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Chapter 5

Chapter 5. Gases. Chapter 5: Gases. 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Dalton’s Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion

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Chapter 5

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  1. Chapter 5 Gases

  2. Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Dalton’s Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions 5.10 Real Gases 5.11 Chemistry in the Atmosphere

  3. Hurricanes, such as this one off the coast of Florida, are evidence of the powerful forces present in the earth's atmosphere.

  4. Important Characteristics of Gases 1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have low viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible Gases mix in any proportion such as in air, a mixture of many gases.

  5. Substances that are Gases under Normal Conditions Substance Formula MM(g/mol) • Helium He 4.0 • Neon Ne 20.2 • Argon Ar 39.9 • Hydrogen H2 2.0 • Nitrogen N2 28.0 • Nitrogen Monoxide NO 30.0 • Oxygen O2 32.0 • Hydrogen Chloride HCL 36.5 • Ozone O3 48.0 • Ammonia NH3 17.0 • Methane CH4 16.0

  6. Some Important Industrial Gases Name - Formula Origin and use Methane (CH4) Natural deposits; domestic fuel Ammonia (NH3) From N2 + H2 ; fertilizers, explosives Chlorine (Cl2) Electrolysis of seawater; bleaching and disinfecting Oxygen (O2) Liquified air; steelmaking Ethylene (C2H4) High-temperature decomposition of natural gas; plastics

  7. Pressure of the Atmosphere • Called “Atmospheric pressure,” or the force exerted upon us by the atmosphere above us. • A measure of the weight of the atmosphere pressing down upon us. • Measured using a Barometer! - A device that can weigh the atmosphere above us! Force Area Pressure =

  8. Figure 5.1: A torricellian barometer.

  9. Construct a Barometer using Water! • Density of water = 1.00 g/cm3 • Density of Mercury = 13.6 g/cm3 • Height of water column = Hw • Hw = Height of Hg x Density of Mercury • Hw = 760 mm Hg x 13.6/1.00 = 1.03 x 104 mm • Hw = 10.3 m = 33.8 ft HeightWater HeightMercury DensityMercury DensityWater = Density of Water

  10. Common Units of Pressure Unit Atmospheric Pressure Scientific Field Used Pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal (kPa) 101.325 kPa Chemistry Atmosphere (atm) 1 atm Chemistry Millimeters of mercury 760 mmHg Chemistry, medicine (mmHg) biology Torr 760 torr Chemistry Pounds per square inch 14.7 lb/in2 Engineering (psi or lb/in2) Bar 1.01325 bar Meteorology, Chemistry

  11. 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. Plan: The pressure is in mmHg, so we use the conversion factors from Table 5.2(p.178) to find the pressure in the other units. Solution: converting from mmHg to torr: 1 torr 1 mm Hg PCO2 (torr) = 341.6 mm Hg x = 341.6 torr converting from torr to atm: 1 atm 760 torr PCO2( atm) = 341.6 torr x = 0.4495 atm converting from atm to kPa: 101.325 kPa 1 atm PCO2(kPa) = 0.4495 atm x = 45.54 kPa

  12. Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container.

  13. Figure 5.3: A J-tube similar to the one used by Boyle.

  14. Boyle’s Law : P - V relationship • Pressure is inversely proportional to Volume • P = or V = or PV=k • Change of Conditions Problems if n and T are constant ! • P1V1 = k P2V2 = k’ k = k’ • Then : P1V1 = P2V2 k V k P

  15. Figure 5.4: Plotting Boyle’s data from Table 5.1.

  16. Figure 5.5: Plot of PV versus P for several gases.

  17. Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm3, what will be the volume if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm3 to ml and then to liters, then we do the pressure change to obtain the final volume! Solution: V1 (cm3) P1 = 1.23 atm P2 = 3.16 atm V1 = 15.8 cm3V2 = unknown T and n remain constant 1cm3 = 1 mL V1 (ml) 1 mL 1 cm3 1 L 1000mL 1000mL = 1L V1 = 15.8 cm3 x x = 0.0158 L V1 (L) x P1/P2 P1 P2 1.23 atm 3.16 atm V2 = V1 x = 0.0158 L x = 0.00615 L V2 (L)

  18. 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 Final Conditions V 1 = 1.00 cm3 V 2 = ? P 1 = ? P 2 = 1.00 atm

  19. Calculation Continued 0.3048 m 1 ft 100 cm 1 m 1.026 g SH2O 1 cm3 Pressure at depth = 6 x 103 ft x x x Pressure at depth = 187,634.88 g pressure from SH2O For a Mercury Barometer: 760 mm Hg = 1.00 atm, assume that the cross-section of the barometer column is 1 cm2. The mass of Mercury in a barometer is: 1.00 atm 760 mm Hg 10 mm 1 cm Area 1 cm2 1.00 cm3 Hg 13.6 g Hg Pressure = x x x x 187,635 g = Pressure = 182 atmDue to the added atmospheric pressure = 183 atm! V1 x P1 P2 1.00 cm3 x 183 atm 1.00 atm V2 = = = 183 cm3= 0.183 liters

  20. 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

  21. Figure 5.6: PV plot verses P for 1 mole of ammonia

  22. T2 T1 T2 T1 = V1 x = V2 V1 V2 Charles Law - V - T- relationship • Temperature is directly related to volume • T proportional to Volume : T = kV • Change of conditions problem: • Since T/V = k or T1 / V1 = T2 / V2 or: Temperatures must be expressed in Kelvin to avoid negative values!

