Problem Set 1 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001

Size: px
Start display at page:

Download "Problem Set 1 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001"

Transcription

1 LEEL ROBLEMS rblem Set Slutins. MI ressr Gerbrand Ceder Fall rblem. Gas is heating in a rigid cntainer rm 4 C t 5 C U U( ) U( ) W + Q (First Law) (a) W Since nly wrk is pssible & since the cntainer is rigid. Q U (. + C ) (. + C) Q U. ( ) 46.5 k kg (b) Entrpy change: since W r,. d ds. d S. ln ds Q Q du. d. k kgk (r reversible heating)

2 rblem. Calculate entrpy change by ging rm t accrding t: ( ) A reversible isthermal cmpressin ( ) Fllwed by an isbaric heating Entrpy change Entrpy change U ( U U( ) ) nr Q W r Q pd d Q nr ds d S nrln nr ln 6. 5 mle K Q c d Cp n R Q cd p ds nr d S nrln mle K p (isbaric heating) Nw r the ttal prcess tt S S + S. 94 mle K

3 rblem. (a) 9 C C H c d + H + c d C mlek j mlek 9C HFe 4 (9 C C) ( mle K mle mle K C 9 C ) (b) System Fe+ice water is adiabatic H H ( m H ), m mass ice transrmed m Fe H H p, H H F e Fe H F e ice melting ice ice Fe melting icewater 495 g 648 ice -4.95k m g p, ml 56g ml 8g rblem.4 Diatmic ideal gas c R and c R p v 5

4 Assume the prcess is reversible. here are tw methds ne can use t ind the wrk dne alng this path. Fr bth appraches it will be useul t ind p,, r the starting and ending pints. We are given the starting pressure, vlume and temperature and the inal pressure. Frm this we can calculate the inal vlume and temperature. Methd I Start with cp cv p p 6. 99m he inal temperature can be und with a similar relatin c c cv R c c p p S, d w pd p ) w ( ( ) ) ( ) ( ) ( ) Inserting in the values r,, we get w. M Methd II 5 p p (with ) Start with the knwn act the the internal energy an ideal gas is nly a unctin temperature. U Q+ W nc d w nc ( ) v (with c ) K It will als be useul t calculate the number mles rm p nr giving n 65 mles. In summary the cnditins at each pint are with p ( a) 595 ( m ) ( K) 98 p Hwever, since this is an adiabatic path, Q and the wrk dne is simple nc d p w. M v v cp cv 5 5 4

5 his prcess is at cnstant pressure, s the wrk is given by w pd p d w p ( ) We just have t ine. hat can be dne easily by using the equatin state an ideal gas alng an istherm S, w a m 5m w 89k he ttal wrk dne is then w tt k rblem.5 Functin : dy ( x + y ) dx + x( x + y ) dy () Integrate alng y x ( dy dx) x( x + x ) dx+ x( x + x ) dx 4 x (4 x + x ) dx x dx 4 () Integrate alng y x ( dy xdx) 4 4 x ( x + x ) dx + x( x + x )xdx x (5 x + 5 x ) dx x + 5 Functin : dy ( x + y) dx + x( x + y) dy () Integrate alng y x ( dy dx) x( x + y) dx+ x( x + x) dx 4 x x (4x + x) dx () Integrate alng y x ( dy xdx) x ( x + x ) dx + x( x + x )xdx 4 x xdx Since unctin is path independent it is an exact dierential. 5

6 rblem.6 du w + Q his is an ideal gas s du alng an isthermal path. (a) R w pd, p R w ln ln d R p R p w R ln(5) 654 (b) (c) du Q w Q 654 H U + H U + Fr an ideal gas at cnstant temperature ( ) U U Als, when herere H H (d) Q pd nr ds d S Rln R ln S. 5 K 6

7 rblem. (a) Wrk dne in the prcess is: Deine water as the system S we write w w inal initial vapr liq vapr w p d (Since p is cnstant) liq 5 pd w p( ) vap p a w p vap vapr. 59 his is abut.5% the heat evaprtatin (b) g, Initial State Liquid Final State apr kg m pd Since the speciic vlume a vapr is much larger than r a liquid liq U H p U g g kg H U + p H U + ( p ) U + p Cnclusin: he heat ne has t transer t water t evaprate it is partly used r increasing the internal energy water (94 breaking bnds) but asl r the wrk required by the vapr expansin. g g rblem.8

