Final Examination ( )
|
|
- Harvey Scott
- 6 years ago
- Views:
Transcription
1 Ministry of high education& scientific Research University of echnology Materials engineering department Class: Second Year Subject: Chemical Metallurgy Examiner: Final Examination (0-0) وزارة التعليم العالي و البحث العلمي الجامعت التكنولوجيت قسم هندست المواد Date: 7/05/0 Allowed time: 3 hrs Note: Attempt only Five Questions. Q / (A) Determine two parameters used to predict spontaneity of the process, then determine when each one from them is useful in dealing with different conditions of the process. State their relationship with enthalpy change. (B) Is the reaction spontaneous under standard conditions? 4KClO 3 (s) 3KClO 4 (s) + KCl (s) i H f (kj/mol) S o KClO KClO KCl (0 Marks) Q / (A) Explain in details the Reversible and Irreversible processes, give an example. (B) Water at 33 C is pumped from storage tank at the rate of 0. m 3 /sec. he motor for the pump supplied work at the rate of 60 KJ/Kg. he water passes through heat exchanger of 000 KJ/Kg, and is delivered to elevation of 0 m above the first tank at the velocity 5 m/sec. What is the temperature of the water delivered to this elevation? (0.94 Btu = 0 3 J). Density of water=000 Kg/m 3. Sketch Flow Diagram. Enthalpy, Btu/kg emperature C Enthalpy, Btu/kg emperature C (0 Marks) Q 3 / (A) For closed system adiabatic processes for an ideal gas, by law of hermodynamic prove that: beginning from the differential first P P (B) One Kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and bar until its volume triple. Calculate Q, W, U and H for the process. Assume that air obeys the relation PV/=83.4 bar cm 3 mol K, Cp=9 J mol K, and Molecular Weight of air=9 Kg/Kmol. (0 Marks)
2 Pressure Q 4 / For the following Figure, calculate for each process U, H, Q & W. Given that: = =600 K P = 0 bar, P = 3 bar, P 3 = bar Cp=3.5R, Cv=.5R R=8.48 J/mol. K Volume (0 Marks) Q 5 / Sulfur dioxide gas is oxidized in 90 percent excess air with 80 % conversion to Sulfur trioxide. he gases enter the reactor at 40 o C and leave at 460 o C. How much heat must be transfer from the reactor on the basis mole of entering gas? 4 3 SO + O SO 3 Given data; Standard heat of formation at 5 C, and Constants. i H 98 J/mole A 0 3 B 0 6 C 0-5 D SO O SO N (0 Marks) Q 6 / (A) One mole of an ideal gas, Cp= (7/)R, Cv=(5/)R, is compressed adiabatically in a piston/cylinder device from bar and 40 C to 5 bar. he process is irreversible and requires 35 percent more work than a reversible adiabatic compression from the initial state to the same final pressure. What is the final temperature and entropy change of the gas? (B) For an ideal gas with constant heat capacities show that: For a pressure change from P to P, the sign of S of an isothermal change is opposite that for a constant volume change. (0 Marks) Good Luck
3 Q / (A) Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department he major parameters to predict spontaneity of the process are Entropy change ( S) and Free Energy ( G), the first one is useful in dealing with different conditions of the process, but usually the second ( G) be more useful for certain conditions of temperature and pressure. Note that ΔG is composite of both ΔH and ΔS: If ΔH < 0 and ΔS > 0.spontaneous at all A reaction is spontaneous if ΔG < 0. Such that: If ΔH > 0 and ΔS < 0.not spontaneous at any If ΔH < 0 and ΔS < 0.spontaneous at low If ΔH > 0 and ΔS > 0.spontaneous at high Q / (B) ΔG = ΔH -ΔS ΔH rxn = ᵧi ΔH i = 3( 43.8kJ) + ( 436.7kJ) 4( 397.7kJ) = 44kJ ΔS rxn = ᵧi S o i = 3(5.0J/K) + (8.6 J/K) 4(43.J/K) = 36.8J/K ΔG rxn = ΔH rxn ΔS rxn = 44kJ (98K)( 36.8J/K)(kJ/000J)= 33kJ ΔG rxn < 0; therefore, reaction is spontaneous under standard conditions. Q / (A) A process is reversible when its direction can be reversed at any point by an infinite change in external conditions. Once the process is initiated, no infinite change in external conditions can reverse its direction; the process is irreversible. he apparatus is shown in figure (a gas in piston/cylinder). Expansion processes result when mass is removed from the piston (we assume the piston without friction), then the piston will rise and new balance of the piston level will be achieved. he oscillation of the peiton level due to that mass m is suddenly removed to a shelf. If we change the mass by powder the piston gradually rise and the other hand will return to initial level in the same path. m Figure : Gas Expasion l
