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**Entropy Contributions by:**

John L. Falconer & Will Medlin Department of Chemical and Biological Engineering University of Colorado Boulder, CO Supported by the National Science Foundation

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**You have 4 balls: red, green blue, and yellow**

You have 4 balls: red, green blue, and yellow. You distribute them all into two cans. How many different ways can you do this? 4 8 12 16 None of the above ANSWER: D. 16

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**A gas is compressed isothermally**

A gas is compressed isothermally. The entropy of the surroundings ________. increases decreases remains the same. Insufficient information. ANSWER: A increases. The total entropy of a system plus its surroundings greater than or equal to zero, so the entropy of the surroundings must increase as the entropy of the gas decreases.

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What external pressure should be used in the WEC term when calculating the entropy change for isothermal compression of an ideal gas? A constant external pressure An external pressure that is always equal to the system pressure Either (A) or (B) Neither (A) or (B) ANSWER: B An external pressure that is always equal to the system pressure. A reversible path, i.e. one that avoids gradients in pressure, must be chosen to evaluate entropy changes.

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Air flows steadily at constant pressure through a pipe that is heated by a furnace. As air flows through the pipe, the entropy of the air _______. Si So increases remains the same decreases Insufficient information ANSWER: A increases. Adding heat to the fluid at constant pressure will raise the temperature, and higher temperature corresponds to higher entropy.

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**The ideal gases are initially at 1 atm.**

Which of the following systems will have the largest entropy change when the blue partition is removed? The ideal gases are initially at 1 atm. N2 Vacuum O2 A C ANSWER: C In this case, the partial pressure of each gas decreases so that the total entropy change (based on contributions from each gas) is the highest. B D

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Which of the following systems will have the largest entropy change when the blue partition is removed? The gases are ideal. 2 atm N2 1 atm O2 3 atm Vacuum A C ANSWER: A The N2 and O2 system. In this case, the partial pressure of each gas decreases so that the total entropy change (based on contributions from each gas) is the highest. Note that the partial pressure of O2 will still go down in case A, even though the pressure on the opposite side of the partition is higher. B D

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**DS = 3Rln(2). What is most correct expression for DG?**

When the blue partition is removed between the two ideal gases below, the change is entropy is DS = 3Rln(2). What is most correct expression for DG? ∆H-3RTln(2) ∆H-3Rln(2) -3RTln(2) 3RTln(2) ∆H+3RTln(2) 1 atm 1 mol Ar 2 atm 2 mol O2 ANSWER: C. - 3RTln(2). ∆G=∆H-T∆S and ∆H = 0.

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**Which process has the larger entropy change for the gas?**

An ideal gas undergoes two irreversible processes between state 1 and state 2. For process A, 100 kJ of work was done by the gas. For process B, 200 kJ of work was done by the gas. Which process has the larger entropy change for the gas? 1 2 100 kJ Process A Process B Both process have the same change in entropy of the gas 1 2 200 kJ ANSWER: C The entropy change of the fluid is the same for both processes, because entropy is a state function and is always the same in state 2. However, the entropy change for surroundings will be less in process B because more work is done (more efficient process).

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A gallon of hot water is mixed with a gallon of cold water in a perfectly insulated container. When the waters are mixed, the entropy _______. increases decreases remains the same Insufficient information ANSWER: A increases. Lots of ways to justify this – for example, now you will have to do work to create the original system once they are mixed, whereas mixing will average the temperatures spontaneously.

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**1. 0 mol of hot water mixes adiabatically with 1. 0 mol of cold water**

1.0 mol of hot water mixes adiabatically with 1.0 mol of cold water. The entropy change of the cold water is 0.21 J/(mol•K). The entropy change of the hot water is _______. positive < J/(mol•K) < 0.21 J/(mol•K) > 0.21 J/(mol•K) > J/(mol•K) ANSWER: E. Because overall entropy change must be positive. ΔS = 0.21 J/(mol•K) ΔS = ?

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**Consider a completely insulated system**

Consider a completely insulated system. A reaction takes place inside the system. Therefore the system’s entropy __________. remains the same since Q = 0 increases decreases Need more information Insulation System ANSWER: D Depends on what the entropy change of reaction is. The fact that Q = 0 is not relevant because what you need to know is Q for the reversible process.

