Gasses • Zumdahl (6th Ed) Chapter 5 Sections 1-3 • The nature of gasses (the air we breathe) • Scale of gasses in atmosphere. • The ideal Gas (I.G.) – How it behaves - Solve Gas Problems - Subsumes Boyle, Charles, Avogadro Laws • Problems 5.21-24, 5.26, 5.29-32, 5.35, 5.37-38. How big are molecules? Water has a density (weighs) 1 gram/cc. Boil the water and the density is 1 gram/liter. Convert that to moles per liter: What is the expansion factor? What does this tell us about what a gas is? How big are molecules? Liquid water has a density (weighs) 1 gram/cc. What is the volume and diameter of a water molecule? (Assume it is a sphere). What is the molar concentration? Water vapor at one Atm pressure: density is 1.2 gram/liter. What is the number of moles per liter? What is the average distance between vapor molecules. Compare that to the distance between water molecules in liquid water. Compare the volume that N water molecules occupy in the gas state with that in the liquid state? (i.e. What is the expansion factor? Why do gas molecules stay as a gas? • Consider liquid N2 or solid CO2. – What makes them condense? • Temperature is the best predictor of the state of matter; it is a direct measure of the kinetic energy inherent in molecules. • So the trade off is the molecular attraction to cause matter to condense with the kinetic energy inherent in matter (at 300K) to overcome adhesion. • Consider the “gedanken” experiment of an isolated gas next to a solid object in contact with a gas at Room Temp. • Temperature makes molecules move that overcomes the “weaker” attractions tending to make the solid. Gas Molecules in a Box: Lots of Empty Space Trace shows the path of one particle, includes collisions with other particles. Average distance between particles (at 1STP) is about 10 times particle diameter. Wbsite for the kinetic demo: http://ccl.northwestern.edu/netlogo/models/run.cgi?GasLabGasinaBox.919.607 Hot Air Balloon Why does a hot air balloon rise? How hot does the gas have to be to carry the people and the basket below? How may units of Pressure are there? • • • • • • • The Torr (or Torricelli) Mm Hg 1Torr=1mm(Hg) Atmospheres (Atm) 1 Atm = 760 Torr Pounds per square inch 1Atm = 14.7 lb/in2 Bars SI units ~ 1.01 Atm The SI unit itself: the Pascal. 1 Bar=105Pa How are they all connected???? Barometer Open to atmosphere What is pressure; how do we measure it? Closed The mercury column exerts a force over the cross‐sectional area of the tube. Vacuum P = h⋅ g ⋅d pressure 1 atm 1 atm 1 atm height accel. due to gravity density 1 atm The pressure exerted by the mercury column is exactly balanced by the pressure of the atmosphere. A torricellian barometer Relation of Pressure, mass, force, Area mass _ air = mass _ mercury mass _ air = density _ air ⋅ Volume _ air = density _ air ⋅ Area ⋅ Height _ air mass _ Hg = density _ Hg ⋅Vol _ Hg = density _ Hg ⋅ Area ⋅ h _ Hg Force = g ⋅ ( mass _ air ) = g ⋅ ( mass _ mercury ) Force g ⋅ ( mass _ air ) g ⋅ ( mass _ mercury ) = = Pressure = Area Area Area Hg is very dense, density is 13. 6 g/cc; Air is sparse, and is equivalent to a column of air of a density of 1.2 g/l up to 8.6 km. The air does not have uniform density but falls off exponentially due to the force of gravity and its own weight. A torricellian barometer Mass of the Hg in the tube must equal the effective mass of the air pushing down. . P = dgh kg PAir = d ⋅ g ⋅ h = 1.2 ⋅ 9.8 ⋅ 8.6 ⋅10 m = 101kPa 3 3 m kg 3 l ⋅10 PHg = d ⋅ g ⋅ h = 13.6 3 ⋅ 9.8 ⋅ 0.760m = 101kPa l m P = 2.28 pounds / cm2 = 14.7 pounds / in2 = 760 mm Hg = 1.00 atm. Why do we have pressure from the atmosphere? Gas in the air is held by gravity: How thick is the earth’s atmosphere? What is pressure? Why don’t we “feel” 1 Atm of pressure? Why is it hard to walk on grass in high-heeled shoes? How deep would the liquid be if the air were to liquefy (which it does when the temperature is below 80K). Why do we have pressure from the atmosphere? There