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Vol. 3 , No. 1
The Chemical Educator © 1998 Springer-Verlag New York, Inc. |
ISSN 1430-4171
S 1430-4171(98)01184-3
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Chemical Thermodynamics: Revision and Worked Examples by H. P. Stadler. The Royal Society of Chemistry, London, 1990. Softcover, ISBN 0 85186 273 X, xiv + 132 pages. Price 11.95 pounds sterling.
This book offers the chemical engineering and chemistry student a review of the basic concepts of thermodynamics, presented primarily in the form of examples. The examples have detailed solutions rather than just answers. This approach can be very useful to a student who is currently studying the topic, or who has already studied it, and would find detailed examples that have been worked through to be a useful way to put theory into practice. One important feature of the solutions is that they frequently remind the reader of the limitations of particular models, something that is becoming especially important with the extensive use of computers. The topics covered include the first and second laws of thermodynamics (the third law is covered very briefly); equilibria, including phase, ideal gas, and chemical equilibria; partial molar quantities and open systems; and nonideal systems including real gases and nonideal mixtures.
Each chapter begins with a brief review of the appropriate relationships and equations, followed by a series of increasingly complex worked examples. The examples typically include both derivations and numerical exercises, and frequently both. As an example, in the nonideal gas chapter, one of the exercises starts with the composition of an equilibrium mixture of N2, H2, and NH3 at a specified temperature and total pressure. The first part is relatively simple—just the calculation of Kp and its comparison with the observed standard free energy of formation of NH3 . The next part of the exercise gives the student an approximate form for the equation-of-state in the form PV = RT + BP + CP2 + DP3 and asks the student to show that this leads to an expression for the fugacity f of the form log(f/P) = aP + bP2 + cP3 . The solution reminds the student of the appropriate choice of equation and of the conversion from ln to log. Finally, values for a, b, and c for all three gases are given at the appropriate temperature and the student asked to verify the value of Kp. Thus three levels of review are given: the basic calculation of the equilibrium constant that would be expected of a beginning chemistry student, the derivation of a correction for nonideality from an approximate equation of state, and the verification that this approximation produces an correct estimate of this correction.
The one weak spot in an otherwise impressive text is that in the solutions to numerical examples, only the number portion of the physical quantities is shown with the "appropriate" units added to the final answer. For example, in this same chapter the student is asked to calculate the Joule–Thomson inversion temperature Tinv = 2a/Rb where a and b are the van der Waals constants for the gas. Values of a = 4 x 106 cm6 atm mol-2 and b = 36 cm3 mol-1 are given. In the solution part of the chapter however, the solution is written in the form
Tinv = 2a/Rb = 2 x 4 x 106/(36 x 82.06) = 2708 K
There are several things wrong with this solution. First of all, there are no units to check for correct cancellation; for example, the student who used R = 8.314 J mol-1 K-1 rather than R = 82.06 cm3 atm mol-1 K-1 would have no chance to check for the incorrect units, which would lead to an error of a factor of 10 in the answer. Second, although this may be considered a quibble, it is generally recommended that numbers be written in the solution in the exact order that the corresponding variables appear in the original equation, since this makes it easier to check the work. Third, there are problems with significant figures, four are given but only one or two are appropriate. A correct presentation would be as follows, where the units are shown and the correct significant figures given in the answer.
Tinv = 2a/Rb = 2 x 4 x 106 cm6 atm mol-2/(82.06 cm3 atm mol-1 K-1 x 36 cm3 mol-1 ) = 2700 K
Actually, to the correct number of significant figures, one can only say that Tinv = 3 x 103 K, since the value of the van der Waals a constant is given only to one significant digit. In a text aimed at chemical engineering students as well as chemistry students, these lapses from proper technique are particularly significant since engineers make use of a wide range of units and must be able to trace back their answer to the original data. Only following correct units arithmetic will permit this.
My overall impression of the book is highly favorable, although I would probably require my students to follow correct units arithmetic in any solutions, and would have to remind them that the solutions given in the book are lacking units.