Chapter 15: Solutions
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Modelling and Numerical Simulation of Phase Separation in Polymer Modified Bitumen by Phase- Field Method
http://www.diva-portal.org Postprint This is the accepted version of a paper published in Materials & design. This paper has been peer- reviewed but does not include the final publisher proof-corrections or journal pagination. Citation for the original published paper (version of record): Zhu, J., Lu, X., Balieu, R., Kringos, N. (2016) Modelling and numerical simulation of phase separation in polymer modified bitumen by phase- field method. Materials & design, 107: 322-332 http://dx.doi.org/10.1016/j.matdes.2016.06.041 Access to the published version may require subscription. N.B. When citing this work, cite the original published paper. Permanent link to this version: http://urn.kb.se/resolve?urn=urn:nbn:se:kth:diva-188830 ACCEPTED MANUSCRIPT Modelling and numerical simulation of phase separation in polymer modified bitumen by phase-field method Jiqing Zhu a,*, Xiaohu Lu b, Romain Balieu a, Niki Kringos a a Department of Civil and Architectural Engineering, KTH Royal Institute of Technology, Brinellvägen 23, SE-100 44 Stockholm, Sweden b Nynas AB, SE-149 82 Nynäshamn, Sweden * Corresponding author. Email: [email protected] (J. Zhu) Abstract In this paper, a phase-field model with viscoelastic effects is developed for polymer modified bitumen (PMB) with the aim to describe and predict the PMB storage stability and phase separation behaviour. The viscoelastic effects due to dynamic asymmetry between bitumen and polymer are represented in the model by introducing a composition-dependent mobility coefficient. A double-well potential for PMB system is proposed on the basis of the Flory-Huggins free energy of mixing, with some simplifying assumptions made to take into account the complex chemical composition of bitumen. -
Solutes and Solution
Solutes and Solution The first rule of solubility is “likes dissolve likes” Polar or ionic substances are soluble in polar solvents Non-polar substances are soluble in non- polar solvents Solutes and Solution There must be a reason why a substance is soluble in a solvent: either the solution process lowers the overall enthalpy of the system (Hrxn < 0) Or the solution process increases the overall entropy of the system (Srxn > 0) Entropy is a measure of the amount of disorder in a system—entropy must increase for any spontaneous change 1 Solutes and Solution The forces that drive the dissolution of a solute usually involve both enthalpy and entropy terms Hsoln < 0 for most species The creation of a solution takes a more ordered system (solid phase or pure liquid phase) and makes more disordered system (solute molecules are more randomly distributed throughout the solution) Saturation and Equilibrium If we have enough solute available, a solution can become saturated—the point when no more solute may be accepted into the solvent Saturation indicates an equilibrium between the pure solute and solvent and the solution solute + solvent solution KC 2 Saturation and Equilibrium solute + solvent solution KC The magnitude of KC indicates how soluble a solute is in that particular solvent If KC is large, the solute is very soluble If KC is small, the solute is only slightly soluble Saturation and Equilibrium Examples: + - NaCl(s) + H2O(l) Na (aq) + Cl (aq) KC = 37.3 A saturated solution of NaCl has a [Na+] = 6.11 M and [Cl-] = -
Chapter 2: Basic Tools of Analytical Chemistry
Chapter 2 Basic Tools of Analytical Chemistry Chapter Overview 2A Measurements in Analytical Chemistry 2B Concentration 2C Stoichiometric Calculations 2D Basic Equipment 2E Preparing Solutions 2F Spreadsheets and Computational Software 2G The Laboratory Notebook 2H Key Terms 2I Chapter Summary 2J Problems 2K Solutions to Practice Exercises In the chapters that follow we will explore many aspects of analytical chemistry. In the process we will consider important questions such as “How do we treat experimental data?”, “How do we ensure that our results are accurate?”, “How do we obtain a representative sample?”, and “How do we select an appropriate analytical technique?” Before we look more closely at these and other questions, we will first review some basic tools of importance to analytical chemists. 