8 Given: N p atm m liter. m liters. m K Need t knw hw many mles gas: (isthermal) (isbaric) w p n. mles R N p p p atm 565 m nr pd d nr ln w nr ln. 9 p p cp 5 with c r mnatmic ideal gas v p. 5 liters. 5m p N w p m m w 4. (reversible adiabatic) w pd w p p p p [ ] p p d w. S the ttal wrk dne is ttal w w + w + w ttal w 9. 5 p p rblem.9 (i) he amunt wrk dne by the gas is zer, since the gas des n wrk n the surruindings utside the chamber. he expressin w pd cannt be applied since the prcess is nt reversible. (ii) he walls the chamber are insulating, thus Q and rm (i) w. hus U Q+ W (iii) Fr an ideal gas U is nly a unctin temperature. Since r a ree expansin in an insulated chamber U initital inal (iv) Fr a nn-ideal gas where U (, ), will change ater a ree expansin in an insulated chamber since changes and U U (, ) U (, ) inal inal initial initital 8

9 (Mst gasses at lw pressures can be well apprximated as being ideal gasses) (v) I nw the walls the chambers cnduct heat, r an ideal gas U U( ) is still true. he initial and inal states are in equilibrium with the envirnment. hus initial inal envirnment (vi) Fr nn-ideal gases U since U U(,) and changes. LEEL ROBLEMS rblem. (a) S (b) S> (c) S> rblem. Given inrmatin i i m? i 5 atm atm 98 K 98K (a) Find the inal vlume ater the expansin i i i i 5 5. m Finding the number mles, n, will be useul r parts b & c L atm R. 85, m L K mle 5 n R n 6. 5 mles (b) Find the wrk dne i the prcess is isthermal w pd. w nr ln ln 5 i w 6 k (c) Find the wrk in the multi-step prcess 9

10 w parts: (I) Adiabatic expansin rm t (II) Heating at cnstant rm t ( wi + wii)+( QI + QII) Usy s Since state and are at the same temperature and U r an ideal gas is nly a unctin temperature, Usys and ( wi + wii) ( QI + QII) S we can calculate w s r Q s. Since QI (adiabatic prcess) let s calculat QII QII ncp ncp( ) n 6. 5 mles, cp R 9 mle K Need t ind R cp 98 68K 5 Nw r QII Q II 6. mles 9 (98 68 K) mle K Q k S, II w ( Q + Q ) ( + k) ttal I II w ttal k rblem. U Ap ( A psitive cnstant) du w + Q Alng a reverisble path, W pd when nly p- wrk is pssible. Alng a reversible adiabatic path, Q and therere du w (First Law)

11 Ging back t equatin Ater intergratin U U du dp + d p p U Ap p U Ap p ( Ap ) dp + ( Ap ) d pd A dp ( Ap +) d dp d ( Ap +) A ln( Ap + ) ln + cnstant [ A ] A ln ( Ap + ) new cnstant () ( Ap +) even newer cnstant rblem.4 (a) Using the irst law Usys Q(5 C) + Q( C) + W + Q(5 C). 5 MW +. 5MW MW + Q(5 C) Q(5 C) MW (b) Chack t see i the secnd law is satisied ds machine Q But ds machine has t be zer since we are perating at steady state Q(5 C) Q C Q C + ( ) + (5 ) MW. MW MW K 5K 98K K s Ks 54. 6

12 rblem.5 (a) Since the prcess is adiabatic, Q (b) U Q+ w U ncv w 5 ( mles ) mle K 85K 88K S S +S S since this prcess is reversible and adiabatic S S isat cnstant pressure, temperatre varies p What is thugh? he temperatre that wuld be reached i expansin was reversible. R cp p 5K p 88 S c d p ln 5 5 S. 4 mle K rblem.6 In case (b) extra wrk is dne n the gas s that its internal energy will remain higher than in (a) > b a rblem. (a) tal amunt energy required Necessary heat input is the ttal enthalpy change the material (since p is cnstat) Determined by the inal and initial state

13 Enthalpy change is nly due t the L (the 5 L remains in exactly the same state) dh nc d g cal H L 5K L g K H 5 kcal 4k H. 8kWh (b) Entrpy change Again,entrpy change is sley determined by the inal and initial state ds p nc d p g cal S L L g K S 999 ln 5 8 mlk 8. kcal K rblem.8 d w pd R Rln 4 w R ln he secnd step where Q gives du dw and the urth step hus and Bd pd R d B ln R ln B ln R ln 4 r 4 4

