DOCSLIB.ORG

Ideal (Solid) Solutions

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Ideal (Solid) Solutions Ideal (solid) solutions M total sites Entopy (=entropy of mixing) Two component system ΔS = −kB (xA ln xA + xB ln xB ) M xA + xB =1 € Multi component system ΔS A € = = −kB ∑ x j ln x j M j = B constraint ∑ x j =1 € j € € € Ideal (solid) solutions M total sites Entopy (=entropy of mixing) Two component system ΔS = −kB (xA ln xA + xB ln xB ) M xA + xB =1 € Multi component system ΔS A € = = −kB ∑ x j ln x j M j = B constraint ∑ x j =1 € j € € € Free energy of mixing for ideal solutions ΔG mix = RT(xA ln xA + xB ln xB ) €Δ G mix Chemical potentials RT 0 ∂ΔG mix µB − µB = ΔG mix (xB ) + xB ∂xB = RT ln xB € 0 µA − µA = RT ln xA = RT ln(1− x ) € B Ideal solution behaves Roaultian! xB € € Regular solution theory ΔG mix = ΔHmix − TΔSmix Bragg‐Williams approximation (or mean‐field approximation) € Washed‐out mixture Enthalpy of mixing 0 0 ΔHmix ≈ ΔE mix = E mix − xA E A − xB E B = wxA xB 2 2 E mix = wAA xA + wBB xB + 2wAB xA xB € € Regular solution theory ΔG mix = wxA xB + RT(xA ln xA + xB ln xB ) 1 w = Z w − (w + w ) AB 2 AA BB € ΔHmix € ΔG mix RT € −TΔSmix € xB € € Regular solution theory ΔG mix = wxA xB + RT(xA ln xA + xB ln xB ) Chemical potentials 0 2 2 µA − µA = w(1− xA ) + RT ln xA = w(xB ) + RT ln(1− xB ) € 0 2 µB − µB = w(1− xB ) + RT ln xB € Activity € 0 w w µB − µB = RT lnaB (1−x )2 +ln x (1−x )2 RT B B RT B w aB = e = xBe lna = (1− x )2 + ln x B RT B B € Fundamental flaw in Regular Solution Theory: The approximation in R. S. T. is that the probability of observing different configurations is the same. This is not true in reality! € € Entropy induced order (not disorder) Consider a system composed of long rod‐like particles and “solvent” (small particles). What is the entropy if mixing in this case? Assume that you have NR rods and Nm solvent molecules, so that the total volume is Vtot = NR VR + Nm Vm . Solvent Rod Hints in the next page Hints: you should consider two systems. One in which your “lattice” boxes are big enough so that the rods can freely rotate, and calculate the total entropy of the system. The boxes should look like the area in the upper left corner of the sketch. Remember that you want to include the rotational part also. The second “lattice” you have to consider is a lattice where you impose the direction of the rods, such as that shown in the lower right corner, and calculate the entropy of this system. What you will find is two different forms for the Free Energy of the system, and above a certain rod length and density of rods the order phase will be preferred. Hints: you should consider two systems. One in which your “lattice” boxes are big enough so that the rods can freely rotate, and calculate the total entropy of the system. The boxes should look like the area in the upper left corner of the sketch. Remember that you want to include the rotational part also. The second “lattice” you have to consider is a lattice where you impose the direction of the rods, such as that shown in the lower right corner, and calculate the entropy of this system. What you will find is two different forms for the Free Energy of the system, and above a certain rod length and density of rods the order phase will be preferred. This is an example that simple intelligent reasoning and math is much better than complicated proofs. It almost always offers a much simpler and clearer picture of the fundamentals. .
