1 Exercise 4.1b pg 153
|
|
- Tracy Washington
- 7 years ago
- Views:
Transcription
1 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the examle, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic stes. Numbers without underlines (including final answers) are shown with the roer number of sig figs. 1 Exercise 4.1b g 153 Question How many hases are resent at each of the oints marked in Fig. 4.23b? a. 1 - oint is entirely inside a single hase region (not on any boundaries). b. 3 - oint on a boundary where 3 hases meet. c. 3 - (same exlanation as b) d. 2 - oint occurs on boundary between two hases (on a single line). 1
2 2 Exercise 4.5b g 153 Given Iron is heated from T i = 25 C to T f = 100 C. Over this temerature range S m = 53 J K 1 1. In terms of given variables, this is written: T i = 25 C T f = 100 C S m = 53 J K 1 1 Find By how much does its chemical otential change? Strategy First, temeratures are converted to Kelvin. T i = C = 373 K T f = C = 1273 K We can use text book Equation 4.2 (g 143) to relate temerature changes to changes in chemical otential. which gives the differential ( ) µ = Sm dµ = S m Integrating dµ over the temerature range gives the change in chemical otential. Tf µ = Sm T i = Sm (T f T i ) J = 53 (1273 K 373 K) K = J = 47.7 kj where the integral has been easily solved, as we ve assumed S m is constant over this temerature range. 2
3 µ = 50 kj 3
4 3 Exercise 4.11b g 153 Given The vaour ressure of a liquid between 15 C and 35 C fits the exression log(/ torr) = /(T/ K) Find Calculate... (a) the enthaly of vaorization (b) the normal boiling oint of the liquid Strategy We ll start with Equation 4.11 (g 148). Through rearrangement we solve for va H. d ln = vah RT 2 va H = RT 2 d ln To find d ln, the given exression for log(/ torr) can be converted to an exression for ln(/ torr) using the change of base formula. and this gives the equation log b x = log k x log k b Differentiating this exression by T gives ln(/ torr) = ln(10) ( / (T/ K)) ln(d) = ln(10) 1625 K T 2 (The Torr units were discarded as the units of ressure would only lead to constant shift in the exression for ln() and this constant is lost on differentiation.) Lastly we can substitute this exression for ln(d) into our earlier exression for va H. 4
5 va H = R T 2 ln(10) 1625 K T 2 = R ln(10) 1625 K = J ln(10) 1625 K K = J = kj The normal boiling oint can be found by solving the given exression (T ) for the temerature at which (T ) = atmoshere where the atmosheric ressure atmoshere = 1.00 atm = 760 torr. log(760 torr/ torr) = /(T/ K) T = 1625 K log 760 = K (a) va H = kj (b) T = 280 K 5
6 4 Exercise 4.14b g 153 Given On a cold, dry morning after a frost, the temerature was T = 5 C and the artial ressure of water in the atmoshere fell to H2 O = 0.30 kpa. In terms of given variables, this is written: T = 5 C = 0.30 kpa Find (a) Will the frost sublime? (b) What artial ressure of water H2O would ensure that the frost remained? Strategy In this exercise, the second question answers the first question, in that once we know the artial ressure of water H2 O needed to ensure the frost remains, we know any H2 O below this will lead to frost sublimation. The artial ressure needed to revent sublimation is found by determing the solid-vaor ressure for water at this temerature; this is the ressure at this temerature on the coexistance curve for ice and water vaor. We know that if the atmoshere has a lower artial ressure of water than the solid-vaor ressure, then the water vaour will be favored over the solid and the ice will sublime. At higher artial ressures the solid hase is favored. We can find the solid-vaor ressure using Equation 4.12 (g 149) from our text book = e χ χ = subh R ( 1 T 1 ) T where we ve relaced va H in the original exression with sub H as we re concerned with the sublimation coexistence oint instead of the vaorization oint. The sublimation enthaly va H is found from the vaorization enthaly and the fusion enthaly, va H and fus H resectively. sub H = fus H + va H = kj kj 1 = kj 1 = J 1 Using Equation 4.12 we can calculate the ressure at temerature T when we know the reference ressure at the reference temerature T. As we re solving for a (T ) on the solid/gas coexistence curve, we ll need the reference to also fall on this curve. Therefore we ll use the the trile oint of water as our reference giving the following values: T = K 6
