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Gases

Gases. Pressure. Gases evenly mix and expand to fill their containers Exert pressure on the container, as well as objects in the container Pressure changes with several factors, including altitude and temperature Barometer – instrument used to measure atmospheric pressure

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Gases

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  1. Gases

  2. Pressure • Gases evenly mix and expand to fill their containers • Exert pressure on the container, as well as objects in the container • Pressure changes with several factors, including altitude and temperature • Barometer – instrument used to measure atmospheric pressure • Invented in 1643 by Evangelista Torricelli • Higher the level of mercury, the higher the air pressure • Atmospheric pressure is based on the weight of the air • Low pressure usually accompanies a storm • High altitude means less air, so lower pressure

  3. Units of Pressure • Instruments that measure pressure (manometers) usually use a column of mercury, so pressure is often expressed in millimeters of mercury (mm Hg) • This is sometimes referred to as a torr • Pressure is also measured in standard atmospheres (atm) • 1 atmosphere = 760 mm Hg = 760 torr • Since pressure is defined as force per unit area (Pressure = Force/area), the units of pressure in SI are newtons/meter squared (N/m2) or pascals (Pa) • 1 atmosphere = 101325 Pascals

  4. The Gas Laws of Boyle, Charles, and Avogadro • Several gas laws relating various properties of gases to one another • Including pressure, volume, temperature, moles of gas particles

  5. Boyle’s Law • Developed by Robert Boyle who performed the first quantitative experiments • Shows that there is an inverse relationship between the pressure and volume of a gas, given constant temperature • Pressure * Volume = a constant (PV=k) • As instruments and measuring techniques have become more sophisticated, scientists have found that Boyle’s law only precisely holds at low pressures • Gases that follow Boyle’s law are called ideal gases • Unless told differently, consider gases to be ideal • Typically use Boyle’s Law to determine the new volume of a gas when the pressure changes • Pressure1 * Volume1 = Pressure2 * Volume2 • P1V1 = P2V2

  6. Boyle’s Law • Example: Given a 1.53L sample of sulfur dioxide at a pressure of 5.6*103Pa. If the pressure is changed to 1.5*104Pa at a constant temperature, what will be the new volume of gas? • P1 = 5.6*103Pa P2 = 1.5*104Pa • V1 = 1.53L V2 = ? • P1V1 = P2V2 • V2 =P1V1/P2 = 5.6*103Pa(1.53)/1.5*104Pa • V2 = .57L

  7. Charles’s Law • Jacques Charles – first person to fill a balloon with hydrogen gas and make a solo balloon flight • in 1787 found that the volume of gas at constant pressure increases linearly with the temperature of the gas • Found that the volume of all gases equal zero at -273.2°C or absolute zero (0 Kelvin) • Below 0K, gases would have a negative volume, which is impossible, so no temperature lower than absolute zero • Absolute zero has never been reached. The lowest temperature in a laboratory is 0.000001K • Volume = proportionality constant * temperature in Kelvins • V=bT • Temperature must be in Kelvins to use Charles’s Law • Much like Boyle’s law, Charles’s law can be rewritten and combined for two situations • Volume1/Temperature1 = Volume2/Temperature2 • V1/T1=V2/T2

  8. Charles’s Law • Example: A sample of gas at 15°C and 1 atm has a volume of 2.58L. What volume will this gas occupy at 38°C and 1 atm? • V1/T1=V2/T2 • V1 = 2.58L V2 = ? • T1 = 15°C T2 = 38°C • Convert temperatures to Kelvin by adding 273 • T1 = 288K T2 = 311K • V2 = T2V1/T1 = 311(2.58)/288 = 2.79L

  9. Avogadro’s Law • Stated by Amedeo Avogadro in 1811, equal volumes of gases at the same temperature and pressure contain the same number of particles • Volume = proportionality constant * number of moles • V=an • Can be rearranged as V1/n1 = V2/n2