  23. Figure 5.7: Plots of V versus T (c) for several gases.

  24. Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.

  25. Charles Law Problem • A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. • V1 = 3.20 L T1 = 125oC = 398 K • V2 = 1.54 L T2 = ? • T2 = T1 x ( V2 / V1) T2 = 398 K x = 192 K • T2 = 192 KoC = K - 273.15 = 192 - 273 oC = -81oC 1.54 L 3.20 L

  26. 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 V2 = ? • T1 = 75o F 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

  27. Antarctic Balloon Problem - II • V1 / T1 = V2 / T2 V2 = V1 x ( T2 / T1 ) • V2 = 20.0 L x • V2 = 14.6 L • The Balloon shrinks from 20 L to 15 L !!!!!!! • Just by going outside !!!!! 216.4 K 297.0 K

  28. Applying the Temperature - Pressure Relationship (Amonton’s Law) P1 P2 T1 T2 = T2 T1 P2 = P1 x = ? Problem: A copper tank is compressed to a pressure of 4.28 atm at a temperature of 0.185 oF. What will be the pressure if the temperature is raised to 95.6 oC? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, and the pressure, so convert to SI units, and calculate the change in pressure from the Temp.and Pressure change. Solution: T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC T1 = -17.68 oC + 273.15 K = 255.47 K T2 = 95.6 oC + 273.15 K = 368.8 K 368.8 K 255.47 K P2 = 4.28 atm x = 6.18 atm

  29. Avogadro’s Law - Amount and Volume The Amount of Gas (Moles) is directly proportional to the volume of the Gas. n  V or n = kV For a change of conditions problem we have the initial conditions, and the final conditions, and we must have the units the same. n1 = initial moles of gas V1 = initial volume of gas n2 = final moles of gas V2 = final volume of gas n1 V1 n2 V2 V1 V2 = or: n1 = n2 x

  30. V2 V1 n2 = n1 x = 0.0183 mol SF6 x = 3.40 mol SF6 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. Solution: Molar mass SF6 = 146.07 g/mol 2.67g SF6 146.07g SF6/mol = 0.0183 mol SF6 543.9 m3 2.93 m3 mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6

  31. Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it’s volume to 112.46 Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing with T and P constant, so we will use Avogadro’s law, and the change of conditions form. Solution: V1 = 28.75 L V2 = 112.46 L T = constant P = constant n1 = 1.14 moles H2 n2 = 1.14 moles + ? moles n1 n2 V1 V2 V2 V1 112.46 L 28.75 L = n2 = n1 x = 1.14 moles H2 x mass = moles x molecular mass mass = 4.46 moles x 2.016 g / mol mass = 8.99 g H2 gas n2 = 4.4593 moles = 4.46 moles added mass = 8.99g - 2.30g = 6.69g

  32. Change of Conditions, with no change in the amount of gas ! • = constant Therefore for a change of conditions : • T1 T2 P x V T P1 x V1 P2 x V2 =

  33. Change of Conditions :Problem -I • A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure? • P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm • P2 = ? • V1 = 45.9 L V2 = 3.10 ml = 0.00310 L • T1 = 25 oC + 273 = 298 K • T2 = 155 oC + 273 = 428 K

  34. Change of Condition Problem I : continued P1 x V1 P2 x V2 = T1 T2 P2 (0.00310 L) ( 0.978 atm) ( 45.9 L) = ( 298 K) ( 428 K) ( 428 K) ( 0.978 atm) ( 45.9 L) = 2.08 x 104 atm P2 = ( 298 K) ( 0.00310 L)

  35. Change of Conditions : Problem II • A weather balloon is released at the surface of the earth. If the volume was 100 m3 at the surface ( T = 25 oC, P = 1 atm ) what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 oC and the pressure is 15 mm Hg ? • Initial Conditions Final Conditions • V1 = 100 m3 V2 = ? • T1 = 25 oC + 273.15 T2 = -90 oC +273.15 • = 298 K = 183 K • P1 = 1.0 atm P2 = 15 mm Hg 760 mm Hg/ atm P2= 0.0198 atm

  36. Change of Conditions Problem II: continued P1V1T2 • P1 x V1 P2 x V2 V2 = • V2 = = • V2 = 310.148 m3 = 310. m3or 3.1 times the volume !!! = T1 T2 T1P2 ( 1.0 atm) ( 100 m3) (183 K) (298 K) (0.198 atm)

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