7 = kpa This allows us to find the solid-vaor ressure on the solid/gas coexistence curve for the temerature T = 5 C = 268 K. ( 1 T 1 ) T χ = subh R = J J K 1 1 = ( ) K K = e χ = kpa e = kpa Now that we know the solid-vaor ressure = kpa at the given temerature, we know the ice will sublime into any gas system with a artial ressure of water H2O < kpa. This gives us the answer to art b as H2 O = kpa. Additionally, can determine that the ice will sublime into the atmoshere as the artial ressure of water is H2O = 0.30 kpa. (a) Yes. (b) H2 O = kpa 7
8 5 Exercise 4.17b g 154 Question What fraction of the enthaly of vaorization va H of ethanol is sent on exanding its vaour? Strategy We can use the definition of enthaly, H = U + V (Equation 2.18 g 56) to decomose the enthaly of vaorization va H into an internal energy comonent va U and an exansion work comonent va (V ). va H = va U + va (V ) Additionally, we can assume the ressure is constant and the liquid volume V liq is negligible relative to the gas volume such that va (V ) = (V gas V liq ) V gas Next, we can use the ideal gas law V = nrt to relate V gas to RT for a ar quantity of gas by assuming the ethanol vaor is ideal. This gives va (V ) = RT And thereby the ratio of exansion work va (V ) to exansion enthaly va H is va (V ) va H = RT va H Using text book table 2.3 we find that ethanol vaorizes at T = 352 K and its exansion enthaly is 43.5 kj 1. Substituting these values gives the ratio value va (V ) va H RT = va H = J K K 43.5 kj J 1 kj = = 6.73% va (V ) va H = 6.73% 8
9 6 Problem 4.4 g 154 Question Calculate the difference in sloe of the chemical otential against temerature on either side of (a) the normal freezing oint of water and and (b) the normal boiling oint of water. (c) By how much does the chemical otential of water suer cooled to 5.0 C exceed that of ice at that temerature. Strategy To solve arts (a) and (b) we ll use Equation 4.13 (art 2 g 150) that relates how the temerature derivative (sloe) of chemical otential, µ, changes across a coexistence curve. ( ) µ (β) where α and β denote the two hases. ( ) µ (α) = S m (β) + S m (α) = trs S = trsh T trs For art (a) we re interested in the difference between the liquid l and solid s hases, the fusion transition fus H, which gives the following exression. ( ) µ (s) = S m (l) + S m (s) = fus S = fush T f Substituting in the enthaly and temerature of water s fusion transition, fus H = kj 1 and T f = K resectively as found in Table 2.3 of our text book, gives ( ) µ (s) = fush T f kj 1 = K = kj 1 K 1 = J 1 K 1 Likewise for art (b) we re interested in the difference between the gas g and the liquid l hases which is the vaorization transition va H. ( ) µ (g) = S m (g) + S m (l) = va S = vah T v Substituting in va H = kj 1 and T v = K from Table 2.3 gives ( ) µ (g) = vah T v kj 1 = K = kj 1 K 1 = J 1 K 1 9
10 For art (c) we re interested in calculating the difference in chemical otential of liquid water at 5 C, µ(l, 5 C) and the chemical otential of solid water at the same temerature µ(s, 5 C) and we ll call this quantity x. x = µ(l, 5 C) µ(s, 5 C) To calculate x we ll take advantage of the normal freezing oint of water, which imlies that at T = 0 C solid water and liquid water have the same chemical otential: µ(l, 0 C) = µ(s, 0 C) Therefore we can subtract µ(l, 0 C) µ(s, 0 C) (as this quantity is 0) from the exression we re working to solve. x = µ(l, 5 C) µ(s, 5 C) = µ(l, 5 C) µ(s, 5 C) [µ (l, 0 C) µ (s, 0 C)] = [µ (l, 5 C) µ (l, 0 C)] [µ (s, 5 C) µ (s, 0 C)] = µ(l) µ(s) In the third equation above we ve rearranged the righthand side so as to lace the difference in chemical otentials at the two temeratures in brackets for each hase. We call this quantity µ(α) where µ(α) = µ (α, 5 C)µ (α, 0 C) and α secifies either the liquid (l) or solid (s) hase. We can calculate µ(l) and µ(s) using Equation 4.2 (g 143) which gives the change in chemical