  10. Avogadro’s Law • Example: Suppose we have a 12.2L sample containing 0.50mol O2 at a pressure of 1 atm and a temperature of 25°C. If all of this O2 were converted to ozone O3 at the same temperature and pressure, what would be the volume of the ozone? • Start by writing the balanced chemical equation • 3O2 2O3 • Calculate the moles of ozone • .50mol O2 2mol O3 = .33mol O3 3mol O2

  11. Avogadro’s Law • n1 = .50mol n2 = .33mol • V1 = 12.2L V2 = ? • V1/n1 = V2/n2 • V2 = (n2/n1)V1 = (.33/.50)12.2 = 8.1L

  12. Ideal Gas Law • Boyle’s, Charles’s, and Avogadro’s Laws can be combined into one equation, known as the Ideal Gas Law • Pressure * Volume = moles of gas * universal gas constant * Temperature in Kelvins • PV=nRT • R = 0.08206 (L atm)/(K mol) • This law really only works at low pressure and high temperature, or under ideal circumstances • Unless told otherwise, assume gases are ideal

  13. Ideal Gas Law • Example: A sample of hydrogen gas (H2) has a volume of 8.56L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample. • P = 1.5 atm V = 8.56L • T = 0 + 273 = 273K n = ? • R = 0.08206 (L atm)/(K mol) • PV=nRT • n=PV/(RT) = 1.5(8.56)/(0.08206*273) = .57mol

  14. Gas Stoichiometry • We can perform stoichiometry problems using gases • For ideal gases, 1 mole of gas at standard temperature and pressure (273.2K and 1 atm, abbreviated STP) has a volume of 22.42L • If the conditions in a problem are not at STP, we must use the ideal gas law to calculate the volume

  15. Gas Stoichiometry • Example: A sample of nitrogen gas has a volume of 1.75L at STP. How many moles of N2 are present? 1.75L N2 1mol N2 = 7.81*10-2mol N2 22.42L N2

  16. Gas Stoichiometry • Example: Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of 152g CaCO3 by the reaction CaCO3(s) CaO(s) + CO2(g)

  17. Gas Stoichiometry 152g CaCO3 1mol CaCO3 1mol CO2 22.42L CO2 = 34.1L CO2 100.09g CaCO3 1mol CaCO3 1molCO2

  18. Gas Stoichiometry • Example: A sample of methane gas (CH4) having a volume of 2.80L at 25°C and 1.65atm was mixed with a sample of oxygen gas (O2) having a volume of 35.0L at 31°C and 1.25atm. The mixture was then ignited to form carbon dioxide (CO2) and water (H2O). Calculate the volume of CO2 formed at a pressure of 2.50atm and a temperature of 125°C. • Write and balance the chemical equation CH4(g) + O2(g) CO2(g) + H2O(g) CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

  19. Gas Stoichiometry • Convert temperatures to Kelvin by adding 273 • TCH4= 25 + 273=298K • TO2=31 + 273=304K • TF = 125 + 273=398K

  20. Gas Stoichiometry • Determine the limiting reagent by calculating the number of moles of each reactant. Use the ideal gas law. • PV = nRT, so n = (PV)/(RT) nCH4 = (1.65*2.80)/(.08206*298) = .198mol CH4 nO2 = (1.25*35.0)/(.08206*304) = 1.75mol O2

  21. Gas Stoichiometry • Since 1mol CH4 requires 2mol of O2, • 1.75mol O2 requires .875mol CH4 • .198molCH4 would require .378molO2 • CH4 is the limiting reagent • Calculate the mol of CO2 produced .189mol CH4 1mol CO2 = .189mol CO2 1mol CH4

  22. Gas Stoichiometry • Use the ideal gas law to calculate the volume of CO2 produced • PV=nRT, so V= (nRT)/P • V = (.189*.08206*398)/2.50 = 2.47L CO2

  23. Molar Mass of a Gas • To calculate the molar mass of a gas us the following formula: • Molar mass = (density * universal gas constant * Temperature in Kelvins) / Pressure in atm) • MM = dRT/P • Units for molar mass should be in g/mol

