DOCSLIB.ORG

Problem 8.11 from Text Solution to Problem 8.11 Solution to Problem 8.11 Cont

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Problem 8.11 from Text Solution to Problem 8.11 Solution to Problem 8.11 Cont Problem 8.11 from text The normal boiling point of iodomethane, CH3I, is 42.43 ˚C and its vapor pressure at 0 ˚C is 140 torr. Calculate (a) standard enthalpy of vaporization (ΔH°vap) of iodomethane (b) standard entropy of vaporization (ΔS°vap) of iodomethane (c) the vapor pressure of CH3I at 25 ˚C Solution to problem 8.11 (a) The data given in the problem actually supplies the vapor pressure of at two temperatures (0 and 42.43˚C, or 273.15 and 315.58 K), so the easiest equation to use is: ln (P2/P1) ≈ - (ΔH°/R)(1/T2 - 1/T1) 760 !H˚ vap $ 1 1 ln = " # & 140 8.314 J mol -1K-1 % 315.58 K 273.15 K' -1 !H˚ vap= +28.6 kJ mol Solution to problem 8.11 cont. (b) Now we can use the basic expression ΔG° = ΔH° - TΔS° to solve for ΔS°. Since ΔG at boiling point is zero, ΔS°vap = (ΔH°vap/Τb) " 28600 J -1 !S˚ = $ = + 90.6 J K vap # 315.58 K% Solution to problem 8.11 (c) Use again: ln (P2/P1) ≈ - (ΔH°/R)(1/T2 - 1/T1) P 28600 J 1 1 ln 298 K = ! " # ! % 1 atm 8.314 J mol -1K-1 $ 298.15 K 315.58 K& P298 K = 0.529 atm = 402 torr Sample Problem CaCO3(s) CaO(s) + CO2(g) limestone lime Given: -23 ΔH° = + 178 kJ; at 25 ˚C, KP = 1.39 × 10 Give an estimate for KP at 800 ˚C. Solubility Equilibrium (Chap. 11) The process is in equilibrium when rate of ions (or molecules) leaving the solid = rate returning to the solid. That is, when dissolution rate = precipitation rate Solubility Product, Ksp (Sec. 11.10) y+ x- AxXy(s) x X (aq) + y X (aq) 2+ - (eg., BaCl2(s) Ba (aq) + 2 Cl (aq)) What is the form for the equilibrium constant? activities of products K = (at equilibrium) eq activities of reactants Write the equilibrium constant expression. Colligative Properties Depend solely on relative numbers of solvent and solute molecules in solutions, not on their chemical identity. The specifics of intermolecular interactions are not important for determining these properties Typical examples: Boiling point elevation and freezing point depression Osmosis and osmotic pressure Osmosis - The process Cross-sectional ∆h ∆V = A∆h area = A π = Osmotic Pressure, h (due to this “hydrostatic head”) Pure Solution Solvent Semi-permeable membrane Cross-sectional ∆h ∆V = A∆h area = A π = Osmotic Pressure, h (due to this “hydrostatic head”) Osmotic Pure Solution Solvent Pressure Semi-permeable membrane ρ = density of solvent ≈ density of solution (if very dilute) hA = Vsolution - Vsolvent = Volume of “hydrostatic head” ρ(Vsolution - Vsolvent) = ρ(hA) = mass of solvent in the “hydrostatic head” π = Osmotic Pressure = Force/Area = (mass × g)/Area (g = gravitational constant) [ρ(hA)] × g π = = ρhg A Osmosis Pure solvent “moves into” a solution when the solution and solvent are separated by a semipermeable membrane Why? Entropy increases when solvent moves through the membrane into the solution because the solute molecules then have a greater volume to move in. (See sections 7.3 and 7.6). Entropy change when solution volume increases by ∆V V V+∆V ∆V n moles of solute molecules expand into available solvent like a gas expands into a vacuum. ΔS = nRln[(V+∆V)/V] = nRln[1+(∆V/V)] Since V >> ∆V, ln[1+(∆V/V)] ≈ ∆V/V ∆S = nR(∆V/V) ⇒ T∆S = (nRT/V)∆V The solvent in the ΔV element must be “lifted” by height h. Cross-sectional ∆h ∆V = A∆h area = A π = Osmotic Pressure, h (due to this “hydrostatic head”) Equivalent Pure work needed Solution Solvent Semi-permeable to do the membrane “lifting” is (ρ∆V)gh. Entropy Gain vs Enthalpy Cost No more solvent will move from the pure solvent into the solution when ∆G for that process is no longer negative: ∆G = ∆H - T∆S The enthalpy cost from “lifting” the solvent” in the volume element ∆V is: ∆H = (ρ∆V)gh, But π = ρgh, ∴ ∆H = π∆V We already found T∆S = (nRT/V)∆V So, when ∆G = 0, π = nRT/V Other Useful Forms n π = nRT/V = [ /V]RT n [ /V] = moles solute per liter solution If we know mass (grams) of solute per liter, cm, then we transform to mol g mol ( /L) ~ ( /L)( /g) n c [ /V] = ( m/M) M = molar mass (“mol. weight”) c π = ( m/M)RT If the solute dissociates, a factor “i” accounts the actual number n of particles: π = i[ /V]RT Example: Determination of Polymer Molecular Weight 7 !/cm 6 atm 10-4 " 5 gm/L 4 3.3 " 10-4 30 2 4 6 8 10 cm , gm/L “Real-life” data for polyvinylchloride dissolved in cyclohexanone at 40 ˚C. What is the average molecular weight?.