4 Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q / (B) H + E p + E k = Q W s Mass flow rate= Density x Volumetric flow rate= 0. m 3 /sec X 000 Kg/m 3 =00 Kg/sec. E p = m Z g= 00 Kg/sec x 0 m x 9.8 m/s = 980 kj/sec. E k =m x u /= 00 Kg/sec (5-0)/= 50 kj/sec. (u could be neglected; tank). Q = 000 kj/kg x 00 Kg/sec = kj/sec. (positive sign assume heat absorbed by the system). W s = 60 kj/kg x 00 Kg/sec = 6000 kj/sec. hen H= Q W s E p E k = = kj/sec 6000 kj/sec 980 kj/sec 50 kj/sec = 8940 kj/sec. = H H By extrapolation method H (at 33 C) = 44 Btu/Kg x ( kj/0.94 Btu) x 00 Kg/sec= 5300 kj/sec. hen H = H + H = 8940 kj/sec kj/sec. = 9840 kj/sec. H = 9840 kj/sec. x (0.94 Btu/ kj) / 00 Kg/sec = 7.5 Btu/kg. By extrapolation method (have H=93.5 Btu/kg ) = 66.8 C. Q 3 / (A) First law of thermodynamic, U = Q W (Adaibatic Q =0) C v d = P dv, P=R/V hen C v d = R(dV/V) d/= R/C v (dv/v) R/C v = (C p C v )/ C v = C p / C v =ᵞ d/ = (ᵞ ) dv/v (Integration) ln( / )=ln(v /V ) -(ᵞ-) V V ( ) P V / = P V /, then V /V = ( / )(P /P ) substitute in the last equation hen: P P
5 Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 3 / (B) No. of moles = /9 Kg/Kmol = mole P V / = P V / = 83.4 bar cm 3 mol K, constant pressure P= P V / = V /, and V / = 3V /, =3 =3 x 300= 900 V = 300 X 83.4 bar cm 3 mol K = 494 cm 3 mol and V = 848 cm 3 V = 3V = 544 cm 3 At constant pressure Q = H=nC p ( )= x 9 ( ) = 600 J W=P V= X 0 5 (544 cm cm 3 ) = 696 x 0 5 N/m. cm 3 =69.6 J U = Q W = 600 J 69.6 J = J. Q 4 / = (Isothermal process), H=0, U=0 P Q=W= R ln 0 =8.34 x 600 ln = 605 J. P 3 3 Constant Volume (W=0) P P 3 3, constant volume. 3 = 600 x /3= K H= C p ( 3 )= 3.5x8.34( )= J U= C v ( 3 )=.5x8.34( )= J Q = U = J 4 Adiabatic Process (Q=0), and P 4 =P 3 = 3 bar 4 P P 4, , and =.4 hen 4 = 34.3 K H= C p ( 4 )= 3.5x8.34( )= J U= C v ( 4 )=.5x8.34( )= J W= - U= J 3 4 Constant pressure (Isoparic), Q= H H= C p ( 4 3 )= 3.5x8.34( )= J U= C v ( 4 3 )=.5x8.34( )= 407. J W = Q U = = 56.8 J
6 Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 5 / Basis: mole of SO entering., = = K, = = K. O required (computed on base completely conversion) According reaction equation ( mole SO Vs 0.5 mole O required) Excess O = 0.5 x 0.9= 0.45 mole otal O entering = = 0.95 mole. 79 N entering= 0.95 x = 3.57 mole. Leaving the reactor: SO = x ( 80% conversion) = 0. mole O reacted = x 80% x 0.5= 0.4 mole. O output = input-reacted = = 0.55 mole. SO 3 produced = No. of mole SO reacted = 0.8 mole. in =40 o C SO mol O 0.95 mol N 3.57 mol H R Q= H H P out =460 o C SO 0. mol SO mol O 0.55 mol N 3.57 mol H= H R + H 98 + H P H 98 H 98 = ᵧi ΔH i = (- ) ( )= J am =( + )/=708.5 K. H R = n i Cp m,hi (98.5 in ) H P = n i Cp m,hi ( out ) n i Cp m,hi =R( n i A i + ( n i B i ) am + ni Di ) n i A i = x x x 3.80= 0.86 n i B i = 3.4, n i D i = n i Cp m,hi = 8.34( x )= x733 he: H R = n i Cp m,hi ( )= J And the same way H P = J Finally H= J + ( J) J = J
7 Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 6 / (A) Compressed adiabatically, Q=0 U = W he first step we find reversible and adiabatic work (i.e S=0) S=0=C p ln P R ln P 7/Rln 4 =Rln, = K 33.5 W= U= C v ( )= 5/R( )= W rev. = 363. J W irrev =.3 x W rev (More 30 %).= 4. J = C v ( irrv. )=5/R( irrv 33.5), irrv. = 5 K 5 4 S=7/R ln R ln 33.5 =.7 J/K > 0 spontaneous process. Q 6 / (B) S=C p ln P R ln, P. Isothermal process = P hen ( S) iso = R ln P. Constant volume P P, and P P P And S=C p ln P R ln P P P = (C p -R)ln P = +Cv ln P P herefore ( S) opposite sign of ( S) V
Chem 1B Dr. White 1 Chapter 17: Thermodynamics. Review From Chem 1A (Chapter 6, section 1) A. The First Law of Thermodynamics
