You MUST sign the honor pledge:
|
|
- Marsha Logan
- 5 years ago
- Views:
Transcription
1 CHEM 3411 MWF 9:00AM Spring 2012 Physical Chemistry I Final Exam, Version A (Dated: May 4, 2012 Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only if the correct answer is in marked boxes, when provided. (Do not write in the small boxes in the right margin; they are for grading purposes only. You MUST sign the honor pledge: Part Problem Max.Pts Actual Score I / II / III / IV / TOTAL: 120 On my honor, I pledge that I will not give or receive aid in examinations; I will not attempt to gain prior access to examinations; I will not represent the work of another as my own; and I will avoid any activity which will encourage others to violate their own pledge of honor. Your Signature: You may find the following information useful: Constant Symbol Value Units Speed of light in vacuum c E+10 cm sec 1 Electronic charge e E 10 esu e E 19 C (Coulombs Avogadro s number N A E+23 molecules mole 1 Gas Constant R J K 1 mole 1 R E 2 L bar K 1 mole 1 R E 2 L atm K 1 mole 1 R E+1 L Torr K 1 mole 1 Boltzmann constant k B E 23 J, K 1 Electron mass m e E 4 amu Proton mass m p amu Neutron mass m n amu Planck Constant h E 27 erg sec h E 34 J sec h E 27 erg sec Bohr radius a E-11 m Atomic mass unit amu E 24 g Electron volt ev E 19 J ev E 12 erg Debye D E 30 C m Calorie cal J Rydberg Constant R E+5 cm 1 A sheet of possibly useful formulas is provided at the end. 1
2 THIS PAGE INTENTIONALLY LEFT BLANK 2
3 Part I: Thermodynamics: First Law 1. Concept Questions. Try to be concise, and correct (5 pts a. What is the equation of state for a gas described by the virial expansion? Provide a qualitative explanation for how this equation corrects for the ideal gas equation. (5 pts b. What is the first law of thermodynamics, and how does it allow you to define internal energy? 3 :
4 2. First Law: The machinery for gases. A sample consisting of 3.3 moles of ideal gas molecules with C P,m = 19.5JK 1 mol 1 is initially at 162 kpa and 323 K. It then undergoes a reversible adiabatic expansion until its temperature is 247 K. (5 pts a. What is the initial volume? (5 pts b. What is q for this process? Answer: Answer: 4
5 (5 pts c. What is the final pressure? (4 pts d. What is w for this process? Answer: Answer: 5 :
6 Part II: Thermodynamics: Second & Third Law 3. Concept Questions. Try to be concise, and correct (5 pts a. What does the phase rule tell you about the phase transition of a typical pure liquid to a solid? Indicate this on a phase diagram being careful to label your axes. (5 pts b. What is the second law of thermodynamics, which thermodynamic variable(s does it help define, and how? 6 :
7 4. Heat Engines. Consider a heat engine that operates between 780.0K and 340.0K. (5 pts a. What is the maximum efficiency of the engine? Answer: (5 pts b. Calculate the maximum work that can be done by each 1.0kJ of heat supplied by the hot source? Answer: 7
8 (5 pts c. How much heat is discharged into the cold sink in a reversible process for each 1.0kJ supplied by the hot source? Answer: (5 pts d. What is the change in entropy in a reversible process for each 1.0kJ supplied by the hot source? Answer: 8 :
9 Part III: The Thermodynamics of Chemical Systems 5. Concept Questions. Try to be concise, and correct (5 pts a. Suppose you have a nonideal mixture of two components, A and B, which satisfies the regular solution model. Let ǫ αβ be the interaction energy for the contact of component α with component β. If 1 2 (ǫ AA +ǫ BB < ǫ AB, draw a sketch of the excess Gibb s free energy of mixing as a function of the composition of component A? (Make sure to label the axes! (5 pts b. Suppose that you find that electrons are flowing from the solution in beaker A to beaker B. What can you say about the chemical potentials in both beakers? What else must also be happening between the two beakers to enable this flow? 9 :