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An endothermic reaction takes place in a completely insulated, closed container of fixed volume. What happens to the entropy of system? A. Increases B. Decreases C. Does not change Insulation System ANSWER: C. Does not change. Temperature will decrease, but the system is adiabatic, so Q=0

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An exothermic reaction takes place in a completely insulated, closed container of fixed volume. What happens to the entropy of system? A. Increases B. Decreases C. Does not change Insulation System ANSWER: A. Increases. Since for an irreversible process, the total entropy must increase, and since the surroundings must change, then the entropy of the system must increase.

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**The entropy of the gas ___________.**

You raise the pressure of a gas at constant temperature in a closed, adiabatic system. The entropy of the gas ___________. does not change since temperature is constant does not change since there is no heat transfer increases decreases ANSWER: D. decreases. For P to increase at constant T, V must decrease. This decreases the configurational entropy.

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This picture shows a DNA molecule on a surface with vertical pillars on half the surface. The DNA will _______. move into the pillars more move out onto the open area stay where it is ANSWER: B move out into the open area because entropy of the system will increase where it is free to take on more configurations.

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**Container A has 2 kg of water, and container B has 1 kg of water**

Container A has 2 kg of water, and container B has 1 kg of water. Both containers are initially at 50oC. To each container, 20 kJ of heat is added. Which container has the greater total (NOT per kg water) entropy change? QA = 20 kJ QB = 20 kJ A B 2 kg water at 50oC 1 kg water at 50oC A B Same entropy change for both Need more information ANSWER: A. Integral of dQ/T is entropy change, and temperature will increase more for smaller container (B). Thus, the denominator will be larger for container B so entropy will increase less for container B. The temperature increases less for container A so the denominater is smaller for integraal dQ/T, and thus the entropy increases more for container A.

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In these piston-cylinder systems, when the red stop is removed, the ideal gas expands, and the piston moves until it hits the black stopper. Each system is adiabatic and starts at 10 atm and 25°C. Which has the largest entropy change? A B C 2 kg Vacuum Gas Piston Block 1 kg Vacuum Gas Vacuum Gas A B C All the same ANSWER: C The system with no weight on the piston has the greatest entropy change. The gas in this system does the most work on the and therefore has the greatest internal energy. The greater the internal energy, the greater the entropy.

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**For which does entropy increase?**

Constant P Constant V Constant P Both Neither ANSWER: B Constant pressure. At constant volume, the configurational entropy of each component is constant. At constant pressure, there is more space over which to arrange configurations. Constant V

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**Which system has the highest entropy if one mole of the same ideal gas is in each piston-cylinder?**

B 2 atm 50°C A B Both have the same entropy 10 atm 50°C ANSWER: A. The lower pressure, for the same temperature, has higher entropy because there is more space in which to arrange gas molecules.

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**A B WA = 0 kJ WB = 50 kJ QA=12 kJ QB=10 kJ**

Which system has the greatest entropy change if 10 mol N2 are in each piston-cylinder, which are initially at 100°C. Both processes are reversible. A B Same change for both A B WA = 0 kJ WB = 50 kJ QA=12 kJ ANSWER: A. Reversible, so integral dQ/T. The temperature will increase more for system B because of work so denominator higher. Also, less heat added. Visualize as such: if you add work adiabatically, reversibly, no entropy change so adding work not criteria for entropy change. QB=10 kJ

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**The entropy of the gas ___________.**

You lower the pressure of a gas at constant temperature in a closed, adiabatic system. The entropy of the gas ___________. does not change since temperature is constant does not change since there is no heat transfer increases decreases ANSWER: C increases. For P to decrease at constant T, V must increase. This increases the configurational entropy.

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**Which is the best approximation to a constant pressure process for a gas on a H-S diagram?**

ANSWER: D. H increases at constant P by temperature increasing; assuming ideal gas bevarior, H increases linearly with T. On the other hand, S increases logarithmically with T, so S will increase less rapidly than H as T increases.

