The Laws of Thermodynamics - Croom Physics€¦ · The Laws of Thermodynamics CLICKER QUESTIONS Question J3.01 ... 558 Chapter 12 Question J3.02 ... Question J3.03

12The Laws of Thermodynamics

CLICKER QUESTIONS

Question J3.01

Description: Relating thermodynamic processes to PV curves: isobar.

Question

A quantity of ideal gas undergoes a thermodynamic process. Which curve represents an isobaric path?

Commentary

Purpose: To develop your ability to interrelate the basic thermodynamic quantities of pressure, tempera-ture, and volume.

Discussion: Isobaric means “constant pressure.” None of the depicted graphs with pressure on one axis show a constant pressure. However, graph (4) displays a linear relationship between volume and tempera-ture; according to the ideal gas law, if volume and temperature are proportional to each other, the pressure must be constant.

Key Points:

• Isobaric means “constant pressure.”

• Be able to relate the behavior of pressure, volume, and temperature via the ideal gas law.

• If pressure is constant for a process, volume, and temperature will be directly proportional to each other.

For Instructors Only

This question provokes students to reason qualitatively with graphical representations and fundamental relationships, which is valuable for developing conceptual understanding and analysis skills.

1)

T

P

2)

P

V

3)

V

T

5)

P

V

4)

V

T

1)

T

P

2)

P

V

3)

V

T

5)

P

V

4)

V

T

557

56157_12_ch12_p557-624.indd 55756157_12_ch12_p557-624.indd 557 1/4/08 1:44:43 PM1/4/08 1:44:43 PM

558 Chapter 12

Question J3.02

Description: Interpreting PV process diagrams.

Question

A vertical cylinder with a movable cap is cooled. The process corresponding to this is:

M

20

10

P (Pa)

V (m3)

1 3

C

B

A

1. CB 2. AB 3. AC 4. CA 5. Not shown

Commentary

Purpose: To link the graphical representation of an isobaric process to the description of such a process.

Discussion: Forces on the moveable cap must always balance, which means that the internal pressure of the gas must always be enough larger than atmospheric pressure to counteract the weight of the cap. Thus, the process described is isobaric (constant pressure).

A cooled gas contracts, so the volume must be decreasing. Only one segment of the PV graph, C to A, depicts decreasing volume at constant pressure; therefore, the correct answer must be (4).


The Laws of Thermodynamics 559

Key Points:

• Interpreting process diagrams and relating them to physical situations is a vital skill to develop.

• A cylinder with a freely moving cap, supported only by the pressure of the contained gas, is a mechanism for insuring constant pressure.


Some follow-up discussion questions you may wish to use: Is work done during this process? (By or on the gas?) How does the temperature at A compare to that at C ? How much heat was extracted during the process?

Question J3.03

Description: Linking graphical representations of thermodynamic processes to quantities: work.

Question

An ideal gas is taken around the process shown. The net work done on the gas is most nearly:

20

10

P (Pa)

V (m3)

1 3

1. 20 J 2. −30 J 3. 15 J 4. −10 J 5. None of the above 6. Cannot be determined

Commentary

Purpose: To hone the concept of “work” for a thermodynamic cycle, and connect it to a graphical represen-tation of a cycle.

Discussion: On a pressure vs. volume graph, the area under a curved segment is the work done by the system during the process represented by that segment: the integral of pressure over a change in volume. (For a seg-ment moving right to left, the work done by the system is the negative of that area.) For a cycle—a process that ends in the same state it began—the net work done by the system must be equal to the area inside the closed PV path if the cycle is traversed clockwise.

The work done on the gas is the negative of the work done by the gas. So, for this situation, the work done on the gas is the negative of the area of the triangle: answer (4).


560 Chapter 12

Key Points:

• The area under a pressure vs. volume curve is the work done by the system.

• The area inside a closed path on a PV diagram is the work done by the system during a cycle.

• The work on a system is the negative of the work by a system.

• Pay attention to signs: work on vs. work by, path traveled left-to-right vs. right-to-left, cycle traveled clockwise vs. counterclockwise, etc.


Students selecting a positive answer – (1) or (3) – should be sensitized to the difference between work on and work by a gas.

Question J4.01a

Description: Relating internal energy to and reasoning with a PV cycle.

Question

Two moles of a perfect gas are taken along the cycle shown below.

Volume, V

Pre

ssur

e, P

A

D C

BP1

V10

Which of the following is true about the internal energy U?

1. U0 = U

A

2. UD < U

A

3. UB > U

C

4. UD = U

B

5. UC = U

A

6. None of these is true.



Commentary

Purpose: To connect the idea of internal energy to a PV cycle and reason with the information contained in a PV diagram.

Discussion: The internal energy of a perfect gas depends only on its temperature and the number of moles. The number of moles is not changing, so, by applying the perfect gas law (PV = nRT), we can fi nd out about how the temperature is changing.

The product PV is smallest in state A, so T is also smallest in state A. In states B and D, the product PV is twice as large as it is in state A, so the temperature in states B and D are the same and twice as large as the temperature in state A. In state C, PV is four times as large as it is in state A, so the temperature is four times as large.

Mathematically, since n = 2 moles:

TA = P

1V

1�2R

TB = P

1(2V

1)�2R = P

1V

1�R = 2T

A

TC = (2P

1)(2V

1)�2R = 2P

1V

1�R = 4T

A

TD = (2P

1)V

1�2R = P

1V

1�R = 2T

A

Looking at the given expressions involving U, we see that #4 is true. Since the temperature of state B is equal to the temperature of state D, the internal energy must be the same too.

Statement (1) is not necessarily true, because we do not need to begin the cycle at state A. (U0 indicates the

initial internal energy.)


Many students will think that statement (1) is true and, not needing to analyze the situation any further, will stop. This distracter answer is a trap for students who assume that all cycles begin with state A. We often begin a cycle at state A, but this is not a requirement; is not even a common convention. (If students ask what the notation “U

0” means, you should tell them it means the “initial internal energy.”)

Students need to make two connections to successfully analyze this situation. First, they must recognize that internal energy depends on temperature. Second, they must recognize that they can compare tempera-tures by looking at the specifi c values of P and V, even though they can’t calculate numerical values.

We intend students to use the graph to scale. From A to B, the volume doubles at constant pressure. From B to C, the pressure doubles at constant volume, etc.


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Question J4.01b

Description: Interrelating energy and ideal gas ideas for a PV cycle.

Question

An ideal monatomic gas is taken around the cycle ABCDA as shown below.

Volume, V

Pre

ssur

e, P

A

D C

BP1

V10

What is the amount of energy removed by heat during one cycle?

1. 0 2. P

1V

1

3. 2P1V

1

4. 3P1V

1

5. 4P1V

1

6. 5P1V

1

7. Impossible to determine

Commentary

Purpose: To interrelate energy and ideal gas ideas within the context of a PV cycle.

Discussion: The internal energy of a perfect gas is a “state function,” which means that any time the gas is in a particular state (a point on the PV diagram), it must have the same internal energy. If the gas goes around a cycle and returns to a state, it must return to the same internal energy. This means the change in internal energy ∆U is equal to zero for this cycle.

Since the internal energy is not changing, the work done on the gas must be equal to the amount of energy removed from the gas by heat, according to the fi rst law of thermodynamics.

When a gas expands, it does work on its environment, and when it is compressed, the environment does work on it. At constant pressure, the work done by a gas is P V∆ . From A to B, the gas is expanding, so the work done by the gas is P

1V

1, and the work done on the gas is –P

1V

1. (Since the graph is drawn to scale,

VB = 2V

A = 2V

1.) From B to C, the volume is constant, so no work is done on the gas, even though the tem-

perature is increasing. From C to D, work is done on the gas to compress it at constant pressure. The work done on the gas is 2P

1V

1. From D to A, the work done is zero.



The total amount of work done on the gas is P1V

1. The amount of energy removed by heat from the gas to

the environment is therefore also P1V

1: answer (2).

Note that the area of the enclosed path in the PV diagram is the work done by the gas when the cycle is counterclockwise (as above), and the area is the work done on the gas when the cycle is clockwise.

Note also that the “amount of energy removed by heat” is a net amount. During two legs of the cycle, energy is added to the gas by heat, and during the other two, energy is removed by heat. The net amount is the difference between these two amounts.

Key Points:

• Internal energy is a “state function.” If a gas returns to a state (a point on the PV diagram) that is has been in previously, it returns to the same internal energy.