Load more

Recommended publications
  • Using the Vapor Pressure of Pure Volatile Organic Compounds to Predict the Enthalpy of Vaporization and Computing the Entropy of Vaporization
    Open Access Library Journal Using the Vapor Pressure of Pure Volatile Organic Compounds to Predict the Enthalpy of Vaporization and Computing the Entropy of Vaporization Shawn M. Abernathy, Kelly R. Brown Department of Chemistry, Howard University, Washington DC, USA Email: [email protected] Received 5 September 2015; accepted 21 September 2015; published 28 September 2015 Copyright © 2015 by authors and OALib. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/ Abstract The objective of this investigation was to develop a vapor pressure (VP) acquisition system and methodology for performing temperature-dependent VP measurements and predicting the en- thalpy of vaporization (ΔHvap) of volatile organic compounds, i.e. VOCs. High quality VP data were acquired for acetone, ethanol, and toluene. VP data were also obtained for water, which served as the system calibration standard. The empirical VP data were in excellent agreement with its ref- erence data confirming the reliability/performance of the system and methodology. The predicted values of ΔHvap for water (43.3 kJ/mol, 1.0%), acetone (31.4 kJ/mol; 3.4%), ethanol (42.0 kJ/mol; 1.0%) and toluene (35.3 kJ/mol; 5.4%) were in excellent agreement with the literature. The com- puted values of ΔSvap for water (116.0 J/mol∙K), acetone (95.2 J/mol∙K), ethanol (119.5 J/mol∙K) and toluene (92.0.J/mol∙K) compared also favorably to the literature. Keywords Vapor Pressure (VP), Enthalpy of Vaporization, Entropy of Vaporization, Volatile Organic Compounds (VOCs), Predict Subject Areas: Physical Chemistry 1.
    [Show full text]
  • THE ENTROPY of VAPORIZATION AS a MBANS of DIS- Tinguismg NORIKAL LIQUIDS
    970 JOEL H. HILDEBRAND. iodide solution are shown to give a constant e. m. f. when measured against an unpolarizable electrode. Freshly cast rods of cadmium are shown to attain this value with time. 4. The behavior of cadmium electrodes in concentration cells is ex- plained as due to an allotropic change. 5. Crystalline electrolytic deposits of cadmium on cadmium rods or platinum are shown to give a reproducible e. m. f. when measured against an unpolarizsMe ekctrode. This e. m. f. is, however, slightly higher than that given by the gray cadmium electrodes. Amalgamated electrodes are not reproducible. 6. Photomicrographs indicate a change in crystalline form and con- firm the conclusions drawn from the e. m. f. measurements. 7. Attempts to inoculate the surface of the freshly cast cadmium with the stable rnaatian indicate that the action is extremely slow. 8. Phobmicrogtsphs indicate that in all cases the change is due to a deposit upon the surface htead of to a pitting of the surface. LHEMICALLAJJOUT~BT, BSYN MAWRCOLLBGE, BlYN MAWR. PA [CONTRIBUTION FROM THE CHEMICAL LABOBATORYOF TaS UNIVERSITY OF cALlmRNIA. I THE ENTROPY OF VAPORIZATION AS A MBANS OF DIS- TINGUISmG NORIKAL LIQUIDS. BY JOSL H. HILDSBRAND. Received February 23. 1915. In a series of studies which we are making on the theory of solutions, it has become necessary to distinguish between deviations from Raoult’s Law, whose cause may be considered physical, and those due to chemical changes. As has been pointed out it1 a previous paper,’ the choice be- tween the alternatives in a given case is possible if we know whether or not the pure liquids are associated.