is a lot of gas in the air and the gas is held by gravity. The force of gravity brings the molecules to the earth’s surface but they do not just lay on the ground, they have too much energy for that and they bounce around. The mass of the atmosphere is And the gravitational constant is M = 5.2 ⋅1018 kg g = 9.8 m sec 6 2 F = Mg The earth is a ball with diameter d o = 12.6 ⋅10 m So the earths surface area is A = π ⋅ d o2 = 5.0 ⋅1014 m 2 Pressure is due to all that mass spread over the entire surface of the earth, which generates a force due to gravity: F Mg P= = = A A 5.2 ⋅1018 kg ⋅ 9.8 m 5 ⋅1014 m 2 sec 2 = 1.01⋅105 Pa = 1 Atm Half of the earth’s atmosphere is contained in the first 4.3 km. So the density of the 1M air is: 18 5.2 ⋅10 kg kg g = = = d= 1.2 1.2 3 14 3 3 A m A ⋅ h 2 ⋅ 5 ⋅10 ⋅ 4.3 ⋅10 m 2 The density and the pressure drop exponentially as one goes away from the earth A simple manometer, a device for measuring the pressure of a gas in a container. P = hgd or ΔP = ( gd ) ⋅ Δh How does a car tire work? • A car tire is usually inflated to 35 psi (or 2 to 3 Atm over atmospheric pressure). • What fraction of the tire is filled with air? At one atmosphere 0.1% are molecules, at 10 Atmosphere 1% is actually air. 10 Atmospheres is 140 psi; which is above the high pressure bike racing tires. And only 1% of the space in the tire is taken up by the air!!!! • Why does air in a tire work? The more air, the more molecules hit the inside of the tire and push out. • How well would a tire work below 80K? (Moon tires?) • Air molecules hit the wall of the tire more often on the inside than the outside. The molecules hitting the wall of the tire push on the wall. They don’t loose kinetic energy. Why? Tire Pressure: Volume decreases, pressure goes up This is Boyle’s law (developed in the 1650s before tires) Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO3) in a closed end manometer, the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals, and bars. Plan: The pressure is in mmHg, so we use the conversion factors to find the pressure in the other units. Solution: converting from mmHg to torr: 1 torr = 341.6 torr 1 mm Hg converting from torr to atm: 1 atm = 0.4495 atm PCO2( atm) = 341.6 torr x 760 torr converting from atm to kPa: 101.325 kPa = 45.54 kPa PCO2(kPa) = 0.4495 atm x 1 atm PCO2 (torr) = 341.6 mm Hg x Gas Laws: an Overview • • • • Picture: Molecules bounce around in a box. Parameters needed to describe the situation: Number of gas molecules; Size of the box; Temperature The outcome is that the molecules produce pressure by hitting the sides of the box. We can measure pressure. • The summary relation among all of these quantities is PV = nRT • Demonstrate the relations by various experiment – Boyle’s Law – Charles’ Law – Avogadro’s Law P ⋅V = k k = k (T , n) V = T ⋅b b = b ( P, n ) V = n⋅a a = a (T , P ) Boyle’s Law : P - V inversely proportional when n and T are constant in a gas sample • Pressure is inversely proportional to Volume P ⋅V = k k = k ( n, T ) k P= V k V= P • Change of Conditions if n and T are constant ! • P1V1 = k P2V2 = k P1V 1 = P 2 V 2 • Then : • When the volume decreases and the molecules move at the same speed the time between wall-hits is less, so the pressure is higher. Plotting Boyle’s data from Table 5.1. Syringe Demo Boyle’s Law : Balloon • A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant(which obviously is not true), what is the final volume of the balloon? • P1 = 1.0 atm P2 = 0.40 atm • V1 = 0.55 L V2 = ? • V2 = Boyle’s Law : Balloon • A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant(which obviously is not true), what is the final volume of the balloon? • P1 = 1.0 atm P2 = 0.40 atm • V1 = 0.55 L V2 = ? • V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm) • V2 = 1.4 L Determine the pressure of water • The mass of water in a column h units m = d ⋅V = d ⋅ h ⋅ A high and with a foot print of area A is related to the density: F • The Pressure