otential with temerature. ( ) µ = S m Hence the change in chemical otential for the α hase can be calculated as where S m is assumed temerature indeendent. (α) = = Tf T i Tf T i ( ) µ S m = S m (T f T i ) Using T f T i = T = 5 K and the standard entroies of liquid water S m (l) and solid water S m (s) we get the following exressions for chemical otential changes (l) = S m (l) T (s) = S m (s) T Substituting these exression into our equation for x gives x = µ(l) µ(s) = [S m (l) S m (s)] T The above term in brackets, the difference in entroy of liquid and solid water, is just the oosite of the entroy of fusion fus S which we know from art (a). 10
11 x = [S m (l) S m (s)] T = fus S T ( = J 1 K 1 ) 5 K = 109 J 1 We are reminded that x is just the difference in chemical otentials that we were solving. µ(l, 5 C) µ(s, 5 C) = x = 109 J 1 This ositive difference in chemical otential between liquid and solid hases imlies higher free energy for the liquid hase relative to the solid hase (ice) and exlains why ice is favorable at this temerature. (a) (b) (c) ( ) µ (g) ( ) µ (s) = J 1 K 1 = J 1 K 1 µ(l, 5 C) µ(s, 5 C) = x = 100 J 1 11
1.3 Saturation vapor pressure. 1.3.1 Vapor pressure
1.3 Saturation vaor ressure Increasing temerature of liquid (or any substance) enhances its evaoration that results in the increase of vaor ressure over the liquid. y lowering temerature of the vaor we
More informationFugacity, Activity, and Standard States
Fugacity, Activity, and Standard States Fugacity of gases: Since dg = VdP SdT, for an isothermal rocess, we have,g = 1 Vd. For ideal gas, we can substitute for V and obtain,g = nrt ln 1, or with reference
More information1 Exercise 2.19a pg 86
In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit,
More informationFirst Law, Heat Capacity, Latent Heat and Enthalpy
First Law, Heat Caacity, Latent Heat and Enthaly Stehen R. Addison January 29, 2003 Introduction In this section, we introduce the first law of thermodynamics and examine sign conentions. Heat and Work
More informationThermodynamics worked examples
An Introduction to Mechanical Engineering Part hermodynamics worked examles. What is the absolute ressure, in SI units, of a fluid at a gauge ressure of. bar if atmosheric ressure is.0 bar? Absolute ressure
More informationIdeal Gas and Real Gases
Ideal Gas and Real Gases Lectures in Physical Chemistry 1 Tamás Turányi Institute of Chemistry, ELTE State roerties state roerty: determines the macroscoic state of a hysical system state roerties of single
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 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 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 information- The value of a state function is independent of the history of the system. - Temperature is an example of a state function.
First Law of hermodynamics 1 State Functions - A State Function is a thermodynamic quantity whose value deends only on the state at the moment, i. e., the temerature, ressure, volume, etc - he value of
More informationPrice Elasticity of Demand MATH 104 and MATH 184 Mark Mac Lean (with assistance from Patrick Chan) 2011W
Price Elasticity of Demand MATH 104 and MATH 184 Mark Mac Lean (with assistance from Patrick Chan) 2011W The rice elasticity of demand (which is often shortened to demand elasticity) is defined to be the
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 information4. Influence of Temperature and Pressure on Chemical Changes
4. Influence of Temerature and Pressure on Chemical Changes Toic: The influence of temerature and ressure on the chemical otential and drive and therefore the behaviour of substances. 4.1 Introduction
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 informationOUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS
UNI 6: ENGINEERING HERMODYNAMICS Unit code: D/60/40 QCF level: 5 Credit value: 5 OUCOME UORIAL No. HERMODYNAMIC SYSEMS. Understand the arameters and characteristics of thermodynamic systems Polytroic rocesses:
More informationStudy the following diagrams of the States of Matter. Label the names of the Changes of State between the different states.
