H C p = T Enthalpy can be obtained from the Legendre transform of the internal energy as follows, U + T = P T (RT )
|
|
- Lisa McCormick
- 7 years ago
- Views:
Transcription
1 roblem Set 2 Solutions 3.20 MI rofessor Gerbrand Ceder Fall 2003 roblem S(U, ) : ds du + d 1 ariables here are U and and intensive variables are and. o go to 1 as a natural variables take the Legendre transform by subtracting U J S U from S So, 1 dj U d + d Note that J is equal to F (F is the Helmholtz energy) Similarly, go from S(U, ) to B(U, ) by taking B S 1 db du d A Legendre transform with respect to both U and defines the lanck function (Yes, you too can be famous by defining your own Legendre transform) U Y S 1 dy U d d One can show that these functions have etrema properties just like the Legendre transforms of energy. Massieu functions are maimum under constant value of their natural variables. roblem 1.2 We know that C p is defined as
2 H C p Enthalpy can be obtained from the Legendre transform of the internal energy as follows, H U + We know that for an ideal gas U is a function of temperature only and the R (for one mole), thus H U + (R ) U is only a function of temperature and this is not a function of pressure. Also, R is not a function of pressure. So C p of an ideal gas must be independent of pressure.
3 roblem 1.3 F F (, ) Compare this with F F df d + d F U S herefore, df Sd d F Using the given formula for F, solve for by taking the derivative w.r.t at constant. F a R f + m b Since f( ) is only a function of, this term drops out and the solution is: F R a m b 2 roblem 1.4 (a) We can write the differential form of the entropy as a function of and 2 m S S ds d + d Multiplying by to get ds (which is equal to dq) S S ds d + d Since we are told the process is adiabatic, ds q 0, so, Or, S S d m d
4 S d α S Cp d C p (b) Using the above relationship, we can get a value for the final temperature, d α d C p f α ln Δ i C p α f i ep C p Δ For this case we can assume we have 1kg of material which means the volumes is m 3. Remember that α 3α L K m 999atm a/atm f 300 ep 2.3J/g K 1000g f K he only trick here is to be careful with your units and the definition of α L and α. roblem 1.5 (a) Start with the internal energy du ds d + HdM Using a Legendre transformation get G(,, H) G U S + HM dg Sd + d MdH And since we know that the order of differentiation does not matter 2 G 2 G H H G G H H H, H,, M H,, H,
5 (b) One possibility is S H, M H, roblem 1.6 (a) U ds + τdl G U S Which yields the solution: dg Sd + τdl ds + Sd (b)we also know if GG(L,) that hus, dg τdl Sd G G dg dl + d L G G τ and S L Since we know that dg is a perfect differential, then the second derivatives must be equal independent of the order which they are taken. So, (c) 2 G 2 G L L G G L L τ S L L L L du ds + τdl L L U L S + τ L
6 Using the relation from part (b), U L (d) If U is only a function of temperature, then LEEL 2 ROBLEMS roblem 2.1 τ + τ U τ 0 τ L L 1 τ 1 τ (a) Start by constructing the differential form of S(,) [In Zemansky this is called the second ds equation] S S ds d + d S S ds d + d But we know S C p and from a Mawell relation we know S.So, ds C p d For an isothermal compression, L L d C p d αd Q ds αd In the case of a solid or liquid, neither or α is very sensitive to pressure. So, Q α d αδ Now we can use the data given to calculate Q (Replacing with M/ρ) Q MαΔ ρ 298K 0.5kg K kg/m a Q 416J
7 (b)we now need to calculate the work during the compression W W d W d β β d β d f i W f Mβ 0.5kg a a 2 2 2ρ kg/m 3 W 44J (c) We know the first law, ΔU Q + W 416J + 44J ΔU 372J hus it can be seen that the etra amount of energy in the form of heat comes from the storage of internal energy. (d) Now we can go back to the relationship we obtained for ds (which we know is 0 since the process is adiabatic and reversible) ds 0 C p d αd d α d C p f α ln Δ i α αm f i ep Δ i ep Δ C p C p ρ 0.5kg K atm a/atm f 298 ep kg/m 3 385J/kg K 0.5kg C p f 300K Δ K roblem 2.2
