Lecture 2 2024 PDF - Chemistry

Lecture #2, p. 1 Lecture 2: Announcements Today: Brown 10.1 Characteristics of Gases 10.2 Pressure 10.3 The Gas Laws 10.4 The Ideal Gas Equation...

Lecture #2, p. 1 Lecture 2: Announcements Today: Brown 10.1 Characteristics of Gases 10.2 Pressure 10.3 The Gas Laws 10.4 The Ideal Gas Equation 10.5 Gas Mixtures and Partial Pressures 10.6 The Kinetic-Molecular Theory of Gases 10.7 Molecular Effusion and Diffusion 10.8 Real Gases: Deviations from Ideal Behavior Problem Set 1: Due by exercise tomorrow (upload on Moodle as pdf) Tutorials: After Exercise 2, tutorials are finished Useful Info Sheets: What are they? Pages of information given to you at exam Where are they? Posted on Moodle Chemistry  Lecture #2, p. 2 Lecture 3 Next Week: Brown 4.1 General Properties of Aqueous Solutions 4.2 Precipitation Reactions 4.3 Acids, Bases, and Neutralization Reactions 4.4 Oxidation–Reduction Reactions 4.5 Concentrations of Solutions Chemistry  Lecture #2, p. 3 Review Nucleus In Lecture 1, we introduced a lot of definitions and nomenclature protons and neutrons ∼10−4 Å Elements, compounds, mixtures Atoms, atomic structure Atomic number, mass number, atomic mass Isotopes, atomic weight, periodic table Molecules, formulas, structural formulas, representations Ions, ionic compounds Electron cloud Then, we started to discuss chemical reactions 1–5 Å Change one substance into another 1 Å = 1×10−10 m Chemical equations, stoichiometry, balancing equations Formula weights, mole, Avogadro’s number, molar mass Chemistry  Lecture #2, p. 4 Example Chemical Reaction Important for “hydrogen economy” shownwith model 1 mole is 6.02 x 1023 “items” Avogadro’s Number, NA Note Mass is conserved as itmustb Molar mass can be determined from formula weight Chemistry  Lecture #2, p. 5 Types of Chemical Reactions Combustion Reactions Produce flame typically by reacting with O! In addition to hydrogen combustion above, we can burn hydrocarbons Combustion reactions power the world, but produce CO2 Example: CH4 + 2 O2 CO2 + 2 H 2O Chemistry 1 Combustion of methane CitynaturalgasErdgas  Lecture #2, p. 6 Types of Chemical Reactions C # + O! (&) → CO! (&) Burning coal N! & + 3 H! (&) → 2 NH" (&) Fixation of nitrogen 2 H! O(/) → 2 H! & + O! (&) Electrolysis of water CaCO" # → CaO # + CO! (&) Cement / concrete Chemistry can late if concrete is secondmostusedsubstanceonEarthbehindHad Concreteproduction on ofconcrete  Lecture #2, p. 7 Importanttopicforindustryapplicationsandeverydaylife Today’s Topic: Gases Important topic for engineers t Three states of matter: solid, liquid, vapor Example: H2O Gases have similar physical properties Even if they have different chemical properties Physical properties of gases: Expand to fill their container Can be compressed Form homogeneous mixtures asoline Molecules occupy tiny fraction of volume I Chemistry a  Lecture #2, p. 8 Common Gases at Room Temperature Chemistry  Lecture #2, p. 9 Parameters to Describe Gases? How can we describe physical state of gas? For simple gas, we need only 4 parameters: Parameter Variable Volume ! Extensive properties Amount " (number of moles) Depend on size of sample Temperature # Intensive properties Pressure $ Do not depend on size of sample Another example: density ' N (mass/volume) $ = = ≡ Pa ( m! Chemistry SI unit of force Newton N SI unit of pressure Pasial Pa  Lecture #2, p. 10 Origin of Pressure? Gas in a Molecules randomly moving container Hit any surface and cause force per unit area Atmospheric pressure? Pressure due to Earth’s gravity Pulling molecules gas Consider column of air, 1 m2 in area owardEarth Molecules $ ≈ 10,000 kg % = 9.8 m2/s strike surface ! = $% ≈ 1×10! N +"#$ ≈ 1×10! N/m% ≈ 100 kPa +"#$ ≈ 1 bar Estimate (see below) IIx Chemistry Itar 105Pa g duet acceleration Earth'sgravity  Lecture #2, p. 11 Measuring Atmospheric Pressure? Galileo Barometer barometer tool to measure !!"# $ &' ()' !!" = = = ) =%ℎ % % % !