Fugacity, Activity, and Standard States



Similar documents
1.3 Saturation vapor pressure Vapor pressure

1 Exercise 4.1b pg 153

Chapter 7 : Simple Mixtures

Thermodynamics of Mixing

Ideal Gas and Real Gases

vap H = RT 1T 2 = kj mol kpa = 341 K

Gibbs Free Energy and Chemical Potential. NC State University

Phase Equilibrium: Fugacity and Equilibrium Calculations. Fugacity

Problem Set 3 Solutions

- The value of a state function is independent of the history of the system. - Temperature is an example of a state function.

a) Use the following equation from the lecture notes: = ( J K 1 mol 1) ( ) 10 L

EXERCISES. 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?

Thermodynamics worked examples

Review of Chemical Equilibrium Introduction

The Ideal Solution. ChemActivity T15

The Reduced van der Waals Equation of State

HEAT UNIT 1.1 KINETIC THEORY OF GASES Introduction Postulates of Kinetic Theory of Gases

Final Exam CHM 3410, Dr. Mebel, Fall 2005

48 Practice Problems for Ch Chem 1C - Joseph

Equilibrium. Ron Robertson

(1) The size of a gas particle is negligible as compared to the volume of the container in which the gas is placed.

Thermodynamics. Chapter 13 Phase Diagrams. NC State University

IDEAL AND NON-IDEAL GASES

First Law, Heat Capacity, Latent Heat and Enthalpy

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Chapter 13 - Chemical Equilibrium

Everest. Leaders in Vacuum Booster Technology

Boyles 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

The International Standard Atmosphere (ISA)

Boyle s law - For calculating changes in pressure or volume: P 1 V 1 = P 2 V 2. Charles law - For calculating temperature or volume changes: V 1 T 1

Problem Set 4 Solutions

= 800 kg/m 3 (note that old units cancel out) J 1000 g = 4184 J/kg o C

Lecture 1: Physical Equilibria The Temperature Dependence of Vapor Pressure

= atm. 760 mm Hg. = atm. d. 767 torr = 767 mm Hg. = 1.01 atm

Chapter 13. Properties of Solutions

Thermodynamics of Adsorption

Pressure Drop in Air Piping Systems Series of Technical White Papers from Ohio Medical Corporation

EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor

CHEM 36 General Chemistry EXAM #1 February 13, 2002

Chemistry 13: States of Matter

Gases. States of Matter. Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large Chaotic (random)

Sample Test 1 SAMPLE TEST 1. CHAPTER 12

OUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING

ORGANIC LABORATORY TECHNIQUES NEVER distill the distillation flask to dryness as there is a risk of explosion and fire.

THE BAROMETRIC FALLACY

) and mass of each particle is m. We make an extremely small

Compressible Fluids. Faith A. Morrison Associate Professor of Chemical Engineering Michigan Technological University November 4, 2004

Gas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.

STABILITY OF PNEUMATIC and HYDRAULIC VALVES

Chapter 18 Homework Answers

THE IDEAL GAS LAW AND KINETIC THEORY

Fugacity Models. I. Model Framework

PV (0.775 atm)( L) n = = = mol RT -1-1

11 Thermodynamics and Thermochemistry

Chapter 8 Maxwell relations and measurable properties

10.7 Kinetic Molecular Theory Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory

Determination of Molar Mass by Freezing-Point Depression

Organic Chemistry Calculations

4. Influence of Temperature and Pressure on Chemical Changes

FUNDAMENTALS OF ENGINEERING THERMODYNAMICS

Performing Calculatons

Experiment 12E LIQUID-VAPOR EQUILIBRIUM OF WATER 1

Experiment 1: Colligative Properties

k 2f, k 2r C 2 H 5 + H C 2 H 6

Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT).

Distillation vaporization sublimation. vapor pressure normal boiling point.

AS1 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

Physics 176 Topics to Review For the Final Exam

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10

Chem 338 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21

Lecture 3: Models of Solutions

momentum change per impact The average rate of change of momentum = Time interval between successive impacts 2m x 2l / x m x m x 2 / l P = l 2 P = l 3

F321 MOLES. Example If 1 atom has a mass of x g 1 mole of atoms will have a mass of x g x 6.02 x = 7.

Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.

Chapter 11 Properties of Solutions

CHAPTER 12. Gases and the Kinetic-Molecular Theory

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6

be 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

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

So T decreases. 1.- Does the temperature increase or decrease? For 1 mole of the vdw N2 gas:

CHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section The Gas Laws The Ideal Gas Law Gas Stoichiometry

Chapter 13 Properties of Solutions

Gases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6

n molarity = M = N.B.: n = litres (solution)

Chemical Equilibrium by Gibbs Energy Minimization on Spreadsheets*

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.

Study the following diagrams of the States of Matter. Label the names of the Changes of State between the different states.