is related to the Force of P = A the water. • The force of the water is proportional to F = g ⋅ m the mass: m • Solve for Pressure in terms of height of P = g = g ⋅d ⋅h the water: A • Water density is about 1g/cc=1kg/liter • Atmospheric pressure is 101.3kPa • Solve for the height in meters using SI units of pressure. P 1Atm 101.3 ⋅103 Pa = = = 10.07 m h= 3 liter kg kg m m g ⋅ d 9.8 sec2 ⋅1.026 liter 9.8 sec2 ⋅1.026 liter ⋅1⋅10 m3 Boyle’s Law - A gas bubble in the ocean! A bubble of gas is released by the submarine “Alvin” at a depth of 6000 ft in the ocean, as part of a research expedition to study under water volcanism. Assume that the ocean is isothermal( the same temperature through out ),a gas bubble is released that had an initial volume of 1.00 cm3, what size will it be at the surface at a pressure of 1.00 atm? (We will assume that the density of sea water is 1.026 g/cm3, and use the mass of Hg in a barometer for comparison!) Initial Conditions V 1 = 1.00 cm3 P1 = ? Final Conditions V2 = ? P 2 = 1.00 atm Each 10 meters of sea water generates a force of 1 Atmosphere; This is a depth of 1840 meters. So 185 Atmosphere of pressure. Ocean Bubble Calculation For every 10 meters or 32 feet there is an additional atmosphere of pressure. Therefore: 32 feet water = 1 Atmosphere pressure 1 Atm x = 6000 ft = 6000 ft ⋅ = 176.5 Atm − water 34 ft The total pressure on the bubble (add one for air) is P1 = x + 1 = 188.5 Atm The final volume then: PV 1 1 = PV 2 2 177.5 Atm ⋅1cc = 1Atm ⋅ V2 V2 = 177.5cc = 0.18l Diving Questions A diver can dive with SCUBA equipment to 40 meters. A deep but doable “recreation” depth, 132 ft. Start to get N2 narcosis without special precautions. What is the pressure at that depth? If the diver took a 1 liter balloon with him to that depth, how large would the balloon be at 40 meters? Charles’ Law: Plots of V versus T (C) for several gases. Charles Law - V - T- proportional • At constant pressure and for a fixed amount (# moles) of gas: Volume is proportional to Temperature : V= constant x Absolute Temperature V= Txb • Change of conditions problems: • Since V = T x b or V1 / T1 = V2 / T2 or: T1 V1 = T2 or T1 = V1 x V2 Temperatures must be expressed in Kelvin! T2 V2 Volume -- Temperature • The box has a piston on the top that is at 1 Atm and can move up and down to keep the pressure constant and keep the gas in the box. The temperature drops to from 117C to 58.5C, the new volume will be as a percentage of the original volume? • A balloon in Antarctica in a building is at room temperature ( 75o F ) and has a volume of 20.0 L . What will be its volume outside where the temperature is -70oF ? Charles Law Problem - I • A balloon in Antarctica in a building is at room temperature ( 75o F ) and has a volume of 20.0 L . What will be its volume outside where the temperature is -70oF ? • V1 = 20.0 L • T1 = 75o F V2 = ? T2 = -70o F • o C = ( o F - 32 ) 5/9 • T1 = ( 75 - 32 )5/9 = 23.9o C • K = 23.9o C + 273.15 = 297.0 K • T2 = ( -70 - 32 ) 5/9 = - 56.7o C • K = - 56.7o C + 273.15 = 216.4 K Antarctic Balloon Problem - II • V1 / T1 = V2 / T2 • V2 = 20.0 L x • V2 = 14.6 L V2 = V1 x ( T2 / T1 ) 216.4 K 297.0 K • The Balloon shrinks from 20 L to 15 L !!!!!!! • Just by going outside !!!!! Combo Problem • Typical car tire is at 35 psi at 100F (or 40C). If the temperature drops to -40F, what is the tire pressure? • First Assume the tire volume does not change • Then redo assuming the tire volume drops by 5%. Variations on Ideal Gas Equation • During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed. • Thus, PV=nRT can be rearranged for the fixed variables: – for a fixed amount at constant temperature • P V = nRT = constant Boyle’s Law • P2V2=nRT=P1V1 – for a fixed amount at constant pressure • V / T = nR / P = constant Charles’ Law – for a fixed pressure and temperature • V = n (RT/P) or V/n = constant Avogadro’s Law Use I.G. E.o.S. : PV=nRT and rearrange as needed Avogadro’s Law - Moles and