Load more

Recommended publications
  • Solution Is a Homogeneous Mixture of Two Or More Substances in Same Or Different Physical Phases. the Substances Forming The
    CLASS NOTES Class:12 Topic: Solutions Subject: Chemistry Solution is a homogeneous mixture of two or more substances in same or different physical phases. The substances forming the solution are called components of the solution. On the basis of number of components a solution of two components is called binary solution. Solute and Solvent In a binary solution, solvent is the component which is present in large quantity while the other component is known as solute. Classification of Solutions Solubility The maximum amount of a solute that can be dissolved in a given amount of solvent (generally 100 g) at a given temperature is termed as its solubility at that temperature. The solubility of a solute in a liquid depends upon the following factors: (i) Nature of the solute (ii) Nature of the solvent (iii) Temperature of the solution (iv) Pressure (in case of gases) Henry’s Law The most commonly used form of Henry’s law states “the partial pressure (P) of the gas in vapour phase is proportional to the mole fraction (x) of the gas in the solution” and is expressed as p = KH . x Greater the value of KH, higher the solubility of the gas. The value of KH decreases with increase in the temperature. Thus, aquatic species are more comfortable in cold water [more dissolved O2] rather than Warm water. Applications 1. In manufacture of soft drinks and soda water, CO2 is passed at high pressure to increase its solubility. 2. To minimise the painful effects (bends) accompanying the decompression of deep sea divers. O2 diluted with less soluble.
    [Show full text]
  • Physical Chemistry
    Subject Chemistry Paper No and Title 10, Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No and Title 10, Free energy functions and Partial molar properties Module Tag CHE_P10_M10 CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 10: Free energy functions and Partial molar properties TABLE OF CONTENTS 1. Learning outcomes 2. Introduction 3. Free energy functions 4. The effect of temperature and pressure on free energy 5. Maxwell’s Relations 6. Gibbs-Helmholtz equation 7. Partial molar properties 7.1 Partial molar volume 7.2 Partial molar Gibb’s free energy 8. Question 9. Summary CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 10: Free energy functions and Partial molar properties 1. Learning outcomes After studying this module you shall be able to: Know about free energy functions i.e. Gibb’s free energy and work function Know the dependence of Gibbs free energy on temperature and pressure Learn about Gibb’s Helmholtz equation Learn different Maxwell relations Derive Gibb’s Duhem equation Determine partial molar volume through intercept method 2. Introduction Thermodynamics is used to determine the feasibility of the reaction, that is , whether the process is spontaneous or not. It is used to predict the direction in which the process will be spontaneous. Sign of internal energy alone cannot determine the spontaneity of a reaction. The concept of entropy was introduced in second law of thermodynamics. Whenever a process occurs spontaneously, then it is considered as an irreversible process.
    [Show full text]
  • Simple Mixtures
    Simple mixtures Before dealing with chemical reactions, here we consider mixtures of substances that do not react together. At this stage we deal mainly with binary mixtures (mixtures of two components, A and B). We therefore often be able to simplify equations using the relation xA + xB = 1. A restriction in this chapter – we consider mainly non-electrolyte solutions – the solute is not present as ions. The thermodynamic description of mixtures We already considered the partial pressure – the contribution of one component to the total pressure, when we were dealing with the properties of gas mixtures. Here we introduce other analogous partial properties. Partial molar quantities: describe the contribution (per mole) that a substance makes to an overall property of mixture. The partial molar volume, VJ – the contribution J makes to the total volume of a mixture. Although 1 mol of a substance has a characteristic volume when it is pure, 1 mol of a substance can make different contributions to the total volume of a mixture because molecules pack together in different ways in the pure substances and in mixtures. Imagine a huge volume of pure water. When a further 1 3 mol H2O is added, the volume increases by 18 cm . When we add 1 mol H2O to a huge volume of pure ethanol, the volume increases by only 14 cm3. 18 cm3 – the volume occupied per mole of water molecules in pure water. 14 cm3 – the volume occupied per mole of water molecules in virtually pure ethanol. The partial molar volume of water in pure water is 18 cm3 and the partial molar volume of water in pure ethanol is 14 cm3 – there is so much ethanol present that each H2O molecule is surrounded by ethanol molecules and the packing of the molecules results in the water molecules occupying only 14 cm3.