• When a gas travels around a closed cycle, net work is done on it and an equal amount of net heat is expelled to the environment. (The net work may be positive or negative, depending on the direction of the cycle.)

• No work is done on a gas during a constant-volume process.

• During a constant-pressure process, the work done on a gas is −P V∆ .


This question requires students to put many ideas together. Some students might not realize that the change in internal energy is zero for a complete cycle. They might not connect work with the amount of energy transferred by heat. Or, they might not realize that they can fi nd the work done using the graph of P vs. V.

A possible misconception is to think that since the system returns to its initial state, the energy transferred by heat is zero. In other words, they might inadvertently think that Q is a state variable. Students might have diffi culty understanding how net heat can be expelled with no net change in internal energy.

Students selecting answer (4) might be considering only the magnitudes of the work done during the steps of the cycle, and neglecting the signs.

Students selecting answer (7), “Impossible to determine,” might think that the phrase “amount of energy removed by heat” is ambiguous: does it refer to the “net” amount, or does it refer only to those legs during which energy is actually removed (as opposed to added)? Alternatively, they may simply be overwhelmed by the complexity of the question.

Students might benefi t from a concrete description of how each leg can be accomplished. For instance, from A to B, we would put a certain weight on the piston to maintain constant pressure, then put the gas in thermal contact with a temperature bath at 2T

A until equilibrium is reached, thus transferring energy to the

gas by heat. From D to A, the piston is held fi xed to maintain constant volume, then energy is removed by heat by placing the gas (at T

D = 2T

A) in thermal contact with a temperature bath at T

A, thus removing energy

from the gas by heat.

It turns out that during the fi rst two legs, 4P1V

1 of energy is added to the gas by heat, and during the last

two, 5P1V

1 of energy is removed.


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Question J5.01a

Description: Understanding free expansion of a gas: change in pressure.

Question

Two moles of an ideal gas fi ll volume V = 10 liters at pressure P = 2.4 atm. The gas is thermally insulated from the surroundings. A membrane is broken which allows the gas to expand into the new volume which is 3 times as large as the old volume. The new pressure is:

1. 0.4 atm 2. 0.8 atm 3. 1.2 atm 4. 1.6 atm 5. 2.4 atm 6. None of the above 7. Cannot be determined

Commentary

Purpose: To explore the application of the ideal gas law to the free expansion of a gas.

Discussion: The temperature of a gas undergoing free expansion does not change. For discussion of that fact, see question b of this question set.

For a constant-temperature process, the ideal gas law may be used to relate the change pressure to the change in volume: P

fV

f = (constant) = P

iV

i. So, if the volume triples, the pressure must drop to 1�3 its origi-

nal value, or 0.8 atm. We don’t actually need to know the temperature of the gas or the number of moles.

The subtlety about this question is recognizing that the temperature is in fact a constant. If the gas had expanded via some different mechanism—for example, if it had pushed a piston to the far wall while thermally insulated—the answer would be different.

Key Points:

• In a constant-temperature process, the ideal gas law allows us to relate changes in volume to changes in pressure.

• In thermodynamics processes, how a system gets from one state to another is frequently important.




This problem is meant to be used alongside Question J5.01b. We recommend presenting the question, collecting answers, and allowing students to voice arguments in favor of their selections, but not resolving any of the disagreements or revealing anything about the correct logic beyond what students articulate themselves. Instead, proceed on to 18b, which addresses the question of temperature head-on. Both questions can then be discussed and resolved together. (If the constancy of the temperature arises as a focal issue, you can say “Let’s look at this next question, which addresses that directly.)

One reason for this approach is to catch students who may apply Boyle’s law or some other form of the ideal gas law blindly, without pausing to consider whether the temperature is in fact constant. We hope that the second question will cause them to stop and ponder whether they approached the fi rst correctly.

One can invert the order of the two questions, which makes them more straightforward for students, but that ordering has less potential for causing students to wake up and think hard.

Question J5.01b

Description: Understanding free expansion of a gas: change in temperature.

Question

Two moles of an ideal gas fi ll a volume of 10 liters with a pressure of 2.4 atm. The gas is thermally insu-lated from the surroundings. A membrane is broken which allows the gas to expand into the new volume which is 3 times as large as the old volume. What is the new temperature?

1. The same as before 2. Lower than before 3. Higher than before 4. Cannot be determined

Commentary

Purpose: To explore how the “free expansion” of a gas affects its temperature.

Discussion: When the membrane is broken, the gas expands into a vacuum. It does no work, since it is not pushing on anything: it exerts zero force through some distance. (If it were doing work, where would the energy be going?)


566 Chapter 12

Also, no heat is absorbed or expelled by the gas, since the container is insulating. Even if it were not insulating, the expansion happens essentially instantaneously, so there is no time for any signifi cant heat transfer to occur.

According to the fi rst law of thermodynamics, if a gas does no work and absorbs no heat, its internal energy does not change. If internal energy does not change, temperature does not change. (Consider the equation of state of a gas, relating internal energy to temperature.) The answer is thus (1).

One way to think about this is to realize that the temperature of a gas describes the average speed of the molecules bouncing around within the gas. When the membrane disappears, the molecules spread out into the new space. When they have fi nished “spreading out” into the larger volume, they have the same veloc-ity distribution as before; there is nothing to slow down the molecules.

(While the “spreading out” is occurring, the gas is not in equilibrium and does not have a well-defi ned temperature. Temperature is defi ned only for a system in equilibrium.)

Many students have learned that a gas cools as it expands, and may therefore choose answer (2). A gas does cool as it expands, if it does work during the process of expansion—for example, by pushing back a piston. In this case, the energy lost by the gas is gained by the piston or by something connected to the piston. In the molecular picture, gas molecules are rebounding from the moving wall of the piston; bouncing from a surface that is receding causes a loss of speed.

Key Points:

• The temperature of a gas does not change when the gas undergoes free expansion.

• Temperature is defi ned only for a system in equilibrium.


This question is meant to follow Question J5.01a, bringing to the fore the issue (implicit in 18a) of whether the temperature of the gas changes. See the commentary for that question for suggestions about using the questions together.

This situation provides a good context for discussion what “temperature” means, and when a system has a well-defi ned temperature. It may be helpful to draw a PV diagram, and indicate that during the gas’s nonequilibrium expansion the system is not represented by any point on the diagram; it has disappeared from the plot at one equilibrium point and reappeared at the other.



Question J5.02a

Description: Developing understanding of adiabatic (thermally isolated) processes.

Question

A gas is placed in a thermally insulated container as shown below.

F

As the piston is slowly lowered, the temperature:

1. Increases 2. Decreases 3. Stays the same 4. The change in the temperature cannot be determined.

Commentary

Purpose: To explore what the phrase “thermally insulated” means.

Discussion: The phrase “thermally insulated” means that the gas is thermally isolated from its environment. No heat can fl ow between the gas and the environment, even if they are at different temperatures. That is, Q = 0 for any process.

However, the force F can do work on the gas, which means its internal energy can change. In this case, as the piston is lowered, positive work is done on the gas, so the internal energy increases. If the internal energy increases the temperature must rise.

Key Points:

• “Thermally insulated” means no heat fl ows into or out of the system.

• The temperature of a thermally insulated system can change if work is done on it.


568 Chapter 12


Students choosing answer (3) might think that “thermally insulated” means that the temperature is constant. Some might know that “thermally insulated” means Q = 0, but interpret that to mean the temperature is constant.

Students choosing answer (2) might assume that the force F is constant, which means the pressure is constant and the temperature must decrease to keep the pressure constant at a smaller volume of the gas. (The next question will tease out this facet even more, so you do not need to deal with the idea explicitly. It is not really relevant, since even a constant force would do positive work on the gas.)

Students choosing answer (1), the “correct” answer,” might be using the ideal gas law (PV = nRT ), assum-ing that P is constant, and concluding (correctly) that since V is getting smaller, T must be getting larger. Since the pressure is not constant, this reasoning is fl awed even though the conclusion is right. Therefore, it is critical that you fi nd out why students chose the answers they did. Even if everyone answers this question “correctly,” you need to discuss the reasoning.

Question J5.02b


Question


F

As the piston is slowly lowered, the applied force F:

1. Increases 2. Decreases 3. Stays the same 4. The change in the applied force cannot be determined.



Commentary

Purpose: To reason with the ideal gas law in an adiabatic (thermally insulated) process.

Discussion: The phrase “thermally insulated” means that Q = 0 for this process. Therefore, according to the First Law of Thermodynamics, the work W done on the gas is equal to the change in internal energy ∆ E. Since the work done on the gas is positive, the internal energy increases. If the internal energy increases, the temperature increases.