8 In terms of and, du can be written U U du d + d We can evaluate the two partials U and U using U(S, ) which we know So du ds d U S Which can be simplified since we know C p S and α Similarly U C p α U S which can be simplified using the Mawell relation S utting this all together, U α + β (β α ) du (C α) d + (β α ) d and β roblem 2.3 Given the definition of dh : dh ds + d, we want first to determine how the enthalpy varies with pressure, at constant temperature: H S + By using the Mawell s Relation S (from the epression for the Gibbs Free Energy) and the definition 1 α
9 , we have: H (1 α ) What we are really interested in is to find how H evap H vap H liq changes with temperature: ΔHevap [H vap H liq ] H vap H liq ΔHevap vap liq vap α vap liq α liq vap (1 α vap ) In general, at relatively low pressures and densities, all gases behave ideally. For an ideal gas, α 1 nr 1 herefore, ΔHevap 0 roblem 2.4 Given U (S, X) X a ep(bs) First find the equations of state U X a ep(bs)b bu (S, X) S X U f ax a 1 au (S, X) ep(bs) X X From these we can derive the following which will be useful later S 1 S ln (1) b bx a b f X (2) a U (S, X) b (3) 1 a f X ep ( bs) a (4) Starting with the first function D(S, f ) we perform a Legendre transform as follows
10 D(S, f) U fx (1 a)u(s, X) Now using (4) to get this in terms of S and f only Similarly for F(,X) (1 a)a f D(S, f) (1 a) ep ( bs) ep(bs) a F (, X) U S U(S, X) bu(s, )S (1 bs)u(s, X) Using (3) and (1) from above F (, X) 1 ln bx a b Now we do a transformation for to both and f K(, f) U S fx Using (3), (1) and (2) a K(, f) ln b b bx a b K(, f) 1 a ln a b a b bf roblem 2.5 (a) Start by using the chain rule on the left side of the equation S S We know that S C and we can epand the second derivative as follows utting those together, C We can replace C by C using the relationship C 1 β α
11 Yielding α 2 C C β S β α 2 C β C α α β α (b) We can epand the derivative as follows S S S he first term is and the second term can be transformed using a Mawell relation S C. his gives (c) Start with the left side Using a Mawell relation, S C α S C 2 2 S 2 S Now the right side S Cv S S C v + he first term in the above is zero since the change in temperature at constant temperature is zero. he second term can be written as S 2 S And since we know the order of differentiation doesn t matter 2 S 2 S
12 roblem 2.6 (a) Start with the differential form of G (b) Start with the differential form of H dg Sd + d G S + G S G 1 S ( α) β G α S β dh ds + d H S + H C H C roblem 2.7 (For this problem Maple is our friend) (a) his part involves taking the indicated derivatives of the given function. he two tricks are knowing what β is and how to change U into something with derivatives of w.r.t or. β 1 and using the differential form of U and a Mawell relation we get, he solutions are then U
13 (b)remember R b 1 1 ( b) 2 2 β R 3 2a 2 + 4a b 2ab 2 U R a b 2 1 α R α Epand H in differential form to try and get H into a form you can solve for dh ds + d H S + H + H R + b Now for C C C C α 2 β R 3 2 C C 2 ( b) 2 roblem 2.8 U 1 k + Nf k 1 N k du ds d For a reversible, adiabatic process,
14 (du) S (d ) S Along a reversible adiabatic process: k g(s) so U Again for a reversible, adiabatic process (du) S Integrate from 0 at constant S Moving things around, g(s) k S g(s) k U d g(s) S d k U(, S) U( 0, S) g(s) 1 (k 1) 1 k 1 0 k 1 g(s) 1 g(s) U(, S) + U( 0, S) + k k 1 (k 1) 0 k 1 s) 1 g(s) U(, S) g( + U( 0, S) + k 1 k 1 (k 1) 0 k 1 he second term above is only a function of S, so we can call it some function w(s) k 1 U(, S) + w(s) +w(s) k 1 k 1 (k 1) w(s) is an etensive function, therefore it can be written as N (the number of moles in the system) times a molar quantity f(s) which is independent of the system size k f N k U(, S) +Nf (k 1) N k roblem 2.9
15 he thermal epansivity is defined as Using a Mawell relation, 1 α S And we are told in the problem that S is independent of crystallite structure and pressure, so And thus α 0 at absolute zero. roblem 2.10 S 0 (a) he key property is that S < 0. his can be proven from the Mawell relation. H S M H (this can be proven from the differential of dφ d(u S + MH) Sd + d MdH ) and since M H 0 H S < 0 < 0 H S( ) as constant H therefore has the following shape. Adiabatic demagnetization consists of two steps: I. Isothermal application of a field II. insulate material and turn off the field isotropic demagnetization. System moves back to state with H 0, hence decreases. (b) We can evaluate M H H S as follows H S H S H S M c H M H c H κ M κ M H H 2 H H