  24. Molar Mass of a Gas • Example: The density of a gas was measured at 1.50atm and 27°C and found to be 1.95g/L. Calculate the molar mass of the gas. • Convert temperature to Kelvin • T = 27 + 273 = 300K • MM = dRT/P = (1.95 * .08206 * 300)/1.50 • MM = 32.0g/mol

  25. Dalton’s Law of Partial Pressures • For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone • Total Pressure = the sum of all pressures • Ptotal = òP • Pressure can also be calculated by adding up all of the moles of gas and using the ideal gas law • Ptotal = ntotal(RT/V)

  26. Dalton’s Law of Partial Pressures • Example: Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46L He at 25°C and 1.0atm and 12L O2 at 25°C and 1.0atm were pumped into a tank with a volume of 5.0L. Calculate the partial pressure of each gas and the total pressure in the tank at 25°C.

  27. Dalton’s Law of Partial Pressures • First calculate the number of moles of each gas using the ideal gas law • n=PV/RT • nHe = (1.0*46)/(.08206*298) = 1.9mol He • nO2 = (1.0*12)/(.08206*298) = .49mol O2

  28. Dalton’s Law of Partial Pressures • Use the ideal gas law to calculate the pressure of each gas in the tank • P=nRT/V • PHe = (1.9*.08206*298)/5.0 = 9.3atm • PO2 = (.49*.08206*298)/5.0 = 2.4atm • Use Dalton’s Law of Partial Pressures to get the total pressure • Ptotal = PHe + PO2= 9.3 + 2.4 = 11.7atm

  29. Dalton’s Law of Partial Pressures • Mole fraction – the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture • Symbolized by the Greek lowercase letter chi (χ) • χ1 = P1/Ptotal = n1/ntotal • The partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure • P1 = χ1 * Ptotal

  30. Dalton’s Law of Partial Pressures • Example: The partial pressure of oxygen was observed to be 156torr in air with a total atmospheric pressure of 743torr. Calculate the mole fraction of O2 present. • χO2 = PO2 /Ptotal = 156/743 = .210

  31. Dalton’s Law of Partial Pressures • Example: The mole fraction of nitrogen is the air is .7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760.torr. • PN2 = χN2 * Ptotal = .7808*760. = 593torr

  32. Gas Collection Over Water • A mixture of gas results whenever a gas is collected by displacement of water • Some water vapor is always present • Water is constantly being cycled between liquid and gaseous water. • The rate of water becoming vapor is the same as the rate of vapor becoming liquid for a given temperature • The pressure of the water vapor in the container remains constant, assuming constant temperature • This is called the vapor pressure of water

  33. The Kinetic Molecular Theory of Gases • Kinetic molecular theory (KMT) – explains the properties of ideal gases • The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero) • The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas • The particles are assumed to exert no forces on each other; they are assumed to neither attract nor repel each other • The average kinetic energy of a collision of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas

  34. The Kinetic Molecular Theory of Gases • Relating to Boyle’s Law • Decrease in volume causes more collisions with the container • Increase in pressure • Relating to the Ideal Gas Law • Increase in temperature causes the particles to move faster, therefore having more collisions with the container • Increase in pressure

  35. The Kinetic Molecular Theory of Gases • Relating to Charles’s Law • Increase in temperature causes the particles to move faster, therefore having more collisions with the container • To keep pressure constant, the container must expand • Relating to Avogadro’s Law • Increase in the number of particles present (number of moles) causes the particles to move faster, therefore having more collisions with the container • To keep pressure constant, the container must expand • Relating to Dalton’s Law • Because the gases are independent of one another (don’t react) and take up no volume, the identities of the gases don’t matter when calculating total pressure from a mixture of gases

  36. The Kinetic Molecular Theory of Gases • The Meaning of Temperature • Temperature in Kelvin serves as an index of the random motions of particles • The higher the temperature, the greater the random motion • The greater the kinetic energy of the particles • Average kinetic energy = 3/2 * universal gas constant * temperature in Kelvins • KE = 3/2RT

