" 1 = # $H vap. Chapter 3 Problems


 Austen Nicholson
 1 years ago
 Views:
Transcription
1 Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x18 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius Clapeyrn equatin n the vapr/liquid bundary: " ln " 1 = $H vap R % $H = R & ln ( ) ln( ) vap 1 1 where is atmspheric pressure, is the triple pint pressure, is the vaprizatin temperature at 1 atm. and is the temperature at the triple pint, which we dn t knw. We als assume that ΔH vap is independent f temperature. On the /L line hwever, d d = "H m where "V l s is the vlume difference between "V l s the slid and liquid phase. his vlume difference is much smaller than the vlume difference between the liquid and the vapr. herefre, the /L line is nearly vertical and as a cnsequence, ~ m. ln( 11325) ln( $ 8.4 %18) As a result "H vap = R $ & 285kJ /ml
2 rblem hase α f species A transfrms int phase β at 55 K and 1 atm. he heat capacities f A,! 5 3,! 4 3 in the structures α and β are, respectively, C = 8.8" 1 and C = 2.38" 1. Using the hird Law and assuming that a and b are pure crystalline elements, calculate the entrpy and enthalpy f transfrmatin at 55 K and K at 1 atm pressure. Frm the third law, we knw that = "! = K; ( if they are pure crystalline elements). " $ = $ C C = ", $, d d = 2.93*1 = 7.95*1! 5! 5 3 = 55K, " $ % % $ =! = 8.37J / ml * K $ G "! ( 55K) = = $ H"! % $ "! " % $! H (55K) = J / ml! " (K) = $ "H $% (K) = "H $% (55K) + & 55 (C,% ' C, )d =117.2J /ml 2
3 rblem in can exist in either the metallic βn phase r in the semicnducting αn phase. Cnsider the phase transitin frm n in the βphase t the αphase. he enthalpy change "H $% assciated with the transitin at 1 atm. is 2.1 kj/ml. βn is in equilibrium with αn at ne atm. pressure and =238.8 K. At 1 atm. the equilibrium temperature is =233.8 K. Estimate the vlume difference between the tw phases. Which ne has the higher density? Explain yur assumptins. Apply the Clausius equatin: d d = " $% "V $% We will assume that Δ and ΔV are independent f temperature and pressure. " $% = "H $% = &8.8 J.K &1.ml &1. Frm the data pints, we have d/d=2x1 6 a/k. hus "V $% = " $% d /d = 4.4 cm 3 /ml. he βphase is denser than the αphase. 3
4 rblem Yu wish t evaprate Au cntacts nt a simple integrated circuit in an evacuated thermal evapratin system. Au has a biling pint f 313K and an entrpy f vaprizatin at 1 atm. f 16.8 J.ml 1.K 1. If the desired vapr pressure f Au in the vacuum chamber fr the depsitin prcess is 17 atm, estimate the temperature t which the Au surce must be heated in the thermal evapratr. tate yur assumptins. Using the Clausius Clapeyrn equatin, d = "H vap d = " vap vap d where we assume R 2 R 2 that the vaprizatin enthalpy (= vap Δ vap ) des nt depend strngly n temperature and pressure ( vap =313K). Integrating the Clausium Clapeyrn equatin, we have: ln( 2 ) " ln( 1 ) = $ vap vap 1 " 1 ' & ). By plugging in this equatin 2 =17 atm, 1 =1 atm, R % 1 2 ( 1 =313K, we btain 2 =1389K 4
5 rblem 1hw that at the triple pint the sublimatin enthalpy f a material is equal t the sum f its melting enthalpy and its vaprizatin enthalpy. 2Belw the triple pint (56 C) the vapr pressure f slid CO 2 is given as: ln( (atm) ) = " with in K. he mlar heat f melting f CO 2 is 8.33 kj. Making sme simple assumptins, calculate the vapr pressure exerted by liquid CO 2 at 25 C and explain why slid CO 2 is referred t as dry ice. ketch the phase diagram f CO 2 and place the =1 atm line and the =25 C line with respect t the triple pint. 1If H is a state functin, we can g frm s t v by tw different rutes with the same ΔH. he tw different rutes are direct sublimatin r melting and then vaprizatin. At the triple pint all the phases can exist in equilibrium and these three prcesses can ccur. herefre "H sv = "H sl + "H l v. 2Using the expressin, we can calculate "H sv = $R % & ln = $R % We btain &1/ "H sv =25.9kJ/ml. Using the infrmatin abut "H sl, we btain "H l v = "H sv $ "H sl =17.6 kj/ml. We nw use the Clausius Clapeyrn equatin t estimate the vapr pressure f the liquid at 25 C using the triple pint as the starting pint: ln & 25C % ( = " )H l *v 1 $ "56C ' R 3K " 1 & % (. $ 217K ' he pressure at the triple pint is btained frm the expressin fr ln() given in the questin: 56 C =5.21 atm. We btain 25C ~75 atm>>1 atm. CO 2 is called dry ice because at rm temperature it sublimates directly frm slid t gas withut prducing a wet liquid phase. he phase diagram is as fllws: 5