Chem 1B Dr. White 1 Chapter 17: Thermodynamics Review From Chem 1A (Chapter 6, section 1) A. The First Law of Thermodynamics 17.1 Spontaneous Processes and Entropy A. Spontaneous Change Chem 1B Dr. White
More informationThermodynamics. Thermodynamics of Chemical Reactions. Enthalpy change
Thermodynamics 1 st law (Cons of Energy) Deals with changes in energy Energy in chemical systems Total energy of an isolated system is constant Total energy = Potential energy + kinetic energy E p mgh
More informationCHAPTER 8 ENTROPY. Blank
CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.
More informationCHEMICAL THERMODYNAMICS. Nature of Energy. ΔE = q + w. w = PΔV
CHEMICAL HERMODYNAMICS Nature of Energy hermodynamics hermochemistry Energy (E) Work (w) Heat (q) Some Definitions Study the transformation of energy from one form to another during physical and chemical
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 informationExam 2 Solutions. for a gas obeying the equation of state. Z = PV m RT = 1 + BP + CP 2,
Chemistry 360 Dr. Jean M. Standard Fall 016 Name KEY 1.) (14 points) Determine # H & % ( $ ' Exam Solutions for a gas obeying the equation of state Z = V m R = 1 + B + C, where B and C are constants. Since
More informationU = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C.
CHAPER LECURE NOES he First Law of hermodynamics: he simplest statement of the First Law is as follows: U = q + w. Here U is the internal energy of the system, q is the heat and w is the work. CONVENIONS
More informationAdvanced Chemistry Practice Problems
Thermodynamics: Review of Thermochemistry 1. Question: What is the sign of DH for an exothermic reaction? An endothermic reaction? Answer: ΔH is negative for an exothermic reaction and positive for an
More informationChemistry 452 July 23, Enter answers in a Blue Book Examination
Chemistry 45 July 3, 014 Enter answers in a Blue Book Examination Midterm Useful Constants: 1 Newton=1 N= 1 kg m s 1 Joule=1J=1 N m=1 kg m /s 1 Pascal=1Pa=1N m 1atm=10135 Pa 1 bar=10 5 Pa 1L=0.001m 3 Universal
More informationHence. The second law describes the direction of energy transfer in spontaneous processes
* Heat and Work The first law of thermodynamics states that: Although energy has many forms, the total quantity of energy is constant. When energy disappears in one form, it appears simultaneously in other
More informationChem 401 Unit 1 (Kinetics & Thermo) Review
KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 1 Δ [H2O] Δ[O 2] = 2 Δt Δt 2. Determine the
More informationSecond law of thermodynamics
Second law of thermodynamics It is known from everyday life that nature does the most probable thing when nothing prevents that For example it rains at cool weather because the liquid phase has less energy
More informationHomework Problem Set 8 Solutions
Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next,
More informationChapter 19 The First Law of Thermodynamics
Chapter 19 The First Law of Thermodynamics The first law of thermodynamics is an extension of the principle of conservation of energy. It includes the transfer of both mechanical and thermal energy. First
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 informationCHAPTER - 12 THERMODYNAMICS
CHAPER - HERMODYNAMICS ONE MARK QUESIONS. What is hermodynamics?. Mention the Macroscopic variables to specify the thermodynamics. 3. How does thermodynamics differ from Mechanics? 4. What is thermodynamic
More information4200:225 Equilibrium Thermodynamics
4:5 Equilibrium Thermodynamics Unit I. Earth, Air, Fire, and Water Chapter 3. The Entropy Balance By J.R. Elliott, Jr. Unit I. Energy and Entropy Chapter 3. The Entropy Balance Introduction to Entropy
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 informationMCAT General Chemistry Discrete Question Set 19: Thermochemistry & Thermodynamics
MCAT General Chemistry Discrete Question Set 19: Thermochemistry & Thermodynamics Question No. 1 of 10 1: A metal with a high heat capacity is put on a hot plate. What will happen? Question #01 A. The
More informationCHEMISTRY 202 Practice Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (40 pts.) 22 (20 pts.)