$%& & Evangelista Torricelli ( ≡ = density = 13.5 g⁄cm# for Hg 1608–1647 ) (wikipedia.org) (%ℎ' !!" = = ('ℎ Hydrostatic pressure % Hg At sea level ℎ = 760 mm Hg !$%& = 760 mm Hg In different units: !$%& = 760 torr = 1 atm = 101.325 kPa = 1.01325 bar Chemistry wewilluserhoe assymbolfordensitytobeconsistent withyourotherclasses ButBrownusesd  Lecture #2, p. 12 Measuring Relative Pressure? Manometer tool to measure ! relative to !$%& !$%& !"$' = !$%& + ( ' ℎ Also used to measure blood pressure “120 over 80” Hg in mm Hg Chemistry systolic over diastolic Imax 1min  Lecture #2, p. 13 Nowhowdothe4 parameters V n T P relatetoeachother Wewillconsiderin pairs P–V Relationship Boyle’s Law Keep ! and " constant If we halve volume " " ⟶ pressure doubles Maksense $% = constant or constant $ = % 0 0 First gas law 0 ! 0 1# ! Law: states what happens I Chemistry Note RobertBoyle16271691 wasfirsttochangeonevariableto determinehowanothervariablechanges i e howthetwovariables arerelated Keyideainscienceengineering Note2 Whenwebreathe ourbody is exploiting Boyle'sLaw  Lecture #2, p. 14 V–T Relationship Charles’s Law Keep ! and " constant # is linear with $ & & I ie Rescale −273.15°C = 0 K ∆$ of 1.0°C = 1.0 K # = constant ( $ or L # = constant 0 0 $ ir −273.15 0 !(°C) 0 !(K) Second gas law Chemistry I 1 Note p units Bit just JacquesCharles 17461823  Lecture #2, p. 15 V–n Relationship Avogadro’s Law Avogadro’s Hypothesis " Keep !, # constant Moreover, constant is approximately the same $ is linear with % for all gases! $ = constant ' % Equal volumes of two different gases at same or !, # contain same $ number of molecules = constant % 0 Third gas law 0 ! Wow! Chemistry AmedeoAvogadro 17761856  Lecture #2, p. 16 Toestablish a normal PT we definethe Standard Temperature and Pressure Ideal Gas STP $ = 0 °C = 273.15 K Hypothetical gas that follows: Threegaslaw We set STP as " = 1 atm = 101.325 kPa 1 #∝ , " !,# Molar volume at STP = 22.41 L !$ Boyle'sLaw #∝$. #∝ " Good approximation for any gas !,$ #∝!. Even though the mass of 1 mole $,# of different gases varies Iii !$ Strictly true for an “ideal gas” #=- " ! ≡ “Gas Constant” Chemistry where wesettheproportionality constantas R  Lecture #2, p. 17 or I def I PORTANT Ideal Gas Law (IGL) Assumptions III as !" =$%& Molecules in gas do not interact Volume that molecules occupy is much - ≡ Gas Constant smaller than the total volume m3⋅Pa - = 8.314 molecule mol⋅K (not to scale) 5 its or J container - = 8.314 mol⋅K L⋅atm Note: Amount of gas depends on number - = 0.08206 mol⋅K of gaseous species present L⋅torr - = 62.36 Ex: A ⟶ B + C ⇒ Pressure doubles mol⋅K if #, $ are constant Chemistry Decomi  Lecture #2, p. 18 Using Ideal Gas Law? Very useful ! !" = %&' ! is constant Many problem variations Common on exams ! Type 1: Given three of four variables; need missing variable Strategy: rearrange ideal gas law for desired variable $%& $%& !' !' ! = ' = $ = & = ' ! %& $% Type 2: Two variables remain constant; two other variables are changing # ! $! Ex1: $, & constant ⟶ !' = $%& is constant ⟶ !! '! = !" '" ⟶ !" = $" # &' #! #" #! %" Ex2: $, ' constant ⟶ = is constant ⟶ = ⟶ !" = % $ %! %" %! Chemistry  Lecture #2, p. 19 Using Ideal Gas Law? !" = %&' Type 3: 1 variable remains constant; three other variables are changing !" !! "! !" "" "! #" Ex: ! constant ⟶ = !$ is constant ⟶ #! = #" ⟶ %$ = %% "" #! # Problem: Balloon of V = 6.0 L at P = 101.3 kPa rises until P = 45.6 kPa. Temperature falls from 22°C to −21°C. What is the final volume? Knowns: ! = fixed Unknown: "" = ? #! = 6.0 L Strategy: !" !! "! !" "" '! = 101.3 kPa # = !$ is constant ⟶ #! = #" '" = 45.6 kPa !! #" Solve for unknown ⟶ &$ = &% ,! = 22 °C !" #! ," = −21 °C Plug in knowns ⟶ &$ = 11 L Chemistry IMPORTANT Temperatur alwaysbeinKelvin must RememberCharles'sL onlymadesensein K  Lecture #2, p. 20 Other Uses of the IGL Density, ! Molecular Weight, #! ! % g Rearranging IGL: " = &' Or molar mass, mol ! % " ( )! = &' ( )! Rearranging density expression: Mw gmol mol g g !