6 rblem he ice f an utdr skating rink is at the ambient temperature f 2.5 C. Calculate in a the minimum pressure (applied fr example by the blade f a skate) required t melt the ice. At C, the specific vlume f water is 1 cm 3 /g and that f ice is 1.9 cm 3 /g. he heat f fusin is J/g Ice is less dense than water therefre increasing pressure lwers the melting temperature. he prblem asks t find the pressure where ice will melt at 2.5 C=27.5 K. We knw that at 1 atm and 273 K water and ice are in equilibrium. We use the Clausius equatin t calculate the pressure variatin f the melting pint: d d = "H sl $ " = "H 27.5 sl d % "V sl "V sl J /g $ 27.5' 1 = + ln& ) = a * 337atm. "9 1 "8 m 3 /g % 273 ( 6
7 rblem ure ice, initially at 2 C, is cmpressed adiabatically and reversibly until it melts. 1) ketch n a / phase diagram hw yu wuld find the slutin (i.e. the path f the cmpressin and shw n the diagram hw yu wuld identify the melting pressure and temperature graphically). (1 pints) 2) Calculate the temperature and pressure at which melting starts? (2 pints) slve this prblem yu will need a few apprximatins:  he temperature ranges are small enugh that the mlar vlume f water is apprximately cnstant (be careful, this is nt the same as stating that α=~)  Remember that ln(1x)~x if x<<1 C =2 kj.kg 1.K 1 ; ριχε=92 kg.m 3 ; αιχε=15 ppm.k 1, L m =334 kj.kg 1, ρ water =1 kg.m 3, β~ " & Hint: t slve questin 2, yu may find useful t calculate % ( $ " ' 1) he adiabatic and reversible cmpressin heats the slid. In a  phase diagram, the cmpressin path will intersect the /L equilibrium line. 2) he simplest way f thinking abut this prblem is t think abut finding the intercept between tw curves: the /L equilibrium curve and the () functin during cmpressin. he /L equilibrium curve is given by the Clapeyrn equatin. Yu already knw frm class and prblem sets that it s very steep and clse t being a line. Yu can use the simplest apprximatin t derive the () functin fr the /L equilibrium: d d = "h L and assume that yu are interested in this functin ver a small range f ( v L $ v ) temperatures therefre n the right side f the equatin is apprximately cnstant and equal t m =273 K. Yu can als integrate this equatin rigrusly and yu can cnvince yurselves that if yu are integrating ver a small interval, assuming ln(1x)~x takes yu t the same result. Hence, the equatin fr the /L equilibrium line is: 7
8 / L L ( ) = + m m " 1 1 L " ( ) m ρ water >ρ ice, the slpe is negative, as it shuld be. ( ) where is the pressure at = m : =1 atm. ince Nw we need t find AC (), which is the trajectry in the / plane during the adiabatic cmpressin. ince the cmpressin is adiabatic and reversible, we can write: d = " & % ( d. $ " ' Integrating this equatin will give us AC (). " & Using a Maxwell relatin, we have % ( = " & % ( $ " ' $ "V ' " & We use the recipe discussed in class t find % ( : $ "V ' d = " & % ( dv + " & % ( d = " & % ( [)Vd * +Vd] + " & % ( d $ "V ' $ " ' $ "V ' $ " ' V d = " & % ( )Vd +... $ "V ' I am nt interested in the term in d as what I wrte is sufficient fr me t identify!! $ &. "!V % " & We have % ( )Vd = C $ "V ' hence " & % ( = " & % ( = C $ "V ' $ " ' )V " & One f yu nticed a much easier way t find % (. We have seen in class that $ " ' d = C d "Vd. herefre, =cnst d=, frm which we quickly btain " & % ( = C $ " ' )V In any case, integrating this equatin using the apprximatins allwed in the prblem and again assuming that the interval ver which we integrate is small, we get AC ( ) = + C "V i he prblem is nw reduced t finding the intersectin between tw lines: L AC ()= /L () m m " 1 1 ( ) = C ( L " i $V i). ( i ) where is the pressure when = i = 271 K: =1 atm. ( ) m lving fr, I get =271.5 K and plugging int any f the expressins fr, I get ~225 atm. V 8
Chem 75 February 16, 2017 Exam 2 Solutions