16 Figure 1: So, κ H H S c H LEEL 3 ROBLEMS roblem 3.1 Given: θ Rθ U S 2 2 R 0 (a) We want to make this the absolute U and not the molar quantity, so multiply by N. 2 2 Nθ NRθ Nθ S NRθ U NU S R 0 R N 0 N 1 θ Rθ U S 2 2 N R 2 0 he equations of state give us 2 U 2 θ S S N R N,
17 U N,S 2 Rθ N U θ Rθ θ Rθ µ 1 S 2 2 S 2 2 N N 2 R 2 R 2,S or (b) Epress µ as a function of and Substitute in to the epression for µ roblem 3.2 µ 1 N 2 µ U N R S 2 θ N Rθ θ N 2 R 2 Rθ N R 4 θ Rθ R µ θ 4 Rθ Start with the epression we used in class y + f z f y y f f z For this problem, this equation can relate the constant l heat capacity to the constant heat capacity as follows, S S S + l l o get heat capacity we need to multiply both sides by, and when we do this we get S C l C p + l he first derivative can be rewritten using the Mawell relation S simplifies to which
18 C l And since we know that C p α C l C p α Now we need to epand the second derivative l l l Since the material is isotropic, l 1 α which now leave us with 3 1 C + 2 α 2 l C p 3 l One more partial to go...if we look at the relationship between stress and strain, σ Eε and thus σ ε where E is the Young s modulus for an isotropic material. o get this in a form we can use, we need to multiply by l to switch from ε to l and then invert. his then will give us 1 2 α 2 E C l C p + 3 l l l E roblem 3.3 he total system is at constant volume and temperature hence its Helmholtz free energy is minimal with respect to the internal degrees of freedom. Figure 2:
19 df tot df α + df β α β df tot S α d α p α d α + µ A dn α A S β d β p β d β + µ dn β A A since d α d β 0 and d α d β and dn α dn β A A we can write µ A µ A df tot p α p β d α + α β α β If d α and dn α A were independent, equilibrium would require that p α p β and µ A µ A. However d α and dn α A are NO independent! df tot d α Kdn α A α β µ A µ A p α p β d α K he bracketed term must equal zero for equilibrium. dn α A α β µ A Kp α µ A Kp β roblem 3.4 he heat capacity c is related to the temperature derivative of the entropy under the condition. ( being the conditions defined in the problem statement) Starting with S(, ) we write S c S S ds d + d We can take the temperature derivative of this holding constant S S (ds) d + d S c p S + (Note: he above can also be obtained using y + f f y f z y f z )
20 We know through a Mawell relation that S α v. So now we need to find. We start from the given information that a a constant, thus d ad. his gives us 1 a Continuing from there, we write out the differential form of (, ) But we know So is given by Replacing we get or utting this all together... So we get finally that d d + d α v and β. We now have d α v d + β d α v β β α v a α v a a + β S c p 1 + ( α v ) a S c p 1 αv a + ( α v ) a a + β S 1 αv a c p + ( α v ) a a + β (α v ) 2 c c p a+β
Problem Set 4 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal
More informationa) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L
hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal
More informationProblem Set 3 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start
More informationGibbs Free Energy and Chemical Potential. NC State University
Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second
More informationProblem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003
LEVEL 1 PROBLEMS Problem Set 1 3.0 MIT Professor Gerbrand Ceder Fall 003 Problem 1.1 The internal energy per kg for a certain gas is given by U = 0. 17 T + C where U is in kj/kg, T is in Kelvin, and C
More informationvap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K
Thermodynamics: Examples for chapter 6. 1. The boiling point of hexane at 1 atm is 68.7 C. What is the boiling point at 1 bar? The vapor pressure of hexane at 49.6 C is 53.32 kpa. Assume that the vapor
More informationGive all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.