13 14 Analytical Chemistry 2.0 2A Measurements in Analytical Chemistry Analytical chemistry is a quantitative science. Whether determining the concentration of a species, evaluating an equilibrium constant, measuring a reaction rate, or drawing a correlation between a compound’s structure and its reactivity, analytical chemists engage in “measuring important chemical things.”1 In this section we briefly review the use of units and significant figures in analytical chemistry. 2A.1 Units of Measurement Some measurements, such as absorbance, A measurement usually consists of a unit and a number expressing the do not have units. Because the meaning of quantity of that unit. We may express the same physical measurement with a unitless number may be unclear, some authors include an artificial unit. It is not different units, which can create confusion. For example, the mass of a unusual to see the abbreviation AU, which sample weighing 1.5 g also may be written as 0.0033 lb or 0.053 oz. -
Page 1 of 6 This Is Henry's Law. It Says That at Equilibrium the Ratio of Dissolved NH3 to the Partial Pressure of NH3 Gas In
CHMY 361 HANDOUT#6 October 28, 2012 HOMEWORK #4 Key Was due Friday, Oct. 26 1. Using only data from Table A5, what is the boiling point of water deep in a mine that is so far below sea level that the atmospheric pressure is 1.17 atm? 0 ΔH vap = +44.02 kJ/mol H20(l) --> H2O(g) Q= PH2O /XH2O = K, at ⎛ P2 ⎞ ⎛ K 2 ⎞ ΔH vap ⎛ 1 1 ⎞ ln⎜ ⎟ = ln⎜ ⎟ − ⎜ − ⎟ equilibrium, i.e., the Vapor Pressure ⎝ P1 ⎠ ⎝ K1 ⎠ R ⎝ T2 T1 ⎠ for the pure liquid. ⎛1.17 ⎞ 44,020 ⎛ 1 1 ⎞ ln⎜ ⎟ = − ⎜ − ⎟ = 1 8.3145 ⎜ T 373 ⎟ ⎝ ⎠ ⎝ 2 ⎠ ⎡1.17⎤ − 8.3145ln 1 ⎢ 1 ⎥ 1 = ⎣ ⎦ + = .002651 T2 44,020 373 T2 = 377 2. From table A5, calculate the Henry’s Law constant (i.e., equilibrium constant) for dissolving of NH3(g) in water at 298 K and 340 K. It should have units of Matm-1;What would it be in atm per mole fraction, as in Table 5.1 at 298 K? o For NH3(g) ----> NH3(aq) ΔG = -26.5 - (-16.45) = -10.05 kJ/mol ΔG0 − [NH (aq)] K = e RT = 0.0173 = 3 This is Henry’s Law. It says that at equilibrium the ratio of dissolved P NH3 NH3 to the partial pressure of NH3 gas in contact with the liquid is a constant = 0.0173 (Henry’s Law Constant). This also says [NH3(aq)] =0.0173PNH3 or -1 PNH3 = 0.0173 [NH3(aq)] = 57.8 atm/M x [NH3(aq)] The latter form is like Table 5.1 except it has NH3 concentration in M instead of XNH3. -
THE SOLUBILITY of GASES in LIQUIDS Introductory Information C
THE SOLUBILITY OF GASES IN LIQUIDS Introductory Information C. L. Young, R. Battino, and H. L. Clever INTRODUCTION The Solubility Data Project aims to make a comprehensive search of the literature for data on the solubility of gases, liquids and solids in liquids. Data of suitable accuracy are compiled into data sheets set out in a uniform format. The data for each system are evaluated and where data of sufficient accuracy are available values are recommended and in some cases a smoothing equation is given to represent the variation of solubility with pressure and/or temperature. A text giving an evaluation and recommended values and the compiled data sheets are published on consecutive pages. The following paper by E. Wilhelm gives a rigorous thermodynamic treatment on the solubility of gases in liquids. DEFINITION OF GAS SOLUBILITY The distinction between vapor-liquid equilibria and the solubility of gases in liquids is arbitrary. It is generally accepted that the equilibrium set up at 300K between a typical gas such as argon and a liquid such as water is gas-liquid solubility whereas the equilibrium set up between hexane and cyclohexane at 350K is an example of vapor-liquid equilibrium. However, the distinction between gas-liquid solubility and vapor-liquid equilibrium is often not so clear. The equilibria set up between methane and propane above the critical temperature of methane and below the criti cal temperature of propane may be classed as vapor-liquid equilibrium or as gas-liquid solubility depending on the particular range of pressure considered and the particular worker concerned. -