  37. The Kinetic Molecular Theory of Gases • Root Mean Square – a special kind of average of the velocities of gas particles • Symbolized by urms • Calculated by multiplying 3 * the universal gas constant * temperature (in Kelvin) divided by the mass of one mole of the gas (in kilograms) and then taking the square root • urms = √(3RT/M) • For these problems, we have a different form of R • R = 8.3145J/(K*mol) • Gives us units of meters per second (m/s)

  38. The Kinetic Molecular Theory of Gases • Example: Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C. • T= 25+273 = 298K • M = 4.00g He 1kg He = .00400kg/mol He mol 1000g He • urms = √(3RT/M) = √(3*8.3145*298/.00400) = 1.36*103m/s

  39. Effusion and Diffusion • Diffusion – describes the mixing of gases as they spread throughout a container • Rate of diffusion is the rate of the mixing • Effusion – describes the passage of gas into an evacuated chamber • Rate of effusion is the speed at which the gas is transferred into the container

  40. Effusion and Diffusion • Graham’s Law of Effusion – the rate of effusion of a gas is inversely proportional to the square root of the mass of it’s particles Rate of effusion for gas 1 = √M2 Rate of effusion for gas 2 = √M1 • M is the molar mass of the gas

  41. Effusion and Diffusion • Example: Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors. Rate of effusion for H2 = √ UF6= √ 352.02 = 13.2 Rate of effusion for UF6 = √ H2 = √ 2.016

  42. Real Gases • No gas is truly an ideal gas • Certain gas come close at low pressure and/or high temperatures • Johannes van der Waals developed an equation that can be used to predict the behavior of real gases in 1873

  43. Real Gases • Van der Waals equation: • [Pobs+a(n/V)2] * (V-nb) = nRT • Pobs is the observed pressure • a is a proportionality constant • b is an empirical (experimental) constant • a(n/V)2 is a pressure correction • nb is a volume correction

  44. Chemistry in the Atmosphere • Atmosphere – the mixture of gases surrounding the Earth’s surface • Mostly N2 and O2 with Ar, CO2, H2O, Ne, CH4, Kr, H2, NO, and Xe • Broken down into several levels based on changes in temperature • Bottom layer is called the troposphere • Influenced by human activities • Lead to air pollution and acid rain • Air pollution is caused primarily by the burning of fossil fuels which produce CO, CO2, NO, and NO2

  45. Chemistry in the Atmosphere • N2 and O2 produce NO in the engine which is oxidized in the air to form NO2 • N2 + O2→ NO → NO2 • This breaks down into NO and free O atoms which combine with O2 to form ozone (O3) • NO2 → NO + O • O + O2 → O3

  46. Chemistry in the Atmosphere • Ozone is highly reactive and can either react with other pollutants or break up to form an excited molecule and an excited O atom. The excited O atom reacts with water to form two hydroxyl radicals (OH). The hydroxyl radicals can then combine with NO2 to produce nitric acid (HNO3) or with unburned hydrocarbons to produce chemicals that cause the eyes to burn and water and are harmful to the respiratory system. This is referred to as photochemical smog • OH + NO2 → HNO3

  47. Chemistry in the Atmosphere • Acid rain is produced from burning coal • Coal contains sulfur (S) which, when burned with oxygen (O2), produces sulfur dioxide (SO2) • S + O2 →SO2 • Sulfur dioxide reacts with oxygen to produce sulfur trioxide (SO3) which reacts with water to form sulfuric acid (H2SO4) • SO2 + O2→SO3 • SO3 + H2O →H2SO4 • Acid rain is corrosive to living things and building materials

  48. Chemistry in the Atmosphere • Anti-pollution measures are designed to reduce unburned fuels and minimize the production of nitric acid or use scrubbers to remove sulfur dioxide from exhaust • Scrubbers blow powdered limestone (calcium carbonate) into the combustion chamber which breaks down into lime (calcium oxide) and carbon dioxide • The lime reacts with the sulfur dioxide to form calcium sulfite which is buried in landfills

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