Chemistry 45/456 0 July, 007 Midterm Examination Professor G. Drobny Universal gas constant=r=8.3j/mole-k=0.08l-atm/mole-k Joule=J= Nt-m=kg-m /s 0J= L-atm. Pa=J/m 3 =N/m. atm=.0x0 5 Pa=.0 bar L=0-3 m 3.
More informationHEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases
UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius
More informationChapter 8 Maxwell relations and measurable properties
Chapter 8 Maxwell relations and measurable properties 8.1 Maxwell relations Other thermodynamic potentials emerging from Legendre transforms allow us to switch independent variables and give rise to alternate
More informationThermodynamics. Chapter 13 Phase Diagrams. NC State University
Thermodynamics Chapter 13 Phase Diagrams NC State University Pressure (atm) Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function
More informationThermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson
Thermochemistry r2 d:\files\courses\1110-20\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy
More informationPhysics 176 Topics to Review For the Final Exam
Physics 176 Topics to Review For the Final Exam Professor Henry Greenside May, 011 Thermodynamic Concepts and Facts 1. Practical criteria for identifying when a macroscopic system is in thermodynamic equilibrium:
More informationThe first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.
The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed
More informationAnswer, Key Homework 6 David McIntyre 1
Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making
More informationChapter 2 Classical Thermodynamics: The Second Law
Chapter 2 Classical hermodynamics: he Second Law 2.1 Heat engines and refrigerators 2.2 he second law of thermodynamics 2.3 Carnot cycles and Carnot engines 2.4* he thermodynamic temperature scale 2.5
More informationAP CHEMISTRY 2007 SCORING GUIDELINES. Question 2
AP CHEMISTRY 2007 SCORING GUIDELINES Question 2 N 2 (g) + 3 F 2 (g) 2 NF 3 (g) ΔH 298 = 264 kj mol 1 ; ΔS 298 = 278 J K 1 mol 1 The following questions relate to the synthesis reaction represented by the
More informationThermodynamics of Mixing
Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What
More information= 800 kg/m 3 (note that old units cancel out) 4.184 J 1000 g = 4184 J/kg o C
Units and Dimensions Basic properties such as length, mass, time and temperature that can be measured are called dimensions. Any quantity that can be measured has a value and a unit associated with it.
More informationPhys222 W11 Quiz 1: Chapters 19-21 Keys. Name:
Name:. In order for two objects to have the same temperature, they must a. be in thermal equilibrium.
More informationWhen the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.
Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs
More informationFinal Exam CHM 3410, Dr. Mebel, Fall 2005
Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At -31.2 C, pure propane and n-butane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture
More informationThermodynamics - Example Problems Problems and Solutions
Thermodynamics - Example Problems Problems and Solutions 1 Examining a Power Plant Consider a power plant. At point 1 the working gas has a temperature of T = 25 C. The pressure is 1bar and the mass flow
More informationThe final numerical answer given is correct but the math shown does not give that answer.
Note added to Homework set 7: The solution to Problem 16 has an error in it. The specific heat of water is listed as c 1 J/g K but should be c 4.186 J/g K The final numerical answer given is correct but
More informationHeat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.
Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical
More informationTo give it a definition, an implicit function of x and y is simply any relationship that takes the form:
2 Implicit function theorems and applications 21 Implicit functions The implicit function theorem is one of the most useful single tools you ll meet this year After a while, it will be second nature to
More informationLecture 3 Fluid Dynamics and Balance Equa6ons for Reac6ng Flows
Lecture 3 Fluid Dynamics and Balance Equa6ons for Reac6ng Flows 3.- 1 Basics: equations of continuum mechanics - balance equations for mass and momentum - balance equations for the energy and the chemical
More informationFluid Mechanics: Static s Kinematics Dynamics Fluid
Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three
More informationScalars, Vectors and Tensors
Scalars, Vectors and Tensors A scalar is a physical quantity that it represented by a dimensional number at a particular point in space and time. Examples are hydrostatic pressure and temperature. A vector
More informationEquilibrium. Ron Robertson
Equilibrium Ron Robertson Basic Ideas A. Extent of Reaction Many reactions do not go to completion. Those that do not are reversible with a forward reaction and reverse reaction. To be really correct we
More informationSTRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION
Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable
More informationEquations, Inequalities & Partial Fractions
Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities
More informationExergy: the quality of energy N. Woudstra
Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if
More information3.1. Solving linear equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Solving linear equations 3.1 Introduction Many problems in engineering reduce to the solution of an equation or a set of equations. An equation is a type of mathematical expression which contains one or
More informationChem 420/523 Chemical Thermodynamics Homework Assignment # 6
Chem 420/523 Chemical hermodynamics Homework Assignment # 6 1. * Solid monoclinic sulfur (S α ) spontaneously converts to solid rhombic sulfur (S β ) at 298.15 K and 0.101 MPa pressure. For the conversion
More informationDefine the notations you are using properly. Present your arguments in details. Good luck!
Umeå Universitet, Fysik Vitaly Bychkov Prov i fysik, Thermodynamics, 0-0-4, kl 9.00-5.00 jälpmedel: Students may use any book(s) including the textbook Thermal physics. Minor notes in the books are also
More informationKinetic Molecular Theory of Matter
Kinetic Molecular Theor of Matter Heat capacit of gases and metals Pressure of gas Average speed of electrons in semiconductors Electron noise in resistors Positive metal ion cores Free valence electrons
More informationSo T decreases. 1.- Does the temperature increase or decrease? For 1 mole of the vdw N2 gas:
1.- One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The gas goes through the free expansion process (Q = 0, W = 0), in which the pressure drops down to the atmospheric
More informationParabolic Equations. Chapter 5. Contents. 5.1.2 Well-Posed Initial-Boundary Value Problem. 5.1.3 Time Irreversibility of the Heat Equation
7 5.1 Definitions Properties Chapter 5 Parabolic Equations Note that we require the solution u(, t bounded in R n for all t. In particular we assume that the boundedness of the smooth function u at infinity
More informationChapter 7 : Simple Mixtures
Chapter 7 : Simple Mixtures Using the concept of chemical potential to describe the physical properties of a mixture. Outline 1)Partial Molar Quantities 2)Thermodynamics of Mixing 3)Chemical Potentials
More informationCompressible Fluids. Faith A. Morrison Associate Professor of Chemical Engineering Michigan Technological University November 4, 2004
94 c 2004 Faith A. Morrison, all rights reserved. Compressible Fluids Faith A. Morrison Associate Professor of Chemical Engineering Michigan Technological University November 4, 2004 Chemical engineering
More information1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes
Arithmetic of Algebraic Fractions 1.4 Introduction Just as one whole number divided by another is called a numerical fraction, so one algebraic expression divided by another is known as an algebraic fraction.