10 6. Reactions. Suppose that a dissociating reaction A(g 2B(g + C(g takes place in a closed container with fixed pressure equal to 1.00 bar. The reaction is described by the chemical equilibrium expression, ln K = A + B/T + C/T 2 where A = 1.13, B = 932 K and C = K 2, at all temperatures between 300K and 500K. (5 pts a. What is the standard entropy of this reaction at K? Answer: (5 pts b. What is the standard enthalpy of this reaction at K? Answer: 10
11 (5 pts c. Suppose that the system is ideal, and initiated with 1.0 moles of A at constant temperature, K. Estimate the chemical composition of A at equilibrium. Answer: (5 pts d. Suppose that the system is nonideal, and initiated with 1.0 moles of A at constant temperature, K. The activity coefficient of components A and B are 1.00 and the activity coefficient of component C is equal to 1.11 times its mole fraction. Estimate the chemical composition of A at equilibrium. Answer: 11 :
12 Part IV: Kinetics & Beyond 7. Concept Questions. Try to be concise, and correct (5 pts a. Suppose that you have a sequential reaction process, A B C. If the rate of the first step is much faster than that for the second step, what can you say about how the concentration of B and C with time? (5 pts b. What does the prefactor in the transition state theory rate forumula have to do with the underlying assumptions? 12 :
13 8. Arrehnius Rates (10 pts a. The rate constant for the decomposition of a certain subtance is dm 3 mol 1 s 1 at 300.0K and dm 3 mol 1 s 1 at 330.0K. Whate are the Arrhenius parameters of the reaction? Answer: Answer: 13
14 (5 pts b. How much would you need to lower the activation energy of this reaction in order to double the rate constant? Answer: (5 pts c. Suppose, instead that a bimolecular gas reaction takes place at constant volume. How do you expect that the prefactor in the Arhenius rate will change (if at all if the temperature is doubled? Justify. 14 :
15 (Possibly Useful Formulas WARNING: Use at your own risk PV = nrt PV m = RTZ x a = na n T ( ( Z = 1 + BP + CP 2 + Z = 1 + B 1 + C V m P = nrt a ( n 2 1 Z = (a/rt V nb V 1 (b/v m V m 2 1 V m + U = q + w du = dq + dw H = q P U = w ad U = q V ( Vf V w = P ex V i dv w = P ex V w = nrt ln f V i H = U + PV P = ( U V S,N ( α 1 V V T V = ( H P S,N du = π T dv + C V dt du = PdV + TdS P ( dh = (µc P dp + C P dt κ T 1 V dh = V dp + TdS V P T C V = ( U C T V P = ( H C T P P C V = nr C V = 3nR 2 V T c = constant c = C V,m R PV γ = constant γ = 1 + R C V,m i νr i R i j νp j P j r H = j νp j f H (P j i νr i f H (R i ds = dqrev T S = T 2 T 1 C p T dt r C P,m = j νp j C P,m (P j i νr i C P,m (R i S vap = Hvap T b 85 J mol S surr = Hsys T ( ǫ w q h = 1 Tc T h A = U TS G = U + PV TS da = dw max dg = dw add,max r S m = j νp j S m(p j i νr i S m(r i r G = j νp j f G (P j i νr i f G (R i ds dq 0 ds T U,V 0 du S,V 0 ds H,P 0 dh S,P 0 da T,V 0 dg T,P 0 ( ( G ( = H P G(P T T T P 2 f P i = V P G(P f P i = NRT ln f P i G(P = G + NRT ln ( f f = φp ln φ = P Z 1 dp P 0 P dp = trss dt trsv µ = G N = G m dµ = S m dt + V m dp p = p e ( Vm P dp dt = trshm T trsv m ln ( ( P P = vaph m ( 1 1 R T T RT ( [ P = P + fus H m fus ln ( ] T V T dw = γdσ P in = P out + 2γ r h = 2γ ρgr cos θ c = γsg γ sl γ lg w ad = γ sg + γ lg γ sl Continued on NEXT Page 15
16 V j = ( V n j P,T,n µ j = ( G n j G = P,T,n i n dµ iµ B i dµ A = n A nb dg = V dp SdT + i µ idn i P A = x A PA P B = x B K B Π = ( n B V RT µ = µ + RT ln ( P (P µ P A (l = µ A (l + RT ln A µ PA A (l = µ A (l + RT ln (x A mix G = nrt(x A ln x A + x B ln x B + βx A x B mix S = nr(x A ln x A + x B ln x B mix G = mix G (id + nrt(x A ln γ A + x B ln γ B µ = µ + RT ln(a a A = P A PA ( b a = γx a = γ B PA B b y A = x A x B = vaph R ( r G G ξ dln(k dt ( T T 2 P,T = rh RT 2 PB +(P A P B x A ( T T 2 T = K b b x B = fush R r G = j νp j µ(p j i νr i µ(r i ln(k 2 /K 1 = rh R ( 1 T 2 1 T 1 a B = P B K B n a l( ad = n b l( db F = C P + 2 T = K f b Q P j P a j νp j P i R a i νr i ln x B = fush R ( 1 1 T T RT ln(k = r G ( solv G = z2 i e2 N A 8πǫ 0 r i 1 1 ǫ r ( µ = µ id + RT ln γ γ ± = (γ + γ 1 2 log(γ ± = Z + Z AI 1 2 A = I = 1 2 i z2 bi i b I k ( b i k(x,m + = 1 k(x,m 2+ = 3 k(x 2,M + = 3 k(x 2,M 2+ = 4 b c 2 = 1 3 v2 ( Z = σ c N rel V r G = νfe E = E RT ln Q F = N νf Ae v 2 e Mv2 /(2RT P( v = ( 3 M 2 e Mv2 /(2RT f(v = 4π ( M 2πRT 2πRT λ c Z λ = kt 2σP Z w = ( N crel V 4 ( T z P = ( 2πRT M 1/2 