Presentation on theme: "Gases CHAPTER 10 Powerpoint from Melissa Brophy’s website (http://teacherweb.com/TX/McNeilHS/brophy/photo7.aspx) and modified for our class."— Presentation transcript:
1 Gases CHAPTER 10Powerpoint from Melissa Brophy’s website (http://teacherweb.com/TX/McNeilHS/brophy/photo7.aspx) and modified for our class
2 Gases (Chapter 10) Resources and Activities Textbook - chapter 10 & ppt fileLab activities(1) Exploring Gas Behavior(with LabQuest Units )(2) Molar volume of a gas (wet lab)POGIL activities:Ideal gas lawPartial PressureChemtour videos from W.W. Norton (chapter 6 : ideal gas law; Dalton’s law; molecular speed)
3 Animation on gases to view: Independent work - view animations & interactive activities (4 in total) and write summary notes on each. These summaries are to be included in your portfolio
4 Activities and Problem set for chapter 10 (due date_______) Complete the chapter 10 packet part (a) Characteristics of Gases ( sample, practice and integrative exercises - show work) and part (b) AP Chemistry Gas Law PracticeDo all chapter 10 GIST and Visualizing concepts problems - write out questions and answers & show work. (NB Photocopy of questions is also acceptable)TextBook ch. 10– all sections required for regents (in part), SAT II and AP examsLab activities: (2)POGILS (2)
5 Measurements on Gases Properties of gases Gases uniformly fill any containerGases are easily compressedGases mix completely with any other gasGases exert pressure on their surroundingsPressure = ForceArea
6 Measurements on Gases Measuring pressure The barometer – measures atmospheric pressureInventor – Evangelista Torricelli (1643)
7 Measurements on GasesThe manometer – measures confined gas pressure
8 Measurements on Gases Units mm Hg (torr) Kilopascal (kPa) Atmospheres 760 torr = standard pressureKilopascal (kPa)kPa = standard pressureAtmospheres1 atmosphere (atm) = standard pressureSTP = 1 atm = 760 torr = 760 mmHg = kPa
9 Measurements on Gases Example: Convert 0.985 atm to torr and to kPa.
10 The Gas Laws of Boyle, Charles and Avogadro Boyle’s Law (Robert Boyle, )The product of pressure times volume is a constant, provided the temperature remains the samePV = k
11 The Gas Laws of Boyle, Charles and Avogadro P is inversely related to VThe graph of P versus V is hyperbolicVolume increases linearly as the pressure decreasesAs you squeeze a zip lock bag filled with air (reducing the volume), the pressure increases making it difficult to keep squeezing
12 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 pressureP1V1 = k = P2V2or P1V1 = P2V2Boyle’s law works best at low pressuresGases that obey Boyle’s law are called Ideal gases
13 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 atmV1 = 27 LP2 = 3.9 atmV2 = ?(1.3atm)(27L) = (3.9atm)V2V2 = 9.0 L
14 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 constantV = bT V = b V1 = b = V2T T T2or V1 = V2T T2
15 The Gas Laws of Boyle, Charles and Avogadro Temperature must be measured in Kelvin( K = °C + 273)0 K is “absolute zero”
16 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 = 0.842LT1 = 30°C (+273 = 303K)V2 = ?T2 = 60°C (+273 = 333K)
18 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, nV = an V/n = aV1 = a = V or V1 = V2n n n n2
19 The Gas Laws of Boyle, Charles and Avogadro Example: A 5.20L sample at 18°C and 2.00 atm contains 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 = 5.20Ln1 = molV2 = ?n2 = 1.27 mol mol = mol
20 The Gas Laws of Boyle, Charles and Avogadro 5.20 L = V20.436 mol molx = 20.3L
21 The Ideal Gas Law Derivation from existing laws P V = kba(Tn) V=k V=bT V=anPV = kba(Tn)
22 The Ideal Gas Law V = nRT or PV = nRT P Constants k, b, a are combined into the universal gas constant (R),V = nRT or PV = nRTP
23 The Ideal Gas Law R = PV using standard numbers will give you R nT STP P = 1 atmV = 22.4Ln = 1 molT = 273KTherefore if we solve for R, R = L• atm/mol •K
24 The Ideal Gas LawExample: A sample containing moles of a gas at 12°C occupies a volume of 12.9L. What pressure does the gas exert?P = ?V = 12.9Ln = molR = L•atm/mol• KT = = 285K
25 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
26 The Ideal Gas Law Combined Law P1V1 = nR = P2V2 or P1V1 = P2V2 Solving for new volumes, temperature, or pressure (n remaining constant)Combined LawP1V1 = nR = P2V or P1V1 = P2V2T T T T2
27 The Ideal Gas LawExample: A sample of methane gas at 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 = atm P2 = 1.52 atmV1 = 7.0L V2 = ?T1 = = 277K T2 = = 284K
28 The Ideal Gas Law(0.848atm)(7.0L) = (1.52atm)(V2)277K KV2 = 4.0L
29 Gas Stoichiometry Molar volume Things to remember volume M One mole of an ideal gas occupies 22.4L of volume at STPThings to rememberDensity = massvolumen = grams of substance = mmolar mass MPV = mRTM