More informationChem 338 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21
Chem 8 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.1, 5.15, 5.17, 5.21 5.2) The density of rhombic sulfur is 2.070 g cm - and that of monoclinic sulfur is 1.957 g cm -. Can
More informationCHAPTER 12. Gases and the Kinetic-Molecular Theory
CHAPTER 12 Gases and the Kinetic-Molecular Theory 1 Gases vs. Liquids & Solids Gases Weak interactions between molecules Molecules move rapidly Fast diffusion rates Low densities Easy to compress Liquids
More informationStatistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004
Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein
More informationLecture 1: Physical Equilibria The Temperature Dependence of Vapor Pressure
Lecture 1: Physical Equilibria The Temperature Dependence of Vapor Pressure Our first foray into equilibria is to examine phenomena associated with two phases of matter achieving equilibrium in which the
More information1. Degenerate Pressure
. Degenerate Pressure We next consider a Fermion gas in quite a different context: the interior of a white dwarf star. Like other stars, white dwarfs have fully ionized plasma interiors. The positively
More informationConstrained optimization.
ams/econ 11b supplementary notes ucsc Constrained optimization. c 2010, Yonatan Katznelson 1. Constraints In many of the optimization problems that arise in economics, there are restrictions on the values
More informationAvailability. Second Law Analysis of Systems. Reading Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.
Availability Readg Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.88 Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of reversible work that can be obtaed
More informationReview of Chemical Equilibrium Introduction
Review of Chemical Equilibrium Introduction Copyright c 2016 by Nob Hill Publishing, LLC This chapter is a review of the equilibrium state of a system that can undergo chemical reaction Operating reactors
More informationLecture 9, Thermal Notes, 3.054
Lecture 9, Thermal Notes, 3.054 Thermal Properties of Foams Closed cell foams widely used for thermal insulation Only materials with lower conductivity are aerogels (tend to be brittle and weak) and vacuum
More informationCHEM 36 General Chemistry EXAM #1 February 13, 2002
CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show
More informationExperiment 12E LIQUID-VAPOR EQUILIBRIUM OF WATER 1
Experiment 12E LIQUID-VAPOR EQUILIBRIUM OF WATER 1 FV 6/26/13 MATERIALS: PURPOSE: 1000 ml tall-form beaker, 10 ml graduated cylinder, -10 to 110 o C thermometer, thermometer clamp, plastic pipet, long
More information) and mass of each particle is m. We make an extremely small
Umeå Universitet, Fysik Vitaly Bychkov Prov i fysik, Thermodynamics, --6, kl 9.-5. Hjälpmedel: Students may use any book including the textbook Thermal physics. Present your solutions in details: it will
More informationSupplementary Notes on Entropy and the Second Law of Thermodynamics
ME 4- hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change
More information4. Introduction to Heat & Mass Transfer
4. Introduction to Heat & Mass Transfer This section will cover the following concepts: A rudimentary introduction to mass transfer. Mass transfer from a molecular point of view. Fundamental similarity
More informationChapter 18 Homework Answers
Chapter 18 Homework Answers 18.22. 18.24. 18.26. a. Since G RT lnk, as long as the temperature remains constant, the value of G also remains constant. b. In this case, G G + RT lnq. Since the reaction
More informationCBE 6333, R. Levicky 1 Review of Fluid Mechanics Terminology
CBE 6333, R. Levicky 1 Review of Fluid Mechanics Terminology The Continuum Hypothesis: We will regard macroscopic behavior of fluids as if the fluids are perfectly continuous in structure. In reality,
More informationTopic 3b: Kinetic Theory
Topic 3b: Kinetic Theory What is temperature? We have developed some statistical language to simplify describing measurements on physical systems. When we measure the temperature of a system, what underlying
More information3 Theory of small strain elastoplasticity 3.1 Analysis of stress and strain 3.1.1 Stress invariants Consider the Cauchy stress tensor σ. The characteristic equation of σ is σ 3 I 1 σ +I 2 σ I 3 δ = 0,
More informationIntegrating algebraic fractions
Integrating algebraic fractions Sometimes the integral of an algebraic fraction can be found by first epressing the algebraic fraction as the sum of its partial fractions. In this unit we will illustrate
More information18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a
First Major T-042 1 A transverse sinusoidal wave is traveling on a string with a 17 speed of 300 m/s. If the wave has a frequency of 100 Hz, what 9 is the phase difference between two particles on the
More informationThermodynamics AP Physics B. Multiple Choice Questions
Thermodynamics AP Physics B Name Multiple Choice Questions 1. What is the name of the following statement: When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium
More informationGases and Kinetic-Molecular Theory: Chapter 12. Chapter Outline. Chapter Outline
Gases and Kinetic-Molecular heory: Chapter Chapter Outline Comparison of Solids, Liquids, and Gases Composition of the Atmosphere and Some Common Properties of Gases Pressure Boyle s Law: he Volume-Pressure
More information01 The Nature of Fluids
01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of
More informationMath 2443, Section 16.3
Math 44, Section 6. Review These notes will supplement not replace) the lectures based on Section 6. Section 6. i) ouble integrals over general regions: We defined double integrals over rectangles in the
More information1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids
1. Fluids Mechanics and Fluid Properties What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids - both liquids and gases.