m A 0 t κ = 1νk 3 Bλ cn κ effusion = Z w A 0 D = 3 cλ 1 κ E = 1λ cc 3 V,M[A] η = 1Nmλτ 3 J z = κ ( T J z z = D ( ( N J = κ T z E J z z = η ( v x z η u η e Ea/(RT G = 1/R κ = (GL/A Λ m κ/c λ ion κ/c ion Λ m = ν + λ + + ν λ Λ m = Λ m κc 1 2 κ = A + BΛ M Λ m = αλ m 1 Λ m = 1 Λ M + ΛmC k a(λ m 2 K a = α2 C 1 α F E = ZeE = Ze φ/l f F = 6πηa s = (Ze/f F E u s/e = (Ze/(6πηa λ = ZuF f F = ( µ = ( C x P,T RT J = D ( C s = Df F D = urt C x P,T x P,T RT Zf F Λ m = ( ν + Z 2 +D + + ν Z 2 D F 2 RT D = kt f F C = J t x C = D 2 C t x 2 C = D 2 C ν C kc t x 2 x C(x,t = n 0 A(πDt 1/2 e x2 /(4Dt C(r,t = n 0 e r2 /(4Dt x 2 1 8(πDt 3/2 2 = (2Dt 1 2 x = st D = λ2 2τ ν(t = [A] t ν(t = 1 [A] a t = + 1 c [C] t ν(t = k[a] n ν(t = k[a] m [B] n k 2 = σ ( 1 8k B T 2 πµ k 2 = ( RT P k K [A] = [A] 0 e akt t 1/2 = ln 2 k = c rel σ K = ( k B T hν k 2 = ( ( kt RT h p NA q C q A q B 1 1 [A] [A] 0 = akt k = Ae E A/(RT e E/(RT k = κν k 2 = σ N A ce E/(RT e S/R e H/(RT 16 k 2 = kak d k a+k d k d = 4πR D
You MUST sign the honor pledge:
Chemistry 3411 MWF 9:00AM Spring 2009 Physical Chemistry I Final Exam, Version A (Dated: April 30, 2009 Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only if the
More informationYou MUST sign the honor pledge:
CHEM 3411 MWF 9:00AM Fall 2010 Physical Chemistry I Exam #2, Version B (Dated: October 15, 2010) Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only if the correct
More informationChemistry 3411 MWF 9:00AM Spring Physical Chemistry I - Exam #2, Version A (Dated: February 27, 2009)
Chemistry 3411 MWF 9:00AM Spring 2009 Physical Chemistry I - Exam #2, Version A (Dated: February 27, 2009) 1 Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only
More informationExam Thermodynamics 12 April 2018
1 Exam Thermodynamics 12 April 2018 Please, hand in your answers to problems 1, 2, 3 and 4 on separate sheets. Put your name and student number on each sheet. The examination time is 12:30 until 15:30.
More informationThe Second Law of Thermodynamics (Chapter 4)
The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made
More informationExam Thermodynamics 2 9 November 2017
1 Exam Thermodynamics 2 9 November 2017 Please, hand in your answers to problems 1, 2, 3 and 4 on separate sheets. Put your name and student number on each sheet. The examination time is 08:30 until 11:30.
More informationUNIVERSITY OF SOUTHAMPTON
UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2014-2015 ENERGY AND MATTER Duration: 120 MINS (2 hours) This paper contains 8 questions. Answers to Section A and Section B must be in separate
More informationChapter 5. Simple Mixtures Fall Semester Physical Chemistry 1 (CHM2201)
Chapter 5. Simple Mixtures 2011 Fall Semester Physical Chemistry 1 (CHM2201) Contents The thermodynamic description of mixtures 5.1 Partial molar quantities 5.2 The thermodynamic of Mixing 5.3 The chemical
More informationLast Name or Student ID
10/06/08, Chem433 Exam # 1 Last Name or Student ID 1. (3 pts) 2. (3 pts) 3. (3 pts) 4. (2 pts) 5. (2 pts) 6. (2 pts) 7. (2 pts) 8. (2 pts) 9. (6 pts) 10. (5 pts) 11. (6 pts) 12. (12 pts) 13. (22 pts) 14.
More informationAdvanced Placement. Chemistry. Integrated Rates
Advanced Placement Chemistry Integrated Rates 204 47.90 9.22 78.49 (26) 50.94 92.9 80.95 (262) 52.00 93.94 83.85 (263) 54.938 (98) 86.2 (262) 55.85 0. 90.2 (265) 58.93 02.9 92.2 (266) H Li Na K Rb Cs Fr
More informationEffect of adding an ideal inert gas, M
Effect of adding an ideal inert gas, M Add gas M If there is no change in volume, then the partial pressures of each of the ideal gas components remains unchanged by the addition of M. If the reaction
More informationPhysical Chemistry I Exam points
Chemistry 360 Fall 2018 Dr. Jean M. tandard October 17, 2018 Name Physical Chemistry I Exam 2 100 points Note: You must show your work on problems in order to receive full credit for any answers. You must
More informationModule 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics
Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics Objectives In this Lecture you will learn the following The need for studying thermodynamics to understand chemical and biological processes.