As the piston is lowered, the temperature is increasing and the volume is decreasing. Therefore, by the Ideal Gas Law (PV = nRT ) the pressure must be increasing. Thus, the force F must increase also in order to keep the piston in equilibrium.

Key Points:


• The ideal gas law relates the pressure, volume, and temperature of a gas.


Students choosing answer (4) might not think of using the ideal gas law, perhaps because the question asks about the force but the ideal gas law refers to pressure. Or, they might be confused by the fact that the pis-ton’s mass is not specifi ed (though this is in fact irrelevant).

Question J5.02c


Question


F

As the piston is slowly raised, the work done on the gas is:

1. Positive 2. Negative 3. Zero 4. The sign of the work done on the gas cannot be determined.


570 Chapter 12

Commentary

Purpose: To check your understanding of the thermodynamic defi nition of work.

Discussion: The thermal insulation of the container is irrelevant. All that matters is the direction of the displacement of the force as compared to the direction of the force.

As the piston is raised, the force F points down, but the displacement of the piston is up, so the work done by F is negative. Thus, the work done “on” the gas (by the piston) is negative. (Note that the work done “by” the gas (on the piston) is positive, because the gas is pushing up on the piston as it moves in the same direction.)

Another way of thinking about this is in terms of pressure. The container holding a gas exerts a pressure (a force per unit area) on the gas on all sides, directed inward. If the gas expands, it is doing positive work against this pressure, so the pressure exerted on it is doing negative work. Any time a gas expands, negative work is done on it. Any time a gas contracts, positive work is done on it.

Key Points:

• The work done by a force is positive if the object acted upon moves in the direction the force pushes, and negative if it moves in the opposite direction.

• Mechanics ideas, defi nitions, and principles apply to thermodynamic systems as well.

• An expanding gas has negative work done on it by its environment.


Some students might still have trouble understanding how the work done can be negative. Depending on how long it has been since students have used work and energy ideas, it might be a good idea to review them in more familiar contexts. This question provides a good opportunity to link and integrate mechanics with thermodynamics.

Students often get confused about the work done “on” the gas (by an external agent) vs. the work done “by” the gas (on the environment).

Students might be unduly distracted by the fact that the container is insulated, and try to apply thermo-dynamic principles to deduce the sign of the work done on the gas rather than simply applying the defi nition of work.



Question J5.02d


Question


F

As the piston is slowly raised:

1. The gas gives off heat to the environment (Q < 0) 2. The gas absorbs heat from the environment (Q > 0) 3. The gas does not exchange any heat with the environment (Q = 0) 4. The heat exchanged with the environment cannot be determined.

Commentary

Purpose: To revisit the concept of “thermally insulated.”

Discussion: “Thermally insulated” means that no heat can be exchanged with the environment. The only way the environment can interact with the gas is through the piston and the applied force F.

This does not mean the temperature of the gas must remain constant. As work is done on the gas by the piston, the energy of the gas changes according to the fi rst law of thermodynamics. If the internal energy changes, the temperature changes.

Key Points:


• The temperature of a thermally insulated system can change if work is done on it.


Though this question may seem redundant with the fi rst in this set, many students will remember only that the temperature can change (a surprising result!) and incorrectly conclude that heat must be exchanged with the environment.


572 Chapter 12

Question J5.03a

Description: Understanding adiabatic processes: pressure.

Question

An ideal gas is allowed to expand slowly. The system is thermally isolated. Which statement regarding the fi nal pressure is true?

F

M

M

Before After

P, T P ′, T ′h′

h

1. ′P < P 2. ′P = P 3. ′P > P 4. Not enough information

Commentary

Purpose: To understand pressure change in an adiabatic process.

Discussion: The system is thermally isolated, so the expansion process is adiabatic (no heat fl ows in or out). For adiabatic processes, PV γ is constant. Thus, if the volume increases, the pressure decreases.

Another way to determine this is to realize that for an adiabatic process, the fi rst law of thermodynamics (energy conservation) requires that if the gas does work on its environment, its internal energy and thus its temperature must drop (since heat transfer is zero). If the temperature drops and the volume increases, the ideal gas law indicates that the pressure must decrease.

A third way to reason to the answer is to realize that the gas won’t expand unless the pressure inside the container is greater than the pressure outside plus the pressure caused by the weight of piston head (the moveable mass). The gas will expand until the forces on the piston head are balanced, so the fi nal internal pressure must be lower than the initial internal pressure.

Key Points:

• For adiabatic processes, PV γ is constant.

• For adiabatic processes, the work done by the gas equals the internal energy lost by the gas.

• There often more than one way to answer a question, and some ways are simpler than others.




This is the fi rst of two related questions.

It is a simple question, but good for introducing or checking students’ understanding of adiabatic processes and the PV γ rule. It is also valuable for demonstrating that a question may be answered through multiple arguments. The third argument, in particular, is elegant and simple, and carries the moral that making physical sense of a situation is as important as remembering and applying abstract mathematical laws.

Question J5.03b

Description: Understanding adiabatic processes: temperature.

Question

An ideal gas is allowed to expand slowly. The system is thermally isolated. Which statement regarding the fi nal temperature is true?

F

M

M

Before After

P, T P ′, T ′h′

h

1. ′T < T 2. ′T = T 3. ′T > T 4. Not enough information

Commentary

Purpose: To understand temperature change in an adiabatic process.

Discussion: The system is thermally isolated, so the expansion process is adiabatic (no heat fl ows in or out). For adiabatic processes, T V γ −1; this can be derived from the ideal gas law and the fact that PV γ is constant. Thus, if the volume increases, the temperature decreases.

Another way to determine this is to realize that for an adiabatic process, the fi rst law of thermodynamics (energy conservation) requires that if the gas does work on its environment, its internal energy must drop (since heat transfer is zero). For an ideal gas, internal energy depends only on temperature, so if the internal energy drops, so also must the temperature.


574 Chapter 12

Key Points:

• For adiabatic processes, TV γ −1 is constant.

• For adiabatic processes, the work done by the gas equals the internal energy lost by the gas.

• There often more than one way to answer a question, and some ways are simpler than others.


This is the second of two related questions. The answer to this question is likely to be implicitly addressed while covering the previous one, making this one good for checking whether students really “got it.”

As with 74a, it is worth making sure students appreciate the multiple ways of determining the answer.

Question J6.01

Description: Sensitizing to assumptions when working with the ideal gas law.

Question

Consider the two systems below, labeled A and B.

F

F

A B

Which gas has the higher pressure?

1. A 2. B 3. Neither; the pressures are the same. 4. Impossible to determine



Commentary

Purpose: To develop your awareness of the role different assumptions play in answering an ideal gas law question.

Discussion: The ice-water mixture keeps the gas in the container at 0°C. Thus, we know the temperatures of the two systems are the same, and the volume of system A is larger than that of system B.

If both systems contain the same amount (number of moles) of gas, then the ideal gas tells us that system B must have the higher pressure. However, we don’t know anything about the amounts of gas, so answer (2) is unjustifi ed.

If the applied force F is the same for both systems, and if both pistons have the same size, shape, and mass, then the pressure applied to the gas must be the same in both cases. The question statement doesn’t say that these conditions are true, but we might assume that the label “F” is a variable that represents the same value for both cases, and the picture seems to show two identical pistons. So answer (3) is defensible, but requires some assumptions.

The only thing we can say for sure is that the volumes are different and the temperatures are the same, which means that either the pressure or the amount of gas must differ. The most conservative answer is therefore (4).

Key Points:

• The ideal gas law relates the pressure, volume, temperature, and amount of a gas. Don’t overlook the amount (number of moles) as a variable.

• Be aware of the assumptions you make, and choose them carefully.


Answer (2), that system B has a higher pressure, tends to be common; this is generally because students implicitly assume both systems contain the same amount of gas.

As always, we recommend focusing on the consequences and reasonableness of various possible assump-tions rather than on the correctness of answers.

Additional Questions:

1. Which system has the higher temperature? 2. Which system has the higher number of moles of gas? 3. Assume now that a thermodynamic process takes system A and turns it into system B. How would

you accomplish this?

Question J6.02a

Description: Understanding gas pressure.

Question

An unknown amount of gas is placed in a container with a moveable piston of negligible mass. The cross-sectional area of the container is A. When a mass M is placed on the piston, the height of the piston above the bottom of the container is h.


576 Chapter 12

M

What is the pressure of the gas in the container?