More informationElasticity Theory Basics
G22.3033-002: Topics in Computer Graphics: Lecture #7 Geometric Modeling New York University Elasticity Theory Basics Lecture #7: 20 October 2003 Lecturer: Denis Zorin Scribe: Adrian Secord, Yotam Gingold
More informationIntegrals of Rational Functions
Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t
More information3.2. Solving quadratic equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Solving quadratic equations 3.2 Introduction A quadratic equation is one which can be written in the form ax 2 + bx + c = 0 where a, b and c are numbers and x is the unknown whose value(s) we wish to find.
More informationMathematics. (www.tiwariacademy.com : Focus on free Education) (Chapter 5) (Complex Numbers and Quadratic Equations) (Class XI)
( : Focus on free Education) Miscellaneous Exercise on chapter 5 Question 1: Evaluate: Answer 1: 1 ( : Focus on free Education) Question 2: For any two complex numbers z1 and z2, prove that Re (z1z2) =
More informationChapter 1 Classical Thermodynamics: The First Law. 1.2 The first law of thermodynamics. 1.3 Real and ideal gases: a review
Chapter 1 Classical Thermodynamics: The First Law 1.1 Introduction 1.2 The first law of thermodynamics 1.3 Real and ideal gases: a review 1.4 First law for cycles 1.5 Reversible processes 1.6 Work 1.7
More informationStandard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K
1 Thermodynamics There always seems to be at least one free response question that involves thermodynamics. These types of question also show up in the multiple choice questions. G, S, and H. Know what
More informationChapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57
Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity
More informationAPPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES
APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES INTRODUCTION This tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more
More informationMaximum Likelihood Estimation
Math 541: Statistical Theory II Lecturer: Songfeng Zheng Maximum Likelihood Estimation 1 Maximum Likelihood Estimation Maximum likelihood is a relatively simple method of constructing an estimator for
More informationis identically equal to x 2 +3x +2
Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3
More informationTHE KINETIC THEORY OF GASES
Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure
More informationHeterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi
Heterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi Module - 03 Lecture 10 Good morning. In my last lecture, I was
More informationFUNDAMENTALS OF ENGINEERING THERMODYNAMICS
FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant
More informationASCII CODES WITH GREEK CHARACTERS
ASCII CODES WITH GREEK CHARACTERS Dec Hex Char Description 0 0 NUL (Null) 1 1 SOH (Start of Header) 2 2 STX (Start of Text) 3 3 ETX (End of Text) 4 4 EOT (End of Transmission) 5 5 ENQ (Enquiry) 6 6 ACK
More informationKINETIC THEORY OF GASES AND THERMODYNAMICS SECTION I Kinetic theory of gases
KINETIC THEORY OF GASES AND THERMODYNAMICS SECTION I Kinetic theory of gases Some important terms in kinetic theory of gases Macroscopic quantities: Physical quantities like pressure, temperature, volume,
More informationMechanical Properties - Stresses & Strains
Mechanical Properties - Stresses & Strains Types of Deformation : Elasic Plastic Anelastic Elastic deformation is defined as instantaneous recoverable deformation Hooke's law : For tensile loading, σ =
More informationHigh Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur
High Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 06 One-dimensional Gas Dynamics (Contd.) We
More informationCHAPTER 14 THE CLAUSIUS-CLAPEYRON EQUATION
CHAPTER 4 THE CAUIU-CAPEYRON EQUATION Before starting this chapter, it would probably be a good idea to re-read ections 9. and 9.3 of Chapter 9. The Clausius-Clapeyron equation relates the latent heat
More informationEXERCISES. 16. What is the ionic strength in a solution containing NaCl in c=0.14 mol/dm 3 concentration and Na 3 PO 4 in 0.21 mol/dm 3 concentration?