More informationEquations: q trans = 2 mkt h 2. , Q = q N, Q = qn N! , < P > = kt P = , C v = < E > V 2. e 1 e h /kt vib = h k = h k, rot = h2.
Constants: R = 8.314 J mol -1 K -1 = 0.08206 L atm mol -1 K -1 k B = 0.697 cm -1 /K = 1.38 x 10-23 J/K 1 a.m.u. = 1.672 x 10-27 kg 1 atm = 1.0133 x 10 5 Nm -2 = 760 Torr h = 6.626 x 10-34 Js For H 2 O
More informationChem 6 Sample exam 1 (150 points total) NAME:
Chem 6 Sample exam 1 (150 points total) @ This is a closed book exam to which the Honor Principle applies. @ The last page contains equations and physical constants; you can detach it for easy reference.
More informationConcentrating on the system
Concentrating on the system Entropy is the basic concept for discussing the direction of natural change, but to use it we have to analyze changes in both the system and its surroundings. We have seen that
More informationwhere R = universal gas constant R = PV/nT R = atm L mol R = atm dm 3 mol 1 K 1 R = J mol 1 K 1 (SI unit)
Ideal Gas Law PV = nrt where R = universal gas constant R = PV/nT R = 0.0821 atm L mol 1 K 1 R = 0.0821 atm dm 3 mol 1 K 1 R = 8.314 J mol 1 K 1 (SI unit) Standard molar volume = 22.4 L mol 1 at 0 C and
More informationThe underlying prerequisite to the application of thermodynamic principles to natural systems is that the system under consideration should be at equilibrium. http://eps.mcgill.ca/~courses/c220/ Reversible
More informationPractice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set.
Practice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set. The symbols used here are as discussed in the class. Use scratch paper as needed. Do not give more than one answer for any question.
More information7 Simple mixtures. Solutions to exercises. Discussion questions. Numerical exercises
7 Simple mixtures Solutions to exercises Discussion questions E7.1(b For a component in an ideal solution, Raoult s law is: p xp. For real solutions, the activity, a, replaces the mole fraction, x, and
More informationChapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.
Chapter 3 Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Concepts Energy functions F and G Chemical potential, µ Partial Molar properties
More informationPractice Questions Placement Exam for Exemption from Chemistry 120
Practice Questions Placement Exam for Exemption from Chemistry 120 Potentially Useful Information Avogadro's number = 6.0221420 10 23 h = 6.6260688 10 34 J s c = 2.9979246 10 8 m/s 1amu = 1.6605387 10
More information1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant
1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v lnt + RlnV + cons tant (1) p, V, T change Reversible isothermal process (const. T) TdS=du-!W"!S = # "Q r = Q r T T Q r = $W = # pdv =
More informationOCN 623: Thermodynamic Laws & Gibbs Free Energy. or how to predict chemical reactions without doing experiments
OCN 623: Thermodynamic Laws & Gibbs Free Energy or how to predict chemical reactions without doing experiments Definitions Extensive properties Depend on the amount of material e.g. # of moles, mass or
More informationPhysical Chemistry Physical chemistry is the branch of chemistry that establishes and develops the principles of Chemistry in terms of the underlying concepts of Physics Physical Chemistry Main book: Atkins
More informationUNIVERSITY OF KWAZULU-NATAL WESTVILLE CAMPUS DEGREE/DIPLOMA EXAMINATIONS: NOVEMBER 2006 CHEMISTRY CHEM230W: PHYSICAL CHEMISTRY 2
UNIVERSITY OF KWAZULU-NATAL WESTVILLE CAMPUS DEGREE/DIPLOMA EXAMINATIONS: NOVEMBER 006 CHEMISTRY CHEM30W: PHYSICAL CHEMISTRY TIME: 180 MINUTES MARKS: 100 EXAMINER: PROF S.B. JONNALAGADDA ANSWER FIVE QUESTIONS.
More informationPhysics 404: Final Exam Name (print): "I pledge on my honor that I have not given or received any unauthorized assistance on this examination.
Physics 404: Final Exam Name (print): "I pledge on my honor that I have not given or received any unauthorized assistance on this examination." May 20, 2008 Sign Honor Pledge: Don't get bogged down on
More informationUNIVERSITY OF SOUTHAMPTON
UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2014-2015 ENERGY AND MATTER Duration: 120 MINS (2 hours) This paper contains 8 questions. Answers to Section A and Section B must be in separate
More informationEnthalpy and Adiabatic Changes
Enthalpy and Adiabatic Changes Chapter 2 of Atkins: The First Law: Concepts Sections 2.5-2.6 of Atkins (7th & 8th editions) Enthalpy Definition of Enthalpy Measurement of Enthalpy Variation of Enthalpy
More informationThermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes
Thermodynamics Chem 36 Spring 2002 Thermodynamics The study of energy changes which accompany physical and chemical processes Why do we care? -will a reaction proceed spontaneously? -if so, to what extent?