1. Mg 2. MgA 3. MA 4. MA�h 5. Mg�h 6. Mh�A 7. Mg�A 8. M�A 9. None of the above 10. Impossible to determine from the given information

Commentary

Purpose: To develop your understanding of the “pressure” of a gas.

Discussion: This is a perfectly static situation. We are not yet discussing any processes; we simply want to make sure we understand the connection between the mass on the piston and the pressure of the gas inside the container.

Focusing on the piston and block as a single system is particularly useful. It is at rest, so its acceleration is zero, and thus the net force on it must be zero.

Pressure is defi ned as force per unit area (P = F�A), so the gas pushes up on the piston with a force of PA, where A is the cross-sectional area of the piston. This makes intuitive sense: the larger the area or the larger the pressure, the larger the force should be.

Gravitation is pulling down on the block and piston with a force of Mg. Since the piston and block are at rest, the total force up must balance the total force down. So, ignoring the atmospheric pressure pushing down on the piston and block, PA = Mg, or P = Mg�A.

The outside air pushes down on the top of the block and piston with one atmosphere of pressure (Patm

), for an additional downward force of P

atmA. If we include this, we fi nd that the pressure of the gas in the

cylinder must be P = Patm

+ Mg�A.

Note that the pressure does not depend on the height h.

Key Points:

• Pressure is force exerted per unit surface area. For a fl at surface that exerts a force on a gas, P = F�A.

• A gas contained by a “fl oating” (movable) piston will expand or contract until its pressure has the right value to make the net force on the piston zero.




This set of four questions introduces constant-pressure processes, beginning with the concept of gas pressure.

Students might not fully appreciate the balance that occurs to keep the piston and block at rest, especially if their understanding of Newtonian mechanics is rusty. This question provides an opportunity to review elementary statics.

Students might think that the height h should be part of the answer. They might think that the larger the pressure, the smaller the height h. (The next question in this set should help them sort out this feature.)

Students can use proportional reasoning. A larger mass would create a larger pressure, so the pressure should depend on M in the numerator. A larger area would require a smaller pressure to produce the same force. And so on . . .

Question J6.02b

Description: Understanding gas pressure.

Question

Consider the two systems below, labeled A and B.

M

M

A B

Which gas has the higher pressure?

1. A 2. B 3. Neither; the pressures are the same. 4. Impossible to determine


578 Chapter 12

Commentary

Purpose: To check your understanding of gas pressure.

Discussion: Assuming the piston is free to move, the force pushing up due to pressure in the gas must exactly balance the forces pushing down on the piston. The forces down are the same for the two situations: the weight of the piston and block and atmospheric pressure from above. Therefore, the pressures inside the two cylinders must be the same.

How can the pressures be the same while the volumes are clearly different? According to the ideal gas law (PV = nRT ), if pressure is constant, a larger V means a larger nRT. So the larger volume in situation A could be due to a higher temperature in the gas, or due to a larger quantity of gas, or both. We are not told enough to determine which.

Key Points:

• For a “fl oating” piston supported by a gas under pressure, the upward force due to the gas pressure must balance all other forces acting on the piston.

• You cannot necessarily infer anything about a gas’ pressure from information about its volume, since temperature and number of moles are also relevant variables.


Students choosing answer (A) might be thinking that A must have a greater pressure because we can see it has pushed the piston higher. Students choosing answer (B) might be thinking that B must have a greater pressure because it has been compressed more. Both groups are imagining a dynamic process, rather than considering a static balance of forces.

Even if students recognize at some point that the pressures inside the containers are the same, they will likely think that the temperature is higher for A than B, because they will implicitly assume that the amounts of gas in the containers are the same. They should be made aware of this assumption.

Some students might think that the answer is impossible to determine, perhaps because they do not know if atmospheric pressure should be included or not or because they do not know its value. This means only that a value for the pressure cannot be computed; it does not mean that they cannot compare the pressures.

Discussion Questions

1. Which system has the higher temperature? 2. Which system has the higher number of moles of gas? 3. Assume now that a thermodynamic process takes system A and turns it into system B. How would

you accomplish this?

Question J6.02c

Description: Understanding isobaric (constant pressure) processes.

Question

An unknown amount of gas is placed in a container with a moveable piston as shown below. The cross- sectional area of the container is A. When a mass M is placed on the piston, the height of the piston above the bottom of the container is h.



M

What happens when an amount of energy Q is transferred by heat to the gas? (Q > 0.)

1. T increases. 2. h increases. 3. T and h increase. 4. T and P increase. 5. h and P increase. 6. T, h, and P increase. 7. T decreases. 8. h increases; T decreases. 9. None of the above 10. Impossible to determine

Commentary

Purpose: To understand constant pressure processes.

Discussion: Q is positive, so thermal energy is being added to the gas. According to the fi rst law of thermodynamics, the added heat must either increase the internal energy and thus the temperature of the gas, cause the gas to expand and do work on the piston, or both. The energy has to go somewhere.

The pressure must remain constant to balance the downward forces on the piston and mass. According to the ideal gas law, if the temperature rises with constant pressure, the volume must increase, and vice-versa. So, the added heat must raise the gas’ temperature and cause it to expand. The best answer is therefore (3).

Key Points:

• When heat is added to a gas at constant pressure, the gas’ temperature and volume increase.

• The fi rst law of thermodynamics helps you keep track of energy that gets added to or removed from a thermodynamic system.

• The ideal gas law relates the pressure, volume, temperature, and quantity of a gas.


Some students might intuit that the gas will expand, but not be sure what happens to the temperature. Since the expanding gas does work on the piston, they might not be sure if this “uses up” all of the added heat: whether the work is larger or smaller than Q, and therefore whether the temperature falls or rises. The fi rst law of thermodynamics is not enough to resolve this; the ideal gas law is required as well.


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Discussion Questions

1. What is the relationship between T and V for this process? 2. What is the relationship between T and h for this process? 3. Which is larger, Q or W? 4. If the same amount of energy Q is transferred to the gas, but this time the piston is held fi xed, how

would the change in temperature compare to the change in this situation?

Question J6.02d

Description: Understanding isobaric (constant pressure) processes.

Question

An unknown amount of gas is placed in a container with a moveable piston as shown below. The cross- sectional area of the container is A. When a mass M is placed on the piston, the height of the piston above the bottom of the container is h.

M

An amount of energy Q is transferred by heat to the gas. (Q > 0.) Put the magnitudes of the energy Q, work W, and change in internal energy ∆ E in order from smallest to largest.

1. Q < |W | < |∆ E| 2. Q < |∆E| < |W | 3. |W | < |∆E| < Q 4. |W | < |∆E| = Q 5. |W | < Q < |∆E| 6. |∆E| < Q < |W | 7. |∆E| < Q = |W | 8. |∆E| < |W | < Q 9. None of the above 10. Impossible to determine

Commentary

Purpose: To understand constant pressure processes.

Discussion: Since Q is positive, some of the energy raises the temperature of the gas and some of the energy does work to raise the piston. We can see this directly from the fi rst law, ∆ E = Q + W. The tempera-ture increases, so ∆ E is positive. The work W done on the gas by the piston is negative. The only way all



of these statements can be true is if both ∆ E and W are smaller than Q. Thus, we know how |∆ E| and |W | compare to Q, but we do not know yet how they compare to each other.

To compare |∆ E| and |W |, we need to dig a little deeper. The internal energy E of a gas is proportional to its temperature. For a monatomic gas, the relationship is E =3

2nRT. For all other gases, the proportionality constant is even larger. Thus, the change in internal energy may be written ∆ E ≥ 3

2nR∆T.

For a constant pressure process such as this one, the work done on the gas is –P∆V. (A positive change in volume means the work done on the gas is negative.) Assuming the gas in the container is ideal, it obeys the ideal gas law, PV = nRT, so P∆V = nR ∆ T. This means the magnitude of the work done on the gas is |W | = nR ∆ T. The change in internal energy is always larger than this, so the work done is smallest. Thus, |W | < |∆ E| < Q.


Many students will likely be able to reason that Q is larger than W and ∆ E, but unable to sort out why W is smaller than ∆ E.

Some students choosing the correct answer, (4), may be doing so erroneously: even though the question asks for a comparison between magnitudes, they may think that W is “smallest” since it is negative.

Some students might think that the change in internal energy is always the largest of the three, perhaps because ∆ E = Q + W. In other words, they are not thinking that either Q or W could be negative.

If students assume the gas is monatomic, challenge them to answer the question and justify their answer without that assumption.