    [Show full text]
  • Chapter 9 Ideal and Real Solutions
    2/26/2016 CHAPTER 9 IDEAL AND REAL SOLUTIONS • Raoult’s law: ideal solution • Henry’s law: real solution • Activity: correlation with chemical potential and chemical equilibrium Ideal Solution • Raoult’s law: The partial pressure (Pi) of each component in a solution is directly proportional to the vapor pressure of the corresponding pure substance (Pi*) and that the proportionality constant is the mole fraction (xi) of the component in the liquid • Ideal solution • any liquid that obeys Raoult’s law • In a binary liquid, A-A, A-B, and B-B interactions are equally strong 1 2/26/2016 Chemical Potential of a Component in the Gas and Solution Phases • If the liquid and vapor phases of a solution are in equilibrium • For a pure liquid, Ideal Solution • ∆ ∑ Similar to ideal gas mixing • ∆ ∑ 2 2/26/2016 Example 9.2 • An ideal solution is made from 5 mole of benzene and 3.25 mole of toluene. (a) Calculate Gmixing and Smixing at 298 K and 1 bar. (b) Is mixing a spontaneous process? ∆ ∆ Ideal Solution Model for Binary Solutions • Both components obey Rault’s law • Mole fractions in the vapor phase (yi) Benzene + DCE 3 2/26/2016 Ideal Solution Mole fraction in the vapor phase Variation of Total Pressure with x and y 4 2/26/2016 Average Composition (z) • , , , , ,, • In the liquid phase, • In the vapor phase, za x b yb x c yc Phase Rule • In a binary solution, F = C – p + 2 = 4 – p, as C = 2 5 2/26/2016 Example 9.3 • An ideal solution of 5 mole of benzene and 3.25 mole of toluene is placed in a piston and cylinder assembly.
    [Show full text]
  • 9.2 Raoult's Law and Henry's
    Chapter 9 The Behavior of Solutions 1 9.2 Raoult’s Law and Henry’s Law (1) Initially evacuated vessel at T liq. A (2) spontaneously evaporate until P in the vessel ° reaches the saturated vapor pressure of liq. A, p A, at T (3) a dynamic equilibrium established between the rate of evaporation of liq. A & the rate of condensation of vapor A. 2 9.2 Raoult’s Law and Henry’s Law • re(A) α the # of A atoms in the vapor phase, ° thus, rc(A) = kp A , and at equilibrium (9.1) • Similarly, pure liquid B (9.2) re(A) rc(A) liq. A 3 9.2 Raoult’s Law and Henry’s Law (A) the effect of the small addition of liquid B to liquid A Assume (1) the composition of the surface of the liquid = that of the bulk liquid, XA, (2) the sizes of A & B atoms comparable (9.3) (9.4) 4 9.2 Raoult’s Law and Henry’s Law • From (9.1) and (9.3) gives (9.5) and from (9.2) and (9.4) gives (9.6) • Raoult’s law: the vapor pressure exerted by a component i in a solution = (the mole fraction of i in the solution) X (the saturated vapor pressure of pure liquid i ) at T of the solution. 5 9.2 Raoult’s Law and Henry’s Law • Requirements: the intrinsic rates of evaporation of A and B ≠ f(comp. of the solution) i.e., the magnitudes of the A-A, B-B, and A-B bond energies in the solution are identical 6 9.2 Raoult’s Law and Henry’s Law (B) If A-B >> A-A and B-B and a solution of A in B is sufficiently dilute: A atom at the surface surrounded only by B, A in a deeper potential E well than are the A atoms at the surface of pure liq… difficult to be lifted to vapor, evaporation rate is decreased to re(A) to re’(A) 7 9.2 Raoult’s Law and Henry’s Law (9.7) re(A) > re’(A) (9.8) ’ (9.9) • A similar consideration of dilute solutions of B in A gives ’ (9.10) • (9.9),(9.10) : Henry’ law 8 9.2 Raoult’s Law and Henry’s Law Beyond a critical value of XA, re’(A) becomes composition- dependent.
