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## Gases CHAPTER 10

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**Gases(Chapter 10)**Resources and Activities • Textbook - chapter 10 • Online practice quiz • Lab activities (1) Molar volume of a gas (wet lab) • POGIL activities: Ideal gas law Partial Pressure • Comprehensive tutorial and animations on Intermolecular forces : http://www.chem.purdue.edu/gchelp/liquids/imf2.html**Animation on gases to view in class:**http://glencoe.com/sites/common_assets/advanced_placement/chemistry_chang9e/animations/chang_7e_esp/gam2s2_6.swf Independent work - students to view animations & interactive activities (4 in total) and write summary notes on each. These summaries are to be included in your notebook http://glencoe.mcgraw-hill.com/sites/0035715985/student_view0/chapter5/animations_center.html# Virtual lab activity (To complete at home and to put in notebook) • Boyle’s Law Objective: Explore the effects of V on P at constant T (for a fixed quantity of gas molecules ). Task: Generate, record and graph data for three gases. Print and Discuss your results. Go to: http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html • Charles Law Objective: Explore the effect of T on V at constant P (for a fixed quantity of gas molecules). Task: generate, record and graph data. Print and discuss your results. Go to: http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/charles_law.html**Measurements on Gases**• Properties of gases • Gases uniformly fill any container • Gases are easily compressed • Gases mix completely with any other gas • Gases exert pressure on their surroundings • Pressure = Force Area**Measurements on Gases**• Measuring pressure • The barometer – measures atmospheric pressure • Inventor – Evangelista Torricelli (1643)**Measurements on Gases**• The manometer – measures confined gas pressure**Measurements on Gases**• Units • mm Hg (torr) • 760 torr = standard pressure • Kilopascal (kPa) • 101.325 kPa = standard pressure • Atmospheres • 1 atmosphere (atm) = standard pressure • STP = 1 atm = 760 torr = 760 mmHg = 101.325 kPa**Measurements on Gases**• Example: Convert 0.985 atm to torr and to kPa. 0.985 atm 760 torr = 749 torr 1 atm 0.985 atm 101.325 kPa = 98.3 kPa 1 atm**The Gas Laws of Boyle, Charles and Avogadro**• Boyle’s Law (Robert Boyle, 1627-1691) • The product of pressure times volume is a constant, provided the temperature remains the same • PV = k**The Gas Laws of Boyle, Charles and Avogadro**• P is inversely related to V • The graph of P versus V is hyperbolic • Volume increases linearly as the pressure decreases • As you squeeze a zip lock bag filled with air (reducing the volume), the pressure increases making it difficult to keep squeezing**The Gas Laws of Boyle, Charles and Avogadro**• At constant temperature, Boyle’s law can be used to find a new volume or a new pressure • P1V1 = k = P2V2 or P1V1 = P2V2 • Boyle’s law works best at low pressures • Gases that obey Boyle’s law are called Ideal gases**The Gas Laws of Boyle, Charles and Avogadro**• Example: A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? • P1 = 1.3 atm • V1 = 27 L • P2 = 3.9 atm • V2 = ? (1.3atm)(27L) = (3.9atm)V2 V2 = 9.0 L**The Gas Laws of Boyle, Charles and Avogadro**• Charles’ Law (Jacques Charles, 1746 – 1823) • The volume of a gas increase linearly with temperature provided the pressure remains constant V = bTV = b V1 = b = V2 T T1 T2 or V1 = V2 T1 T2**The Gas Laws of Boyle, Charles and Avogadro**• Temperature must be measured in Kelvin ( K = °C + 273) • 0 K is “absolute