CHEMISTRY 202 Practice Hour Exam II Fall 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 7 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationWhere F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1
In order to force the fluid to flow across the boundary of the system against a pressure p1, work is done on the boundary of the system. The amount of work done is dw = - F1.dl1, Where F1 is the force
More informationTHERMODYNAMICS. Dr. Sapna Gupta
THERMODYNAMICS Dr. Sapna Gupta FIRST LAW OF THERMODYNAMICS Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes. First Law of Thermodynamics Energy cannot
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 informationOAT General Chemistry Problem Drill 15: Thermochemistry & Thermodynamics
OAT General Chemistry Problem Drill 15: Thermochemistry & Thermodynamics Question No. 1 of 10 1. A metal with a high heat capacity is put on a hot plate. What will happen? Question #01 (A) The temperature
More information(g) + 2H 2. (g) CH [1] (g) H 2. Explain, with a calculation, whether this reaction is feasible at 25 C [3]
1 This question looks at two reactions involving sulfur compounds (a) Hydrogen reacts with carbon disulfide as shown below 4H 2 + CS 2 CH 4 + 2H 2 S For this reaction, ΔH = 234 kj mol 1 and ΔS = 164 J
More informationtemperature begins to change noticeably. Feedback D. Incorrect. Putting an object on a hot plate will always cause the temperature to increase.
SAT Chemistry - Problem Drill 22: Thermodynamics No. 1 of 10 1. A metal with a high heat capacity is placed on top of a hot plate that is turned on. What will happen to the temperature of the piece of
More informationTwo mark questions and answers UNIT II SECOND LAW 1. Define Clausius statement. It is impossible for a self-acting machine working in a cyclic process, to transfer heat from a body at lower temperature
More informationThis follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore. (for reversible process only) (22.
Entropy Clausius inequality can be used to analyze the cyclic process in a quantitative manner. The second law became a law of wider applicability when Clausius introduced the property called entropy.
More informationChapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.)
CHEMISTRY 202 Hour Exam II October 27, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 32 questions on 11 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationChapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
More informationUnit 12. Thermochemistry
Unit 12 Thermochemistry A reaction is spontaneous if it will occur without a continuous input of energy However, it may require an initial input of energy to get it started (activation energy) For Thermochemistry
More informationTHERMODYNAMICS Lecture 5: Second Law of Thermodynamics
HERMODYNAMICS Lecture 5: Second Law of hermodynamics Pierwsza strona Second Law of hermodynamics In the course of discussions on the First Law of hermodynamics we concluded that all kinds of energy are
More informationChapter 19 Chemical Thermodynamics Entropy and free energy
Chapter 19 Chemical Thermodynamics Entropy and free energy Learning goals and key skills: Explain and apply the terms spontaneous process, reversible process, irreversible process, and isothermal process.
More informationMAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI-621213. Department: Mechanical Subject Code: ME2202 U N IT - 1 Semester: III Subject Name: ENGG. THERMODYNAMICS 1. 1 kg of gas at 1.1 bar, 27 o C is compressed
More informationAP CHEMISTRY SCORING GUIDELINES
Mean 5.64 out of 9 pts AP CHEMISTRY Question 1 CO(g) + 1 2 O 2 (g) CO 2 (g) 1. The combustion of carbon monoxide is represented by the equation above. (a) Determine the value of the standard enthalpy change,
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (30 pts.) 22 (30 pts.)