30 Gas StoichiometryExample: 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?
31 Gas Stoichiometry Moles of CO2(s): 15g 1 mol = 0.341 mol CO2(s) 44 g 0.341 mol CO2(s) 1 CO2(g) = mol CO2(g)1 CO2(s) = n
32 Gas Stoichiometry P = 1.04 atm V = ? n = 0.341 mol R = L•atm/mol• KT = 22C = 295KV = 7.94L
33 Gas StoichiometryLimiting Reactant Example: L 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?
35 Gas StoichiometryH2:(1 atm)(V) = (0.0223mol)( L•atm/mol• K)(623K)V = 1.14LO2:(1atm)(V) = (0.0536mol)( L•atm/mol• K)(623K)V = 2.74LLimiting reactant and how much volume is produced!
36 Gas StoichiometryDensity/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 = gV LP = 1.75 atm R = L•atm/mol• KV = 1L T = = 307Km = 3.40 g M = x
37 Gas Stoichiometry (1.75atm)(1L) = (3.4g)(0.0821 L•atm/mol• K)(307K) M = molar massM = g/mol
38 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.
39 Dalton’s Law of Partial Pressures DerivationPtotal = P1 + P2 + P3 …P1 = n1RT P2 = n2RT P3 = n3RT ….V V VPtotal = n1RT + n2RT + n3RTV V VPtotal = (n1+n2+n3+…) (RT)VPtotal = ntotal (RT)
40 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 + PH2O5.5 = x + 1.2X = 4.3 atm
41 Dalton’s Law of Partial Pressures Mole FractionThe ratio of the number of moles of a given component in a mixture to the total number of moles in the mixtureFor an ideal gas, the mole fraction (x):x1 = n = P1ntotal Ptotal
42 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 = torrPair torr= 0.036
43 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?
44 Dalton’s Law of Partial Pressures Calculate the total number of moles= 9.5 molesUse mole fractions for each individual gasPO2: 800mmHg x 1.5 mol = 126 mmHg9.5 molPN2: 800mmHg x 7.5 mol = 632 mmHgPCO2: 800mmHg x 0.5 mol = 42 mmHg
45 Kinetic Molecular Theory of Gases Postulates of the KMT Related to Ideal GasesThe particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be zeroThe 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 otherThe average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas
46 Kinetic Molecular Theory of Gases Explaining Observed Behavior with KMTP 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
47 Kinetic Molecular Theory of Gases P and T (V = constant)As T increase, P increasesThe 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.
48 Kinetic Molecular Theory of Gases V and T (P = constant)As T increases, V also increasesIncreased T creates more frequent, more forceful collisions. V must increase proportionally to increase the surface area, and maintain P
49 Kinetic Molecular Theory of Gases V and n (T and P constant)As n increases, V must increaseIncreasing the number of particles increases the number of collisions. This can be balanced by an increase in V to maintain constant P
50 Kinetic Molecular Theory of Gases Dalton’s Law of partial pressuresP is independent of the type of gas moleculeKMT states that particles are independent, and V is assumed to be zero. The identity of the molecule is therefore unimportant
51 Kinetic Molecular Theory of Gases Root Mean Square VelocityVelocity of a gas is dependent on mass and temperatureVelocity of gases is determined as an averageM = mass of one mole of gas particles in kgR = J/K•moljoule = kg•m2/s2urms = 3RTM
52 Kinetic Molecular Theory of Gases Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0°C.MO2 = 32g/mol32g 1 kg = kg1000gT = = 273KR = J/K•mol
54 Kinetic Molecular Theory of Gases Mean Free Path – distance between collisionsAverage distance a molecule travels between collisions1 x m for O2 at STP
55 Effusion and Diffusion Movement of a gas through a small opening into an evacuated container (vacuum)Graham’s law of effusionRate of effusion for gas 1 = √M2Rate of effusion for gas 2 √M1
56 Effusion and Diffusion Example: How many times faster than He would NO2 gas effuse?MNO2 = g/mol larger massMHe = g/mol effuses slowerRate He = = He effuses 3.39Rate NO times faster
57 Real Gases and van der Waals Equation (Johannes van der Waals, 1873) VolumeReal gas molecules do have volumeVolume available is not 100% of the container volumen= number of molesb = is an empirical constant, derived from experimental results
58 Real Gases and van der Waals Equation Ideal: P = nRTVReal: P = nRTV-nb
59 Real Gases and van der Waals Equation PressureMolecules of real gases do experience attractive forcesa = proportionality constant determined by observation of the gasPobs = Pideal – a(n/V)2
60 Real Gases and van der Waals Equation Combining to derive van der Waals equationPobs = nRT - n2aV-nb V2And then rearranging…(Pobs + n2a/V2)(V-nb) = nRT
61 Chemistry in the Atmosphere Composition of the TroposphereThe atmosphere is composed of 78% N2, 21% O2, 0.9%Ar, and 0.03% CO2 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.
62 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 radiationManufacturing 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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