    [Show full text]
  • The Dissolution Process Page 1 of 2 Chemistry 2E 11: Solutions And
    OpenStax Chemistry 2e 11.1: The Dissolution Process Chemistry 2e 11: Solutions and Colloids 11.1: The Dissolution Process 1. How do solutions differ from compounds? From other mixtures? Solution A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. 2. Which of the principal characteristics of solutions are evident in the solutions of K2Cr2O7 shown in Figure 11.2? Solution The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K2Cr2O7 in water can vary. 3. When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally. (a) Is the dissolution of KNO3 an endothermic or an exothermic process? (b) What conclusions can you draw about the intermolecular attractions involved in the process? (c) Is the resulting solution an ideal solution? Solution (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and NO3 ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. 4. Give an example of each of the following types of solutions: (a) a gas in a liquid (b) a gas in a gas (c) a solid in a solid Solution (a) CO2 in water; (b) O2 in N2 (air); (c) bronze (solution of tin or other metals in copper) 5.
    [Show full text]
  • Calculation of the Gibbs Free Energy of Solvation and Dissociation of Hcl in Water Via Monte Carlo Simulations and Continuum
    PCCP View Article Online PAPER View Journal | View Issue Calculation of the Gibbs free energy of solvation and Cite this: Phys. Chem. Chem. Phys., 2013, dissociation of HCl in water via Monte Carlo 15, 13578 simulations and continuum solvation models Matthew J. McGrath,*ab I-F. Will Kuo,c Brice F. Ngouana W.,de Julius N. Ghogomu,d Christopher J. Mundy,f Aleksandr V. Marenich,b Christopher J. Cramer,b Donald G. Truhlarb and J. Ilja Siepmannbg The Gibbs free energy of solvation and dissociation of hydrogen chloride in water is calculated through a combined molecular simulation/quantum chemical approach at four temperatures between T = 300 and 450 K. The Gibbs free energy is first decomposed into the sum of two components: the Gibbs free energy of transfer of molecular HCl from the vapor to the aqueous liquid phase and the standard-state Gibbs free energy of acid dissociation of HCl in aqueous solution. The former quantity is calculated Received 24th April 2013, using Gibbs ensemble Monte Carlo simulations using either Kohn–Sham density functional theory or a Accepted 21st June 2013 molecular mechanics force field to determine the system’s potential energy. The latter Gibbs free energy DOI: 10.1039/c3cp51762d contribution is computed using a continuum solvation model utilizing either experimental reference data or micro-solvated clusters. The predicted combined solvation and dissociation Gibbs free energies www.rsc.org/pccp agree very well with available experimental data. 1 Introduction Studies of the thermophysical properties of HCl dissolution in water were performed early in the last century using electro- The interaction between hydrogen chloride and water has been chemical cells,7,8 and the structures of both dilute and concen- studied for many years due to the importance of HCl as a strong trated HCl solutions have been probed in neutron and X-ray acid (including acid dissociation in very cold nanoclusters1) 9,10 Published on 21 June 2013.
    [Show full text]
  • Articles Plays Relative Humidity (RH), and Hygroscopic Growth Factors at an Important Role in Numerous Atmospheric Processes Such As 90% RH Ranged from 1.08 to 1.17
    Atmos. Chem. Phys., 4, 35–50, 2004 www.atmos-chem-phys.org/acp/4/35/ Atmospheric SRef-ID: 1680-7324/acp/2004-4-35 Chemistry and Physics Hygroscopic properties of water-soluble matter and humic-like organics in atmospheric fine aerosol M. Gysel1, E. Weingartner1, S. Nyeki1, D. Paulsen1, U. Baltensperger1, I. Galambos2, and G. Kiss3 1Laboratory of Atmospheric Chemistry, Paul Scherrer Institut, CH-5232 Villigen, Switzerland 2Department of Earth and Environmental Sciences, University of Veszprem,´ 8201 Veszprem,´ Hungary 3Air Chemistry Group of the Hungarian Academy of Sciences, University of Veszprem,´ 8201 Veszprem,´ Hungary Received: 27 July 2003 – Published in Atmos. Chem. Phys. Discuss.: 1 October 2003 Revised: 15 December 2003 – Accepted: 28 December 2003 – Published: 22 January 2004 Abstract. Ambient continental-rural fine aerosol (K-puszta, of WSM), the WSOM accounted for a significant fraction of Hungary, PM1.5) was sampled on quartz fibre filters in win- particulate water. At 90% RH, according to model predic- ter and summer 2001. Water-soluble matter (WSM) was ex- tions and measurements, about 80–62% of particulate water tracted in MilliQ-water, and, in a second step, solid phase in the samples are associated with inorganic salts and about extraction was used to isolate the less hydrophilic fraction 20–38% with WSOM. The relative contributions of both dis- (ISOM) of the water-soluble organic matter (WSOM) from tinguished WSOM fractions, ISOM and MHOM, remains remaining inorganic salts and “most” hydrophilic organic uncertain since MHOM was not available in isolated form, matter (MHOM). This approach allowed ISOM, which con- but the results suggest that the less abundant MHOM is also stitutes the major fraction of WSOM, to be isolated from am- important due to its presumably larger hygroscopicity.
