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Gases CHAPTER 10 Powerpoint from Melissa Brophy’s website (http://teacherweb.com/TX/McNeilHS/brophy/ photo7.aspx) and modified for our classhttp://teacherweb.com/TX/McNeilHS/brophy/ photo7.aspx

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Gases (Chapter 10) Resources and Activities Textbook - chapter 10 & ppt file Lab activities (1) Exploring Gas Behavior (with LabQuest Units ) (2) Molar volume of a gas (wet lab) POGIL activities: Ideal gas law Partial Pressure Chemtour videos from W.W. Norton (chapter 6 : ideal gas law; Dalton’s law; molecular speed) http://www.wwnorton.com/college/chemistry /chemistry3/ch/06/chemtours.aspxhttp://www.wwnorton.com/college/chemistry /chemistry3/ch/06/chemtours.aspx

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Animation on gases to view: http://glencoe.com/sites/common_assets/advanced_placement/chemistry_ chang9e/animations/chang_7e_esp/gam2s2_6.swf Independent work - view animations & interactive activities (4 in total) and write summary notes on each. These summaries are to be included in your portfolio http://glencoe.mcgraw- hill.com/sites/0035715985/student_view0/chapter5/animations_center.html#

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Activities and Problem set for chapter 10 (due date_______) TextBook ch. 10– all sections required for regents (in part), SAT II and AP exams Lab activities: (2) POGILS (2) Complete the chapter 10 packet part (a) Characteristics of Gases ( sample, practice and integrative exercises - show work) and part (b) AP Chemistry Gas Law Practice Do all chapter 10 GIST and Visualizing concepts problems - write out questions and answers & show work. (NB Photocopy of questions is also acceptable)

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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

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Measuring pressure –The barometer – measures atmospheric pressure Inventor – Evangelista Torricelli (1643) Measurements on Gases

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–The manometer – measures confined gas pressure Measurements on Gases

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–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

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Example: Convert 0.985 atm to torr and to kPa. Measurements on Gases 0.985 atm760 torr 1 atm = 749 torr 0.985 atm101.325 kPa 1 atm = 98.3 kPa

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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

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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

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–At constant temperature, Boyle’s law can be used to find a new volume or a new pressure P 1 V 1 = k = P 2 V 2 or P 1 V 1 = P 2 V 2 –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

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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? P 1 = 1.3 atm V 1 = 27 L P 2 = 3.9 atm V 2 = ? The Gas Laws of Boyle, Charles and Avogadro (1.3atm)(27L) = (3.9atm)V 2 V 2 = 9.0 L

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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 = bT V = b V 1 = b = V 2 T T 1 T 2 or V 1 = V 2 T 1 T 2

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The Gas Laws of Boyle, Charles and Avogadro Temperature must be measured in Kelvin ( K = °C + 273) –0 K is “absolute zero”

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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? V 1 = 0.842L T 1 = 30°C (+273 = 303K) V 2 = ? T 2 = 60°C (+273 = 333K) The Gas Laws of Boyle, Charles and Avogadro

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0.842L = V 2 303K 333K V 2 = 0.925L

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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 V 1 = a = V 2 or V 1 = V 2 n 1 n 2

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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? V 1 = 5.20L n 1 = 0.436 mol V 2 = ? n 2 = 1.27 mol + 0.436 mol = 1.706 mol

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The Gas Laws of Boyle, Charles and Avogadro 5.20 L = V 2 0.436 mol 1.706 mol x = 20.3L

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The Ideal Gas Law Derivation from existing laws V=k V=bT V=an P V = kba(Tn) P

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The Ideal Gas Law Constants k, b, a are combined into the universal gas constant (R), V = nRT or PV = nRT P

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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

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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 Latm/mol K T = 12 + 273 = 285K

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The Ideal Gas Law (P)(12.9L) = (0.614mol)(0.0821 Latm/mol K)(285K) Hint: Rearrange and re-write (watch out for units in numerator and denominator) P =1.11 atm

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The Ideal Gas Law Solving for new volumes, temperature, or pressure (n remaining constant) Combined Law P 1 V 1 = nR = P 2 V 2 or P 1 V 1 = P 2 V 2 T 1 T 2

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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? P 1 = 0.848 atmP 2 = 1.52 atm V 1 = 7.0LV 2 = ? T 1 = 4+273 = 277KT 2 = 11+273 = 284K

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The Ideal Gas Law (0.848atm)(7.0L) = (1.52atm)(V 2 ) 277K284K V 2 = 4.0L

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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

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Gas Stoichiometry Example: A sample containing 15.0g of dry ice (CO 2 (s)) is put into a balloon and allowed to sublime according to the following equation: CO 2 (s) CO 2 (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?