zero”**The Gas Laws of Boyle, Charles and Avogadro**• Example: A gas at 30°C and 1.00 atm has a volume of 0.842L. What volume will the gas occupy at 60°C and 1.00 atm? • V1 = 0.842L • T1 = 30°C (+273 = 303K) • V2 = ? • T2 = 60°C (+273 = 333K)**0.842L = V2**303K 333K V2 = 0.925L**The Gas Laws of Boyle, Charles and Avogadro**• Avogadro’s Law (Amedeo Avogadro, 1811) • For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles, n • V = an V/n = a V1 = a = V2 or V1 = V2 n1 n2n1 n2**The Gas Laws of Boyle, Charles and Avogadro**• Example: A 5.20L sample at 18°C and 2.00 atm contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? • V1 = 5.20L • n1 = 0.436 mol • V2 = ? • n2 = 1.27 mol + 0.436 mol = 1.706 mol**The Gas Laws of Boyle, Charles and Avogadro**5.20 L = V2 0.436 mol 1.706 mol x = 20.3L**The Ideal Gas Law**• Derivation from existing laws V=k V=bT V=an P V = kba(Tn) P**The Ideal Gas Law**Constants k, b, a are combined into the universal gas constant (R), V = nRT or PV = nRT P**The Ideal Gas Law**R = PV using standard numbers will give you R nT STP • P = 1 atm • V = 22.4L • n = 1 mol • T = 273K • Therefore if we solve for R, R = 0.0821 L• atm/mol •K**The Ideal Gas Law**• Example: A sample containing 0.614 moles of a gas at 12°C occupies a volume of 12.9L. What pressure does the gas exert? P = ? V = 12.9L n = 0.614 mol R = 0.0821 L•atm/mol• K T = 12 + 273 = 285K**The Ideal Gas Law**(P)(12.9L) = (0.614mol)(0.0821 L•atm/mol• K)(285K) Hint: Rearrange and re-write (watch out for units in numerator and denominator) P =1.11 atm**The Ideal Gas Law**Solving for new volumes, temperature, or pressure (n remaining constant) • Combined Law P1V1 = nR = P2V2 or P1V1 = P2V2 T1 T2T1 T2**The Ideal Gas Law**• Example: A sample of methane gas at 0.848 atm and 4.0°C occupies a volume of 7.0L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11.0°C? • P1 = 0.848 atm P2 = 1.52 atm • V1 = 7.0L V2 = ? • T1 = 4+273 = 277K T2 = 11+273 = 284K**The Ideal Gas Law**(0.848atm)(7.0L) = (1.52atm)(V2) 277K 284K V2 = 4.0L**Gas Stoichiometry**• Molar volume • One mole of an ideal gas occupies 22.4L of volume at STP • Things to remember • Density = mass volume • n = grams of substance = m molar mass M • PV = mRT M**Gas Stoichiometry**• Example: A sample containing 15.0g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following equation: CO2(s) CO2(g) How big will the balloon be (i.e. what is the volume of the balloon) at 22°C and 1.04 atm, after all of the dry ice has sublimed?**Gas Stoichiometry**• Moles of CO2(s): 15g 1 mol = 0.341 mol CO2(s) 44 g 0.341 mol CO2(s) 1 CO2(g) = 0.341 mol CO2(g) 1 CO2(s) = n**Gas Stoichiometry**P = 1.04 atm V = ? n = 0.341 mol R = 0.0821 L•atm/mol• K T = 22C + 273 = 295K V = 7.94L**Gas Stoichiometry**• Limiting Reactant Example: 0.500L of H2(g) are reacted with 0.600L of O2(g) at STP according to the equation 2H2(g) + O2(g) 2H2O(g) • What volume will the H2O occupy at 1.00 atm and 350°C?**Gas Stoichiometry**• H2: 0.5L 1 mol 2 H2O = 0.0223 mol H2O 22.4L 2 H2 • O2: 0.6L 1 mol 2 H2O = 0.0536 mol H2O 22.4L 1 O2**Gas Stoichiometry**• H2: (1 atm)(V) = (0.0223mol)(0.0821 L•atm/mol• K)(623K) V = 1.14L • O2: (1atm)(V) = (0.0536mol)(0.0821 L•atm/mol• K)(623K) V = 2.74L Limiting reactant and how much volume is produced!