14 w 4 w R ln R ln w( ) w( ) Q ( ) Q( ) w w w w + w rblem.9 Q By drawing the heat Q rm a heat suce at ne may ths prduce mechanical wrk in the amunt i the rest the energy, can be dispsed as heat t a heat sink at Q( ) Suniverse Ssys +Senvirnment here is n heat lw int the envirnment, s Senvirnment But r the system (the gas) he gas ges rm state at i, i, i t a state at,, ( i) with < i cmpute the Sgas we need t hw the entrpy changes with pressure (r vlume) at cnstant S (Maxwell Relatin) and r an ideal gas R Sgas R Rln i i since i > i R ln > Since S > the prcess is irreversible universe rblem. calculate U(,) r any (,) given a value U(,) at sme (,) it is sitable r this system t llw the path sketched belw. 4

15 First g rm,, A cnstant clume, i nly - wrk is pssible du Q U U(, ) U(, ) Q A( ) is unknwn, but we d knw because the path we chse rm (,) (,) that (, ) and (,) are cnnected by a reversible adiabat. herere, r S, (i.e. cnstant vlume) Secnd g rm,, Alng the reversible adiabat Q and therere U U(,) U(, ) d U, ( ) U, ( ) Ar ( ) (with r ) with and representing values alng the adiabat ging thrugh (, ). calaculate this integral use S, Cmbining the irst and secnd step U d U ( ( ) ) ( ) ( ) ( ) U U(,) U(, ) r (with r ) ( ) U U(,) U(, ) A( r )+ r (with r ) tt ( ) LEEL ROBLEMS rblem. he reactin at K is nr NH N +H, and are cnstant but nincreases by a actr in the reactin. herere atm i (b) Heat lw he system is undercnstant vlume ( d ) s W and U Q We can cmpute U rm H because U H 5

16 U H ( ) H ( nr ) H R n k H 8 and n mles mle U 8 () 8. 4 ( K) mle mle K (c) System is adiabtic and under cnstant vlume U S, use c c R p v a U + ( c, + c, ) d reactin a U Q 646 a v N v H U c, +c, vn vh a 5K mle reactin mle mlek he system cls dwn which is expected given that the reactin is endthermic. rblem. We can deine the system as the gas in the tank ( n mles) and the gas that will be squeezed int the tank ( n mles). It will helpul t deine sme variables c he initial pressure in the tank he external pressure pushing gas in he initial temperature the gas he inal temperature the gas (what we want t slve r) he vlume the tank he vlume the gas pushed in We can write the llwing relatinships n R n + nc R c nr c Since this is an ideal gas we knw that the internal energy change is nly a unctin temperature, given by Givem that the prcess is adiabatic (islated), U ( n + n ) c ( ) c v But r this case we knw that U Q+ W W ( Q ) W n R c c 6

17 r the external pressure (which is cnstant) times the vlume gas pushed in. Nw we can just wrk thrugh eliminating variables t ind the inal temperature. nr ( n + n) c( ) c c v nr c v ( ) R c R v ( ) R R R c R R c ( ) v ( R+ cv) ( R+ cv) ( R+ c ) R v + c v reating air as a diatmic ideal gas, we can use cv R and slve r 98 ( R + R ) 98 (. 5 R+ cv) K 5 rblem. Assume that the vlume change resulting rm the reactin A + B AB is and the heat reactin is Q. During the reactin, the system will perrm wrk against the envirnment which has a cnstant pressure. Since the pressure the envirnment is cnstant, the external wrk dne by the system is w Using the irst law U Q Q U + In the initial and inal states the reactin (bth equilibrium states), the pressure the gas has t equal (therwise they wuldn t be equilibrium states). herere i is the pressure the gas r rblem.4 Q U +( ) Q H where H is the dierence enthalpies in the initial and inal states. Clsed system slutin System is gas lwing int the tank Fr an ideal gas, U U Q+ w w ( Q (adiabatic)) U U U U nc ( ) v nr

18 S, Open system slutin System is the tank U nc ( ) nr v c ( R+ c) c c v v p p c v Hm U U U m initial i i i because m m in m U U initial H U U + i in in in c ( ) R cp c i i in in v i i v rblem.5 (a) Gas in tank I is underging an adiabatic reversible expansin R cp R 5 R 46K (b) ry t get II rm the act that Utt UI + UII because the ttal system is isthermal. UI is the internal energy change gas that remains in I and UII is the internal energy change gas that ends up in tank II. U n c ( ) need t ind n I nw rm U + U I II II II v II U n c ( ) I I v I n R ni I atm K 5 6. n I atm 46K n n n (. 6). 9n II I UII UI. 9 n c ( ). 6 n c (46 ) v II v II 84K (c) Entrpy change S varies with and 8