Introduction to the Solubility of Liquids in Liquids
INTRODUCTION TO THE SOLUBILITY OF LIQUIDS IN LIQUIDS The Solubility Data Series is made up of volumes of comprehensive and critically evaluated solubility data on chemical systems in clearly defined areas. Data of suitable precision are presented on data sheets in a uniform format, preceded for each system by a critical evaluation if more than one set of data is available. In those systems where data from different sources agree sufficiently, recommended values are pro posed. In other cases, values may be described as "tentative", "doubtful" or "rejected". This volume is primarily concerned with liquid-liquid systems, but related gas-liquid and solid-liquid systems are included when it is logical and convenient to do so. Solubilities at elevated and low 'temperatures and at elevated pressures may be included, as it is considered inappropriate to establish artificial limits on the data presented. For some systems the two components are miscible in all proportions at certain temperatures or pressures, and data on miscibility gap regions and upper and lower critical solution temperatures are included where appropriate and if available. TERMINOLOGY In this volume a mixture (1,2) or a solution (1,2) refers to a single liquid phase containing components 1 and 2, with no distinction being made between solvent and solute. The solubility of a substance 1 is the relative proportion of 1 in a mixture which is saturated with respect to component 1 at a specified temperature and pressure. (The term "saturated" implies the existence of equilibrium with respect to the processes of mass transfer between phases) • QUANTITIES USED AS MEASURES OF SOLUBILITY Mole fraction of component 1, Xl or x(l): ml/Ml nl/~ni = r(m.IM.) '/. -
Homework 5 Solutions
SCI 1410: materials science & solid state chemistry HOMEWORK 5. solutions Textbook Problems: Imperfections in Solids 1. Askeland 4-67. Why is most “gold” or “silver” jewelry made out of gold or silver alloyed with copper, i.e, what advantages does copper offer? We have two major considerations in jewelry alloying: strength and cost. Copper additions obviously lower the cost of gold or silver (Note that 14K gold is actually only about 50% gold). Copper and other alloy additions also strengthen pure gold and pure silver. In both gold and silver, copper will strengthen the material by either solid solution strengthening (substitutional Cu atoms in the Au or Ag crystal), or by formation of second phase in the microstructure. 2. Solid state solubility. Of the elements in the chart below, name those that would form each of the following relationships with copper (non-metals only have atomic radii listed): a. Substitutional solid solution with complete solubility o Ni and possibly Ag, Pd, and Pt b. Substitutional solid solution of incomplete solubility o Ag, Pd, and Pt (if you didn’t include these in part (a); Al, Co, Cr, Fe, Zn c. An interstitial solid solution o C, H, O (small atomic radii) Element Atomic Radius (nm) Crystal Structure Electronegativity Valence Cu 0.1278 FCC 1.9 +2 C 0.071 H 0.046 O 0.060 Ag 0.1445 FCC 1.9 +1 Al 0.1431 FCC 1.5 +3 Co 0.1253 HCP 1.8 +2 Cr 0.1249 BCC 1.6 +3 Fe 0.1241 BCC 1.8 +2 Ni 0.1246 FCC 1.8 +2 Pd 0.1376 FCC 2.2 +2 Pt 0.1387 FCC 2.2 +2 Zn 0.1332 HCP 1.6 +2 3. -
Self-Assembly of a Colloidal Interstitial Solid Solution with Tunable Sublattice Doping: Supplementary Information
