04.03 The Macroscopic View of Entropy

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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.

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Molecular Nature of S: Micro to Macro of Thermal Entropy (uakron.edu, 20min) We can explain configurational entropy by studying particles in boxes, but only at constant temperature. How does the entropy change if we change the temperature? Why should it change if we change the temperature? The key is to recognize that energy is quantized, as best exemplified in the Einstein Solid model. We learned in Chapter 1 that energy increases when temperature increases. If we have a constant number of particles confined to lattice locations, then the only way for the energy to increase is if some of the molecules are in higher energy states. These "higher energy states" correspond to faster (higher frequency) vibrations that stretch the bonds (Hookean springs) to larger amplitudes. We can count the number of molecules in each energy state similar to the way we counted the number of molecules in boxes. Then we supplement the formula for configurational entropy changes to arrive at the following simple relation for all changes in entropy for ideal gases: ΔS = Cv ln(T2/T1) + R ln(V2/V1). Note that we have related the entropy to changes in state variables. This observation has two significant implications: (1) entropy must also be a state function (2) we can characterize the entropy by specifying any two variables. For example, substituting V = RT/P into the above equation leads to: ΔS = Cp ln(T2/T1) - R ln(P2/P1).

Comprehension Questions:
1. Show the steps required to derive ΔS = Cp ln(T2/T1) - R ln(P2/P1) from ΔS = Cv ln(T2/T1) + R ln(V2/V1).
2. We derived a memorable equation for adiabatic, reversible, ideal gases in Chapter 2. Hopefully, you have memorized it by now! Apply this formula to compute the change in entropy for adiabatic, reversible, ideal gases as they go through any change in temperature and pressure.

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You might better understand the macroscopic definition of entropy (uakron, 9min) if you consider isothermal reversible expansion of an ideal gas. Note the word "isothermal" is different from "adiabatic." If the expansion was an adiabatic and reversible expansion of an ideal gas, then we know from Chapter 2 that the temperature would go down, ie. T2/T1=(P2/P1)^(R/Cp)=(V1/V2)^(R/Cv). Therefore, holding the temperature constant must require the addition of heat. We can calculate the change in entropy for this isothermal process from the microscopic balance, then show that the amount of heat added is exactly equal to the change in entropy (of this reversible process) times the (isothermal) temperature. Studying the energy and entropy balance for the irreversible process helps us to appreciate how entropy is a state function. As suggested by the hint at the end of this video, you can turn this perspective around and infer the relation of entropy to volume by starting with the macroscopic definition and calculating exactly how much heat must be added after adiabatic, reversible expansion in order to recover the original (isothermal) temperature. Through this thought process, you should start to appreciate that the micro and macro definitions are really interchangeable expressions of the same quantity.

Comprehension Questions: (Hint: entropy is a state function.)

1. Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 450K and 4cm3/mol.

2. Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 4cm3/mol to 258.46K and 4cm3/mol.

3. Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 258.46K and 4cm3/mol.

4. Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 300K and 3 cm3/mol.

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Once we establish equations relating macroscopic properties to entropy changes, it becomes straightforward to compute entropy changes for all sorts of situations. To begin, we can compute entropy changes of ideal gases (learncheme, 3 min). Entropy change calculations may also take a more subtle form in evaluating reversibility (learncheme, 3min). 

Comprehension Questions: 

1. Nitrogen at 298K and 2 bars is adiabatically compressed to 375K and 5 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
2. Nitrogen at 350K and 2 bars is adiabatically compressed to 575K and 15 bars in a piston/cylinder. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
3. Steam at 450K and 2 bars is adiabatically compressed to 575K and 15 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
4. Steam at 450K and 2 bars is isothermally compressed to 8 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?

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