10.9 - 10.12 Mixture Properties Overview (6:53) (msu.edu)

Why does Raoult's law work sometimes? Why does it fail sometimes? How can we hope to understand why it fails?

This section of the text is thick with lots of equations. It may help to filter out the most important equations and results so that you have the perspective of the overall objectives of this section. There are a lot of equations in this section to show that Raoult's law is a equlity of an ideal gas component fugacity with an ideal solution liquid fugacity! By understanding the assumptions used in the development of the equation, we can begin to understand the limitations of Raoult's law. This screencast goes on to preview the methods developed in the next sections of the textbook to deal with deviations in fugacities from ideal solutions and the ideal gas law.

Stream enthalpies for the DME process (uakron, 7min) can be estimated using the "heat of reaction" pathway (Fig 3.5a) or the "heat of formation" pathway (Fig 3.5b). This presentation is based on Fig 3.5b, which is very similar to Fig 2.6c. The main difference is the inclusion of the heat of formation for each compound relative to its elements. Including the heat of formation puts the reference state for each compound on the same basis of comparison (ie. the elements). If one stream (e.g. "products") possesses more enthalpy than another stream (e.g. "reactants") then the energy difference between the streams (e.g. "heat of reaction") would be accounted for by simply subtracting the two stream enthalpies. Reactions inherently involve multiple components, so including the heats of formation in the stream enthalpies, as well as the other enthalpic contributions represented in Fig 2.6c, is inevitable. These sample calculations are illustrated for all the streams appearing in the DME process. The presentation follows up on the discussion of Fig 2.6c for pure fluids. Once you understand the calculations for each pure fluid, the mixture property simply involves taking the molar average, so: H ≈ ∑(x_i*Hf_i+Cp_i^igΔT+(q_i-1)*H_i^vap). In this equation, (q_i-1)*H_i^vap accounts crudely for departures from ideal gas behavior. For example, if a stream is a vapor, then q=1 and H^vap doesn't matter. If q=0, then the stream is a liquid and H^vap must be subtracted. We will study more accurate models of ideal gas departures in Unit II.

Comprehension Questions:

1. Compute the enthalpy, H(J/mol), of methanol at 250C and 2 bars relative to its ideal gas standard state elements.

2. Compute the enthalpy, H(J/mol), of DME at 250C and 2 bars relative to its ideal gas standard state elements.

3. Compute the enthalpy, H(J/mol), of water at 250C and 2 bars relative to its ideal gas standard state elements.

4. Compute the enthalpy, H(J/mol), of a stream that is 50% methanol, 25% DME, and 25% water at 250C and 2 bars relative to its ideal gas standard state elements.

Shortcut estimation of thermodynamic properties (sample calculation) can be very quick and sometimes reasonably accurate.(6min, uakron.edu) As a follow-up exercise, it is suggested to adapt the shortcut vapor pressure equation in combination with Eqn. 2.45 and the pathway of Fig. 2.6c to rapidly estimate stream properties. Briefly, all you need is an "IF" statement that checks whether the T is less than T^sat at the given P. If so, then H=H^ref+CpΔT+H_vap. If not, then H=H^ref+CpΔT. This can be a quick and convenient method to estimate stream attributes of a process flow diagram. One equation per cell and you're done. This sample calculation illustrates the process for the heat duty of a butane vaporizer and compares the PREOS to the methods of Chapter 2 (ie. Eq. 2.45 etc.)

Comprehension Questions: Suppose you want to tabulate the entropy (S) of your stream attributes by this approach.
1. How would you compute the S^ig(T,P)-S^ig(T^ref,P^ref) contribution?
2. How would you compute ΔS_vap?
3. Compute "S" for propane at 355K and 3MPa relative to the liquid at 230K and 0.1MPa by this approach.
4. Compute "H" for propane at 355K and 3MPa relative to the liquid at 230K and 0.1MPa by this approach.
5. Compute H and S for the same conditions/reference using the PREOS.
6. Explain the discrepancies between the two approaches. e.g. compare the H_vap values and the (H^V-H^ig) values, where H^V represents the enthalpy of the vapor phase, not the heat of vaporization (H_vap).

Heat and entropy in a glass of water (uakron, 9min) Taking a glass from the refrigerator causes heat to flow from the room to the water. The temperature of the water slowly rises while the temperature of the (relatively large) room remains fairly constant. Applying the macroscopic definition of entropy makes it easy to compute the entropy changes, but is one larger than the other? Are all entropy changes greater than zero? What does the second law mean exactly?

Comprehension Questions:

1. Describe your own example of a process with an entropy decrease and explain why it doesn't violate the second law.