Chapter 6 - Classical Thermodynamics - Generalization to any Fluid

By Elliott, 5 March, 2013

06.1 The Fundamental Property Relation

Energy balance to state functions

From the physical world to the realm of mathematics (uakron.edu, 15min) In Unit I, students develop the skills to infer simplified energy and entropy balances for various physical situations. In order to facilitate that approach for applications involving chemicals other than steam and ideal gases, we need to transform that approach into a realm of pure mathematics. In this context it suffices to apply the energy and entropy balance of a very simple system (piston/cylinder) then focus on the state functions that are involved (U,H,S,...). The mathematical realm is relatively abstract, but it is ideally suited for the generalizations required to extend our principles from steam and ideal gases to any chemical.

Comprehension Questions:

1. In example 4.16, we noted that the estimated work to compress steam was less when treated with the steam tables than when treated as an ideal gas. Explain why while referring to the molecular perspective.

2. In Chapter 5, we noted that the temperature drops when dropping the pressure across a valve when treating steam or a refrigerant with thermodynamic tables, but the energy balance suggests that the temperature drop for an ideal gas should be zero. Explain how these two apparently contradictory observations can both be true while referring to the molecular perspective.

3. What is the relation of the state variable dU to the state variables S and V according to the fundamental property relation?

4. What is the relation of the state variable dH to the state variables S and P according to the fundamental property relation?

5. What is the significance of writing changes of state variables in terms of changes in other state variables?

6. Why is the compressibility factor (Z=PV/RT) less than one sometimes?

7. Is it possible for Z to be greater than one? Explain.

8. What is the significance of having a relation for P = P(V,T)? How will that help us to solve problems involving chemicals other than steam and ideal gases?

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Recognizing Incorrect Arrangements of State Variables

State Functions Equation Check (LearnChemE.com, 3min) determines which form of various state functions is incorrect.

Comprehension Questions:

1. Which of the following relations is not valid (may be more than one)?
dA = dH - d(PV) - d(TS); dA = dG - d(PV); dG = -SdT + VdP; dA = dH - d(PV) - TdS

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06.2 Derivative Relations

Derivative Toolbox

Exact Differentials and Partial Derivatives (LearnChemE.com, 5min) This math review puts into context the discussion of exact differentials in Section 6.2 of the textbook using an example related to the volume of a cylinder.

Comprehension Questions:

1. Given that dU = TdS - PdV, what derivative relation comes from setting ∂²U/(∂S∂P) = ∂²U/(∂P∂S)?

2. Given that dA = -SdT - PdV, what derivative relation comes from setting ∂²A/(∂T∂V) = ∂²A/(∂V∂T)?

3. Given that dG = -SdT + VdP, what derivative relation comes from setting ∂²G/(∂T∂P) = ∂²G/(∂P∂T)?

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Derivative Toolbox

Assembling your derivative toolbox including the triple product rule, (uakron.edu, 13min) Beginning with the fundamental property relation, substitutions lead to Eqns. 6.4-6.7. Differentiating these and equating through exact differentials leads to Eqns. 6.29-6.32 (aka. Maxwell's Relations). Combining Maxwell's Relations with Eqns. 6.4-6.7 leads to Eqns. 6.37-6.41. With these tools in hand, and Eqn. 6.15 (aka. Triple Product Rule), you have all the tools you need to quickly transform any derivative into "expressions involving Cp, Cv, P, V, T, and their derivatives." This capability is fundamental to obtaining expressions for U, H, and S from any given equation of state for any chemical of interest. Four sample derivations are illustrated: (∂U/∂V)_T, (∂T/∂S)V, (∂T/∂V)_S, (∂S/∂V)A,

Comprehension Questions:

1. Transform the following into "expressions involving Cp, Cv, P, V, T, and their derivatives:" (∂T/∂V)_S.

2. Transform the following into "expressions involving Cp, Cv, P, V, T, and their derivatives." Your expression may involve absolute values of S as long as they are not associated with any derivative. (∂T/∂U)_P.

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Pressure Dependence of Heat Capacity: (dCp/dP)T

Heat Capacity Pressure Dependence (LearnChemE.com, 7min) This example derives how the heat capacity of the gas depends on pressure, ie. (∂Cp/∂P)_T.

Comprehension Questions:

1. Derive the relation for (∂Cv/∂V)_T

2. The van der Waals (vdW) equation of state (EOS) is: P = RT/(V-b) - a/V². Evaluate (∂Cp/∂P)_T for the vdW EOS.

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Heat Capacity Volume Dependence:(dCv/dV)T

Heat Capacity Volume Dependence (uakron.edu, 10min) This example derives how the heat capacity of the gas depends on volume, ie. (∂Cv/∂V)_T. It may seem paradoxical that a quantity defined at constant volume can change with respect to volume. The discussion here shows how to solve this puzzle. The sample derivation presented here follows an alternative approach to what is illustrated in Example 6.9 of the textbook.

Comprehension Questions:

1. The van der Waals (vdW) equation of state (EOS) is: P = RT/(V-b) - a/V². Evaluate (∂Cv/∂V)_T for the vdW EOS.
2. The Soave-Redlich-Kwong (SRK) EOS is: P = RT/(V-b) - a/[V(V+b)]
where a=[1+K*(1-sqrt(T/Tc))]². Evaluate (∂Cv/∂V)_T for the SRK EOS.
3. Comment on the differences between the results for 1 and 2 above. Do these results change the way you look at the vdW EOS?

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