Dept. of Materials Science & Engineering. Problem Set 2 Solutions

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1 Problem Set 2 Solutions 1. Using dv and ds as a function of T and P that we derived in class, fully derive the rest of toolbox 3 (i.e., du, dh, df, and dg). See DeHoff pages 73 74, equations

2 2

3 2. Derive the relationship that describes the dependence of Helmholtz free energy upon entropy and temperature. Substituting M and N into df, we have: df MdS NdT P P ST ds Cv dt T P P 1 ST Cv dt ds T P 3

4 3. Consider the situation in which you were asked to compare the numerical values of the coefficient: F S V for a monatomic ideal gas, for a solid, or a liquid. To do such a comparison, derive the state function in differential form and the coefficient as a function of materials parameters that you would use to make the comparison. Then, discuss how the coefficient would compare between the following: a monatomic ideal gas, for a solid, or a liquid. Hint: to start this problem, think about what variables are sought (dependent variable) and given (independent variables). For the last part of the problem, think in terms of gas phases versus condensed phases. 4

5 Matrix algebra methods in Mathematica can be used to solve these linear equations. In Mathematica, we can use LinearSolve[ ] to do so. See the extra credit assignment on matrix algebra posted on the course website. cv Tv2 M1,v,, ; T M2 s pv, p ; MNAnswer LinearSolveM1, M2; Print"M,N ", MNAnswer FullSimplify M,N st st, p cv cv We can writing it in matrix algebra form. We can also use Cramer s rule which is essentially the same approach, but is only good for two equations and two unknowns. 5

6 Here, we discuss how the coefficient would compare between the following: a monatomic ideal gas, for a solid, or a liquid. Hence, reasonable assumptions are sought for comparing or implementing the coefficient for a monatomic ideal gas, 6

7 for a solid, or a liquid. The latter two are considered condensed phases and may, in many cases, be treated the same. The fact that the change in Helmholtz free energy with respect to entropy at constant volume is negative suggests that a reaction would spontaneously occur. The entropy of a gas is, in general, larger than the entropy for a condensed phase (liquid or solid). The capacity to hold heat is, in general, greater for a condensed phase than it is for a gas phase. This is observed when comparing Cv values for condensed versus liquid phases. Hence, the coefficient for a gas would be more negative than for condensed matter and thus more apt to react. Mathematically, this can be described in the following way: F F F F Sgas Scondensed & Cp, gas Cp, condensed thus S S S S v, gas v, condensed v, gas v, condensed 7

8 4. Give an example of how to use 4.30 on page 68 of DeHoff to manipulate the following partial derivative: P T V to develop a thermodynamic relationship with materials parameters. 8

9 5. Consider yourself the instructor of a materials thermodynamics course and you need to create a homework problem that follows the general topics covered in determining state functions. Using tables in DeHoff to obtain data need for the problem (or you can use other sources of data make sure to cite your source), pose a question (i.e., problem) in which you need to derive a state function (in differential form) for a solid or liquid using the methodology we covered in class in which we derived a state function with respect to any two variables (DeHoff Ch , p. 75) using toolbox 3. The state function must not just be a function of T or P (i.e., only one of them or none). Then, use that state function to solve the problem you created (i.e., integrate). Chose the simplest path to integrate. Plot your state function as a function of each independent variable while holding the other variable constant. You should end up with two plots. Use a mathematical program (not Excel) to generate your plot. Comment on interesting aspects of the results of your plot. Hint: An example to help you is the combination of DeHoff s examples 4.6 and 4.9 combined except example 4.9 gives one numerical answer rather than two plots and it considers an ideal gas rather than a solid or liquid. 9

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