More informationI affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class.
Chem340 Physical Chemistry for Biochemists Exam Mar 16, 011 Your Name _ I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade
More informationR = J/mole K = cal/mole K = N A k B ln k = ( E a /RT) + ln A
Chemistry 134 Prof. Jason Kahn Your Name: University of Maryland, College Park Your SID #: General Chemistry and Energetics Exam I (100 points) You have 53 minutes for this exam. Your Section # or time:
More informationSolutions to Problem Set 9
Solutions to Problem Set 9 1. When possible, we want to write an equation with the quantity on the ordinate in terms of the quantity on the abscissa for each pf the labeled curves. A B C p CHCl3 = K H
More informationChemical Kinetics. Chapter 13. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chemical Kinetics Chapter 13 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chemical Kinetics Thermodynamics does a reaction take place? Kinetics how fast does
More informationFIRST PUBLIC EXAMINATION SUBJECT 3: PHYSICAL CHEMISTRY
CCHE 4273 FIRST PUBLIC EXAMINATION Trinity Term 2005 Preliminary Examination in Chemistry SUBJECT 3: PHYSICAL CHEMISTRY Wednesday June 8 th 2005, 9:30am Time allowed: 2 ½ hours Candidates should answer
More informationTHERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system
THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system and its surroundings. a. System = That part of universe
More informationProblem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are
Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are perfectly insulated from the surroundings. Is this a spontaneous
More informationSome properties of the Helmholtz free energy
Some properties of the Helmholtz free energy Energy slope is T U(S, ) From the properties of U vs S, it is clear that the Helmholtz free energy is always algebraically less than the internal energy U.
More informationln( P vap(s) / torr) = T / K ln( P vap(l) / torr) = T / K
Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Number 9 Solutions 1. McQuarrie and Simon, 9-4. Paraphrase: Given expressions
More informationChapter 3 - First Law of Thermodynamics
Chapter 3 - dynamics The ideal gas law is a combination of three intuitive relationships between pressure, volume, temp and moles. David J. Starling Penn State Hazleton Fall 2013 When a gas expands, it
More informationChemistry 5350 Advanced Physical Chemistry Fall Semester 2013
Chemistry 5350 Advanced Physical Chemistry Fall Semester 2013 Mierm Examination: Thermodynamics and Kinetics Name: October 10, 2013 Constants and Conversion Factors Gas Constants: 8.314 J mol 1 K 1 ; 8.314
More informationCHEMICAL THERMODYNAMICS
DEPARTMENT OF APPLIED CHEMISTRY LECTURE NOTES 6151- ENGINEERING CHEMISTRY-II UNIT II CHEMICAL THERMODYNAMICS Unit syllabus: Terminology of thermodynamics - Second law: Entropy - entropy change for an ideal
More informationLecture 4. The Second Law of Thermodynamics
Lecture 4. The Second Law of Thermodynamics LIMITATION OF THE FIRST LAW: -Does not address whether a particular process is spontaneous or not. -Deals only with changes in energy. Consider this examples:
More informationAP Chemistry. Free-Response Questions
2018 AP Chemistry Free-Response Questions College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. AP Central is the official online
More informationPhysics 408 Final Exam
Physics 408 Final Exam Name You are graded on your work (with partial credit where it is deserved) so please do not just write down answers with no explanation (or skip important steps)! Please give clear,
More informationWhat is thermodynamics? and what can it do for us?
What is thermodynamics? and what can it do for us? The overall goal of thermodynamics is to describe what happens to a system (anything of interest) when we change the variables that characterized the
More informationI PUC CHEMISTRY CHAPTER - 06 Thermodynamics
I PUC CHEMISTRY CHAPTER - 06 Thermodynamics One mark questions 1. Define System. 2. Define surroundings. 3. What is an open system? Give one example. 4. What is closed system? Give one example. 5. What
More information4. All questions are NOT ofequal value. Marks available for each question are shown in the examination paper.
THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF PHYSICS \1111~11\llllllllllllftllll~flrllllllllll\11111111111111111 >014407892 PHYS2060 THER1\1AL PHYSICS FINAL EXAMINATION SESSION 2 - NOVEMBER 2010 I. Time
More informationFinal Exam, Chemistry 481, 77 December 2016
1 Final Exam, Chemistry 481, 77 December 216 Show all work for full credit Useful constants: h = 6.626 1 34 J s; c (speed of light) = 2.998 1 8 m s 1 k B = 1.387 1 23 J K 1 ; R (molar gas constant) = 8.314
More informationLiquids and Solutions Crib Sheet
Liquids and Solutions Crib Sheet Determining the melting point of a substance from its solubility Consider a saturated solution of B in a solvent, A. Since the solution is saturated, pure solid B is in
More informationCHAPTER 3 LECTURE NOTES 3.1. The Carnot Cycle Consider the following reversible cyclic process involving one mole of an ideal gas:
CHATER 3 LECTURE NOTES 3.1. The Carnot Cycle Consider the following reversible cyclic process involving one mole of an ideal gas: Fig. 3. (a) Isothermal expansion from ( 1, 1,T h ) to (,,T h ), (b) Adiabatic
More informationWeek 1 Temperature, Heat and the First Law of Thermodynamics. (Ch. 19 of Serway&J.)