    [Show full text]
  • Jgt3 R:. A) -2ZJ(6 .' .'
    2 1. (24) a. In the diagram below to the to the right, helium atoms are represented by unshaded spheres, neon atoms by gray spheres, and argon atoms by black spheres. If the total pressure in the container is 900 mm Hg, wh at is the parti~ressure of helium? A) 90 mm Hg ~ 180 mm HgC) 270 mm Hg D) 450 mm Hg • 0 o b. Which of the following is not a ~ e function? O. A) enthalpy ® heat C) internal energy D) volume .0.o c. Which one of the followin g~ ses will have the high est rate of effusion? A) N02 ~ N20 C) N204 D) N03 SF(" '$ ,..""I"' DI A-r:: Q09/mol d. At what temperature will sulfur hexafluoride molecules have th e same average speed as argon atoms at 20°C? JGt3 r:. A) -2ZJ(6 B) 73.2° C C) 381 °C @ 799°C e. Which substance in each of the following pairs is expected to have the larger dispersion forces? I BrZ o r I Z '" ""o'o....r~SS ~ A) Br2 in set I and n-butane in set" H H H H B) Br2 in set 1 and isobutane in set II I I I I © 12 in set I and n-butane in set" I I E-t-C----,:::-C-C-H o r H I I I I I D) 12 in set I and isobutane in set II H H H H H-C- C---C-H n- bu t ane I I I H H H l.sDbut a na f.
    [Show full text]
  • Entropy and Free Energy
    Spontaneous Change: Entropy and Free Energy 2nd and 3rd Laws of Thermodynamics Problem Set: Chapter 20 questions 29, 33, 39, 41, 43, 45, 49, 51, 60, 63, 68, 75 The second law of thermodynamics looks mathematically simple but it has so many subtle and complex implications that it makes most chemistry majors sweat a lot before (and after) they graduate. Fortunately its practical, down-to-earth applications are easy and crystal clear. We can build on those to get to very sophisticated conclusions about the behavior of material substances and objects in our lives. Frank L. Lambert Experimental Observations that Led to the Formulation of the 2nd Law 1) It is impossible by a cycle process to take heat from the hot system and convert it into work without at the same time transferring some heat to cold surroundings. In the other words, the efficiency of an engine cannot be 100%. (Lord Kelvin) 2) It is impossible to transfer heat from a cold system to a hot surroundings without converting a certain amount of work into additional heat of surroundings. In the other words, a refrigerator releases more heat to surroundings than it takes from the system. (Clausius) Note 1: Even though the need to describe an engine and a refrigerator resulted in formulating the 2nd Law of Thermodynamics, this law is universal (similarly to the 1st Law) and applicable to all processes. Note 2: To use the Laws of Thermodynamics we need to understand what the system and surroundings are. Universe = System + Surroundings universe Matter Surroundings (Huge) Work System Matter, Heat, Work Heat 1.
    [Show full text]
  • Physical Chemistry Laboratory I CHEM 445 Experiment 6 Vapor Pressure of a Pure Liquid (Revised, 01/10/03)
    1 Physical Chemistry Laboratory I CHEM 445 Experiment 6 Vapor Pressure of a Pure Liquid (Revised, 01/10/03) The vapor pressure of a pure liquid is an intensive property of a compound. That is, the vapor pressure of a liquid is independent of the amounts of the two phases as long as both phases are present. This phenomenon is shown in many physical chemistry textbooks in plots of PV isotherms at temperatures below the critical temperature. The horizontal part of the P vs. V isotherm corresponds to the constant vapor pressure of the liquid. The vapor pressure of every liquid is a strongly increasing function of temperature and a very slightly increasing function of applied pressure. The equation for the variation of vapor pressure of a liquid with temperature is derived in standard texts and often called the Clausius - Clapeyron equation. (Tinoco, Jr., Sauer, and Wang, Chap. 5; and Noggle, Chap.4). Reasonable approximations have been made to achieve this equation. 1 dP ∆HVap = (1) P dT RT 2 In Eq. (1), P is the vapor pressure of the liquid, T is the absolute temperature in K, and ∆HVap is the heat of vaporization of the liquid. The vapor pressure of a pure liquid is often represented by Eq. (2), the integrated form of Eq. (1) with the additional assumption that the heat of vaporization is independent of temperature. ∆H ln{P} = A − Vap (2) RT Although the heat of vaporization of a liquid is not independent of temperature, Eq. (2) will give an accurate fit to vapor pressure data over a wide range of temperatures, as long as the temperature is well below the critical temperature.