Volume The volume of gas is directly proportional to the amount of gas (in moles), when measured at the same P and T: V ∝ n or V = a ⋅ n where a = a ( P, T ) This is the most amazing and puzzling law. Why does the mass of the gas molecules not matter? He, CO2 and propane obey exactly the same gas law. Why? For problems where P,T are fixed go from state 1 to state 2: V = a⋅n V2 n2 n1 n2 = or C= = V1 n1 V1 V2 Concentration, C, of the gas does not change, @ fixed P and T Standard Molar Volumes; density depends on gas One mole of gas in familiar object Avogadro’s Law: Volume and Amount of Gas Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in the atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC is 2.93 m3, what will be the mass of SF6 in a container whose volume is 543.9 m3 at 1.143 atm and 28.5 oC? Plan: Since the temperature and pressure are the same it is a V - n problem, so we can use Avogadro’s Law to calculate the moles of the gas, then use the molecular mass to calculate the mass of the gas. M W ( SF6 ) = 146.07 g mol n m n C= and m = n ⋅ M W therefore d = = ⋅ M W = C ⋅ M W V V V Calculation: Avogdaro’s Law Reduced Problem: 2.67 g of SF6 has volume 2.93 m3; What is mass of SF6 in volume 543.9 m3 Solution m1 m2 V2 543.9 d= = ∴ m2 = m1 = 2.67 ⋅ = 496.g V1 V2 V1 2.93 Alternate Solution: n2 = n1 x V2 V1 2.67g SF6 = 0.0183 mol SF6 146.07g SF6/mol 3 543.9 m = 0.0183 mol SF6 x = 3.40 mol SF6 3 2.93 m mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6 Use Avogadro’s Principle to Obtain the Gas Constant, R • One mole occupies 22.4 liters at STP PV = nRT PV 1Atm ⋅ 22.4A = = 0.0820 A − Atm R= mol − K nT 1mole ⋅ 273.15 K • How many Joules of energy are in 1 literAtmosphere (PV has units of energy!) 8.314 J R mol − K 1= = = 101.3 J A − Atm R 0.0820 A − Atm mol − K 1Atm = 101.3kPa and 1Bar = 100kPa Summary: I.G. • The ideal gas equation combines both Boyle’s, Charles’ and Avogadro’s law into one easy-to-remember law: PV=nRT • • • • n = number of moles of gas in volume V R = Ideal gas constant R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1 In SI units R = 8.314 Pa-m3/ (mol K) = 8.314 J mol-1 K-1 An ideal gas is one for which both the volume of molecules and forces between the molecules are so small that they have insignificant effect on its P-V-T behavior. Independent of substance, in the limit that n/V →0, all gases behave ideally. Usually true below 2 atm. Hot Air Balloon Why does a hot air balloon rise? Typical hot air balloons displace 100,000 cu ft (2,800 m3). The air in the balloon is typically heated to around 100C (max op temp is 120C). Air is 78% N2, 21% O2 (<1% Ar, other). Temperature is 17C. What is the total mass that can be carried? Hot Air Balloon Need the number of moles inside the balloon at the higher temperature. The density of air is M W ( Air ) = 0.78 ⋅ 28 + 0.21 ⋅ 32 + 0.01⋅ 40 = 28.96 g d = M W ( Air ) ⋅ d Air = 1.218 g mol n1 1Atm P = M W ( Air ) ⋅ = 28.96 g mol 0.082 ⋅ 290 l − Atm V RT1 mol l Use: P,V are constant, so as T goes up n goes down. PV = nRT or n1T1 = n2T2 or d Balloon T1 P 290 = M W ( Air ) ⋅ = d Air = 1.218 ⋅ = 0.9468 g l RT2 T2 373 The difference in mass between the mass of the air and the balloon is what you can carry. Δm = V ⋅ ( d Air − d Balloon ) = (1.218 − 0.9468 ) g Δm = 0.76 ⋅106 g = 760kg l = 2.8 ⋅106 l ⋅ (1.218 − 0.9468 ) g l Hot Air Balloon: Weight Distribution component kilogra ms pounds 100,000 cu ft (2,800 m3) envelope 250 113.4 5-passenger basket 140 63.5 50 22.7 3 20-gallon (75.7-liter) fuel tanks full of propane 3 × 135 = 405 183.7 5 passengers 5 × 150 = 750 340.2 sub total 1595 723.5 100,000 cu ft (2,800 m3) of heated air 5922 2686.2 (3.76 tons) 7517 3409.7 double burner total using a density of 0.9486 kg/m³ for dry air heated to 210 °F (99 °C). Inflated Dual Airbags How do we store the gas so well? The primary chemical reactant is NaN3(S). How much powder do we need in the air bag?
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