Describe the strength of attractive forces between particles. Describe the amount of space between particles. Can the particles in this state be compressed? Do the particles in this state have a definite
More informationEXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor
EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor Purpose: In this experiment you will use the ideal gas law to calculate the molecular weight of a volatile liquid compound by measuring the mass,
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 informationPythagorean Triples and Rational Points on the Unit Circle
Pythagorean Triles and Rational Points on the Unit Circle Solutions Below are samle solutions to the roblems osed. You may find that your solutions are different in form and you may have found atterns
More informationThe Reduced van der Waals Equation of State
The Redued van der Waals Equation of State The van der Waals equation of state is na + ( V nb) n (1) V where n is the mole number, a and b are onstants harateristi of a artiular gas, and R the gas onstant
More informationSTABILITY OF PNEUMATIC and HYDRAULIC VALVES
STABILITY OF PNEUMATIC and HYDRAULIC VALVES These three tutorials will not be found in any examination syllabus. They have been added to the web site for engineers seeking knowledge on why valve elements
More informationbe the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out
Chater 4 4. Energy Balances on Nonreactive Processes he general energy balance equation has the form Accumulation Inut Outut Heat added = + of Energy of Energy of Energy to System Work by done System Let
More informationGases. States of Matter. Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large Chaotic (random)
Gases States of Matter States of Matter Kinetic E (motion) Potential E(interaction) Distance Between (size) Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large
More informationChapter 4 Practice Quiz
Chapter 4 Practice Quiz 1. Label each box with the appropriate state of matter. A) I: Gas II: Liquid III: Solid B) I: Liquid II: Solid III: Gas C) I: Solid II: Liquid III: Gas D) I: Gas II: Solid III:
More informationCalorimetry: Heat of Vaporization
Calorimetry: Heat of Vaporization OBJECTIVES INTRODUCTION - Learn what is meant by the heat of vaporization of a liquid or solid. - Discuss the connection between heat of vaporization and intermolecular
More information5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C
1. The average kinetic energy of water molecules increases when 1) H 2 O(s) changes to H 2 O( ) at 0ºC 3) H 2 O( ) at 10ºC changes to H 2 O( ) at 20ºC 2) H 2 O( ) changes to H 2 O(s) at 0ºC 4) H 2 O( )
More informationWe will study the temperature-pressure diagram of nitrogen, in particular the triple point.
K4. Triple Point of Nitrogen I. OBJECTIVE OF THE EXPERIMENT We will study the temperature-pressure diagram of nitrogen, in particular the triple point. II. BAKGROUND THOERY States of matter Matter is made
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 informationChemistry 13: States of Matter
Chemistry 13: States of Matter Name: Period: Date: Chemistry Content Standard: Gases and Their Properties The kinetic molecular theory describes the motion of atoms and molecules and explains the properties
More informationScience Department Mark Erlenwein, Assistant Principal
Staten Island Technical High School Vincent A. Maniscalco, Principal The Physical Setting: CHEMISTRY Science Department Mark Erlenwein, Assistant Principal - Unit 1 - Matter and Energy Lessons 9-14 Heat,
More informationTest 5 Review questions. 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will
Name: Thursday, December 13, 2007 Test 5 Review questions 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will 1. decrease 2. increase 3. remain the same 2. The graph below
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 informationCoordinate Transformation
Coordinate Transformation Coordinate Transformations In this chater, we exlore maings where a maing is a function that "mas" one set to another, usually in a way that reserves at least some of the underlyign
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 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 information7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.
CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,
More informationGas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.
Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.
More informationStates of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided.