    [Show full text]
  • 3. Ideal Solution Dissolution Equilibrium of Gas in Dissolution Equilibrium State of Gas Into Water, A(G)  A(Aq)
    2014 Reaction Systems Engineering II (Complex Reaction Systems Analysis) 1. Chemical Equilibrium (A. Miyoshi) 5/8 3. Ideal Solution Dissolution Equilibrium of Gas In dissolution equilibrium state of gas into water, A(g) A(aq): (g) = (aq) (3.1) For an ideal gas and an ideal solution [from (0.6) and (0.7)]: b / b –solG° = °(g) – °(aq) = RT ln Ksol where Ksol (3.2) p / p Solution Phase Chemical Equilibrium The CO2(aq) dissociates in solution phase as, + – CO2(aq) + H2O(l) H(aq) + HCO3 (aq) [a1] – + 2– HCO3 (aq) H(aq) + CO3 (aq) [a2] The respective acid-dissociation constants are, + [H ][HCO3 ] –a1G° = RT ln Ka1 where Ka1 (3.3) [CO2 (aq)] [H+2 ][CO ] – G° = RT ln K where K 3 (3.4) a2 a2 a2 [HCO3 ] where [H+] b[H+(aq)] / b°, etc. [CO (aq)] [HCO ] [CO2 ] The"apparent" dissolution equilibrium constant, K 233, for a sol, app pp(CO2 ) / given [H+] is, KK KKa211 a1 (3.5) sol, app sol [H ] [H ] Exercise 3.1 1) Compute the dissolution equilibrium constant of CO2 into water, Ksol, and subsequent acid-dissociation constants, Ka1 and Ka2, from the following standard Gibbs energies. –1 fG°(298 K) / kJ mol CO2(g) –394.4 CO2(aq) –386.0 H2O(l) –237.1 – HCO3 (aq) –586.8 H+(aq) 0 * * zero by definition 2– CO3 (aq) –527.8 2) Calculate Ksol, app / Ksol for pH = 4 and 8. Solution to Exercise 3.1 –1 1) solG° = –386.0 – (–394.4) = 8.4 kJ mol . –2 –1 –1 Ksol = exp(–solG° / RT) = exp[–8400 / (8.3145298)] = 3.3710 [-] (or mol kg bar ) –1 –7 a1G° = –586.8 – (–386.0–237.1) = 36.3 kJ mol Ka1 = 4.3410 [-] –1 –11 a2G° = –527.8 – (–586.8) = 59.0 kJ mol Ka2 = 4.5610 [-] K 4.56 1011 4.34 10 7 2) pH = 4: sol, app 11 = 1.004 [-].