    [Show full text]
  • Thermodynamics of the Solution of Mercury Metal James Nelson Spencer Iowa State University
    Iowa State University Capstones, Theses and Retrospective Theses and Dissertations Dissertations 1967 Thermodynamics of the solution of mercury metal James Nelson Spencer Iowa State University Follow this and additional works at: https://lib.dr.iastate.edu/rtd Part of the Physical Chemistry Commons Recommended Citation Spencer, James Nelson, "Thermodynamics of the solution of mercury metal " (1967). Retrospective Theses and Dissertations. 3182. https://lib.dr.iastate.edu/rtd/3182 This Dissertation is brought to you for free and open access by the Iowa State University Capstones, Theses and Dissertations at Iowa State University Digital Repository. It has been accepted for inclusion in Retrospective Theses and Dissertations by an authorized administrator of Iowa State University Digital Repository. For more information, please contact [email protected]. This dissertation has been microfilmed exactly as received 67-8937 SPENCER, James Nelson, 1941- THERMGDYNAMICS OF THE SOLUTION OF MERCURY METAL. Iowa State University of Science and Technology, Ph.D., 1967 Chemistry, physical University Microfilms, Inc., Ann Arbor, Michigan THERPDDYi^AMCS OF THE SOLUTION OF MERCURY feT/" by James Nelson Spencer A Dissertation Submitted to the Graduate Faculty in Partial Fulfillment of The Requirements for the Degree of DOCTOR OF PHILOSOPHY Major Subject; Physical Chemistry Approved: Signature was redacted for privacy. In Charge of Major Work Signature was redacted for privacy. Head of hajor Department Signature was redacted for privacy. (|3aduate ^ollege
    [Show full text]
  • Boiling Point
    AccessScience from McGraw-Hill Education Page 1 of 2 www.accessscience.com Boiling point Contributed by: Ira N. Levine Publication year: 2014 The boiling point of a liquid is the temperature at which the liquid and vapor phases are in equilibrium with each other at a specified pressure. Therefore, the boiling point is the temperature at which the vapor pressure of the liquid is equal to the applied pressure on the liquid. The boiling point at a pressure of 1 atmosphere is called the normal boiling point. For a pure substance at a particular pressure P , the stable phase is the vapor phase at temperatures immediately above the boiling point and is the liquid phase at temperatures immediately below the boiling point. The liquid-vapor equilibrium line on the phase diagram of a pure substance gives the boiling point as a function of pressure. Alternatively, this line gives the vapor pressure of the liquid as a function of temperature. The vapor ◦ ◦ pressure of water is 1 atm (101.325 kilopascals) at 100 C (212 F), the normal boiling point of water. The vapor ◦ ◦ ◦ pressure of water is 3.2 kPa (0.031 atm) at 25 C (77 F), so the boiling point of water at 3.2 kPa is 25 C. The liquid-vapor equilibrium line on the phase diagram of a pure substance begins at the triple point (where solid, liquid, and vapor coexist in equilibrium) and ends at the critical point, where the densities of the liquid and vapor phases have become equal. For pressures below the triple-point pressure or above the critical-point pressure, the boiling point is meaningless.