CHAPTER 10 REVIEW States of Matter SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Identify whether the descriptions below describe an ideal gas or a real gas. ideal gas
More informationThe International Standard Atmosphere (ISA)
Nomenclature The International Standard Atmoshere (ISA) Mustafa Cavcar * Anadolu University, 2647 Eskisehir, Turkey a = seed of sound, m/sec g = acceleration of gravity, m/sec 2 h = altitude, m or ft =
More informationChapter 12 - Liquids and Solids
Chapter 12 - Liquids and Solids 12-1 Liquids I. Properties of Liquids and the Kinetic Molecular Theory A. Fluids 1. Substances that can flow and therefore take the shape of their container B. Relative
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 informationCHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)
CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26
More informationMaterials 10-mL graduated cylinder l or 2-L beaker, preferably tall-form Thermometer
VAPOR PRESSURE OF WATER Introduction At very low temperatures (temperatures near the freezing point), the rate of evaporation of water (or any liquid) is negligible. But as its temperature increases, more
More informationName Date Class STATES OF MATTER. SECTION 13.1 THE NATURE OF GASES (pages 385 389)
13 STATES OF MATTER SECTION 13.1 THE NATURE OF GASES (pages 385 389) This section introduces the kinetic theory and describes how it applies to gases. It defines gas pressure and explains how temperature
More information5 Answers and Solutions to Text Problems
Energy and States of Matter 5 Answers and Solutions to Text Problems 5.1 At the top of the hill, all of the energy of the car is in the form of potential energy. As it descends down the hill, potential
More informationCHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank
CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS Blank SONNTAG/BORGNAKKE STUDY PROBLEM 7-1 7.1 A car engine and its fuel consumption A car engine produces 136 hp on the output shaft with a thermal efficiency
More informationChapter 3. Phase transitions. 3.1 Introduction. 3.2 Thermodynamics of phase transitions
Chater 3 Phase transitions 3.1 Introduction Phase transitions are ubiquitous in Nature. We are all familiar with the different hases of water (vaour, liquid and ice), and with the change from one to the
More informationIn order to solve this problem it is first necessary to use Equation 5.5: x 2 Dt. = 1 erf. = 1.30, and x = 2 mm = 2 10-3 m. Thus,
5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Solution (a) With vacancy diffusion,
More informationA n = 2 to n = 1. B n = 3 to n = 1. C n = 4 to n = 2. D n = 5 to n = 2
North arolina Testing Program EO hemistry Sample Items Goal 4 1. onsider the spectrum for the hydrogen atom. In which situation will light be produced? 3. Which color of light would a hydrogen atom emit
More informationWarm-Up 9/9. 1. Define the term matter. 2. Name something in this room that is not matter.
Warm-Up 9/9 1. Define the term matter. 2. Name something in this room that is not matter. Warm-Up 9/16 1. List the three most important rules of lab safety. 2. Would you classify jello as a solid or a
More informationTHE BAROMETRIC FALLACY
THE BAROMETRIC FALLACY It is often assumed that the atmosheric ressure at the surface is related to the atmosheric ressure at elevation by a recise mathematical relationshi. This relationshi is that given
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 informationBoyles Law. At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 P = P
Boyles Law At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 or k 1 Boyles Law Example ressure olume Initial 2.00 atm 100 cm 3
More informationPressure Drop in Air Piping Systems Series of Technical White Papers from Ohio Medical Corporation
Pressure Dro in Air Piing Systems Series of Technical White Paers from Ohio Medical Cororation Ohio Medical Cororation Lakeside Drive Gurnee, IL 600 Phone: (800) 448-0770 Fax: (847) 855-604 info@ohiomedical.com
More informationDistillation vaporization sublimation. vapor pressure normal boiling point.