Week 1 Temperature, Heat and the First Law of Thermodynamics. (Ch. 19 of Serway&J.) (Syllabus) Temperature Thermal Expansion Temperature and Heat Heat and Work The first Law Heat Transfer Temperature Thermodynamics:
More informationLecture 3 Clausius Inequality
Lecture 3 Clausius Inequality Rudolf Julius Emanuel Clausius 2 January 1822 24 August 1888 Defined Entropy Greek, en+tropein content transformative or transformation content The energy of the universe
More informationOutline Review Example Problem 1. Thermodynamics. Review and Example Problems: Part-2. X Bai. SDSMT, Physics. Fall 2014
Review and Example Problems: Part- SDSMT, Physics Fall 014 1 Review Example Problem 1 Exponents of phase transformation : contents 1 Basic Concepts: Temperature, Work, Energy, Thermal systems, Ideal Gas,
More informationName (Print) Section # or TA. 1. You may use a crib sheet which you prepared in your own handwriting. This may be
Name (Print) Section # or TA 1. You may use a crib sheet which you prepared in your own handwriting. This may be one 8-1/2 by 11 inch sheet of paper with handwriting only on one side. 2. You may use a
More informationChemical standard state: 1 M solutes, pure liquids, 1 atm gases Biochemical standard state: ph 7, all species in the ionic form found at ph 7
Chemistry 271, Section 22xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Exam II (100 points total) Your Section #: November 4, 2009 You
More informationChemistry 1A, Spring 2011 Midterm 1 February 7, 2011 (90 min, closed book)
Name: SID: TA Name: Chemistry 1A, Spring 2011 Midterm 1 February 7, 2011 (90 min, closed book) There are 40 multiple choice questions worth 3 points each. There is only one correct answer for each question
More informationCHAPTER 9 LECTURE NOTES
CHAPTER 9 LECTURE NOTES 9.1, 9.2: Rate of a reaction For a general reaction of the type A + 3B 2Y, the rates of consumption of A and B, and the rate of formation of Y are defined as follows: Rate of consumption
More informationPhysics 360 Review 3
Physics 360 Review 3 The test will be similar to the second test in that calculators will not be allowed and that the Unit #2 material will be divided into three different parts. There will be one problem
More informationMME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition
MME 2010 METALLURGICAL THERMODYNAMICS II Fundamentals of Thermodynamics for Systems of Constant Composition Thermodynamics addresses two types of problems: 1- Computation of energy difference between two
More informationAdvanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS
Experimental Kinetics and Gas Phase Reactions Advanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS Professor Angelo R. Rossi http://homepages.uconn.edu/rossi Department of Chemistry, Room
More informationTransition Theory Abbreviated Derivation [ A - B - C] # E o. Reaction Coordinate. [ ] # æ Æ
Transition Theory Abbreviated Derivation A + BC æ Æ AB + C [ A - B - C] # E A BC D E o AB, C Reaction Coordinate A + BC æ æ Æ æ A - B - C [ ] # æ Æ æ A - B + C The rate of reaction is the frequency of
More informationPhysical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points
Physical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points Name: KEY Gas constant: R = 8.314 J mol -1 K -1 = 0.008314 kj mol -1 K -1. Boltzmann constant k = 1.381 10-23 J/K = 0.6950 cm -1 /K h =
More informationCHEM 4641 Fall questions worth a total of 32 points. Show your work, except on multiple-choice questions. 1 V α=
Physical Chemistry I Final Exam Name: KEY CHEM 4641 Fall 017 15 questions worth a total of 3 points. Show your work, except on multiple-choice questions. 1 V 1 V α= κt = V T P V P T Gas constant R = 8.314
More informationIntroduction into thermodynamics
Introduction into thermodynamics Solid-state thermodynamics, J. Majzlan Chemical thermodynamics deals with reactions between substances and species. Mechanical thermodynamics, on the other hand, works
More informationChemistry. Lecture 10 Maxwell Relations. NC State University
Chemistry Lecture 10 Maxwell Relations NC State University Thermodynamic state functions expressed in differential form We have seen that the internal energy is conserved and depends on mechanical (dw)
More informationFoundations of Chemical Kinetics. Lecture 12: Transition-state theory: The thermodynamic formalism
Foundations of Chemical Kinetics Lecture 12: Transition-state theory: The thermodynamic formalism Marc R. Roussel Department of Chemistry and Biochemistry Breaking it down We can break down an elementary
More informationLecture 6. NONELECTROLYTE SOLUTONS
Lecture 6. NONELECTROLYTE SOLUTONS NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRATION UNITS percent
More informationPhysics 4230 Final Exam, Spring 2004 M.Dubson This is a 2.5 hour exam. Budget your time appropriately. Good luck!