    [Show full text]
  • P-Eq Lec 1.Pptx
    ∆G & phase changes: ! Example: H2O(l ) ⇌ H2O(g) Physical Equilibrium ! For P = 1 atm; T < 373 K (100 ˚C) ∆G > 0 ∆G & phase changes (we know water doesn’t boil below 100 ˚C) For P = 1 atm; T > 373 K ∆G < 0 For P = 1 atm; T = 373 K ∆G = 0 CHEM 102 At the boiling point, the liquid and vapor T. Hughbanks are in equilibrium. The boiling point is defined as the temp. at which the liquid and vapor phases are at equilibrium at 1 atm. ∆G : H2O(l ) ⇌ H2O(g) — more ∆G : H2O(l ) ⇌ H2O(g) — more From Appendix 2A: H2O(g) H2O(l ) We saw that at 298 K, ∆G > 0 ∆Hf˚ (kJ/mol) -241.8 -285.8 (∆H298/∆S298) = (44.0/0.119) K = 370 K this is close S˚ (J/mol K) 188.8 69.9 to 373 K, the boiling point! Why? For the vaporization process at 298 K: rearrange: ∆H298 = (370 K) ∆S298 ∆H298 = ∆Hf˚(gas) - ∆Hf˚(liq.) = ∆H - (370 K) ∆S = 0 –241.8 – (–285.8) = +44.0 kJ 298 298 recall: ∆H -(373 K)∆S = ∆G = 0 (why?) ∆S298 = S˚(gas) – S˚(liq.) 373 373 373 = 188.8 - 69.9 = 118.9 J/K = 0.1189 kJ/K if ∆H373 ≈ ∆H298 and ∆S373 ≈ ∆S298 ∆G298 = ∆H298 – (298 K)∆S298 = +8.5 kJ then ∆G373 = 0 ≈ ∆H298 - (373 K)∆S298 Generalization Ex: Phase transitions for NH3 ∆H and ∆S often don’t change much for a The molar enthalpy and entropy of melting and vaporization for ammonia are given process or reaction from ∆H298 and ∆S298 below.
    [Show full text]
  • Hypothetical Thermodynamic Properties: the Boiling and Critical 2 Temperatures of Polyethylene and Polytetrafluoroethylene
    BATCH: je3a28 USER: cmh69 DIV: @xyv04/data1/CLS_pj/GRP_je/JOB_i03/DIV_je030211u DATE: March 31, 2004 1 Hypothetical Thermodynamic Properties: The Boiling and Critical 2 Temperatures of Polyethylene and Polytetrafluoroethylene 3 James S. Chickos* 4 Department of Chemistry and Biochemistry, University of MissourisSt. Louis, St. Louis, Missouri 63121 5 6 The normal (p ) 101.325 kPa) boiling-temperature behavior of a homologous series, TB, as a function of 7 the number of repeat units, N, is fit by a hyperbolic function whose limiting value asymptotically 8 approaches TB(∞) ) (1217 ( 246) K for series related to polyethylene and approximately TB(∞) ) 915 K 9 for those related to polytetrafluoroethylene. Normal boiling temperatures have been fit to the function 10 TB ) TB(∞)[1-1/(aBN +bB)] where aB and bB are constants characteristic of the series investigated. Similar 11 behavior is observed for the critical temperature, TC. As the number of repeat units, N, approaches infinity, 12 TB and TC are found to converge. This provides an indirect method of determining TC(∞). Combined with 13 a few other experimentally determined critical temperatures, the TC of an entire series can be predicted. 14 Consistent with the convergence of TB and TC, the limiting critical pressure, PC(∞), is found to approach 15 1 atm as N approaches infinity. The vaporization enthalpy and entropy at the boiling temperature, g g 16 ∆l Hm(TB) and ∆l Sm(TB), used initially to suggest hyperbolic behavior of TB, show complex behavior with 17 increasing N. This behavior is discussed in terms of a simple mathematical model proposed previously. 18 19 Introduction temperature approaches a finite limiting value, TB(∞).
