What volume in milliliters will g H2 gas occupy at STP?
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Ideal gas law Describes the physical behavior of an ideal gas in terms of all four variables Pressure (P) Volume (V) Temperature (T) Amount of gas (n)
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Ideal gases vs. Real gases
Described by Kinetic Molecular Theory Particles have no volume, no intermolecular attractive forces, and perfectly elastic collisions Follow all gas laws under all conditions of temperature and pressure Behave less like ideal gases at extremely high pressures and low temperatures Polar gases do not behave as ideal gases Larger gas molecules behave less like ideal gases than smaller gas molecules IDEAL GASES REAL GASES
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Ideal gas law Combines: Boyle’s Law Charles’ Law Gay-Lussac’s Law
Avogadro’s Law Combined Gas Law If number of moles (n) is constant → Combined Gas Law If number of moles (n) is not constant, and At STP → Avogadro’s Law Not at STP → Ideal Gas Law
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How will you know which value to use?
Ideal gas constant (r) Experimentally determined Three possible numerical values How will you know which value to use?
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Using the ideal gas law A sample of nitrogen gas (N2) has a volume of liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present? Rearrange to solve for unknown variable, n VARIABLES P = atm V = L n = ? mol R = T = 273 K
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Using the ideal gas law A sample of nitrogen gas (N2) has a volume of liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present? VARIABLES P = atm V = L n = ? mol R = T = 273 K n = (1.50 atm)(5.56 L) ( ( )(273 K) n = mol
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U12-3 PRACTICE/EXAMPLES p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33
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Which gas law? Avogadro’s Law 22.4 L = 1 mol
What volume does 9.45 g of C2H2 occupy at STP? L = 9.45 g C2H2 L = 8.14 L 1 mol C2H2 22.4 L C2H2 26.0 g C2H2 1 mol C2H2 P #29, #30, #31, #32, #33
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Which gas law? Ideal Gas Law PV = nRT
What volume does 9.45 g of C2H2 occupy at STP? Rearrange to solve for unknown variable, V VARIABLES P = kPa V = ? L n = 9.45 g = mol 26.0 g/mol R = T = 273 K
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Which gas law? What volume does 9.45 g of C2H2 occupy at STP?
VARIABLES P = kPa V = ? L n = 9.45 g = mol 26.0 g/mol R = T = 273 K
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Which gas law? V = (0.3634615385 mol)( )(273 K) (101.325 kPa)
What volume does 9.45 g of C2H2 occupy at STP? V = ( mol)( )(273 K) ( kPa) V = 8.14 L
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Which gas law? You decide to go on a long hot air balloon ride, and pack a bottle of shampoo for your trip. There is some gas inside the shampoo bottle when you climb into the basket at the beginning of your journey. Being the good scientist that you are, you decide to take constant measurements of your surroundings. The shampoo bottle contains 435 mL of gas, is under a pressure of 1.10 atm, and is at a temperature of 30.0°C. As you climb high into the air, the bottle starts to expand and eventually explodes, covering you and your companions with Pert Plus®. Eager to explain this phenomenon, you take some quick measurements, noting that the pressure has dropped to atm and the temperature is now 5.00°C. To what new volume did the gas inside the shampoo bottle expand? VARIABLES P1 = V1 = T1 = 1.10 atm P2 = V2 = T2 = 0.734 atm 435 mL ? = 598 mL ? 30.0°C = 303 K 5.00°C = 278 K
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U12-1 Gas Laws Notes p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33
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Gas stoichiometry Stoichiometric calculations always require a balanced equation N2(g) H2(g) → NH3(g) Coefficients indicate ratios of reactants and products 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas OR 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gas to produce 2 volumes of ammonia gas Volume ratios only work when all reactants and products are gases (1) 3 2
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Gas stoichiometry N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas? L NH3 = 5.00 L H2
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Gas stoichiometry N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas? L NH3 = 5.00 L H2 L H2
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Gas stoichiometry N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas? L NH3 = 5.00 L H2 L NH3 L H2 = L NH3 2 3
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U12-10 Volume-Volume WS
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U12-1 Gas Laws Notes p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33
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Gas stoichiometry Gas stoichiometry problems can involve mass and volume Zn(s) + HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP? 2 L H2 = 6.54 g Zn