1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml
More informationRecitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol 1 K 1 * 338 K) = mol
Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is 0.175 atm and that f O 2 (g) is 0.250 atm in a mixture f the tw gases. a. What is the mle fractin f each
More informationREVIEW QUESTIONS Chapter 18. H = H (Products)  H (Reactants) H (Products) = (1 x 125) + (3 x 271) = 938 kj
Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)
More informationALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?
Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S
More informationChemistry 114 First Hour Exam
Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)
More informationChapter Outline 4/28/2014. PV Work. PV Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w
Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin
More informationMore Tutorial at
Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,
More informationThermodynamics and Equilibrium
Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,
More informationAutumn 2012 CHEM452B Bruce H. Robinson 322 Gould Hall HW 10(A) Homework 10A KEY (there will not be a 10B) 2
Autumn 0 CHEM45B Bruce H. Rbinsn Guld Hall HW 0(A) Hmewrk 0A KEY (there will nt be a 0B) QA) Let c be the speed f sund in air. he square f the speed f sund, () f the gas with respect t the change in the
More informationCHM112 Lab Graphing with Excel Grading Rubric
Name CHM112 Lab Graphing with Excel Grading Rubric Criteria Pints pssible Pints earned Graphs crrectly pltted and adhere t all guidelines (including descriptive title, prperly frmatted axes, trendline
More informationExperiment #3. Graphing with Excel
Experiment #3. Graphing with Excel Study the "Graphing with Excel" instructins that have been prvided. Additinal help with learning t use Excel can be fund n several web sites, including http://www.ncsu.edu/labwrite/res/gt/gt
More informationCHEM443, Fall 2013, Section 010 Midterm 2 November 4, 2013
CHEM443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal
More informationAP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY
AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy energy can be cnverted frm ne frm t anther but it can be neither
More informationUnit 14 Thermochemistry Notes
Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm
More information1/2 and e0 e s ' 1+ imm w 4 M s 3 πρ0 r 3 m. n 0 ktr. .Also,since n 0 ktr 1,wehave. 4 3 M sπρ 0 r 3. ktr. 3 M sπρ 0
Chapter 6 6.1 Shw that fr a very weak slutin drplet (m 4 3 πr3 ρ 0 M s ), (6.8) can be written as e 0 ' 1+ a r b r 3 where a σ 0 /n 0 kt and b imm w / 4 3 M sπρ 0. What is yur interpretatin f thecnd and
More informationLecture 17: Free Energy of Multiphase Solutions at Equilibrium
Lecture 17: 11.07.05 Free Energy f Multiphase Slutins at Equilibrium Tday: LAST TIME...2 FREE ENERGY DIAGRAMS OF MULTIPHASE SOLUTIONS 1...3 The cmmn tangent cnstructin and the lever rule...3 Practical
More informationSection 5.8 Notes Page Exponential Growth and Decay Models; Newton s Law
Sectin 5.8 Ntes Page 1 5.8 Expnential Grwth and Decay Mdels; Newtn s Law There are many applicatins t expnential functins that we will fcus n in this sectin. First let s lk at the expnential mdel. Expnential
More informationChapter 4 Thermodynamics and Equilibrium
Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 16. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume
More informationCHEM 103 Calorimetry and Hess s Law
CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the
More informationChapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes
Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3
More informationPart One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)
CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationChem 116 POGIL Worksheet  Week 3  Solutions Intermolecular Forces, Liquids, Solids, and Solutions
Chem 116 POGIL Wrksheet  Week 3  Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces
More informationThermodynamics Partial Outline of Topics
Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)
More informationCHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK
CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature
More informationChem 111 Summer 2013 Key III Whelan
Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general
More informationChapters 29 and 35 Thermochemistry and Chemical Thermodynamics
Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany
More informationFall 2013 Physics 172 Recitation 3 Momentum and Springs
Fall 03 Physics 7 Recitatin 3 Mmentum and Springs Purpse: The purpse f this recitatin is t give yu experience wrking with mmentum and the mmentum update frmula. Readings: Chapter.3.5 Learning Objectives:.3.
More informationNAME TEMPERATURE AND HUMIDITY. I. Introduction
NAME TEMPERATURE AND HUMIDITY I. Intrductin Temperature is the single mst imprtant factr in determining atmspheric cnditins because it greatly influences: 1. The amunt f water vapr in the air 2. The pssibility
More information( ) kt. Solution. From kinetic theory (visualized in Figure 1Q91), 1 2 rms = 2. = 1368 m/s
.9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the HeNe gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at
More informationMaterials Engineering 272C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion
Materials Engineering 272C Fall 2001, Lecture 7 & 8 Fundamentals f Diffusin Diffusin: Transprt in a slid, liquid, r gas driven by a cncentratin gradient (r, in the case f mass transprt, a chemical ptential
More informationGeneral Chemistry II, Unit II: Study Guide (part 1)
General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt
More informationv , Michael E. Hanyak, Jr., All Rights Reserved Page 718
v07.08.04 007, Michael E. Hanyak, Jr., All Rights Reserved age 78 Energy Balance Intrductin Functinal frm: Hw t evaluate hmix,,? [ X H 0.60* + 0.40* + Δ () mix Ĥ is mlar enthalpy (kj/gml) f mixture at
More informationExamples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?
NOTES: Thermchemistry Part 1  Heat HEAT TEMPERATURE  Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic
More informationCHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25
CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing
More information4F5 : Performance of an Ideal Gas Cycle 10 pts
4F5 : Perfrmance f an Cycle 0 pts An ideal gas, initially at 0 C and 00 kpa, underges an internally reversible, cyclic prcess in a clsed system. The gas is first cmpressed adiabatically t 500 kpa, then
More informationMath 105: Review for Exam I  Solutions
1. Let f(x) = 3 + x + 5. Math 105: Review fr Exam I  Slutins (a) What is the natural dmain f f? [ 5, ), which means all reals greater than r equal t 5 (b) What is the range f f? [3, ), which means all
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationO C S polar  greater force. H polar greater force. H polar. polar Hbond
hapter 10 : 29, 30, 31, 33, 36, 40, 46, 48, 50, 72, 87, 91, 93, 110 29. a. Lndn e. Lndn b. diplediple f. diplediple c. bnding g. inin d. inin 30. a. inin e. bnding b. Lndn f. diplediple c. Lndn
More informationCHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review
Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system
More informationChapter 3 Homework Solutions
Chapter Hmewrk Slutins. n = ml = 5 C = 98 K = 5 C = 98 K p = atm p = 5 atm. Cpm = (5/)R he entrpy changes fr the heating and cmpressin can e calculated separately and added. Heat at cnstant pressure S
More informationAP Physics Kinematic Wrap Up
AP Physics Kinematic Wrap Up S what d yu need t knw abut this mtin in twdimensin stuff t get a gd scre n the ld AP Physics Test? First ff, here are the equatins that yu ll have t wrk with: v v at x x
More informationChemistry 1A Fall 2000
Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur
More informationCompressibility Effects
Definitin f Cmpressibility All real substances are cmpressible t sme greater r lesser extent; that is, when yu squeeze r press n them, their density will change The amunt by which a substance can be cmpressed
More information18:0021:00, 15 April, 2016 (Total Score: 106 points)