19 S S ds d + d S cp S R and cp ds d R d Fr the gas remaining in tank I dsi as it is underging an adibatic reversible expansin. Fr the gas ending up in tank II II II S nii cp ln R ln atm, K LEEL 4 ROBLEMS n 9. (per mle gas initial in tank I) II II 84K niirii nii II II 5atm n 9 ln 84 5 S. n R Rln S 5. 5n Kmle rblem 4. his is a very tricky questin which is why it is alne in Level 4. First let s determine the equilibrium cnditins. Assume that the system has sme hw reached equilibirum with inal values,, r cmpartment and,, r cmpartment. We can ssk what criteria must there variable satisy r the system t be in equilibrium. Well we knw that r any reversible change arund equilibrium du du + du i.e. U,S ( ) must be minimal at equilibrium, and ds ds + ds i.e. SU,S ( ) must be maximal at equilibrium. Since there is n heat lw between cmpartments and between the chamber and the envirnment (since all walls are adiabatic) ds ds d S at equilibrium, the nly reversible changes we can make are in the vlume the cmpartments, i.e. hen we can write (r a reversible change d du d + d du d 9

20 d du du+ du ( ) d since this shuld hld r arbitrary d, i.e. equilibrium is characterized by mechanical equilibrium. NOE: Since ds and ds, the equilibrium cnditin that du des nt yield any inrmatin abut the inal temperatures and. his means that the inal temperatures this system depend n the path llwed by the system t attain equilibrium. he questin remains: can we btain equilibrium rm the initial state,,, and n n by llwing a reverible path? he answer is: prbably nt. One candidate reversible path is t allw the pistn t very slw mve until the equilibrium pressures are attained (i.e. ). his path, hwever, cannt prceed reversible r the llwing reasn: Assume that is can prceed reversibly. hen an ininitesimal change in the vlume cmpartment by d will result in an increase the internal energy cmpartment ne by du he change in internal energy cmpartment is du d d But since bth cmpartments are adiabatically sealed rm the envirnment, du du du r d d r But this is nt true except in the inal equilibrium state. In the initial state by assumptin. herere the suggested path cannt prceed reversibly. he suggest path must be accmpanied by dissipatin (thrugh rictin between the pistn and the walls r example). his dissipatin will be accmpanied by a prductin heat. We cannt predict a-priri hw much that heat evlves t cmpartment r cmpartment thugh.

" 1 = # $H vap. Chapter 3 Problems

 1 = # $H vap. Chapter 3 Problems Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius

More information

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred

More information

Thermochemistry. Thermochemistry

Thermochemistry. Thermochemistry Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk

More information

4F-5 : Performance of an Ideal Gas Cycle 10 pts

4F-5 : Performance of an Ideal Gas Cycle 10 pts 4F-5 : Perfrmance f an Cycle 0 pts An ideal gas, initially at 0 C and 00 kpa, underges an internally reversible, cyclic prcess in a clsed system. The gas is first cmpressed adiabatically t 500 kpa, then

More information

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3

More information

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,

More information

A) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N

A) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N Q1. A transverse sinusidal wave travelling n a string is given by: y (x,t) = 0.20 sin (2.5 x 80 t) (SI units). The length f the string is 2.0 m and its mass is 1.5 g. What is the magnitude f the tensin

More information

Work and Heat Definitions

Work and Heat Definitions Wrk and eat Deinitins FL- Surrundings: Everything utside system + q -q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces - eat, q: transer energy resulting rm

More information

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)

More information

Thermodynamics Partial Outline of Topics

Thermodynamics Partial Outline of Topics Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)

More information

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat

More information

More Tutorial at

More Tutorial at Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,

More information

Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics

More information

Unit 14 Thermochemistry Notes

Unit 14 Thermochemistry Notes Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm

More information

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change? Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S

More information

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany

More information

Solution to HW14 Fall-2002

Solution to HW14 Fall-2002 Slutin t HW14 Fall-2002 CJ5 10.CQ.003. REASONING AND SOLUTION Figures 10.11 and 10.14 shw the velcity and the acceleratin, respectively, the shadw a ball that underges unirm circular mtin. The shadw underges

More information

Compressibility Effects

Compressibility Effects Definitin f Cmpressibility All real substances are cmpressible t sme greater r lesser extent; that is, when yu squeeze r press n them, their density will change The amunt by which a substance can be cmpressed

More information

CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS 16. REASONING AND SOLUTION A trapeze artist, starting rm rest, swings dwnward n the bar, lets g at the bttm the swing, and alls reely t the net. An assistant,

More information

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013 CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal

More information

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin

More information

Prof. Dr. I. Nasser Phys530, T142 3-Oct-17 Fermi_gases. 0 f e. and fall off exponentially like Maxwell-Boltzmaan distribution.