CHEMISTRY 202 Hour Exam II November 2, 2017 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationI. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.
I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit. Both the Kelvin and Fahrenheit scales are absolute temperature scales. Specific volume, v, is an intensive property,
More informationChapter 19 Chemical Thermodynamics
Chapter 19 Chemical Thermodynamics Kinetics How fast a rxn. proceeds Equilibrium How far a rxn proceeds towards completion Thermodynamics Study of energy relationships & changes which occur during chemical
More informationEntropy. ΔS = q rev T. CLASSICAL THERMODYNAMICS study of macroscopic/thermodynamic properties of systems: U, T, V, P,
Entropy Rudolf Clausius 1 st law: ΔU = q - w A new State Func,on: ΔS = q rev CLASSICAL HERMODYNAMICS study of macroscopic/thermodynamic properties of systems: U,, V, P, Ludvig Boltzmann 1844-1906 S = k
More informationChapter 27. Energy and Disorder
Chapter 27 Energy and Disorder Why Reactions Occur Exothermic Rxns - Take place spontaneously Go from high energy to low energy Downhill Endothermic Rxns. - Not usually spontaneous Go from low energy to
More informationChapter 4. Energy Analysis of Closed Systems
Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat
More informationHence. The second law describes the direction of energy transfer in spontaneous processes
Heat and Work The first law of thermodynamics states that: Although energy has many forms, the total quantity of energy is constant. When energy disappears in one form, it appears simultaneously in other
More informationLecture #13. Chapter 17 Enthalpy and Entropy
Lecture #13 Chapter 17 Enthalpy and Entropy First Law of Thermodynamics Energy cannot be created or destroyed The total energy of the universe cannot change Energy can be transferred from one place to
More informationOutline. 9. The Second Law of Thermodynamics: Entropy. 10.Entropy and the Third law of thermodynamics 11.Spontaneous change: Free energy
hermochemistry opic 6. hermochemistry hermochemistry Outline. Getting Started: Some terminology. State functions 3. Pressure-Volume Work 4. he First Law of hermodynamics: Heat, work and enthalpy 5. Heat
More informationUNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING TERM TEST 2 17 MARCH First Year APS 104S
UNIERSIY OF ORONO Please mark X to indicate your tutorial section. Failure to do so will result in a deduction of 3 marks. U 0 U 0 FACULY OF APPLIED SCIENCE AND ENGINEERING ERM ES 7 MARCH 05 U 03 U 04
More informationEnergy is the capacity to do work
1 of 10 After completing this chapter, you should, at a minimum, be able to do the following. This information can be found in my lecture notes for this and other chapters and also in your text. Correctly
More informationSecond Law of Thermodynamics
Second Law of Thermodynamics First Law: the total energy of the universe is a constant Second Law: The entropy of the universe increases in a spontaneous process, and remains unchanged in a process at
More informationClass XI Chapter 6 Thermodynamics Chemistry
Class XI Chapter 6 Chemistry Question 6.1: Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used
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 informationChemical thermodynamics the area of chemistry that deals with energy relationships
Chemistry: The Central Science Chapter 19: Chemical Thermodynamics Chemical thermodynamics the area of chemistry that deals with energy relationships 19.1: Spontaneous Processes First law of thermodynamics
More informationThermodynamics (XI) Assignment(Solution)
SYLLABUS CUM COM./XI/03 4 hermodynamics (XI) Assignment(Solution) Comprehension ype Questions aragraph for Question -5 For an ideal gas, an illustration of three different paths A, (B + C) and (D + E)
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 informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (30 pts.) 22 (30 pts.)
CHEMISTRY 202 Hour Exam II October 25, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationChemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry
Recall the equation. w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 10 2 J -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate. 8.3145 J/mol K 0.082057
More informationChapter 6. Using Entropy
Chapter 6 Using Entropy Learning Outcomes Demonstrate understanding of key concepts related to entropy and the second law... including entropy transfer, entropy production, and the increase in entropy
More informationPrevious lecture. Today lecture
Previous lecture ds relations (derive from steady energy balance) Gibb s equations Entropy change in liquid and solid Equations of & v, & P, and P & for steady isentropic process of ideal gas Isentropic
More information1 A reaction that is spontaneous.