Additional Questions:

• An amount of energy Q is added by heat such that the volume exactly doubles. In terms of known quantities (M, A, h, etc.), what are (a) Q, (b) W, and (c) ∆ E for this process?

Question J7.01a

Description: Introducing isothermal processes, linking to latent heat and phase change.

Question

A gas is placed in a container, which is immersed in an ice-water mixture as shown below.

F


582 Chapter 12

As the piston is slowly lowered, the temperature:

1. Increases 2. Decreases 3. Stays the same 4. The change in the temperature cannot be determined.

Commentary

Purpose: To introduce the idea of an isothermal process.

Discussion: A mixture of ice and water remains at 0°C until all the ice is melted or until all the water has turned into ice. If you add energy (heat) to the mixture, some of the ice melts, but the temperature remains constant. The energy added literary goes into melting the ice rather than changing the temperature of the mixture. When all the ice has melted, then adding more energy would increase the temperature of the water. Likewise, removing energy from the mixture causes some water to turn into ice, and again, the temperature remains constant. When all of the water has turned into ice, then removing more energy would lower the temperature of the ice. Thus, an ice-water mixture is a mechanism for creating a “constant temperature” environment.

The gas chamber is not insulated, so we will assume that heat can fl ow freely between the gas and the ice-water mixture. Since the piston is lowered slowly, the gas must stay at 0°C: if the temperature of the gas rises (or drops) infi nitesimally, heat will fl ow to (or from) the mixture, bringing it back into thermal equilib-rium. We call such a process isothermal (meaning “constant temperature”).

If the piston were pushed down rapidly, the temperature of the gas might rise suddenly, and then slowly fall back to 0°C as heat fl ows through the cylinder wall and the system returns to thermal equilibrium.

Key Points:

• An isothermal process has a constant temperature.

• An ice-water “bath” is a mechanism for maintaining a system at approximately constant temperature.


Depending on students’ understanding of latent heat and phase changes, they might not understand how the temperature of an ice-water mixture can remain constant when energy is added or removed. A demon-stration might be helpful. This is a good question for linking students’ understanding of thermodynamic processes with their understanding of latent heat and phase change.

Students might have trouble understanding how ice is formed by “removing” energy from the water; adding energy and melting ice seem much more easily accomplished. An apparent counterexample that may confuse students is a freezer, which requires electrical energy to form ice. A refrigerator or freezer is in fact a more complicated system. Heat only fl ows spontaneously from warm to cold; to pump heat from a cold object to a warmer one, additional energy is required. A discussion of this can be used as a forward reference to entropy and the second law of thermodynamics.

Students might select answer (1) because they do not recognize that heat will fl ow out of the gas into the ice-water bath, or because they think heat cannot fl ow rapidly enough out of the gas to maintain constant temperature.



Students selecting answer (2) might be misinterpreting the situation, not realizing that the gas has already equilibrated to 0°C when the piston is lowered.

One possible reason students might give for claiming the temperature change cannot be determined (answer 4) is that they do not know whether the mixture contains enough ice.

Question J7.01b

Description: Developing understanding of the ideal gas law in an isothermal context.

Question


F

As the piston is slowly lowered, the applied force F:

1. Increases 2. Decreases 3. Stays the same 4. The change in the applied force cannot be determined.

Commentary

Purpose: To understand the ideal gas law for an isothermal process.

Discussion: As the piston is lowered, the temperature of the gas remains constant at 0°C, but the volume decreases. According to the ideal gas law (PV = nRT ), if the temperature is constant but the volume is decreasing, the pressure must increase. Therefore, as the piston is lowered, the force must increase to keep it in equilibrium.

Key Points:

• For an ideal gas at constant temperature, pressure and volume have an inverse relationship. That is, if one decreases, the other must increase.


584 Chapter 12


One possible source of confusion here is that students might not appreciate the connection between the pressure of the gas and the force applied to keep the piston from moving. Constructing a free-body diagram for the piston may help.

Students occasionally assume that the letter “F” indicates a constant value for the force, and choose answer (3).

Question J7.01c

Description: Exploring thermodynamic work in an isothermal context.

Question


F

As the piston is slowly raised, the work done on the gas is:

1. Positive 2. Negative 3. Zero 4. The sign of the work done on the gas cannot be determined.

Commentary

Purpose: To develop your understanding of thermodynamic work for an isothermal process.

Discussion: As the piston is raised, the work done by the gas on the piston is positive: the force is up, and the displacement of the piston is up also. It does not matter whether the force is constant or changing.

However, the work done on the gas is negative, because the displacement is again up, but the force F��

is down.

Key Points:

• The work done by a force is positive if the object acted upon moves in the same direction as the force, and negative if it moves in the opposite direction.

• The work done on a gas and the work done by the gas have opposite signs. Be careful not to confuse them.

• If a gas expands, it does positive work on its environment.




Students often get confused about the work done “on” the gas (by an external agent) vs. the work done “by” the gas (on the environment).

Some students might still have trouble understanding how the work done can be negative. Depending on how long it has been since students have used work and energy ideas, it might be a good idea to review them in more familiar contexts.

Students might be unduly distracted by the isothermal process, not able to ignore it even though it is irrel-evant. They might try to apply thermodynamic principles to deduce the sign of the work done on the gas, rather than simply apply the defi nition of work. For instance, they might conclude that the work done must be zero, since both the temperature and internal energy are constant.

Question J7.01d

Description: Developing understanding of heat transfer during an isothermal process.

Question


F

As the piston is slowly raised:

1. The gas gives off heat to the environment (Q < 0) 2. The gas absorbs heat from the environment (Q > 0) 3. The gas does not exchange any heat with the environment (Q = 0) 4. The heat exchanged with the environment cannot be determined.

Commentary

Purpose: To understand heat exchange during an isothermal process.

Discussion: An ice-water mixture maintains a constant temperature of 0°C unless all the ice melts or all the water freezes. A gas in good thermal contact with such a mixture remains at 0°C; if energy is added to (or removed from) the gas, heat fl ows out of (or into) the gas so its temperature stays constant. If temperature is constant, internal energy E is also (∆ E = 0).

As the piston is raised, the work W done on the gas is negative. According to the fi rst law of thermo-dynamics, the heat Q absorbed by the gas must be positive (Q + W = 0). Thus, the gas absorbs heat from the environment.


586 Chapter 12

Key Points:

• An isothermal system has constant internal energy (∆ E = 0).

• According to the fi rst law of thermodynamics, any work done on (or by) an isothermal system must cause an equal amount of heat to be expelled (or absorbed).


Students might not realize that when a gas is at a constant temperature, the internal energy is constant too.

Students might not fully appreciate that work and heat cause changes in internal energy.

Students might not believe that the work done on a gas can be negative. You need to create suffi cient confl ict to help them challenge this belief. For instance, when the gas expands it does work on the environ-ment. Where does it get the energy to do this work? Its internal energy is constant. The only way to get energy is to absorb energy from the ice-water mixture, so Q > 0. If Q > 0 and ∆ E = 0, then W < 0. That is, the work done on the gas is negative.

Students might not be sure exactly what is meant by the “environment.” In this case, it is the ice-water mixture, since that is what it is in thermal contact with. In other words, there are two “environments.” The gas is in thermal contact with an environment with which it exchanges heat (microscopic) energy, and it is in mechanical contact with an environment with which it exchanges mechanical (macroscopic) energy. In this case, the ice-water mixture is the “thermal” environment, and the piston and applied force F is the “mechanical” environment.

Question J7.01e

Description: Honing the concept of a thermodynamic system’s “environment.”

Question


F

As the piston is slowly raised, is it possible for the temperature of the gas to be different from the tempera-ture of the environment?

1. Yes 2. No



Commentary

Purpose: To develop your understanding of what the “environment” of a thermodynamic system refers to.

Discussion: In physics, we typically separate the world into a system and its environment. Sometimes the environment is poorly defi ned. Generally, the environment refers to that part of the world the system is in contact with or that affects its behavior.

In this case, the gas is the system, and it is in thermal contact with the ice-water mixture and mechanical contact with the piston. We usually assume that the piston is insulating, so that we do not need to worry about heat exchange with it. Since this question asks about temperatures, we can interpret “environment” as referring to what the gas is in thermal contact with: the ice-water mixture. Therefore, the best answer is “No,” the gas cannot have a different temperature than the environment (as long as the piston is raised suffi ciently slowly).

Key Points:

• The “environment” of a “system” is usually that part of the rest of the world that contacts or otherwise directly affects the system.