Prof. Dr. I. Nasser Phys530, T142 3-Oct-17 Fermi_gases. 0 f e. and fall off exponentially like Maxwell-Boltzmaan distribution. Pr. Dr. I. Nasser Phys, -Oct-7 FERMI_DIRAC GASSES Fermins: Are particles hal-integer spin that bey Fermi-Dirac statistics. Fermins bey the Pauli exclusin principle, which prhibits the ccupancy an available

More information

Chemistry 114 First Hour Exam

Chemistry 114 First Hour Exam Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)

More information

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review) CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal

More information

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O WYSE Academic Challenge Sectinal Chemistry Exam 2008 SOLUTION SET 1. Crrect answer: B. Use PV = nrt t get: PV = nrt 2. Crrect answer: A. (2.18 atm)(25.0 L) = n(0.08206 L atm/ml K)(23+273) n = 2.24 ml Assume

More information

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither

More information

Solutions to the Extra Problems for Chapter 14

Solutions to the Extra Problems for Chapter 14 Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +

More information

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs

More information

Chem 75 February 16, 2017 Exam 2 Solutions

Chem 75 February 16, 2017 Exam 2 Solutions 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml

More information

Chemistry 1A Fall 2000

Chemistry 1A Fall 2000 Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur

More information

ChE 471: LECTURE 4 Fall 2003

ChE 471: LECTURE 4 Fall 2003 ChE 47: LECTURE 4 Fall 003 IDEL RECTORS One f the key gals f chemical reactin engineering is t quantify the relatinship between prductin rate, reactr size, reactin kinetics and selected perating cnditins.

More information

Chapter 9: Quantization of Light

Chapter 9: Quantization of Light Chapter 9: Quantizatin Light 9.1 Planck s Quantum Thery 9.1.1 Distinguish between Planck s quantum thery and classical thery energy The undatin the Planck s quantum thery is a thery black bdy radiatin.

More information

Lecture 4. The First Law of Thermodynamics

Lecture 4. The First Law of Thermodynamics Lecture 4. The First Law f Thermdynamics THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and

More information

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals f Diffusin Diffusin: Transprt in a slid, liquid, r gas driven by a cncentratin gradient (r, in the case f mass transprt, a chemical ptential

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis

More information

General Chemistry II, Unit II: Study Guide (part 1)

General Chemistry II, Unit II: Study Guide (part 1) General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt

More information

Differentiation Applications 1: Related Rates

Differentiation Applications 1: Related Rates Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm

More information

Examiner: Dr. Mohamed Elsharnoby Time: 180 min. Attempt all the following questions Solve the following five questions, and assume any missing data

Examiner: Dr. Mohamed Elsharnoby Time: 180 min. Attempt all the following questions Solve the following five questions, and assume any missing data Benha University Cllege f Engineering at Banha Department f Mechanical Eng. First Year Mechanical Subject : Fluid Mechanics M111 Date:4/5/016 Questins Fr Final Crrective Examinatin Examiner: Dr. Mhamed

More information

Phys102 First Major-122 Zero Version Coordinator: Sunaidi Wednesday, March 06, 2013 Page: 1

Phys102 First Major-122 Zero Version Coordinator: Sunaidi Wednesday, March 06, 2013 Page: 1 Crdinatr: Sunaidi Wednesday, March 06, 2013 Page: 1 Q1. An 8.00 m lng wire with a mass f 10.0 g is under a tensin f 25.0 N. A transverse wave fr which the wavelength is 0.100 m, and the amplitude is 3.70

More information

Matter Content from State Frameworks and Other State Documents

Matter Content from State Frameworks and Other State Documents Atms and Mlecules Mlecules are made f smaller entities (atms) which are bnded tgether. Therefre mlecules are divisible. Miscnceptin: Element and atm are synnyms. Prper cnceptin: Elements are atms with

More information

Heat Effects of Chemical Reactions

Heat Effects of Chemical Reactions eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = -9690

More information

Chapter 3 Homework Solutions

Chapter 3 Homework Solutions Chapter Hmewrk Slutins. n = ml = 5 C = 98 K = 5 C = 98 K p = atm p = 5 atm. Cpm = (5/)R he entrpy changes fr the heating and cmpressin can e calculated separately and added. Heat at cnstant pressure S

More information

Spontaneous Processes, Entropy and the Second Law of Thermodynamics

Spontaneous Processes, Entropy and the Second Law of Thermodynamics Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer

More information

General Chemistry II, Unit I: Study Guide (part I)

General Chemistry II, Unit I: Study Guide (part I) 1 General Chemistry II, Unit I: Study Guide (part I) CDS Chapter 14: Physical Prperties f Gases Observatin 1: Pressure- Vlume Measurements n Gases The spring f air is measured as pressure, defined as the

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Computational modeling techniques