%& Units? ⋅ = (! = L mol L $ = density $ ⇒ ! = ' (! Thus, we can use IGL to %& determine the density if we know molecular weight Chemistry i or vice versa density canget of thegas wen a s at Insistent Note depends on dgf.it fmqas  Lecture #2, p. 21 Consequence of IGL for Gaseous Mixtures Partial pressure Consider mixture containing i different gaseous species IGL ⇒ Total pressure, !!"! , just depends on total number of moles, "!"! , of gas totalnumber "!"! = "# + "$ + ⋯ + "% me '( '( '( '( '( HE samecontai !!"! = "!"! = "# + "$ + ⋯ + "% = "# + "$ + ⋯ + "% ) ) ) ) ) !!"! = !# + !$ + ⋯ + !% Dalton’s law of partial pressures !% ≡ partial pressure of species i Chemistry Makessense behaves En onlythenumberofgasmoleculesmatters  Lecture #2, p. 22 Mixtures and Mole Fraction Note: This implies that "% !% = ! = +% !!"! "!"! !"! &! where +% ≡ &"#" Mole fraction Fraction of total moles in mixture that are species i Chemistry  Lecture #2, p. 23 But Why Does IGL Explain Gas Behavior? Need theory ⇒ Kinetic-Molecular Theory of Gases Model ⇒ Gaseous species are in constant motion with kinetic energy !k 1 EI fiin !! = &' " & ≡ mass of molecule ' ≡ velocity of molecule 2 Ek Kite Exercise are randomly moving occupy negligible volume Gaseous species collide with each other but do not interact have average !k that depends on T Chemistry kinetictheorystatesthat  Lecture #2, p. 24 Kinetic-Molecular Theory of Gases Gas species transfer energy upon collisions A specific molecule in the gas may move faster or slower... But average !k is constant for a given absolute temperature Does not even depend on type of molecule, only T(K). Explains P,T relationship at molecular level! P caused by molecular collisions with container walls Depends on number and force of such collisions If T(K) is doubled, average !k doubles Iii volume Would expect that P is also doubled ⇒ " ∝ $(&) Meaning of T? ⇒ Average kinetic energy, !k , of molecules Chemistry  Lecture #2, p. 25 Butsofar wehavejustdiscussed average velocities Howfastdoindividual moleculesmove Velocity Distributions Important: molecules move with a range of velocities Distribution depends on T and molecular mass that T constant IMPORTAN con p e i Chemistry fast it iii I Limited resource  Lecture #2, p. 26 To describethe velocitydistribution wecandefineseveralphysicallymeaningfulvelocities Special Velocities !"& !$% !!"# 3+$ (!"# = ,$ iii I Chemistry Why is vomstotherightofthepeak inthevelocitydistribution Because kineticenergy Ek isImy  Lecture #2, p. 27 Diffusion Spread of substance through space Ums 500m Gases at atmospheric pressure diffuse much slower than !!"# Because molecules have 1010 collisions per second At Mean free path Average distance between collisions Air at sea level (" = 1 atm) ⇒ 60 nm! This reduces the actual distance traveled But for air at 100 km altitude ⇒ 10 cm! Chemistry ffg.gg gf gggggyggeyggggg Collisions randomize its directionof motion ggggggggqggqgq.gg gggµ  Lecture #2, p. 28 Nonidealities in Gases IGL incredibly useful! But real gases not always ideal At high ! ⇒ Volume occupied by molecules no longer negligible At low " ⇒ Molecules start to interact (i.e. “stick” to each other) Chemistry  Lecture #2, p. 29 Nonidealities in Gases: Van der Waals Equation canfind nonifalities gas Chemistry meaning tointeractions sowe add nawterm to P Them.ie suraifraP.Ie Ier sIhetE www.prgistsf.rd.ee thanthemoleculesexperience Someofthevolumeis nowoccupiedbyothermolecules Thy s fern if gfr  Lecture #2, p. 30 What We Learned Variables needed to describe gases Pressure, barometers, and manometers Gas laws (Boyle’s law, Charles’s law, Avogadro’s law) Given P and T, equal volume of any gas contains same number of molecules Standard temperature and pressure (STP) Ideal gas law (IGL) Partial pressures, mole fraction Kinetic-molecular theory of gases Velocities and velocity distributions Molecular diffusion, collisions, and mean free path Nonidealities in gases (van der Waals equation) Chemistry HS22 NEXT TIME Reactions in Water