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**Which is the correct plot of entropy of the system (S) vs**

Which is the correct plot of entropy of the system (S) vs. mole fraction of component i (yi) for a binary mixture of ideal gases? yi S yi S yi S A B C yi S yi S ANSWER: C. Entropy of a mixture is greater than the sum of the entropy of the pure components. D E

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**A gas goes from states A to B in a reversible adiabatic process**

A gas goes from states A to B in a reversible adiabatic process. It then goes from B back to A by a different pathway that is irreversible and not adiabatic. The entropy change for the gas for the irreversible pathway is _______ zero. V P A B greater than less than equal to Irrev. ANSWER: C equal to zero. Consider first the reversible process. The entropy of the universe doesn’t change because it’s reversible, and the entropy of the surroundings doesn’t change because it’s adiabatic. Therefore the entropy change of the system is zero too. Because entropy is a state function, any path from state A to state B will also results in an entropy change of zero, including the irreversible path. (However, the entropy change of the surroundings will be greater than zero for the irreversible process.) Rev.

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**Which line most likely corresponds to a constant entropy process?**

B A Pressure ANSWER: C. In order to maintain constant entropy as the pressure is increased, the fraction of the fluid that is in the more disordered, vapor phase must be increased. E is not correct because we are both lowering the pressure and increasing the temperature moving from left-to-right, so the entropy would not be constant since both of those processes increase the entropy of gases. D E Enthalpy

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One mol of CH4 at 1 bar and 50°C is mixed with 1 mol of O2 at 1 bar and 50°C. The final mixture is at 2 bar and 50°C. Assume ideal gases. The Gibbs free energy change is ___________. positive negative zero ANSWER: C zero. Both non-interacting gases are at the same temperature and partial pressure, and thus have the same state functions.

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**Which is the most correct plot of Gibbs free energy versus temperature for a single component?**

Gibbs (J/g) T-sat B Gibbs (J/g) T-sat C Gibbs (J/g) T(K) T-sat T(K) T(K) D Gibbs (J/g) T(K) T-sat E Gibbs (J/g) T(K) T-sat ANSWER: D.

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**Which is the most correct plot of Gibb’s free energy versus pressure for a single component?**

Gibbs (J/g) Gibbs (J/g) Gibbs (J/g) A B C P-sat P(atm) P-sat P-sat P(atm) P(atm) ANSWER: E. Gibbs (J/g) Gibbs (J/g) D E P-sat P(atm) P-sat P(atm)

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**Two of these plots represent enthalpy and entropy vs**

Two of these plots represent enthalpy and entropy vs. temperature for a pure component going from liquid to vapor. Identify the correct axes labels. T(K) W T(K) X T(K) Y Y = S ; Z = H W = S ; X = H W = H ; X = S Z = H ; X = S R = H ;Z = S ANSWER: B. Both entropy and enthalpy will increase in both phases with temperature. The rapid increase will be associated with the phase change. T(K) Z T(K) R

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For an ideal gas, if the pressure is increased while keeping the entropy constant, the enthalpy ________. increases decreases remains the same Need more information. ANSWER: A. To keep the entropy constant, the temperature must also increase; for an ideal gas, enthalpy is directly proportional to temperature. (Entropy increases as T increases and P decreases.)

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Two different paths can be used for a process that changes the state of a fluid from 300 K and 1 bar to 500 K and 9 bar. Path A is adiabatic and reversible, and Path B is non-adiabatic and irreversible. Which of the following statements is true? The change in entropy of the fluid will be greater for path A The change in entropy of the fluid will be greater for path B Both paths will have the same change in entropy of the fluid, but the change will be positive The compression in path A will require less work ANSWER: D The compression in path A will require less work. In both paths, the change in entropy of the fluid is the same because entropy is a state function. The entropy change of the fluid is zero because one of the paths is adiabatic (entropy change of surroundings is zero) and reversible (entropy change of universe is zero). However, the less efficient, irreversible process will require more work because of a greater entropy change of the surroundings (corresponding to waste heat).

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Hydrogen storage often involves the sorption of hydrogen from the gas phase into a solid material, where it can be stored more densely. For the adsorption/desorption process involved in H2 storage, which process should be run at a lower temperature? Adsorption of hydrogen into material. Desorption of hydrogen from material. It does not matter. ANSWER: A Adsorption of hydrogen into material. A lower temperature will facilitate adsorption because this is an ordered process and entropy is less at colder temperatures. Desorption would be facilitated at higher temperatures because greater entropy results in disorder

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