Self-assembly of a colloidal interstitial solid solution with tunable sublattice doping: Supplementary information L. Filion,∗ M. Hermes, R. Ni, E. C. M. Vermolen,† A. Kuijk, C. G. Christova,‡ J. C. P. Stiefelhagen, T. Vissers, A. van Blaaderen, and M. Dijkstra Soft Condensed Matter, Debye Institute for NanoMaterials Science, Utrecht University, Princetonplein 1, NL-3584 CC Utrecht, the Netherlands I. CRYSTAL STRUCTURE LS6 In the main paper we use full free-energy calculations to determine the phase diagram for binary mixtures of hard spheres with size ratio σS/σL = 0.3 where σS(L) is the diameter of the small (large) particles. The phases we considered included a structure with LS6 stoichiometry for which we could not identify an atomic analogue. Snapshots of this structure from various perspectives are shown in Fig. S1. FIG. S1: Unit cell of the binary LS6 superlattice structure described in this paper. Note that both species exhibit long-range crystalline order. The unit cell is based on a body-centered-cubic cell of the large particles in contrast to the face-centered-cubic cell associated with the interstitial solid solution covering most of the phase diagram (Fig. 1). II. PHASE DIAGRAM AT CONSTANT VOLUME 3 In the main paper, we present the xS − p representation of the phase diagram where p = βPσL is the reduced pressure, xS = NS/(NS + NL), NS(L) is the number of small (large) hard spheres, β = 1/kBT , kB is the Boltzmann constant, and T is the absolute temperature. This is the natural representation arising from common tangent construc- tions at constant pressure and the representation required for further simulation studies such as nucleation studies. -
Solid Solution Softening and Enhanced Ductility in Concentrated FCC Silver Solid Solution Alloys
UC Irvine UC Irvine Previously Published Works Title Solid solution softening and enhanced ductility in concentrated FCC silver solid solution alloys Permalink https://escholarship.org/uc/item/7d05p55k Authors Huo, Yongjun Wu, Jiaqi Lee, Chin C Publication Date 2018-06-27 DOI 10.1016/j.msea.2018.05.057 Peer reviewed eScholarship.org Powered by the California Digital Library University of California Materials Science & Engineering A 729 (2018) 208–218 Contents lists available at ScienceDirect Materials Science & Engineering A journal homepage: www.elsevier.com/locate/msea Solid solution softening and enhanced ductility in concentrated FCC silver T solid solution alloys ⁎ Yongjun Huoa,b, , Jiaqi Wua,b, Chin C. Leea,b a Electrical Engineering and Computer Science University of California, Irvine, CA 92697-2660, United States b Materials and Manufacturing Technology University of California, Irvine, CA 92697-2660, United States ARTICLE INFO ABSTRACT Keywords: The major adoptions of silver-based bonding wires and silver-sintering methods in the electronic packaging Concentrated solid solutions industry have incited the fundamental material properties research on the silver-based alloys. Recently, an Solid solution softening abnormal phenomenon, namely, solid solution softening, was observed in stress vs. strain characterization of Ag- Twinning-induced plasticity In solid solution. In this paper, the mechanical properties of additional concentrated silver solid solution phases Localized homologous temperature with other solute elements, Al, Ga and Sn, have been experimentally determined, with their work hardening Advanced joining materials behaviors and the corresponding fractography further analyzed. Particularly, the concentrated Ag-Ga solid so- lution has been discovered to possess the best combination of mechanical properties, namely, lowest yield strength, highest ductility and highest strength, among the concentrated solid solutions of the current study. -
Lecture 3. the Basic Properties of the Natural Atmosphere 1. Composition
Lecture 3. The basic properties of the natural atmosphere Objectives: 1. Composition of air. 2. Pressure. 3. Temperature. 4. Density. 5. Concentration. Mole. Mixing ratio. 6. Gas laws. 7. Dry air and moist air. Readings: Turco: p.11-27, 38-43, 366-367, 490-492; Brimblecombe: p. 1-5 1. Composition of air. The word atmosphere derives from the Greek atmo (vapor) and spherios (sphere). The Earth’s atmosphere is a mixture of gases that we call air. Air usually contains a number of small particles (atmospheric aerosols), clouds of condensed water, and ice cloud. NOTE : The atmosphere is a thin veil of gases; if our planet were the size of an apple, its atmosphere would be thick as the apple peel. Some 80% of the mass of the atmosphere is within 10 km of the surface of the Earth, which has a diameter of about 12,742 km. The Earth’s atmosphere as a mixture of gases is characterized by pressure, temperature, and density which vary with altitude (will be discussed in Lecture 4). The atmosphere below about 100 km is called Homosphere. This part of the atmosphere consists of uniform mixtures of gases as illustrated in Table 3.1. 