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Gas stoichiometry Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP? L H2 = 6.54 g Zn 1 mol Zn 65.4 g Zn
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Gas stoichiometry Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP? L H2 = 6.54 g Zn 1 mol Zn 1 mol H2 65.4 g Zn 1 mol Zn
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Gas stoichiometry Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP? L H2 = 6.54 g Zn 1 mol Zn 1 mol H L H2 65.4 g Zn 1 mol Zn 1 mol H2 2.24 L H2
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Gas stoichiometry Gas stoichiometric calculations only involving volume do not take the temperature and pressure conditions into consideration After mixing, each gas is at the same temperature and pressure Gas stoichiometric calculations involving volume and mass must take temperature and pressure conditions into account volume-volume relationships from balanced chemical equation are combined with the Ideal Gas Law
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Gas stoichiometry When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102) Determine volume of water produced (volume-volume relationships). C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g) C3H8 (g) + O2 (g) → H2O (g) + CO2 (g) L H2O = L C3H8 4 L H2O = 12.0 L H2O 1 L C3H8
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C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
Gas stoichiometry When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102) Determine number of moles under given conditions (Ideal Gas Law). C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g) n = mol (0.990 atm)(12.0 L H2O) ( )(623 K) Latm molK =
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C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
Gas stoichiometry When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102) C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g) g H2O = mol H2O 18.0 g H2O 1 mol H2O g H2O = 4.2 g H2O g H2O = g H2O
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U12-10 Gas Stoichiometry WS
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p #100, #103, #104 U12-5 Ch. 14 homework g C3H8 103. 2KClO3 (s) → 2KCl (s) + 3O2 (g) 5.70 L O2 104. a. 2CO (g) + 2NO (g) → N2 (g) + 2CO2 (g) b. 2 volumes CO 2 volumes CO2 c L N2 volume = L, mL or cm3
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U12-3 PRACTICE/EXAMPLES
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U12-3: p. 437, #41 If the pressure exerted by a gas at 25°C in a volume of L is 3.81 atm, how many moles of gas are present? Variable(s) involved? Gas law/application? P: V: n: R: T: P, V, T, n Ideal Gas Law 3.81 atm 0.044 L n = mol ? Latm molK 0.0821 298 K
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U12-3: p. 433, #33 Calculate the volume that 4.5 kg of ethylene gas (C2H4) will occupy at STP. Variable(s) involved? Gas law/application? V, n (mass) @STP Avogadro’s Law 3600 L C2H4
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U12-3: p. 425, #7 The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0°C to 30.0°C. What will be the resulting volume of this gas? Variable(s) involved? Gas law/application? V, T Combined Gas Law VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = V2 = L
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U12-3: p. 441, #57 Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen gas to form water. Variable(s) involved? Gas law/application? Write the balanced chemical equation. 2H2 (g) + O2 (g) → 2H2O (g) Unknown: Given: Answer: V volume-volume stoichiometry L H2 5.00 L O2 10.0 L H2
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U12-3: p. 437, #43 Calculate the volume that a mol sample of a gas will occupy at 265 K and a pressure of atm. Variable(s) involved? Gas law/application? P: V: n: R: T: P, V, T, n Ideal Gas Law 0.900 atm ? V = 7.81 L 0.323 mol Latm molK 0.0821 265 K
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U12-3: p. 437, #44 What is the pressure in atmospheres of a mol sample of helium gas at a temperature of 20.0°C if its volume is L? Variable(s) involved? Gas law/application? P: V: n: R: T: P, V, T, n Ideal Gas Law ? 0.505 L P = 5.14 atm 0.108 mol Latm molK 0.0821 293 K
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U12-3: p. 427, #13 A rigid plastic container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0°C. How much more pressure will the gas exert if the temperature is raised to 44.6°C? Variable(s) involved? Gas law/application? P, T Combined Gas Law VARIABLES P1 = V1 = T1 = P2 = V2 = T2 = P2 = 711 torr Additional pressure of 51 torr will be exerted.
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U12-3: p. 441, #56 What volume of oxygen is needed to react with solid sulfur to form 3.5 L SO2? Variable(s) involved? Gas law/application? Write the balanced chemical equation. S (s) + O2 (g) → SO2 (g) Unknown: Given: Answer: V volume-volume stoichiometry L O2 3.5 L SO2 3.5 L O2
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