Computational modeling techniques Cmputatinal mdeling techniques Lecture 4: Mdel checing fr ODE mdels In Petre Department f IT, Åb Aademi http://www.users.ab.fi/ipetre/cmpmd/ Cntent Stichimetric matrix Calculating the mass cnservatin relatins

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

MBN 305 Phase Diagrams & Transformations

MBN 305 Phase Diagrams & Transformations MBN 35 Phase Diagrams & ransfrmatins Dr. Ersin Emre Oren Department f Bimedical Engineering Department f Materials Science & Nantechnlgy Engineering OBB University f Ecnmics and echnlgy Ankara - URKEY

More information

Autumn 2012 CHEM452B Bruce H. Robinson 322 Gould Hall HW 10(A) Homework 10A KEY (there will not be a 10B) 2

Autumn 2012 CHEM452B Bruce H. Robinson 322 Gould Hall HW 10(A) Homework 10A KEY (there will not be a 10B) 2 Autumn 0 CHEM45B Bruce H. Rbinsn Guld Hall HW 0(A) Hmewrk 0A KEY (there will nt be a 0B) QA) Let c be the speed f sund in air. he square f the speed f sund, () f the gas with respect t the change in the

More information

1/2 and e0 e s ' 1+ imm w 4 M s 3 πρ0 r 3 m. n 0 ktr. .Also,since n 0 ktr 1,wehave. 4 3 M sπρ 0 r 3. ktr. 3 M sπρ 0

1/2 and e0 e s ' 1+ imm w 4 M s 3 πρ0 r 3 m. n 0 ktr. .Also,since n 0 ktr 1,wehave. 4 3 M sπρ 0 r 3. ktr. 3 M sπρ 0 Chapter 6 6.1 Shw that fr a very weak slutin drplet (m 4 3 πr3 ρ 0 M s ), (6.8) can be written as e 0 ' 1+ a r b r 3 where a σ 0 /n 0 kt and b imm w / 4 3 M sπρ 0. What is yur interpretatin f thecnd and

More information

PHYS 314 HOMEWORK #3

PHYS 314 HOMEWORK #3 PHYS 34 HOMEWORK #3 Due : 8 Feb. 07. A unifrm chain f mass M, lenth L and density λ (measured in k/m) hans s that its bttm link is just tuchin a scale. The chain is drpped frm rest nt the scale. What des

More information

Lim f (x) e. Find the largest possible domain and its discontinuity points. Why is it discontinuous at those points (if any)?

Lim f (x) e. Find the largest possible domain and its discontinuity points. Why is it discontinuous at those points (if any)? THESE ARE SAMPLE QUESTIONS FOR EACH OF THE STUDENT LEARNING OUTCOMES (SLO) SET FOR THIS COURSE. SLO 1: Understand and use the cncept f the limit f a functin i. Use prperties f limits and ther techniques,

More information

Chapter 4 Thermodynamics and Equilibrium

Chapter 4 Thermodynamics and Equilibrium Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume

More information

Chapter 2 GAUSS LAW Recommended Problems:

Chapter 2 GAUSS LAW Recommended Problems: Chapter GAUSS LAW Recmmended Prblems: 1,4,5,6,7,9,11,13,15,18,19,1,7,9,31,35,37,39,41,43,45,47,49,51,55,57,61,6,69. LCTRIC FLUX lectric flux is a measure f the number f electric filed lines penetrating

More information

Instructions: Show all work for complete credit. Work in symbols first, plugging in numbers and performing calculations last. / 26.

Instructions: Show all work for complete credit. Work in symbols first, plugging in numbers and performing calculations last. / 26. CM ROSE-HULMAN INSTITUTE OF TECHNOLOGY Name Circle sectin: 01 [4 th Lui] 02 [5 th Lui] 03 [4 th Thm] 04 [5 th Thm] 05 [4 th Mech] ME301 Applicatins f Thermdynamics Exam 1 Sep 29, 2017 Rules: Clsed bk/ntes

More information

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C? NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic

More information

Phys102 Final-061 Zero Version Coordinator: Nasser Wednesday, January 24, 2007 Page: 1

Phys102 Final-061 Zero Version Coordinator: Nasser Wednesday, January 24, 2007 Page: 1 Crdinatr: Nasser Wednesday, January 4, 007 Page: 1 Q1. Tw transmitters, S 1 and S shwn in the figure, emit identical sund waves f wavelength λ. The transmitters are separated by a distance λ /. Cnsider

More information

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium Lecture 17: 11.07.05 Free Energy f Multi-phase Slutins at Equilibrium Tday: LAST TIME...2 FREE ENERGY DIAGRAMS OF MULTI-PHASE SOLUTIONS 1...3 The cmmn tangent cnstructin and the lever rule...3 Practical