Chapter 2 - Porosity PIA NMR BET

David A. Katz Department of Chemistry Pima Community College

atm = 760 torr = 760 mm Hg = kpa = psi. = atm. = atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

Calorimetry: Heat of Vaporization

10.7 Kinetic Molecular Theory Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory

MODEL OF THE PNEUMATIC DOUBLE ACTING CYLINDER COMPILED BY RHD RESISTANCES

5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C

Lecture 24 - Surface tension, viscous flow, thermodynamics

To calculate the value of the boiling point constant for water. To use colligative properties to determine the molecular weight of a substance.

Transcription:

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 to a standard ressure,, we may write G G = nrt ln,or G m = G m + RT ln. Now, consider a real gas that obeys the equation of state which would give ( + a)(v m b)=rt, G m = G m + RT ln + a + a b( ). More comlicated equations of state will give rise to more comlicated exressions for G. In order to reserve the simlicity of the exression for G even for real gases, the concet of fugacity, f, was introduced by G.N. Lewis so that G m = G m + RT ln f. The defining equation for fugacity is d = RTd ln f with the condition that Lim f @0 = 1. 1

A working exression to calculate fugacity: For a constant temerature rocess, dg m = V m d, or,g m = G,m G 1,m = 1 V m d Let us add and subtract RT/ to the right hand side:,g m = RT 1 P + V m RT d = RT ln P 1 + P1 V m RT d. Now, by definition, for all gases, G m = Rtln(f /f 1 ). Therefore, we get RT ln f / P = f 1 / 1 P1 V m RT d. We now set 1 = 0 (recall that f = as 0), f = f and =, where is the ressure of interest, to get RT ln f = 0 V m RT d. The right hand side can be evaluated if V m can be conveniently exressed as a function of R, T and. However, this is often very difficult (consider the van der Waals equation of state, for examle). An alternative is to exress V m in terms of the comressibility factor Z, to get RT ln f = ZRT 0 ln f = P Z 1 0 V m = ZRT/, RT d, or d. The last exression can be conventiently and fairly accurately evaluated in most cases using Virial exansions.

Fugacity of Pure Condensed Phases The defining equation for fugacity of ure condensed hases is, like that of gases, with the condition that d c = RTd ln f c Lim f c @0 = 1, where µ c is the chemical otential of the condensed hase. Alying the criterion for hase equilibrium, dµ c = dµ g (where g stands for gas hase), we get RTd ln f c = RTd ln f g, from which we conclude that f c = kf g. Now, in theory, any solid or liquid taken to zero ressure would eventually become a gas for which f =, so that k = 1. The conclusion is that fugacity of a condensed hase is the same as the fugacity of the gas hase in equilibrium with it. Variation of fugacity with ressure The above statement (italics) and the relationshi ln f = RT 1 T G m = V m RT, T is sufficient to derive the Gibbs equation ln f = V m f 1 RT 1, where the left hand side refers to vaor fugacities (which could be relaced with vaor ressures) while the right hand side refers to total ressures acting on the liquid surface. Therefore, this equation rovides a way to calculate the variation of vaor ressures with changes in the total ressure. 3

Activity The defining equation for fugacity of the i-th comonent in a mixture is d i = RTd ln f i. This equation can be integrated to give i ln f d = RT i ln fi d ln f i i = i + RT ln f i f i, where and i f i are the chemical otential and fugacity in a reference or standard state. The ratio f i /f i is defined as the activity a i, so that the above equation becomes = i + RT ln a i. Standard States The standard state for a gas The standard state for a gas is taken to be the ideal gas at 1 bar (100 kpa or 0.100 Ma) ressure. For such a gas, we have f = = 1 bar. For a real gas, in general, f 1 bar when = 1 bar but, with the above definition, we can define activity for a gas as a i = f i /, and set = 1 bar. Standard states for ure solids and liquids The ure solid or liquid under 1 bar external ressure is taken to be the standard state. With this definition, we have f = f (where f* is the fugacity at = 1 bar). Activity is still defined as a = f f = f f (1 bar), which means that for a ure substance at standard ressure, a = 1. Standard state of a solvent in a solution The convention is the same as above, with a i = f i /f i (1 bar). Raoult s law (to be covered later) states that for a volatile comonent of a solution, f i = x i f i. Therefore, we get a i = x i for solutions that obey Raoult s law. 4

In general, we may write a i =? i x i where γ i is called the activity coefficient. For solutions that obey Raoult s law, obviously, γ i = 1. Sometimes, the activity coefficient defined here is given the subscrit R to indicate that this refers to a Raoult s law standard state. Standard state of solute in a solution The standard state of a solute in a solution is defined as its activity in a solution whose concentration is 1 molal, i.e., 1 mole of solute in 1 kg of solvent. In general, a i =? H,i m with the condition that? H,i @ 1asm @ 0. 5