Distillation Distillation is an important commercial process that is used in the purification of a large variety of materials. However, before we begin a discussion of distillation, it would probably be
More informationPROPERTIES OF NATURAL GAS
1 PROPERIES OF NAURAL GAS Jón Steinar Guðmundsson PG4140 Naturgass Setember 16, 2008 Introduction Equations and roerties used in course Real gas law, z-factor, density and FVF Corresonding states Viscosity
More informationUNIT (1) MEASUREMENTS IN CHEMISTRY
UNIT (1) MEASUREMENTS IN CHEMISTRY Measurements are part of our daily lives. We measure our weights, driving distances, and gallons of gasoline. As a health professional you might measure blood pressure,
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 informationMeasuring relative phase between two waveforms using an oscilloscope
Measuring relative hase between two waveforms using an oscilloscoe Overview There are a number of ways to measure the hase difference between two voltage waveforms using an oscilloscoe. This document covers
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 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 informationThe International Association for the Properties of Water and Steam
he International Association for the Proerties of ater and team London, United Kindom etember 2013 Advisory Note No. 5: Industrial Calculation of the hermodynamic Proerties of eawater 2013 International
More informationEffect Sizes Based on Means
CHAPTER 4 Effect Sizes Based on Means Introduction Raw (unstardized) mean difference D Stardized mean difference, d g Resonse ratios INTRODUCTION When the studies reort means stard deviations, the referred
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 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 informationFree Software Development. 2. Chemical Database Management
Leonardo Electronic Journal of Practices and echnologies ISSN 1583-1078 Issue 1, July-December 2002. 69-76 Free Software Develoment. 2. Chemical Database Management Monica ŞEFU 1, Mihaela Ligia UNGUREŞAN
More information= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm
Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by
More informationNet ionic equation: 2I (aq) + 2H (aq) + H O (aq) I (s) + 2H O(l)
Experiment 5 Goals To determine the differential rate law for the reaction between iodide and hydrogen peroxide in an acidic environment. To determine the activation energy and pre-exponential factor for
More informationThe Magnus-Derek Game
The Magnus-Derek Game Z. Nedev S. Muthukrishnan Abstract We introduce a new combinatorial game between two layers: Magnus and Derek. Initially, a token is laced at osition 0 on a round table with n ositions.
More informationEXPERIMENT 13: THE IDEAL GAS LAW AND THE MOLECULAR WEIGHT OF GASES
Name Section EXPERIMENT 13: THE IDEAL GAS LAW AND THE MOLECULAR WEIGHT OF GASES PRE-LABORATORY QUESTIONS The following preparatory questions should be answered before coming to lab. They are intended to
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 informationChemistry 212 VAPOR PRESSURE OF WATER LEARNING OBJECTIVES
Chemistry 212 VAPOR PRESSURE OF WATER LEARNING OBJECTIVES The learning objectives of this experiment are to explore the relationship between the temperature and vapor pressure of water. determine the molar
More informationCHEM 120 Online Chapter 7
CHEM 120 Online Chapter 7 Date: 1. Which of the following statements is not a part of kinetic molecular theory? A) Matter is composed of particles that are in constant motion. B) Particle velocity increases
More informationIB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
More informationPhysical Transformations of Pure Substances
Physical Transformations of Pure Substances Chapter 6 of Atkins Sections 6.1-6.6 (6th, 7th Eds.), Sections 4.1-4.6 (8th Ed.) Phase Diagrams Stabilities of Phases Phase Boundaries Three Typical Phase Diagrams
More informationTemperature Scales. The metric system that we are now using includes a unit that is specific for the representation of measured temperatures.
Temperature Scales INTRODUCTION The metric system that we are now using includes a unit that is specific for the representation of measured temperatures. The unit of temperature in the metric system is
More informationChapter 8. Phase Diagrams
Phase Diagrams A phase in a material is a region that differ in its microstructure and or composition from another region Al Al 2 CuMg H 2 O(solid, ice) in H 2 O (liquid) 2 phases homogeneous in crystal
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 informationProblem 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 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 informationVapor Pressure of Liquids
Vapor Pressure of Liquids Experiment 10 In this experiment, you will investigate the relationship between the vapor pressure of a liquid and its temperature. When a liquid is added to the Erlenmeyer flask
More informationGAS TURBINE PERFORMANCE WHAT MAKES THE MAP?