More information

Advanced Chemistry Practice Problems

Advanced Chemistry Practice Problems Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with

More information

Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W

Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W Eample 1 rbt has a mass f 60 kg. Hw much des that rbt weigh sitting n the earth at sea level? Given: m Rbt = 60 kg ind: Rbt Relatinships: Slutin: Rbt =589 N = mg, g = 9.81 m/s Rbt = mrbt g = 60 9. 81 =

More information

Semester 2 AP Chemistry Unit 12

Semester 2 AP Chemistry Unit 12 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved

More information

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string? Term: 111 Thursday, January 05, 2012 Page: 1 Q1. A string f length L is fixed at bth ends. Which ne f the fllwing is NOT a pssible wavelength fr standing waves n this string? Q2. λ n = 2L n = A) 4L B)

More information

lecture 5: Nucleophilic Substitution Reactions

lecture 5: Nucleophilic Substitution Reactions lecture 5: Nuclephilic Substitutin Reactins Substitutin unimlecular (SN1): substitutin nuclephilic, unimlecular. It is first rder. The rate is dependent upn ne mlecule, that is the substrate, t frm the

More information

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry

More information

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia: University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,

More information

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied Phsics 141H lutins r Hmewrk et #5 Chapter 5: Multiple chice: 8) (a) he maimum rce eerted b static rictin is µ N. ince the blck is resting n a level surace, N = mg. the maimum rictinal rce is ( ) ( ) (

More information

Plan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations

Plan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations STAPLE Physics 201 Name Final Exam May 14, 2013 This is a clsed bk examinatin but during the exam yu may refer t a 5 x7 nte card with wrds f wisdm yu have written n it. There is extra scratch paper available.

More information

Pipetting 101 Developed by BSU CityLab

Pipetting 101 Developed by BSU CityLab Discver the Micrbes Within: The Wlbachia Prject Pipetting 101 Develped by BSU CityLab Clr Cmparisns Pipetting Exercise #1 STUDENT OBJECTIVES Students will be able t: Chse the crrect size micrpipette fr

More information

Chemistry/ Biotechnology Reference Sheets

Chemistry/ Biotechnology Reference Sheets Cmmn Metric Prefixes: Giga (G) = 1,000,000,000 = Kil (k) = 1,000 = Deci (d) =.1 = Milli (m) =.001 = Nan (n) =.000000001 = 9 6 1 10 Mega (M) = 1,000,000 = 1 10 0 1 10 Basic unit = meter, gram, liter, secnd

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

What determines how matter behaves? Thermodynamics.

What determines how matter behaves? Thermodynamics. What determines hw matter ehaves? hermdynamics.. What determines hw matter ehaves? Final Stale State Equilirium State Gis free energy: Minimum (at specific, ). Classical thermdynamics Macrscpic phenmena.

More information

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

x x

x x Mdeling the Dynamics f Life: Calculus and Prbability fr Life Scientists Frederick R. Adler cfrederick R. Adler, Department f Mathematics and Department f Bilgy, University f Utah, Salt Lake City, Utah

More information

Chem 111 Summer 2013 Key III Whelan

Chem 111 Summer 2013 Key III Whelan Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Chapter 4. Unsteady State Conduction

Chapter 4. Unsteady State Conduction Chapter 4 Unsteady State Cnductin Chapter 5 Steady State Cnductin Chee 318 1 4-1 Intrductin ransient Cnductin Many heat transfer prblems are time dependent Changes in perating cnditins in a system cause

More information

2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS

2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS 2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS 6. An electrchemical cell is cnstructed with an pen switch, as shwn in the diagram abve. A strip f Sn and a strip f an unknwn metal, X, are used as electrdes.

More information

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature

More information

Electrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent.

Electrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent. Electrchemistry Review: Reductin: the gaining f electrns Oxidatin: the lss f electrns Reducing agent (reductant): species that dnates electrns t reduce anther reagent. Oxidizing agent (xidant): species

More information

Study Guide Physics Pre-Comp 2013

Study Guide Physics Pre-Comp 2013 I. Scientific Measurement Metric Units S.I. English Length Meter (m) Feet (ft.) Mass Kilgram (kg) Pund (lb.) Weight Newtn (N) Ounce (z.) r pund (lb.) Time Secnds (s) Secnds (s) Vlume Liter (L) Galln (gal)

More information

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25 CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing

More information

Acids and Bases Lesson 3

Acids and Bases Lesson 3 Acids and Bases Lessn 3 The ph f a slutin is defined as the negative lgarithm, t the base ten, f the hydrnium in cncentratin. In a neutral slutin at 25 C, the hydrnium in and the hydrxide in cncentratins