• How we interpret “environment” depends on the specifi c situation and question asked.


We tend to use a rather narrow defi nition of “environment” in these situations, though students are not always sensitized to its narrow use. It is reasonable for them to think that the temperature of the rest of the world is different than 0°C, so students who answer “Yes” are not necessarily confused about the physics involved. They might understand completely that the gas is always at 0°C. They simply are not using the same defi nition of environment as you are.

We recommended letting students debate this question, explaining their reasoning and assumptions and also challenging each other’s answers. After a few minutes, you can provide the narrow defi nition of environ-ment typically used in thermodynamics.

Question J7.01f


Question


F


588 Chapter 12

As the piston is slowly raised, the work done on the gas is – 400 cal. During this process:

1. The temperature of the gas remains the same 2. 5 grams of ice are formed 3. The heat absorbed by the gas is 400 cal 4. All three of the above are true. 5. Exactly two of the above are true. 6. None of the above is true.

Commentary

Purpose: To integrate several ideas related to isothermal processes.

Discussion: As the piston is raised, the applied force F does negative work W on the gas. Since the temperature of the gas is constant, its internal energy E is constant. So, according to the fi rst law of thermodynamics, Q for this process must be positive. That is, the gas absorbs energy from its environment.

Mathematically, ∆ E = Q + W. For an isothermal process, ∆ E = 0, so Q + W = 0, or Q = ∆ W. Since W = − 400 cal, Q = 400 cal. In other words, the gas absorbs 400 cal of energy from its environment.

In this case, the environment (or “thermal” environment) is the ice-water mixture. We will assume no signifi cant heat transfer occurs between the mixture and the outside atmosphere. Since Q is positive, the ice-water mixture must lose 400 cal of energy. When this occurs, ice is formed. Specifi cally, it takes 80 cal of energy to melt 1 gram of ice, so for each 80 cal of energy removed, 1 gram of ice is formed. Removing 400 cal means that 5 grams of ice are formed.

Thus, the fi rst three statements are all true, and the best answer is (4).

Key Points:

• Total energy is conserved. You should be able to account for how all the energy in a system moves and where it goes to and comes from by considering mechanical work done, heat fl ow, and changes in internal energy (due to both temperature and phase change).

• Understanding a nontrivial physical system often requires putting together multiple processes and ideas: in this case, phase change, heat transfer, the fi rst law of thermodynamics and energy conservation, work, and thermal equilibrium.

• Removing 80 cal of thermal energy from 0°C water causes 1 gram of it to freeze; adding 80 cal causes 1 gram of 0°C ice to melt.


Students will probably have more diffi culty with this than you might expect. If the situation is reversed, with the piston being slowly lowered, students generally have less diffi culty. Often, they just don’t under-stand negative work done on a system.

It might help students to think in terms of the gas doing 400 cal of work on the piston. The temperature is constant, so the internal energy is constant too. Where does this energy come from? It is pulled from the ice-water mixture, which causes ice to be formed, etc.

Students might not remember that the latent heat of fusion for water is 80 cal�g or might not know that the amount of heat needed to melt 1 gram of ice (80 cal) is also the amount of heat removed to freeze 1 gram of water.



Answer (5) is also defensible, on the grounds that if any nonzero heat transfer occurs anywhere except between the ice-water bath and the gas—for example, between the bath and the outside air—the amount of ice formed will not be exactly 5 grams, and statement (2) will not be exactly true.

Question J8.01

Description: Reasoning from graphical representations of thermodynamic processes: internal energy.

Question

An amount of an ideal gas is taken around the process shown. Which of the following statements about the internal energy of the states is true?

20

10

P(Pa)

V (m3)

1 3

C

B

A

1. The internal energy of state B is twice that of state C. 2. The internal energy of state B is equal to that of A and C combined. 3. The internal energy of state A is half that of state C. 4. The internal energy of state B is less than the internal energy of state A. 5. None of the above 6. Cannot be determined

Commentary

Purpose: To link the concepts of pressure, volume, and internal energy by way of a state diagram.

Discussion: The internal energy of an ideal gas depends only on and is directly proportional to its temperature in Kelvin. According to the ideal gas law, temperature is proportional to the product of pressure and volume. State B has twice the pressure and three times the volume of state A, so it must have six times the temperature and six times the internal energy. B has the same volume and twice the pressure as C, so it must have twice the internal energy. Similarly, C has three times the volume and the same pressure as A, so it must have three times the internal energy. Thus, only statement (1) is accurate.

Key Points:

• The internal energy of an ideal gas is proportional to its absolute temperature.

• The ideal gas law relates temperature, pressure, and volume.

• Changes in a system’s internal energy can be inferred from a pressure vs. volume graph.


Qualitative reasoning from information contained in a graph is a powerful tool for developing conceptual understanding and analysis skill.


590 Chapter 12

Question J8.02

Description: Integrating thermodynamic process ideas: PV diagrams, cycles, the First Law, energy, etc.

Question

An amount of a monatomic ideal gas is taken around the process shown. The amount of heat extracted during process BC is:

20

10

P(Pa)

V (m3)

1 3

C

B

A

1. 10 J 2. 20 J 3. 15 J 4. 45 J 5. 60 J 6. None of the above 7. Cannot be determined

Commentary

Purpose: To interconnect several thermodynamics ideas: cycles, processes, PV diagrams, pressure, volume, the fi rst law of thermodynamics, work, heat, and internal energy.

Discussion: BC is a constant volume process, so no work is done. Thus, according to the fi rst law of thermodynamics, the heat extracted from the system is equal to the negative of the system’s change in internal energy. (Pressure is dropping at constant volume, so the gas must be cooling, so a positive amount of heat must be extracted.)

For a monatomic ideal gas, the internal energy U = (3�2)nRT. According to the ideal gas law, PV = nRT. Thus, for monatomic ideal gases, U = (3�2)PV. (For nonmonatomic ideal gases, the factor would be something larger than 3�2.) Since we can read the initial and fi nal pressures and volumes from the graph, we can determine that the change in internal energy is equal to − ⋅ = −45 45 Pa m J3 . The correct answer is therefore (4).

Note that this question can be answered without knowing how many moles of gas are present.

Key Points:

• For a constant-volume process, change in internal energy is equal to the heat absorbed by a system.

• For an ideal gas, internal energy can be expressed in terms of the temperature or in terms of the product of pressure and volume. For a monatomic ideal gas, U = (3�2)PV.




This is a fertile question for interconnecting several different ideas and skills, and helps students organize and structure their knowledge and to develop problem-solving skills.

It is common for students to get stuck or choose “Cannot be determined” because they do not know the number of moles of gas present, and consequently cannot calculate the actual temperatures of the states without that.

Question J8.03

Description: Reasoning with the fi rst law of thermodynamics and PV diagrams.

Question

A system consisting of a quantity of ideal gas has the two isotherms shown. The system, initially at state C, can be taken along path CA to fi nal state A or along path CB to state B.

P(Pa)

A

BC

V (m3)

T1T2

Which of the following is true?

1. QCA < QCB 2. QCA = QCB 3. QCA > QCB 4. Not enough information

Commentary

Purpose: To promote your ability to reason qualitatively about thermodynamics states using the fi rst law of thermodynamics and a PV diagram.

Discussion: Since the internal energy of a gas depends only on the temperature of the gas, states A and B have the same internal energy, so the change in internal energy of the system is the same along paths CA and CB.

According to the fi rst law of thermodynamics, the change in internal energy of a system is equal to the heat absorbed by it plus the work done on it, which is equal to the heat absorbed by it minus the work done by it. Path CA is isovolumetric, so no work is done on or by the gas; thus, the heat absorbed from C to A is equal to the systems’ change in internal energy. The work done by the gas from C to B is equal to the area under the path, so the heat absorbed from C to B must be greater than the change in internal energy.


592 Chapter 12

Key Points:

• The internal energy of an ideal gas depends only on its temperature.

• The fi rst law of thermodynamics relates a system’s change in internal energy to the heat it absorbs and the work done on it.

• If an amount of heat is absorbed by a gas, its temperature will rise more if the volume is fi xed than if it can expand.


Students may fi nd it counterintuitive that the same change in temperature may require different amounts of heat, since they are used to thinking about heat and temperature in the context of heat capacities and latent heats. This question makes a good basis for introducing the concept of specifi c heat and heat capacity for an ideal gas, including the isobaric and isovolumetric expressions.

Question K1.01a

Description: Developing understanding of entropy.