1 Table 3.1. The composition of air. Gases Fraction of air Constant gases Nitrogen, N2 78.08% Oxygen, O2 20.95% Argon, Ar 0.93% Neon, Ne 0.0018% Helium, He 0.0005% Krypton, Kr 0.00011% Xenon, Xe 0.000009% Variable gases Water vapor, H2O 4.0% (maximum, in the tropics) 0.00001% (minimum, at the South Pole) Carbon dioxide, CO2 0.0365% (increasing ~0.4% per year) Methane, CH4 ~0.00018% (increases due to agriculture) Hydrogen, H2 ~0.00006% Nitrous oxide, N2O ~0.00003% Carbon monoxide, CO ~0.000009% Ozone, O3 ~0.000001% - 0.0004% Fluorocarbon 12, CF2Cl2 ~0.00000005% Other gases 1% Oxygen 21% Nitrogen 78% 2 • Some gases in Table 3.1 are called constant gases because the ratio of the number of molecules for each gas and the total number of molecules of air do not change substantially from time to time or place to place. -
Chemistry C3102-2006: Polymers Section Dr. Edie Sevick, Research School of Chemistry, ANU 5.0 Thermodynamics of Polymer Solution
Chemistry C3102-2006: Polymers Section Dr. Edie Sevick, Research School of Chemistry, ANU 5.0 Thermodynamics of Polymer Solutions In this section, we investigate the solubility of polymers in small molecule solvents. Solubility, whether a chain goes “into solution”, i.e. is dissolved in solvent, is an important property. Full solubility is advantageous in processing of polymers; but it is also important for polymers to be fully insoluble - think of plastic shoe soles on a rainy day! So far, we have briefly touched upon thermodynamic solubility of a single chain- a “good” solvent swells a chain, or mixes with the monomers, while a“poor” solvent “de-mixes” the chain, causing it to collapse upon itself. Whether two components mix to form a homogeneous solution or not is determined by minimisation of a free energy. Here we will express free energy in terms of canonical variables T,P,N , i.e., temperature, pressure, and number (of moles) of molecules. The free energy { } expressed in these variables is the Gibbs free energy G G(T,P,N). (1) ≡ In previous sections, we referred to the Helmholtz free energy, F , the free energy in terms of the variables T,V,N . Let ∆Gm denote the free eneregy change upon homogeneous mix- { } ing. For a 2-component system, i.e. a solute-solvent system, this is simply the difference in the free energies of the solute-solvent mixture and pure quantities of solute and solvent: ∆Gm G(T,P,N , N ) (G0(T,P,N )+ G0(T,P,N )), where the superscript 0 denotes the ≡ 1 2 − 1 2 pure component. -
Chemistry 1220 Chapter 17 Practice Exam
Dr. Fus CHEM 1220 CHEMISTRY 1220 CHAPTER 17 PRACTICE EXAM All questions listed below are problems taken from old Chemistry 123 exams given here at The Ohio State University. Read Chapter 17.4 – 17.7 and complete the following problems. These problems will not be graded or collected, but will be very similar to what you will see on an exam. Ksp and Molar Solubility (Section 17.4) The solubility product expression equals the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium expression. For example: PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp=[Pb2+] [Cl-]2 Why isn’t the concentration of PbCl2 in the denominator of the Ksp? Recall from Section 15.4 that whenever a pure solid or pure liquid is in a heterogeneous (multiple phases present) equilibrium, its concentration is not included in the equilibrium expression. 1. The solubility product expression for La2(CO3)3 is Ksp = ? First, write the equilibrium expression. The subscript following each atom in the solid is the number of ions of that atom in solution. The total charges must add to zero: La2(CO3)3(s) 2La3+(aq) + 3CO32-(aq) Given the information above and the equilibrium expression: Ksp=[La3+]2 [CO32-]3 2. The solubility product expression for Zn3(PO4)2 is Ksp = ? Zn3(PO4)2(s) 3Zn2+(aq) + 2PO43-(aq) Ksp=[Zn2+]3 [PO43-]2 3. The solubility product expression for Al2(CO3)3 is Ksp = ? Al2(CO3)3(s) 2Al3+(aq) + 3CO32-(aq) Ksp=[Al3+]2 [CO32-]3 4.