EXERISES 1. The standard enthalpy of reaction is 512 kj/mol and the standard entropy of reaction is 1.60 kj/(k mol) for the denaturalization of a certain protein. Determine the temperature range where
More informationFLUID DYNAMICS. Intrinsic properties of fluids. Fluids behavior under various conditions
FLUID DYNAMICS Intrinsic properties of fluids Fluids behavior under various conditions Methods by which we can manipulate and utilize the fluids to produce desired results TYPES OF FLUID FLOW Laminar or
More information9.2.3.7 Retention Parameters in Column Chromatography
9.2.3.7 Retention Parameters in Column Chromatography Retention parameters may be measured in terms of chart distances or times, as well as mobile phase volumes; e.g., t R ' (time) is analogous to V R
More informationEngineering Problem Solving as Model Building
Engineering Problem Solving as Model Building Part 1. How professors think about problem solving. Part 2. Mech2 and Brain-Full Crisis Part 1 How experts think about problem solving When we solve a problem
More informationDefine conversion and space time. Write the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR.
CONERSION ND RECTOR SIZING Objectives: Deine conversion and space time. Write the mole balances in terms o conversion or a batch reactor, CSTR, PR, and PBR. Size reactors either alone or in series once
More information11 Thermodynamics and Thermochemistry
Copyright ç 1996 Richard Hochstim. All rights reserved. Terms of use.» 37 11 Thermodynamics and Thermochemistry Thermodynamics is the study of heat, and how heat can be interconverted into other energy
More informationChapter 9 Partial Differential Equations
363 One must learn by doing the thing; though you think you know it, you have no certainty until you try. Sophocles (495-406)BCE Chapter 9 Partial Differential Equations A linear second order partial differential
More informationThermodynamics 2nd year physics A. M. Steane 2000, revised 2004, 2006
Thermodynamics 2nd year physics A. M. Steane 2000, revised 2004, 2006 We will base our tutorials around Adkins, Equilibrium Thermodynamics, 2nd ed (McGraw-Hill). Zemansky, Heat and Thermodynamics is good
More informationTHE IDEAL GAS LAW AND KINETIC THEORY
Chapter 14 he Ideal Gas Law and Kinetic heory Chapter 14 HE IDEAL GAS LAW AND KINEIC HEORY REIEW Kinetic molecular theory involves the study of matter, particularly gases, as very small particles in constant
More informationStirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator
Lecture. Real eat Engines and refrigerators (Ch. ) Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator Carnot Cycle - is not very
More informationU = x 1 2. 1 x 1 4. 2 x 1 4. What are the equilibrium relative prices of the three goods? traders has members who are best off?
Chapter 7 General Equilibrium Exercise 7. Suppose there are 00 traders in a market all of whom behave as price takers. Suppose there are three goods and the traders own initially the following quantities:
More informationChapter 2. Parameterized Curves in R 3
Chapter 2. Parameterized Curves in R 3 Def. A smooth curve in R 3 is a smooth map σ : (a, b) R 3. For each t (a, b), σ(t) R 3. As t increases from a to b, σ(t) traces out a curve in R 3. In terms of components,
More informationFluids and Solids: Fundamentals
Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.
More informationUniversity of Maryland Fraternity & Sorority Life Spring 2015 Academic Report
University of Maryland Fraternity & Sorority Life Academic Report Academic and Population Statistics Population: # of Students: # of New Members: Avg. Size: Avg. GPA: % of the Undergraduate Population
More informationAPPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.
APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend
More informationSchooling, Political Participation, and the Economy. (Online Supplementary Appendix: Not for Publication)
Schooling, Political Participation, and the Economy Online Supplementary Appendix: Not for Publication) Filipe R. Campante Davin Chor July 200 Abstract In this online appendix, we present the proofs for
More information