Question

One mole of “o” gas is confi ned to volume V�2 by a thin membrane with a vacuum on the other side.

V/2 V/2

The membrane is broken allowing the gas to fl ow into the full volume. The entropy has:

1. Decreased 2. Stayed the same 3. Increased

Commentary

Purpose: To develop your comprehension of entropy.

Discussion: Entropy can be though of as a measure of disorder: the more disordered the constituents of a system, the larger the entropy. It can also be thought of as a measure of information: the less information we have about the constituents of a system, the more entropy it has.

Before the membrane is broken, we know that all the molecules are in the left half of the box. After the membrane has broken, the molecules spread out, and we know only that they are somewhere in the whole



box. This is less information about their locations, and the system is more disordered. So, the system’s entropy has increased.

Note that work would be required to put the system back into its original state, for example by sliding a new membrane in along one wall of the box and then moving it to the center so that it compresses the gas. Doing this work would require a power source such as an engine, so although the entropy of the system (box and gas) is decreased, the engine pumps exhaust heat into the environment and the global entropy of the universe increases, in accord with the second law of thermodynamics.

Key Points:

• If the disorder of a system increases, and we lose information about the arrangement of its constituents, its entropy has increased.

• Decreasing the entropy of a system takes work, and doing that work will necessarily increase the entropy of the environment so that the total entropy of the universe stays the same or increases.


This is the fi rst of a three-question set leading up to Gibb’s paradox. Most students will answer this correctly, based on a vague notion of entropy as disorder. We suggest progressing onward towards the third question.

Question K1.01b


Question

One mole of “o” gas is confi ned to volume V�2 by a thin membrane with one mole of “x” gas on the other side. Both gases are at the same temperature.

V/2 V/2��

��

��

�

� ��

� � �

�

��

�

The membrane is broken allowing the gases to fl ow into the full volume. The entropy has:



594 Chapter 12

Commentary


Discussion: This question is much like the previous. When the membrane breaks, the “o” molecules spread out, and we lose information about where they are located. The “x” molecules also spread out and we lose information about them. The system has become more disordered, and its entropy has increased. This would be true no matter what the temperature of the two gases was. If the temperatures were different then the distributions of speeds of the molecules would also change when the gases mixed. In this case, because the temperatures are the same, the increase in entropy is due entirely to the increase in the volumes of the gases.

It would take work to restore the system to its original confi guration, work that would necessarily increase the entropy of the external environment. For example, let’s imagine that “x” molecules are slightly larger than “o” molecules. In that case, we could slide a membrane along the left wall that had tiny pores, large enough for “o” molecules to pass through but too small for “x” molecules. We could then push that mem-brane almost all the way to the far side, crunching all the “x” molecules into a very tiny space and letting almost all the “o” molecules pass through and spread out in the rest of the box. Then, we could slide a solid membrane alongside the permeable one (or just close the pores somehow), and move the two-membrane wall back to the center. We now have all the “x” molecules on one side, and almost all the “o” molecules on the other. We have approximately regained the original situation, but this required work to push the membranes against the pressures exerted by the gases.

Key Points:

• If the disorder of a system increases, and we lose information about the arrangement of its constituents, its entropy has increased.

• Decreasing the entropy of a system takes work, and doing that work will necessarily increase the entropy of the environment so that the total entropy of the universe stays the same or increases.


This is the second of a three-question set leading up to Gibb’s paradox. Most students will answer this correctly as well; if they understand the fi rst, this does not pose additional challenges (except for the more convoluted thought experiment for restoring the situation to its initial confi guration). It sets up the third question, however.

Question K1.01c


Question

One mole of “o” gas is confi ned to volume V�2 by a thin membrane with another mole of “o” gas on the other side. Both gases are at the same temperature.



V/2 V/2

The membrane is broken allowing the gases to fl ow into the full volume. The entropy has:


Commentary


Discussion: This situation is similar to the previous one, except for one crucial difference. When the membrane breaks, the “o” molecules in the left side spread out, and we lose information about where they are located. The “o” molecules in the right side also spread out and we lose information about them. However, because the temperatures are the same and all the molecules are indistinguishable, the fi nal arrangement of molecules is statistically identical to the original, and we have lost no information about where “o” molecules are located. They are still, on average, uniformly distributed throughout the box. So, the entropy of the system has stayed the same.

It would take no work to restore the system to its original confi guration, simply by reinserting a new membrane where the old one had been. This means that the original situation could be restored without increasing the entropy of the external environment. Entropy is a state function: if the system is returned to a previous state, its entropy must return to the previous value. If breaking the membrane increased the entropy of the system, then restoring it would have to decrease the entropy; but if the entropy could be decreased without increasing the entropy of the outside environment, the second law of thermodynamics would be violated.

(Note that if the temperatures of the two sides were different, then the entropy would increase and it would require work to be done to restore the system to its original state.)

Key Points:

• Mixing indistinguishable particles at the same temperature does not increase entropy.

• If a system that has changed can be restored to its starting state without doing any work and increasing the entropy of the environment, then the change could not have altered the system’s entropy.

• According to the second law of thermodynamics, the entropy of the entire universe cannot decrease, ever.


This situation is the famous Gibb’s paradox, and is guaranteed to cause argument and confusion. The key to resolving the paradox lies in appreciating the signifi cance of the fact that the molecules are indistinguish-able, fundamentally and in principle. Arguments that begin “If we could watch a particular molecule. . . ” or “If we painted all the left-hand ones blue. . .” must be rejected. Quantum mechanically, when two identical particles collide and scatter, the question of which one went in which direction is not even meaningful.


596 Chapter 12

QUICK QUIZZES

1. (b). The work done on a gas during a thermodynamic process is the negative of the area under the curve on a PV diagram. Processes in which the volume decreases do positive work on the gas, while processes in which the volume increases do negative work on the gas. The work done on the gas in each of the four processes shown is

Wa = − ×4 00 105. J, Wb = + ×3 00 105. J, Wc = − ×3 00 105. J, and Wd = + ×4 00 105. J

Thus, the correct ranking (from most negative to most positive) is a, c, b, d.

2. A is isovolumetric, B is adiabatic, C is isothermal, D is isobaric.

3. (c). The highest theoretical effi ciency of an engine is the Carnot effi ciency given by e T TC c h= −1 . Hence, the theoretically possible effi ciencies of the given engines are

eA

K

1 000 K= − =1

7000 300. , eB

K

800 K= − =1

5000 375. , and eC

K

600 K= − =1

3000 500.

and the correct ranking (from highest to lowest) is C, B, A.

4. (b). ∆S Q Tr= = 0 and Q = 0 in an adiabatic process. If the process was reversible, but not adiabatic, the entropy of the system could undergo a nonzero change. However, in that case, the entropy of the system’s surroundings would undergo a change of equal magnitude but opposite sign, and the total change of entropy in the universe would be zero. If the process was irreversible, the total entropy of the universe would increase.

5. The number 7 is the most probable outcome because there are six ways this could occur: 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. The numbers 2 and 12 are the least probable because they could only occur one way each: either 1-1 or 6-6. Thus, you are six times more likely to throw a 7 than a 2 or 12.

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. The work done by the system on the environment is

W P Venv3 Pa m= + ( ) = ×( ) −( ) = − ×∆ 70 0 10 0 20 14 103 3. . J kJ= −14

and (c) is the correct choice.

2. When volume is constant, the work done on the gas is zero so the fi rst law of thermodynamics gives the change in internal energy as ∆U Q W= + = + =100 0 100 J J, and (d) is the correct answer for this question.

3. For a monatomic ideal gas, the internal energy is U nRT= 32 and the change in internal

energy is ∆ ∆U nR T= 32 ( ) . From the ideal gas law, PV nRT= , observe that

nR T nRT nRT P V PVf i f f i i( )∆ = − = − . When the pressure is constant, P P Pf i= = , this reduces to ∆ ∆U P V V P Vf i= − =3

232[ ( )] ( ), so the change in internal energy for this gas is

∆U = × + = ×32

5 52 00 10 1 50 4 50 10( ( ). . .Pa m J) 3 making (c) the correct answer.



4. The work an ideal gas does on the environment during an isothermal expansion is

W nRT V Vf ienv = ( )ln , so for the given process,

Wenv mol J mol K K= ×( ) ⋅( )( ) ( )3 9 10 8 31 850 22. . ln == ×1 9 106. J

and (a) is the correct choice.