**Gas Stoichiometry**• Density/Molar Mass example: A gas at 34°C and 1.75 atm has a density of 3.40g/L. Calculate the molar mass of the gas. D = mass 3.40 = 3.40g V 1L P = 1.75 atm R = 0.0821 L•atm/mol• K V = 1L T = 34 + 273 = 307K m = 3.40 g M = x**Gas Stoichiometry**(1.75atm)(1L) = (3.4g)(0.0821 L•atm/mol• K)(307K) M M = molar mass M = 48.97 g/mol**Dalton’s Law of Partial Pressures (John Dalton, 1803)**• Statement of law • “for a mixture of gases in a container, the total pressure exerted is the sum of the pressures each gas would exert if it were alone” • It is the total number of moles of particles that is important, not the identity or composition of the gas particles.**Dalton’s Law of Partial Pressures**• Derivation • Ptotal = P1 + P2 + P3 … • P1 = n1RT P2 = n2RT P3 = n3RT …. V VV • Ptotal = n1RT + n2RT + n3RT V VV • Ptotal = (n1+n2+n3+…) (RT) V • Ptotal= ntotal (RT) V**Dalton’s Law of Partial Pressures**• Example: Oxygen gas is collected over water at 28C. The total pressure of the sample is 5.5 atm. At 28C, the vapor pressure of water is 1.2 atm. What pressure is the oxygen gas exerting? • Ptotal = PO2 + PH2O • 5.5 = x + 1.2 • X = 4.3 atm**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 • For an ideal gas, the mole fraction (x): x1 = n1 = P1 ntotalPtotal**Dalton’s Law of Partial Pressures**• Example: The vapor pressure of water in air at 28°C is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28°C and 1.03 atm. • XH2O = PH2O = 28.3 torr Pair 783 torr = 0.036**Dalton’s Law of Partial Pressures**• Example: A mixture of gases contains 1.5 moles of oxygen, 7.5 moles of nitrogen and 0.5 moles of carbon dioxide. If the total pressure exerted is 800 mmHg, what are the partial pressures of each gas in the mixture?**Dalton’s Law of Partial Pressures**• Calculate the total number of moles • 1.5 + 7.5 + 0.5 = 9.5 moles • Use mole fractions for each individual gas • PO2: 800mmHg x 1.5 mol = 126 mmHg 9.5 mol • PN2: 800mmHg x 7.5 mol = 632 mmHg 9.5 mol • PCO2: 800mmHg x 0.5 mol = 42 mmHg 9.5 mol**Kinetic Molecular Theory of Gases**• Postulates of the KMT Related to 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 zero • The particles are in constant motion. Collisions of the particles with the walls of the container cause pressure. • Assume that the particles exert no forces on each other • The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas**Kinetic Molecular Theory of Gases**• Explaining Observed Behavior with KMT • P and V (T = constant) • As V is decreased, P increases: • V decrease causes a decrease in the surface area. Since P is force/area, the decrease in V causes the area to decrease, thus increasing the P**Kinetic Molecular Theory of Gases**• P and T (V = constant) • As T increase, P increases • The increase in T causes an increase in average kinetic energy. Molecules moving faster collide with the walls of the container more frequently, and with greater force.**Kinetic Molecular Theory of Gases**• V and T (P = constant) • As T increases, V also increases • Increased T creates more frequent, more forceful collisions. V must increase proportionally to increase the surface area, and maintain P**Kinetic Molecular Theory of Gases**• V and n (T and P constant) • As n increases, V must increase • Increasing the number of particles increases the number of collisions. This can be balanced by an increase in V to maintain constant P**Kinetic Molecular Theory of Gases**• Dalton’s Law of partial pressures • P is independent of the type of gas molecule • KMT states that particles are independent, and V is assumed to be zero. The identity of the molecule is therefore unimportant**Kinetic Molecular Theory of Gases**• Root Mean Square Velocity • Velocity of a gas is dependent on mass and temperature • Velocity of gases is determined as an average • M = mass of one mole of gas particles in kg • R = 8.3145 J/K•mol • joule = kg•m2/s2 • urms = 3RT M