1 Physics 4230 Final Exam, Spring 2004 M.Dubson This is a 2.5 hour exam. Budget your time appropriately. Good luck! For all problems, show your reasoning clearly. In general, there will be little or no
More informationChapter 14 Kinetic Theory
Chapter 14 Kinetic Theory Kinetic Theory of Gases A remarkable triumph of molecular theory was showing that the macroscopic properties of an ideal gas are related to the molecular properties. This is the
More informationdg = V dp - S dt (1.1) 2) There are two T ds equations that are useful in the analysis of thermodynamic systems. The first of these
CHM 3410 Problem Set 5 Due date: Wednesday, October 7 th Do all of the following problems. Show your work. "Entropy never sleeps." - Anonymous 1) Starting with the relationship dg = V dp - S dt (1.1) derive
More informationVersion 001 HW 15 Thermodynamics C&J sizemore (21301jtsizemore) 1
Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.
More informationRate of Heating and Cooling
Rate of Heating and Cooling 35 T [ o C] Example: Heating and cooling of Water E 30 Cooling S 25 Heating exponential decay 20 0 100 200 300 400 t [sec] Newton s Law of Cooling T S > T E : System S cools
More informationPES 2130 Exam 1/page 1. PES Physics 3 Exam 1. Name: SOLUTIONS Score: / 100
PES 2130 Exam 1/page 1 PES 2130 - Physics 3 Exam 1 Name: SOLUTIONS Score: / 100 Instructions Time allowed or this is exam is 1 hours 15 minutes 10 written problems For written problems: Write all answers
More informationName: Discussion Section:
CBE 141: Chemical Engineering Thermodynamics, Spring 2017, UC Berkeley Midterm 2 FORM B March 23, 2017 Time: 80 minutes, closed-book and closed-notes, one-sided 8 ½ x 11 equation sheet allowed lease show
More informationClassical Thermodynamics. Dr. Massimo Mella School of Chemistry Cardiff University
Classical Thermodynamics Dr. Massimo Mella School of Chemistry Cardiff University E-mail:MellaM@cardiff.ac.uk The background The field of Thermodynamics emerged as a consequence of the necessity to understand
More information140a Final Exam, Fall 2006., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT.
40a Final Exam, Fall 2006 Data: P 0 0 5 Pa, R = 8.34 0 3 J/kmol K = N A k, N A = 6.02 0 26 particles/kilomole, T C = T K 273.5. du = TdS PdV + i µ i dn i, U = TS PV + i µ i N i Defs: 2 β ( ) V V T ( )
More information3.20 Exam 1 Fall 2003 SOLUTIONS
3.0 Exam 1 Fall 003 SOLUIONS Question 1 You need to decide whether to work at constant volume or constant pressure. Since F is given, a natural choice is constant volume. Option 1: At constant and V :
More informationPractice Questions Placement Exam for Entry into Chemistry 120
Practice Questions Placement Exam for Entry into Chemistry 120 Potentially Useful Information Avogadro's number = 6.0221420 10 23 h = 6.6260688 10 34 J s c = 2.9979246 10 8 m/s 1amu = 1.6605387 10 27 kg
More informationME 501. Exam #2 2 December 2009 Prof. Lucht. Choose two (2) of problems 1, 2, and 3: Problem #1 50 points Problem #2 50 points Problem #3 50 points
1 Name ME 501 Exam # December 009 Prof. Lucht 1. POINT DISTRIBUTION Choose two () of problems 1,, and 3: Problem #1 50 points Problem # 50 points Problem #3 50 points You are required to do two of the
More informationName: First three letters of last name
Name: First three letters of last name Chemistry 342 Third Exam April 22, 2005 2:00 PM in C6 Lecture Center Write all work you want graded in the spaces provided. Both the logical solution to the problem
More informationChemical standard state: 1 M solutes, pure liquids, 1 atm gases Biochemical standard state: ph 7, all species in the ionic form found at ph 7
Chemistry 271, Section 22xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Exam II (100 points total) Your Section #: November 4, 2009 You
More informationOutline Review Example Problem 1 Example Problem 2. Thermodynamics. Review and Example Problems. X Bai. SDSMT, Physics. Fall 2013
Review and Example Problems SDSMT, Physics Fall 013 1 Review Example Problem 1 Exponents of phase transformation 3 Example Problem Application of Thermodynamic Identity : contents 1 Basic Concepts: Temperature,
More informationThis is important to know that the P total is different from the initial pressure (1bar) because of the production of extra molecules!!! = 0.