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Gas Stoichiometry Moles of CO 2 (s): 15g 1 mol = 0.341 mol CO 2 (s) 44 g 0.341 mol CO 2 (s) 1 CO 2 (g) = 0.341 mol CO 2 (g) 1 CO 2 (s) = n

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Gas Stoichiometry P = 1.04 atm V = ? n = 0.341 mol R = 0.0821 Latm/mol K T = 22C + 273 = 295K V = 7.94L

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Gas Stoichiometry Limiting Reactant Example: 0.500L of H 2 (g) are reacted with 0.600L of O 2 (g) at STP according to the equation 2H 2 (g) + O 2 (g) 2H 2 O(g) What volume will the H 2 O occupy at 1.00 atm and 350°C?

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Gas Stoichiometry H 2 : 0.5L 1 mol 2 H 2 O = 0.0223 mol H 2 O 22.4L 2 H 2 O 2 : 0.6L 1 mol 2 H 2 O = 0.0536 mol H 2 O 22.4L 1 O 2

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Gas Stoichiometry H 2 : (1 atm)(V) = (0.0223mol)(0.0821 Latm/mol K)(623K) V = 1.14L O 2 : (1atm)(V) = (0.0536mol)(0.0821 Latm/mol K)(623K) V = 2.74L Limiting reactant and how much volume is produced!

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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 Latm/mol K V = 1L T = 34 + 273 = 307K m = 3.40 g M = x

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Gas Stoichiometry (1.75atm)(1L) = (3.4g)(0.0821 Latm/mol K)(307K) M M = molar mass M = 48.97 g/mol

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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.

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Dalton’s Law of Partial Pressures Derivation –P total = P 1 + P 2 + P 3 … –P 1 = n 1 RT P 2 = n 2 RT P 3 = n 3 RT …. VV V –P total = n 1 RT + n 2 RT + n 3 RT V V V –P total = (n 1 +n 2 +n 3 +…) (RT) V –P total = n total (RT) V

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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? P total = P O2 + P H2O 5.5 = x + 1.2 X = 4.3 atm

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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): x 1 = n 1 = P 1 n total P total

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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. X H2O = P H2O = 28.3 torr P air 783 torr = 0.036

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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?

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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 P O2 : 800mmHg x 1.5 mol = 126 mmHg 9.5 mol P N2 : 800mmHg x 7.5 mol = 632 mmHg 9.5 mol P CO2 : 800mmHg x 0.5 mol = 42 mmHg 9.5 mol

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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

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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

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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.

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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

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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

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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

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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/Kmol –joule = kgm 2 /s 2 –u rms = 3RT M

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Kinetic Molecular Theory of Gases Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0°C. M O2 = 32g/mol 32g 1 kg = 0.032 kg 1000g T = 0 + 273 = 273K R = 8.3145 J/Kmol

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Kinetic Molecular Theory of Gases U rms = 3(8.3145 kgm 2 s -2 K -1 mol -1 )(273 K) = 461ms -1 0.032 kgmol -1 Answer 461ms -1 or 461 m/s Recall 1 Joule = 1 J = 1 kg m 2 /s 2 = 1 kg m 2 s -2

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Kinetic Molecular Theory of Gases Mean Free Path – distance between collisions –Average distance a molecule travels between collisions 1 x 10 -7 m for O 2 at STP

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Effusion and Diffusion Effusion –Movement of a gas through a small opening into an evacuated container (vacuum) –Graham’s law of effusion Rate of effusion for gas 1 = √M 2 Rate of effusion for gas 2 √M 1

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Effusion and Diffusion Example: How many times faster than He would NO 2 gas effuse? M NO2 = 46.01 g/mol larger mass M He = 4.003 g/mol effuses slower Rate He = 46.01 = He effuses 3.39 Rate NO 2 4.003 times faster

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Real Gases and van der Waals Equation (Johannes van der Waals, 1873) Volume –Real gas molecules do have volume –Volume available is not 100% of the container volume n= number of moles b = is an empirical constant, derived from experimental results

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Real Gases and van der Waals Equation Ideal: P = nRT V Real: P = nRT V-nb

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Real Gases and van der Waals Equation Pressure –Molecules of real gases do experience attractive forces a = proportionality constant determined by observation of the gas P obs = P ideal – a(n/V) 2

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Real Gases and van der Waals Equation Combining to derive van der Waals equation P obs = nRT - n 2 a V-nb V 2 And then rearranging… (P obs + n 2 a/V 2 )(V-nb) = nRT

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Chemistry in the Atmosphere Composition of the Troposphere –The atmosphere is composed of 78% N 2, 21% O 2, 0.9%Ar, and 0.03% CO 2 along with trace gases. –The composition of the atmosphere varies as a function of distance from the earth’s surface. Heavier molecules tend to be near the surface due to gravity.

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Chemistry in the Atmosphere –Upper atmospheric chemistry is largely affected by ultraviolet, x ray, and cosmic radiation emanating from space. The ozone layer is especially reactive to ultraviolet radiation –Manufacturing and other processes of our modern society affect the chemistry of our atmosphere. Air pollution is a direct result of such processes.

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