Physical Chemistry I

Concordia University - Department of Chemistry & Biochemistry PHYSICAL CHEMISTRY I : Thermodynamics (CHEM 234/2 Section 51) Fall 2014 Instructor: Dr. Guy Paquette Office: SP-201.14. E-mail: guy.paquette@concordia.ca Office hours: Thursday: 10h00-11:30 and 16h00-17h00; please make an appointment GENERAL INFORMATION This 3-credit course intro

Course_Outline_-_Fall_2014.docx-Concordia University - Department of Chemistry

About the book There are numerous features in this edition that are designed to make learning physical chemistry more effective and more enjoyable. One of the problems that make the subject daunting is the sheer amount of information: we have introduced several devices for organizing the material: see Organizing the information . We appreciate that mathematics is often troublesome, and therefore have taken care to give help with this enormously important aspect of physical chemistry: see Mathematics support . Problem solving—especially, ‘where do I start?’—is often a challenge, and we have done our best to help overcome this ﬁrst hurdle: see Problem solving . Finally, the web is an extraordinary resource, but it is necessary to know where to start, or where to go for a particular piece of information; we have tried to indicate the right direction: see About the Book Companion Site . The following paragraphs explain the features in more detail. Organizing the information Key points The Key points act as a summary of the main take-home message(s) of the section that follows. They alert you to the principal ideas being introduced. 1.1 The states of gases Key points Each substance is described by an equation of state. (a) Pressure, force divided by area, provides a criterion of mechanical equilibrium for systems free to change their volume. (b) Pressure is measured with a barometer. (c) Through the Zeroth Law of thermodynamics, temperature provides a criterion of thermal equilibrium. The physical state of a sample of a substance, its physical condition, is deﬁned by its physical properties. Two samples of a substance that have the same physical proper- h h f f l ﬁdb mental fact that each substance is described by an equation of state , an equation that interrelates these four variables. The general form of an equation of state is p = f ( T , V , n ) (1.1) General form of an equation of state These relations are called the Margules equations . Justiﬁcation 5.5 The Margules equations The Gibbs energy of mixing to form a nonideal solution is Δ mix G = nRT { x A ln a A + x B ln a B } This relation follows from the derivation of eqn 5.16 with activities in place of mole fractions. If each activity is replaced by γ x , this expression becomes Δ mix G = nRT { x A ln x A + x B ln x B + x A ln γ A + x B ln γ B } Now we introduce the two expressions in eqn 5.64, and use x A + x B = 1, which gives Δ mix G = nRT { x A ln x A + x B ln x B + ξ x A x 2 B + ξ x B x 2 A } = nRT { x A ln x A + x B ln x B + ξ x A x B ( x A + x B )} = nRT { x A ln x A + x B ln x B + ξ x A x B } as required by eqn 5.29. Note, moreover, that the activity coefﬁcients behave cor- rectly for dilute solutions: γ A → 1 as x B → 0 and γ B → 1 as x A → 0. At this point we can use the Margules equations to write the activity of A as 2 2 Equation and concept tags The most signiﬁcant equations and concepts—which we urge you to make a particular effort to remember—are ﬂagged with an annotation, as shown here. Justiﬁcations On ﬁrst reading it might be sufﬁcient simply to appreciate the ‘bottom line’ rather than work through detailed develop- ment of a mathematical expression. However, mathematical development is an intrinsic part of physical chemistry, and to achieve full understanding it is important to see how a par- ticular expression is obtained. The Justiﬁcations let you adjust the level of detail that you require to your current needs, and make it easier to review material.

CHEM | 234 | Physical Chemistry I | lecture notes | About the book There are numer

• Exercises : – 1.1 a & b ; 1.4 a & b ; 1.8 a & b ;1.10 b ;1.13 a ; 1.15 a ; 1.16 b ; 1.18 a ; 1.21 a • Problems : – 1.2 ; 1.6 ; 1.11 Exercises & problems – Chap. 1 9 th Edion of the textbook

Exercise_Problems_-_Chapter_1_Atkins_9th_Ed.pptx-Exercises : – 1.1 a &

CHEM 234 Fall 2014 Suggested exercises & problems, Chapter 3 Physical Chemistry (Atkins & de Paula) 10th Edition 9th Edition Exer. 3A.3(a) Similar to 3.1(b) Exer. 3A.7(b) 3.2(b) Exer. 3A.10(a) 3.13(a) Exer. 3B.2(a) 3.8(a) Exer. 3C.2(a) Similar to first part of problem 3.12

Exercise_Problems_-_Chapter_3.docx-CHEM 234 Fall 2014 Suggested exercises

Molecular interpretation of U • For molecules, rotations and vibrations are also possible (CO 2 and H 2 O for example) • Each contribution = ½ kT • U m (T) = U m (0) + 5/2 R • Linear molecule; translation & rotation only • U m (T) = U m (0) + Molecular interpretation • From the previous slide : • It takes twice as much energy to undergo the same rise the temperature of a nonl

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