More information

Thermodynamics: Gas Laws

Thermodynamics: Gas Laws hermdynamics: Gas Laws Resurces: Serway emperature & hermal Expansin: 10.1, 10.2, & 10.3 Ideal Gas Law: 10.4 & 10.5 AP Physics B Vides Physics B Lessn 25: Mechanical Equivalent f Heat Physics B Lessn 26:

More information

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions ED- (JEE) Slutins Answer : Optin () ass f the remved part will be / I Answer : Optin () r L m (u csθ) (H) Answer : Optin () P 5 rad/s ms - because f translatin ωr ms - because f rtatin Cnsider a thin shell

More information

GOAL... ability to predict

GOAL... ability to predict THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict

More information

Chapter 5: Diffusion (2)

Chapter 5: Diffusion (2) Chapter 5: Diffusin () ISSUES TO ADDRESS... Nn-steady state diffusin and Fick s nd Law Hw des diffusin depend n structure? Chapter 5-1 Class Eercise (1) Put a sugar cube inside a cup f pure water, rughly

More information

Department of Economics, University of California, Davis Ecn 200C Micro Theory Professor Giacomo Bonanno. Insurance Markets

Department of Economics, University of California, Davis Ecn 200C Micro Theory Professor Giacomo Bonanno. Insurance Markets Department f Ecnmics, University f alifrnia, Davis Ecn 200 Micr Thery Prfessr Giacm Bnann Insurance Markets nsider an individual wh has an initial wealth f. ith sme prbability p he faces a lss f x (0

More information

Appendix I: Derivation of the Toy Model

Appendix I: Derivation of the Toy Model SPEA ET AL.: DYNAMICS AND THEMODYNAMICS OF MAGMA HYBIDIZATION Thermdynamic Parameters Appendix I: Derivatin f the Ty Mdel The ty mdel is based upn the thermdynamics f an isbaric twcmpnent (A and B) phase

More information

Edexcel GCSE Physics

Edexcel GCSE Physics Edexcel GCSE Physics Tpic 10: Electricity and circuits Ntes (Cntent in bld is fr Higher Tier nly) www.pmt.educatin The Structure f the Atm Psitively charged nucleus surrunded by negatively charged electrns

More information

Entropy. Chapter The Clausius Inequality and Entropy

Entropy. Chapter The Clausius Inequality and Entropy Chapter 7 Entrpy In the preceding chapter we btained a number f imprtant results by applying the secnd law t cyclic prcesses assciated with heat engines and reversed heat engines perating with ne and tw

More information

CHM 152 Practice Final

CHM 152 Practice Final CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the

More information

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture Few asic Facts but Isthermal Mass Transfer in a inary Miture David Keffer Department f Chemical Engineering University f Tennessee first begun: pril 22, 2004 last updated: January 13, 2006 dkeffer@utk.edu

More information

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b . REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but

More information

Thermodynamics EAS 204 Spring 2004 Class Month Day Chapter Topic Reading Due 1 January 12 M Introduction 2 14 W Chapter 1 Concepts Chapter 1 19 M MLK

Thermodynamics EAS 204 Spring 2004 Class Month Day Chapter Topic Reading Due 1 January 12 M Introduction 2 14 W Chapter 1 Concepts Chapter 1 19 M MLK Thermdynamics EAS 204 Spring 2004 Class Mnth Day Chapter Tpic Reading Due 1 January 12 M Intrductin 2 14 W Chapter 1 Cncepts Chapter 1 19 M MLK Hliday n class 3 21 W Chapter 2 Prperties Chapter 2 PS1 4

More information

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes. Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied

More information

Preparation work for A2 Mathematics [2017]

Preparation work for A2 Mathematics [2017] Preparatin wrk fr A2 Mathematics [2017] The wrk studied in Y12 after the return frm study leave is frm the Cre 3 mdule f the A2 Mathematics curse. This wrk will nly be reviewed during Year 13, it will

More information

Being able to connect displacement, speed, and acceleration is fundamental to working

Being able to connect displacement, speed, and acceleration is fundamental to working Chapter The Big Three: Acceleratin, Distance, and Time In This Chapter Thinking abut displacement Checking ut speed Remembering acceleratin Being able t cnnect displacement, speed, and acceleratin is undamental

More information

Lab 1 The Scientific Method

Lab 1 The Scientific Method INTRODUCTION The fllwing labratry exercise is designed t give yu, the student, an pprtunity t explre unknwn systems, r universes, and hypthesize pssible rules which may gvern the behavir within them. Scientific

More information