Slide 1 / 55 1 A reaction that is spontaneous. A B C D E is very rapid will proceed without outside intervention is also spontaneous in the reverse direction has an equilibrium position that lies far to
More information1. (25 points) C 6 H O 2 6CO 2 + 7H 2 O C 6 H O 2 6CO + 7H 2 O
MEEBAL Exam 2 November 2013 Show all work in your blue book. Points will be deducted if steps leading to answers are not shown. No work outside blue books (such as writing on the flow sheets) will be considered.
More informationConsequences of Second Law of Thermodynamics. Entropy. Clausius Inequity
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd
More informationUniversity of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2011
Homework Assignment #: Due at 500 pm Wednesday July 6. University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 0 ) he respiratory system uses oxygen to degrade glucose to carbon
More informationHomework 11 - Second Law & Free Energy
HW11 - Second Law & Free Energy Started: Nov 1 at 9:0am Quiz Instructions Homework 11 - Second Law & Free Energy Question 1 In order for an endothermic reaction to be spontaneous, endothermic reactions
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (30 pts.) 22 (30 pts.)
CHEMISTRY 202 Hour Exam II October 25, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
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 informationClassification following properties of the system in Intensive and Extensive
Unit I Classification following properties of the system in Intensive and Extensive Extensive : mass, weight, volume, potential energy, Kinetic energy, Internal energy, entropy, exergy, energy, magnetization
More informationEngineering Thermodynamics. Chapter 6. Entropy: a measure of Disorder 6.1 Introduction
Engineering hermodynamics AAi Chapter 6 Entropy: a measure of Disorder 6. Introduction he second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of
More informationSelected Questions on Chapter 5 Thermochemistry
Selected Questions on Chapter 5 Thermochemistry Circle the correct answer: 1) At what velocity (m/s) must a 20.0 g object be moving in order to possess a kinetic energy of 1.00 J? A) 1.00 B) 100 10 2 C)
More informationConsequences of Second Law of Thermodynamics. Entropy. Clausius Inequity
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd
More informationReversibility, Irreversibility and Carnot cycle. Irreversible Processes. Reversible Processes. Carnot Cycle
Reversibility, Irreversibility and Carnot cycle The second law of thermodynamics distinguishes between reversible and irreversible processes. If a process can proceed in either direction without violating
More informationContent 10 Thermodynamics of gases Objectives Objectives 10.1 Heat capacity
hermodynamics of gases ontent. Heat capacities. ork done by a gas.3 irst law of thermodynamics.4 Isothermal adiabatic changes Objectives (a) define heat capacity, specific heat capacity molar heat capacity
More informationDishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)
HW-1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment
More informationSecond Law of Thermodynamics
Dr. Alain Brizard College Physics II (PY 211) Second Law of Thermodynamics Textbook Reference: Chapter 20 sections 1-4. Second Law of Thermodynamics (Clausius) Heat flows naturally from a hot object to
More informationPhysical Chemistry. Chapter 3 Second Law of Thermodynamic
Physical Chemistry Chapter 3 Second Law of hermodynamic by Izirwan Bin Izhab FKKSA izirwan@ump.edu.my Chapter Description Aims Develop the calculational path for property change and estimate enthalpy and
More informationCHEM Thermodynamics. Entropy, S
hermodynamics Change in Change in Entropy, S Entropy, S Entropy is the measure of dispersal. he natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING 007 5.9 Energy Environment and Society (a Project Based First Year Subject supported by the d'arbeloff Program) ---------------------------------------------------------------------------------------
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE
More informationAnswers to Problem Sheet (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous (e) nonspontaneous
Answers to Problem Sheet 5 1. (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous (e) nonspontaneous 2. (a) Heat will flow from the warmer block of iron to the colder block of iron until
More informationENGR Thermodynamics
ENGR 224 - hermodynamics W #5 Problem : 7.14 - he Increase of Entropy Principle - 2 pts 11-May-11 Will the entropy of steam increase, decrease or remain the same as it flows through a real adiabatic turbine?
More informationUniversity of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2008
University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 008 Midterm Examination Key July 5 008 Blue books are required. Answer only the number of questions requested. Chose wisely!
More informationKNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.
Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity
More informationPhysics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 32: Heat and Work II. Slide 32-1
Physics 1501 Fall 2008 Mechanics, Thermodynamics, Waves, Fluids Lecture 32: Heat and Work II Slide 32-1 Recap: the first law of thermodynamics Two ways to raise temperature: Thermally: flow of heat Energy
More informationc Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd
More informationHence. The second law describes the direction of energy transfer in spontaneous processes
Heat and Work The first law of thermodynamics states that: Although energy has many forms, the total quantity of energy is constant. When energy disappears in one form, it appears simultaneously in other
More informationCHAPTER THERMODYNAMICS
54 CHAPTER THERMODYNAMICS 1. If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction, then ΔHis always greater than ΔE ΔH< ΔE only if the number of moles of
More informationPressure Volume Temperature Relationship of Pure Fluids
Pressure Volume Temperature Relationship of Pure Fluids Volumetric data of substances are needed to calculate the thermodynamic properties such as internal energy and work, from which the heat requirements
More informationChem 401 Unit 1 (Kinetics & Thermo) Review
KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 2. Determine the relative reaction rates of
More informationCHEMICAL ENGINEERING DESIGN & SAFETY
CHEMICAL ENGINEERING DESIGN & SAFETY CHE 4253 Prof. Miguel Bagajewicz Explosions Explained EXPLOSION A rapid and uniform expansion Expanding Shock wave Initial System Boundary EXPLOSION The expansion ends
More informationSummarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0)
Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0) Key Point: ΔS universe allows us to distinguish between reversible and irreversible
More informationFirst Law of Thermodynamics: Closed Systems
utorial # First Law of hermodynamics: Closed Systems Problem -7 A 0.-m tank contains oxygen initially at 00kPa and 7 C. A paddle wheel within the tank is rotated until the pressure inside rise to 50kPa.
More informationw = -nrt hot ln(v 2 /V 1 ) nrt cold ln(v 1 /V 2 )[sincev/v 4 3 = V 1 /V 2 ]
Chemistry 433 Lecture 9 Entropy and the Second Law NC State University Spontaneity of Chemical Reactions One might be tempted based on the results of thermochemistry to predict that all exothermic reactions
More informationS.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks : 100
Total No. of Questions 12] [Total No. of Printed Pages 7 Seat No. [4162]-189 S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks
More informationProblem Set 10 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 10 Solutions 1 Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the triple point occurs at 543 K and 114 torr;
More informationChemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry S H 2 = S H 2 R ln P H2 P NH
N (g) + 3 H (g) NH 3 (g) S N = S H = S NH 3 = S N R ln P N S H R ln P H S NH 3 R ln P NH3 ΔS rxn = (S Rln P NH 3 NH3 ) (S N Rln P N ) 3 (S H Rln P H ) ΔS rxn = S S NH 3 N 3S H + Rln P P 3 N H ΔS rxn =
More informationCh. 19 Entropy and Free Energy: Spontaneous Change
Ch. 19 Entropy and Free Energy: Spontaneous Change 19-1 Spontaneity: The Meaning of Spontaneous Change 19-2 The Concept of Entropy 19-3 Evaluating Entropy and Entropy Changes 19-4 Criteria for Spontaneous
More informationPUT A SQUARE BOX AROUND ALL ANSWERS
UNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING CHE 112F PHYSICAL CHEMISTRY MID-TERM EXAM, OCTOBER 16 2014 6:10 pm to 8:00 pm EXAMINER P.V. Yaneff 1. Answer all questions. The marks add
More informationEnthalpy Chapter 5.3-4,7
Enthalpy Chapter 5.3-4,7 heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM E = q + w w transfer in (+w) w transfer out (-w) Internal Energy at Constant Volume E = E K + E P ΔE
More informationName: Discussion Section:
CBE 141: Chemical Engineering Thermodynamics, Spring 2018, UC Berkeley Midterm 1 February 13, 2018 Time: 80 minutes, closed-book and closed-notes, one-sided 8 ½ x 11 equation sheet allowed Please show
More informationOCR Chemistry A H432
All the energy changes we have considered so far have been in terms of enthalpy, and we have been able to predict whether a reaction is likely to occur on the basis of the enthalpy change associated with
More informationS = S(f) S(i) dq rev /T. ds = dq rev /T
In 1855, Clausius proved the following (it is actually a corollary to Clausius Theorem ): If a system changes between two equilibrium states, i and f, the integral dq rev /T is the same for any reversible
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 information