    [Show full text]
  • Solutions Notes.Notebook February 24, 2015
    Solutions Notes.notebook February 24, 2015 Problem: A solution is prepared by adding 5.84 g of formaldehyde, H2CO, to 100.0 g of H2O. The final volume of the solution was 104.0 mL. Calculate the molarity, molality, mass percent, and mole fraction. Solution: homegeneous mixture Molarity (M): moles solute/liter solution Mass Percent: mass solute/(mass solute + mass solvent) X 100 Mole Fraction of A: moles A/total moles Molality (m): moles solute/kilograms solvent Feb 28­3:01 PM Feb 28­3:02 PM Chapter 11 problems 29, 33, and 35 Normality (N) = equivalents of solute liter of solution An equivalent is the mass of the solute that can furnish or accept one mole of protons (H+ in acid/base reaxns)) or one mole of electrons (in redox reaxns) e.g. H3PO4 e.g. HCl 3 H+ equivalents/mol acid 1 H+ equivalent/mole acid equivalent weight = 98.0 g/mol = 32.7 g/eq equivalent weight = 36.5 g/mol = 36.5 g/eq 3 eq/mol 1 eq/mol Feb 28­3:03 PM Feb 28­3:03 PM Problem: Given the reaction, HEAT OF SOLUTION + 3­ H2SO4 + 2NaOH ­­­­­> 2H2O + 2Na + SO4 "Like dissolves like" If we have 28.42 g H2SO4 in 800. mL of water, what are the normality and molarity of the solution? Solution formation occurs in three steps: 1. Expanding the solute (endothermic, ∆H1 = +) 2. Expanding the solvent (endothermic, ∆H2 = +) 3. Combining solute and solvent (exothermic, ∆H3 = ­) ∆ ∆Hsolution = ∆H1 + ∆H2 + H3 Recall two factors driving reaction: enthalpy and entropy must be considered.
    [Show full text]
  • SOLUTIONS and SOLUBILITY  a Saturated Solution Is One in Which the Solute Is in Equilibrium with the Solid Phase (Solute)
    SOLUTIONS AND SOLUBILITY A saturated solution is one in which the solute is in equilibrium with the solid phase (solute). SoIubiliy is defined in quantitative terms as the concentration of solute in a saturated solution at a certain temperature, and in a qualitative way, it can be defined as the spontaneous interaction of two or more substances to form a homogeneous molecular dispersion. An unsaturated or subsaturated solution is one containing the dissolved solute in a concentration below that necessary for complete saturation at a definite temperature. A supersaturated solution is one that contains more of the dissolved solute than it would normally contain at a definite r r r diss v d s r s tempe atu e, we e the un ol e olute p e ent. 1 Solubility Expressions The solubility of a drug may be expressed in a number of ways. The U.S. Pharnzacopeia and National Formulary lists the solubility of drugs as the number of milliliters of solvent in which 1 g of solute will dissolve. For example, the solubility of boric acid is given in the U.S. Pharmacopeia as follows: 1 g of boric acid dissolves in 18 mL of water, in 18 mL of alcohol, and in 4 mL of glycerin. Solubility is also quantitatively expressed in terms of molality, molarity, and percentage. 2 3 1 g of boric acid dissolves in 18 mL of water 1 g of boric acid dissolves in 4 mL of glycerin 4 SOLVENT²SOLUTE INTERACTIONS Water is a good solvent for salts, sugars, and similar compounds, whereas mineral oil and benzene are often solvents for substances that are normally only slightly soluble in water.
    [Show full text]
  • Raoult's Law Po Po B B Henry's Law
    Solution Behavior • Goal: Understand the behavior of homogeneous systems containing more than one component. • Concepts Activity Partial molar properties Ideal solutions Non-ideal solutions Gibbs’-Duhem Equation Dilute solutions (Raoult’s law and Henry’s law) • Homework: WS2002 1 1 Raoult’s Law • Describes the behavior of solvents with containing a small volume fraction of solute • The vapor pressure of a component A in a solution at temperature T is equal to the product of the mole fraction of A in solution and the vapor pressure of pure A at temperature T. • Assumes: A, B evaporation rates are independent of composition A-B bond strength is the same as A-A and B-B or A-B bond strength is the mean of A-A and B-B • Example of liquid A in evacuated tank Initially, A evaporates at re(A) • At equilibrium Evaporation/condensation same rate WS2002 2 2 Raoult’s Law • Add a small amount of B (solute) to A (solvent) in tank • If the composition of the surface is the same as the bulk, then the fraction of surface sites occupied by A atoms is XA, the mole fraction • The evaporation rate will decrease accordingly along with the vapor pressure of A above the liquid. Condensation should be the same. Ideal Raoultian Solution = o pXp BBB po pp+ B AB o pA p B pA Vapor Pressure A B XB WS2002 3 3 Henry’s Law • Describes the behavior of a solute in a dilute solution • The solvent obeys Raoult’s law in the same region where the solute follows Henry’s law.
    [Show full text]