Chemistry II Midterm Exam 18:0021:00, 15 April, 2016 (Ttal Scre: 106 pints) R = 0.08206 atm L/ml K = 8.314 J/ ml K = 0.08314 bar L/ml K F = 96500 C/ml 1. Please answer the fllwing questins. (ttal 16%)
More informationGOAL... ability to predict
THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict
More informationDifferentiation Applications 1: Related Rates
Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm
More informationLecture 12: Chemical reaction equilibria
3.012 Fundamentals f Materials Science Fall 2005 Lecture 12: 10.19.05 Chemical reactin equilibria Tday: LAST TIME...2 EQUATING CHEMICAL POTENTIALS DURING REACTIONS...3 The extent f reactin...3 The simplest
More informationA.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1
A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat
More informationCHAPTER 24: INFERENCE IN REGRESSION. Chapter 24: Make inferences about the population from which the sample data came.
MATH 1342 Ch. 24 April 25 and 27, 2013 Page 1 f 5 CHAPTER 24: INFERENCE IN REGRESSION Chapters 4 and 5: Relatinships between tw quantitative variables. Be able t Make a graph (scatterplt) Summarize the
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6  Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis
More informationWork and Heat Definitions
Wrk and eat Deinitins FL Surrundings: Everything utside system + q q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces  eat, q: transer energy resulting rm
More informationNUMBERS, MATHEMATICS AND EQUATIONS
AUSTRALIAN CURRICULUM PHYSICS GETTING STARTED WITH PHYSICS NUMBERS, MATHEMATICS AND EQUATIONS An integral part t the understanding f ur physical wrld is the use f mathematical mdels which can be used t
More informationPart a: Writing the nodal equations and solving for v o gives the magnitude and phase response: tan ( 0.25 )
+  Hmewrk 0 Slutin ) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the
More information[COLLEGE ALGEBRA EXAM I REVIEW TOPICS] ( u s e t h i s t o m a k e s u r e y o u a r e r e a d y )
(Abut the final) [COLLEGE ALGEBRA EXAM I REVIEW TOPICS] ( u s e t h i s t m a k e s u r e y u a r e r e a d y ) The department writes the final exam s I dn't really knw what's n it and I can't very well
More information188 CHAPTER 6 THERMOCHEMISTRY
188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.
More informationMODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b
. REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but
More informationChapter 4. Unsteady State Conduction
Chapter 4 Unsteady State Cnductin Chapter 5 Steady State Cnductin Chee 318 1 41 Intrductin ransient Cnductin Many heat transfer prblems are time dependent Changes in perating cnditins in a system cause
More informationSemester 2 AP Chemistry Unit 12
Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved
More informationEdexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.
Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied
More informationSession #22: Homework Solutions
Sessin #22: Hmewrk Slutins Prblem #1 (a) In the cntext f amrphus inrganic cmpunds, name tw netwrk frmers, tw netwrk mdifiers, and ne intermediate. (b) Sketch the variatin f mlar vlume with temperature
More informationAdvanced Chemistry Practice Problems
Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with
More informationSPONTANEITY, ENTROPY, AND FREE ENERGY
CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and
More informationCHM 152 Practice Final
CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the
More informationProcess Engineering Thermodynamics E (4 sp) Exam
Prcess Engineering Thermdynamics 42434 E (4 sp) Exam 9329 ll supprt material is allwed except fr telecmmunicatin devices. 4 questins give max. 3 pints = 7½ + 7½ + 7½ + 7½ pints Belw 6 questins are given,
More informationUniversity Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:
University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6  Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk
More informationLim f (x) e. Find the largest possible domain and its discontinuity points. Why is it discontinuous at those points (if any)?