GAS TURBINE PERFORMANCE WHAT MAKES THE MAP? by Rainer Kurz Manager of Systems Analysis and Field Testing and Klaus Brun Senior Sales Engineer Solar Turbines Incororated San Diego, California Rainer Kurz
More informationIntermolecular and Ionic Forces
Intermolecular and Ionic Forces Introduction: Molecules are attracted to each other in the liquid and solid states by intermolecular, or attractive, forces. These are the attractions that must be overcome
More informationSection 1.4. Difference Equations
Difference Equations to Differential Equations Section 1.4 Difference Equations At this point almost all of our sequences have had explicit formulas for their terms. That is, we have looked mainly at sequences
More information1. The Kinetic Theory of Matter states that all matter is composed of atoms and molecules that are in a constant state of constant random motion
Physical Science Period: Name: ANSWER KEY Date: Practice Test for Unit 3: Ch. 3, and some of 15 and 16: Kinetic Theory of Matter, States of matter, and and thermodynamics, and gas laws. 1. The Kinetic
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 informationFailure Behavior Analysis for Reliable Distributed Embedded Systems
Failure Behavior Analysis for Reliable Distributed Embedded Systems Mario Tra, Bernd Schürmann, Torsten Tetteroo {tra schuerma tetteroo}@informatik.uni-kl.de Deartment of Comuter Science, University of
More informationc 2009 Je rey A. Miron 3. Examples: Linear Demand Curves and Monopoly
Lecture 0: Monooly. c 009 Je rey A. Miron Outline. Introduction. Maximizing Pro ts. Examles: Linear Demand Curves and Monooly. The Ine ciency of Monooly. The Deadweight Loss of Monooly. Price Discrimination.
More informationIDEAL AND NON-IDEAL GASES
2/2016 ideal gas 1/8 IDEAL AND NON-IDEAL GASES PURPOSE: To measure how the pressure of a low-density gas varies with temperature, to determine the absolute zero of temperature by making a linear fit to
More informationError Analysis. Table 1. Capacity Tolerances for Class A Volumetric Glassware.
Significant Figures in Calculations Error Analysis Every lab report must have an error analysis. For many experiments, significant figure rules are sufficient. For a brush up on significant figure rules,
More informationCurrent mirrors are commonly used for current sources in integrated circuit design. This section covers other current sources that are often seen.
c Coyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Comuter Engineering. Current Sources Current mirrors are commonly used for current sources
More informationPopcorn Laboratory. Hypothesis : Materials:
Popcorn Laboratory Problem: Popcorn kernels explode into delightful, edible parcels because of a build-up of pressure inside the kernel during heating. In this experiment you will try to calculate the
More informationFour Derivations of the Black-Scholes Formula by Fabrice Douglas Rouah www.frouah.com www.volopta.com
Four Derivations of the Black-Scholes Formula by Fabrice Douglas Rouah www.frouah.com www.volota.com In this note we derive in four searate ways the well-known result of Black and Scholes that under certain
More informationType: Single Date: Homework: READ 12.8, Do CONCEPT Q. # (14) Do PROBLEMS (40, 52, 81) Ch. 12
Type: Single Date: Objective: Latent Heat Homework: READ 12.8, Do CONCEPT Q. # (14) Do PROBLEMS (40, 52, 81) Ch. 12 AP Physics B Date: Mr. Mirro Heat and Phase Change When bodies are heated or cooled their
More informationExperiment 1: Colligative Properties
Experiment 1: Colligative Properties Determination of the Molar Mass of a Compound by Freezing Point Depression. Objective: The objective of this experiment is to determine the molar mass of an unknown
More informationChapter 2 Measurement and Problem Solving
Introductory Chemistry, 3 rd Edition Nivaldo Tro Measurement and Problem Solving Graph of global Temperature rise in 20 th Century. Cover page Opposite page 11. Roy Kennedy Massachusetts Bay Community
More informationChapter 1: Chemistry: Measurements and Methods
Chapter 1: Chemistry: Measurements and Methods 1.1 The Discovery Process o Chemistry - The study of matter o Matter - Anything that has mass and occupies space, the stuff that things are made of. This
More informationAS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol -1. Example
More information10.7 Kinetic Molecular Theory. 10.7 Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory
The first scheduled quiz will be given next Tuesday during Lecture. It will last 5 minutes. Bring pencil, calculator, and your book. The coverage will be pp 364-44, i.e. Sections 0.0 through.4. 0.7 Theory
More information