Question 1: N O 4 (g) NO (g) Amounts of material at the initial state: n 0 0 Amounts of material at equilibrium: (1 α)n 0 αn 0 Where α equals 0.0488 at 300 K and α equals 0.141 at 400K. Step 1: Calculate
More informationN h (6.02x10 )(6.63x10 )
CHEM 5200 - Final Exam - December 13, 2018 INFORMATION PAGES (Use for reference and for scratch paper) Constants and Conversion Factors: R = 8.31 J/mol-K = 8.31 kpa-l/mol-k = 0.00831 kj/mol-k 1 L-atm =
More informationIntroduction. Statistical physics: microscopic foundation of thermodynamics degrees of freedom 2 3 state variables!
Introduction Thermodynamics: phenomenological description of equilibrium bulk properties of matter in terms of only a few state variables and thermodynamical laws. Statistical physics: microscopic foundation
More informationChapter 3. The Second Law Fall Semester Physical Chemistry 1 (CHM2201)
Chapter 3. The Second Law 2011 Fall Semester Physical Chemistry 1 (CHM2201) Contents The direction of spontaneous change 3.1 The dispersal of energy 3.2 The entropy 3.3 Entropy changes accompanying specific
More informationIntroduction to Chemical Thermodynamics. D. E. Manolopoulos First Year (13 Lectures) Michaelmas Term
Introduction to Chemical Thermodynamics D. E. Manolopoulos First Year (13 Lectures) Michaelmas Term Lecture Synopsis 1. Introduction & Background. Le Chatelier s Principle. Equations of state. Systems
More informationThe Chemistry Maths Book
Solutions for Chapter The Chemistr Maths Book Erich Steiner Universit of Eeter Second Edition 008 Solutions Chapter. Differentiation. Concepts. The process of differentiation. Continuit. Limits.5 Differentiation
More informationAppendix 4. Appendix 4A Heat Capacity of Ideal Gases
Appendix 4 W-143 Appendix 4A Heat Capacity of Ideal Gases We can determine the heat capacity from the energy content of materials as a function of temperature. The simplest material to model is an ideal
More informationChem142 Introduction to Physical Chemistry
Chem4 Introduction to hysical Chemistry Exam # (50 minutes) Name 00 points September 5, 07 Units and Constants R = 8.34 J/K*mol = 0.0834 L*bar/K*mol = 0.0806 L*atm/K*mol dm 3 = L K = 73.5 + C bar = 00
More informationWork and heat. Expansion Work. Heat Transactions. Chapter 2 of Atkins: The First Law: Concepts. Sections of Atkins
Work and heat Chapter 2 of Atkins: The First Law: Concepts Sections 2.3-2.4 of Atkins Expansion Work General Expression for Work Free Expansion Expansion Against Constant Pressure Reversible Expansion
More informationEntropy Changes & Processes
Entropy Changes & Processes Chapter 4 of Atkins: The Second Law: The Concepts Section 4.4-4.7 Third Law of Thermodynamics Nernst Heat Theorem Third- Law Entropies Reaching Very Low Temperatures Helmholtz
More informationContents and Concepts
Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous
More informationContents and Concepts
Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous
More informationCHEMICAL ENGINEERING THERMODYNAMICS. Andrew S. Rosen
CHEMICAL ENGINEERING THERMODYNAMICS Andrew S. Rosen SYMBOL DICTIONARY 1 TABLE OF CONTENTS Symbol Dictionary... 3 1. Measured Thermodynamic Properties and Other Basic Concepts... 5 1.1 Preliminary Concepts
More informationPHY214 Thermal & Kinetic Physics Duration: 2 hours 30 minutes
BSc Examination by course unit. Friday 5th May 01 10:00 1:30 PHY14 Thermal & Kinetic Physics Duration: hours 30 minutes YOU ARE NOT PERMITTED TO READ THE CONTENTS OF THIS QUESTION PAPER UNTIL INSTRUCTED
More informationMore Thermodynamics. Specific Specific Heats of a Gas Equipartition of Energy Reversible and Irreversible Processes
More Thermodynamics Specific Specific Heats of a Gas Equipartition of Energy Reversible and Irreversible Processes Carnot Cycle Efficiency of Engines Entropy More Thermodynamics 1 Specific Heat of Gases
More informationESCI 341 Atmospheric Thermodynamics Lesson 12 The Energy Minimum Principle
ESCI 341 Atmospheric Thermodynamics Lesson 12 The Energy Minimum Principle References: Thermodynamics and an Introduction to Thermostatistics, Callen Physical Chemistry, Levine THE ENTROPY MAXIMUM PRINCIPLE
More informationLecture Notes 2014March 13 on Thermodynamics A. First Law: based upon conservation of energy
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 1 Lecture Notes 2014March 13 on Thermodynamics A. First Law: based upon conservation of energy 1. Work 1 Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 2 (c)
More information