THESE ARE SAMPLE QUESTIONS FOR EACH OF THE STUDENT LEARNING OUTCOMES (SLO) SET FOR THIS COURSE. SLO 1: Understand and use the cncept f the limit f a functin i. Use prperties f limits and ther techniques,
More informationProblem Set 1 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001
LEEL ROBLEMS rblem Set Slutins. MI ressr Gerbrand Ceder Fall rblem. Gas is heating in a rigid cntainer rm 4 C t 5 C U U( ) U( ) W + Q (First Law) (a) W Since nly wrk is pssible & since the cntainer is
More informationWhen a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q
Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: ( We finished using the lecture slides frm lecture 14) In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is
More information, which yields. where z1. and z2
The Gaussian r Nrmal PDF, Page 1 The Gaussian r Nrmal Prbability Density Functin Authr: Jhn M Cimbala, Penn State University Latest revisin: 11 September 13 The Gaussian r Nrmal Prbability Density Functin
More informationUnit 11 Solutions Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?
Name: Perid: Unit 11 Slutins Guided Ntes Mixtures: What is a mixture and give examples? What is a pure substance? What are allys? What is the difference between hetergeneus and hmgeneus mixtures? Slutins:
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationSolutions to the Extra Problems for Chapter 14
Slutins t the Extra Prblems r Chapter 1 1. The H 670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +
More informationChE 471: LECTURE 4 Fall 2003
ChE 47: LECTURE 4 Fall 003 IDEL RECTORS One f the key gals f chemical reactin engineering is t quantify the relatinship between prductin rate, reactr size, reactin kinetics and selected perating cnditins.
More informationFlipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal MassSpring System
Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal MassSpring System A Hrizntal MassSpring System is where a mass is attached t a spring, riented hrizntally, and then placed
More informationEntropy, Free Energy, and Equilibrium
Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs
More informationInstructions: Show all work for complete credit. Work in symbols first, plugging in numbers and performing calculations last. / 26.
CM ROSEHULMAN INSTITUTE OF TECHNOLOGY Name Circle sectin: 01 [4 th Lui] 02 [5 th Lui] 03 [4 th Thm] 04 [5 th Thm] 05 [4 th Mech] ME301 Applicatins f Thermdynamics Exam 1 Sep 29, 2017 Rules: Clsed bk/ntes
More informationES201  Examination 2 Winter Adams and Richards NAME BOX NUMBER
ES201  Examinatin 2 Winter 20032004 Adams and Richards NAME BOX NUMBER Please Circle One : Richards (Perid 4) ES20101 Adams (Perid 4) ES20102 Adams (Perid 6) ES20103 Prblem 1 ( 12 ) Prblem 2 ( 24
More informationCHAPTER 6  ENERGY. Approach #2: Using the component of mg along the line of d:
SlutinsCh. 6 (Energy) CHAPTER 6  ENERGY 6.) The f.b.d. shwn t the right has been prvided t identify all the frces acting n the bdy as it mves up the incline. a.) T determine the wrk dne by gravity
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring
More informationBIT Chapters = =
BIT Chapters 170 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride
More informationCHEM 116 Electrochemistry at NonStandard Conditions, and Intro to Thermodynamics
CHEM 116 Electrchemistry at NnStandard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam
More informationL a) Calculate the maximum allowable midspan deflection (w o ) critical under which the beam will slide off its support.
ecture 6 Mderately arge Deflectin Thery f Beams Prblem 61: Part A: The department f Highways and Public Wrks f the state f Califrnia is in the prcess f imprving the design f bridge verpasses t meet earthquake
More informationReview Problems 3. Four FIR Filter Types
Review Prblems 3 Fur FIR Filter Types Fur types f FIR linear phase digital filters have cefficients h(n fr 0 n M. They are defined as fllws: Type I: h(n = h(mn and M even. Type II: h(n = h(mn and M dd.