    [Show full text]
  • Thermodynamics Chapter 16 • You Are Responsible for Thermo Concepts from CHM Thermodynamics 151
    Thermodynamics Chapter 16 • You are responsible for Thermo concepts from CHM Thermodynamics 151. You may want to review Chapter 8, specifically GCC CHM152 sections 2, 5, 6, 7, 9, and 10 (except “work”). • Thermodynamics: Study of energy changes in chemical reactions and physical processes. • Thermodynamics allows us to determine IF a reaction will occur. • It also tells us the direction and extent of a reaction. • It does not tell us how fast a reaction occurs (kinetics tell us this!) o Thermochemistry Review Predicting sign for H rxn • 1st Law of Thermodynamics: Energy is Predict whether the following processes are endothermic (need heat) or exothermic (release conserved. Energy can be converted from one form heat): to another but it cannot be created or destroyed. Decomposition If a system gives off heat, the surroundings must Acid-Base Neutralization absorb it (and vice versa). Combustion Heat flow (enthalpy, H) is defined with reference to the system Melting • System absorbs heat, H > 0, endothermic Freezing • System gives off heat, H < 0, exothermic Boiling o o Calculating H rxn Heat of formation rxn & H rxn • The o symbol refers to the standard state, 1.00 atm • Write heat of formation reaction for H2O(l). pressure, 25.0oC, 1.00 M for solutions • Product = 1 mol H2O(l) o H f = Standard Molar Enthalpy of Formation • Reactants = elements in most stable form • The enthalpy change for forming 1 mole of a H2(g) + ½ O2(g) H2O(l) compound from its elements in their standard states. o o o Calculate H for the following reaction. Appendix
    [Show full text]
  • Lecture 14: November 4, 2019 1 / 21 More Permutations of Forbidden
    ...Thermodynamics Equation of state Phase Diagrams: stable phase by 2nd Law. System minimises (U,H,F or G) depending on boundary conditions Coexisting phases have same g Second Law prohibits negative Compressibility in equilibrium Second Law prohibits negative Heat Capacity in equilibrium Graeme Ackland Lecture 14: November 4, 2019 1 / 21 More permutations of Forbidden Considering H in place of U gives cp > 0. Using F gives positive isothermal bulk modulus. KT > 0. Note this works for conjugate variables: PV, TS dV 1 There is no similar argument for dT (thermal expansion). All these derivations that quantities must be positive trace back to violations of the Second Law. Negative modulus would allow heat to become work. Negative heat capacity would allow heat to flow from hot to cold. 1There is an assumption about homogeneous samples: for phase coexistence the bulk modulus is zero. Curious behaviour can be obtained for inhomogeneous metamaterials and self-gravitating clusters Graeme Ackland Lecture 14: November 4, 2019 2 / 21 UoE is committed to eliminating negative stereotyping Graeme Ackland Lecture 14: November 4, 2019 3 / 21 Stress-strain relations: Negative elastic moduli \What change of shape do you get for given stress?" Stress σij and strain ekl are 3x3 second rank tensors Elasticity tensor Cijkl (or 6 × 6 matrix) Cij is a material property. σij and eij depend on process. P 1 Elastic Energy = ijkl 2 Cijkl eij ekl σij and eij are conjugate variables, BUT energy is a sum. Second Law means that P and V are scalars, eigenvalues must be +ve =) K is scalar Cijkl may have negative elements.
    [Show full text]
  • Phase Equilibrium and Solutions
    Solutions manual for Burrows et.al. Chemistry3 Third edition 17 Phase equilibrium and solutions Answers to worked examples WE 17.1 Using a phase diagram (on p. 772 in Chemistry3) The normal melting and boiling points for methanol (CH3OH) are 175.7 K and 337.8 K, respectively. The critical temperature and pressure are 512.8 K and 81 bar. The triple point occurs at 175.5 K. Sketch (not to scale) the single- component phase diagram for methanol. Strategy Consider Figure 17.4, which shows a typical phase diagram for a one-component system. Remember that the normal melting and boiling points correspond to a pressure of 1 atm = 1.013 bar. Solution H i g h e r E d u c a t i o n © Oxford University Press, 2017. All rights reserved. Solutions manual for Burrows et.al. Chemistry3 Third edition WE 17.2 Effect of pressure on melting point (on p. 778 in Chemistry3) Calculate the pressure at which the melting temperature of sodium is 373 K. Strategy Apply the Clapeyron equation, Equation 17.4, using the data in the Worked Example to determine the change in the volume on melting and the enthalpy of fusion. Solution The Clapeyron equation, Equation 17.4, has the form d푝 Δ 퐻 = fus d푇 푇Δfus푉 From Worked Example 17.2, −6 3 −1 Δfus푉 = 푉m(liquid) − 푉m(solid) = +0.60 × 10 m mol and −1 3 −1 Δfus퐻 = +3.00 kJ mol = +3.00 × 10 J mol so that, at a temperature of 373 K, the gradient of the pressure–temperature line on the phase diagram is d푝 (+3.00 × 103J mol−1) = d푇 (373 K) × (+0.60 × 10−6m3mol−1) = 13.5 × 106 J m−3K−1 = 13.5 × 106 Pa K−1 Assuming that the solid–liquid equilibrium forms a straight line on the pressure– temperature phase diagram, then Δ푝 = 13.5 × 106 Pa K−1 Δ푇 The normal melting temperature for sodium, which is the temperature at which sodium melts for a pressure of 1 atm = 1.013 × 105 Pa, is 97.8C = 371 K.
    [Show full text]