More informationCHAPTER 13 Temperature and Kinetic Theory. Units
CHAPTER 13 Temperature and Kinetic Thery Units Atmic Thery f Matter Temperature and Thermmeters Thermal Equilibrium and the Zerth Law f Thermdynamics Thermal Expansin Thermal Stress The Gas Laws and Abslute
More informationHeat Effects of Chemical Reactions
eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = 9690
More informationWe can see from the graph above that the intersection is, i.e., [ ).
MTH 111 Cllege Algebra Lecture Ntes July 2, 2014 Functin Arithmetic: With nt t much difficulty, we ntice that inputs f functins are numbers, and utputs f functins are numbers. S whatever we can d with
More informationChapter 17 Free Energy and Thermodynamics
Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics
More informationHeat is energy and is measured in joules (J) or kilojoules (kj). The symbol for heat is H.
Causes f Change Calrimetry Hw Des Energy Affect Change? Heat vs. Temerature HEAT TEMPERATURE Definitin: Deends n: Examles: Heat is energy and is measured in jules (J) r kiljules (kj). The symbl fr heat
More informationDispersion Ref Feynman VolI, Ch31
Dispersin Ref Feynman VlI, Ch31 n () = 1 + q N q /m 2 2 2 0 i ( b/m) We have learned that the index f refractin is nt just a simple number, but a quantity that varies with the frequency f the light.
More information6.3: Volumes by Cylindrical Shells
6.3: Vlumes by Cylindrical Shells Nt all vlume prblems can be addressed using cylinders. Fr example: Find the vlume f the slid btained by rtating abut the yaxis the regin bunded by y = 2x x B and y =
More informationIntermolecular forces Intermolecular Forces van der Waals forces Iondipole forces Dipoledipole forces
Intermlecular frces frces that exist between mlecules determines many f the physical prperties f mlecular liquids and slids lead t deviatins frm ideal gas behavir as well Mlecular Cmparisns f Liquids and
More information**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!**
Tpics lists: UVVis Absrbance Spectrscpy Lab & ChemActivity 36 (nly thrugh 4) I. UVVis Absrbance Spectrscpy Lab Beer s law Relates cncentratin f a chemical species in a slutin and the absrbance f that
More informationPressure And Entropy Variations Across The Weak Shock Wave Due To Viscosity Effects
Pressure And Entrpy Variatins Acrss The Weak Shck Wave Due T Viscsity Effects OSTAFA A. A. AHOUD Department f athematics Faculty f Science Benha University 13518 Benha EGYPT Abstract:The nnlinear differential
More informationhttps://goo.gl/eaqvfo SUMMER REV: HalfLife DUE DATE: JULY 2 nd
NAME: DUE DATE: JULY 2 nd AP Chemistry SUMMER REV: HalfLife Why? Every radiistpe has a characteristic rate f decay measured by its halflife. Halflives can be as shrt as a fractin f a secnd r as lng
More informationFlipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal MassSpring System
Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal MassSpring System A Hrizntal MassSpring System is where a mass is attached t a spring, riented hrizntally, and then placed
More informationFive Whys How To Do It Better
Five Whys Definitin. As explained in the previus article, we define rt cause as simply the uncvering f hw the current prblem came int being. Fr a simple causal chain, it is the entire chain. Fr a cmplex
More informationPhysics 321 Solutions for Final Exam
Page f 8 Physics 3 Slutins fr inal Exa ) A sall blb f clay with ass is drpped fr a height h abve a thin rd f length L and ass M which can pivt frictinlessly abut its center. The initial situatin is shwn
More informationComputational modeling techniques
Cmputatinal mdeling techniques Lecture 2: Mdeling change. In Petre Department f IT, Åb Akademi http://users.ab.fi/ipetre/cmpmd/ Cntent f the lecture Basic paradigm f mdeling change Examples Linear dynamical
More information