5. For an adiabatic process, PV γ = constant, where γ = =c cp v 1 4. for diatomic gases (See Table 12.1

in the textbook.) Thus, P V PVf f i i1 4 1 4. .= , or the fi nal pressure will be

P PV

Vf ii

f

=⎛

⎝⎜⎞

⎠⎟= ×( )

1 4

51 00 101 00

3 50

.

..

. Pa

m3

m Pa3

⎛⎝⎜

⎞⎠⎟

= ×1 4

41 73 10.

.

and the correct response is (e).

6. In a cyclic process, the net work done equals the area enclosed by the process curve in a PV diagram. Thus,

Wnet

3Pa m= −( ) ×⎡⎣ ⎤⎦ −( )⎡⎣4 00 1 00 10 2 00 1 005. . . . ⎤⎤⎦+ −2 00 1 0. . 00 10 3 00 2 00 4 00 105 5( ) ×⎡⎣ ⎤⎦ −( )⎡⎣ ⎤⎦ = ×Pa m3. . . JJ

and (d) is the correct answer.

7. From conservation of energy, the energy input to the engine must be

Q W Qh c= + = + =eng kJ kJ kJ15 37 52

so the effi ciency is

eW

Qc

= = =eng kJ

52 kJor 29%

150 29.

and the correct choice is (b).

8. The coeffi cient of performance of this refrigerator is

COP kJ

18 kJ= = =

Q

Wc 115

6 4. which is choice (d).

9. The maximum theoretical effi ciency (the Carnot effi ciency) of a device operating between absolute temperatures T Tc h< is e T TC c h= −1 . For the given steam turbine, this is

eC = − × =13 0 10

0 332.

.K

450 Kor 33%

and (c) is the correct answer.

10. At a pressure of 1.0 atm, ice melts at absolute temperature TK ° K= + =0 273 15 273 15. . . The thermal energy this block of ice must absorb to fully melt is

Q mLr f= = ( ) ×( ) = ×1 00 3 33 10 3 33 105 5. . . kg J kg J

so the change in entropy of the ice is

∆SQ

Tr= = + × = + × = +

K

J

273.15 KJ K

3 33 101 22 10 1

53.

. ..22 kJ K

and (d) is the correct choice.


598 Chapter 12

11. In an ideal gas, the internal energy is directly proportional to the absolute temperature. Thus, in an isothermal process (constant temperature process) the internal energy of the ideal gas is con-stant. Choice (d) is the only true statement among the listed choices. In the compression, work is done on the gas while the internal energy is constant. The fi rst law of thermodynamics then says that energy must be transferred from the gas by heat. Also, by the ideal gas law, when the volume decreases while temperature is constant, the pressure must increase.

12. By defi nition, in an adiabatic process, no energy is transferred to or from the gas by heat. Thus, (c) is a true statement. All other choices are false. In an expansion process, the gas does work on the environment. Since, there is no energy input by heat, the fi rst law of thermodynamics says that the internal energy of the ideal gas must decrease, meaning the temperature will decrease. Also, in an adiabatic process, PV γ = constant meaning that the pressure must decrease as the volume increases in this adiabatic expansion.

13. In an isobaric process on an ideal gas, pressure is constant while the gas either expands or is compressed. Since the volume of the gas is changing, work is done either on or by the gas, so choice (b) is a true statement. Also, from the ideal gas law with pressure constant, P V nR T∆ ∆( ) = ( ). Thus, the gas must undergo a change in temperature having the same sign as the change in volume. If ∆V > 0, then both ∆ T and the change in the internal energy of the gas are positive ∆U >( )0 . However, when ∆V > 0, the work done on the gas is negative W <( )0 , and the fi rst law of thermodynamics says that there must be a positive transfer of energy by heat to the gas Q U W= − >( )∆ 0 . When ∆V < 0, a similar argument shows that ∆U <( )0 , W >( )0 , and Q U W= − <∆ 0. Thus, all of the other listed choices are false statements.

14. Choice (c) is a statement of the fi rst law of thermodynamics, not the second law, and hence is the correct answer to this question. Choices (a), (b), (d) and (e) are alternative statements of the second law, (a) being the Kelvin-Planck formulation, (b) the Carnot statement, (d) the Clausius statement, and (e) summarizes the primary consequence of all these various statements.

15. First, eliminate choice (d) since the work involved in an isovolumetric process is zero, and by defi nition, one cannot have an isovolumetric process when the volume is changing. In the other processes, the work done on the gas equals the area under the process curve in a PV diagram. In an isobaric process, the pressure is constant, so P Pf i= and the work done is the area under curve 1 in the sketch at the right. For an isother-mal process, the ideal gas law gives P V PVf f i i= ,

so P V V P Pf i f i i= =( ) 2 and the work done is the area under curve 2 in the sketch. Finally, for an

adiabatic process, P V PVf f i iγ γ= = constant, so P V V Pf i f i= ( )γ and P P Pf i i= >2 2γ since γ > 1

for all ideal gases (See Table 12.1 in the textbook). The work done in an adiabatic process is the area under curve 3, which exceeds that done in either of the other processes. Thus, the correct choice is (b), the adiabatic process involves the most work.

16. With this method of using an air conditioner, the average temperature in the room will increase, and choice (a) is the correct answer. The air conditioner operates on a cyclic process so the change in the internal energy of the refrigerant is zero. Then, the conservation of energy gives the thermal energy exhausted to the room as Q Q Wh c= + eng where Qc is the thermal energy the air conditioner removes from the room and Weng is the work done to operate the device. Since

Weng > 0, the air conditioner is returning more thermal energy to the room than it is removing, so the average temperature in the room will increase.

Vf � Vi�2

Pf � Pi

Pf � 2Pi

Pf � 2�Pi

1

2

3

Vi

V

P

Vf � Vi�2

Pf � Pi

Pf � 2Pi

Pf � 2�Pi

1

2

3

Vi

V

P



17. The correct choice is (d). The second law basically says that you must put in some work to trans-fer energy by heat from a lower-temperature to a higher-temperature location. But it can be very little work if the two temperatures are very nearly equal. Still, in the coeffi cient of performance

for a refrigerator, COP = Q Wc , W can never be zero and the result must be fi nite in all cases.

18. The Clausius statement of the second law of thermodynamics says that in any thermodynamics process, reversible or irreversible, the total entropy of the universe must either remain constant (reversible process) or increase (irreversible process). Thus, if in a thermodynamics process, the entropy of a system changes by −6 J K, the entropy of the environment (i.e., the rest of the universe) must increase by +6 J K or more. The correct choice here is (e).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. Shaking opens up spaces between the jelly beans. The smaller ones have a chance of falling down into spaces below them. The accumulation of larger ones on top and smaller ones on the bottom implies an increase in order and a decrease in one contribution to the total entropy. However, the second law is not violated and the total entropy of the system increases. The increase in the inter-nal energy of the system comes from the work required to shake the jar of beans (that is, work your muscles must do, with an increase in entropy accompanying the biological process) and also from the small loss of gravitational potential energy as the beans settle together more compactly.

4. Temperature = A measure of molecular motion. Heat = the process through which energy is trans-ferred between objects by means of random collisions of molecules. Internal energy = energy associated with random molecular motions plus chemical energy, strain potential energy, and an object’s other energy not associated with center of mass motion or location.

6. A higher steam temperature means that more energy can be extracted from the steam. For a con-stant temperature heat sink at Tc and steam at Th, the maximum effi ciency of the power plant goes as

T T

T

T

Th c

h

c

h

− = −1

and is maximized for high Th.

8. Assuming an air temperature of 20°C above the surface of the pond, the difference in temperature between the lower layer of the pond and the atmosphere is

∆T T Th c= − = +( ) − +( ) =100 273 20 273 80° ° K

and the Carnot effi ciency is

e eT T

TCh c

hmax

K

373 K= = − = ≈80

22%

10. Even at essentially constant temperature, energy must be transferred by heat out of the solidifying sugar into the surroundings. This action will increase the entropy of the environment. The water molecules become less ordered as they leave the liquid in the container to mix with the entire atmosphere.

12. A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the fl oor and shatters. Any process is irreversible if it looks funny or frightening when shown in a videotape running backward. At fairly low speeds, air resistance is small and the fl ight of a projectile is nearly reversible.


600 Chapter 12

PROBLEM SOLUTIONS

12.1 (a) The work done on the gas is

W P V= − ( ) = − ×( ) −( ) =∆ 1 50 10 0 250 0 7505. . . Pa m m2 2 77 50 104. × J

(b) The work done by the gas on the environment is

W P Venv

2 2 Pa m m= + ( ) = ×( ) −(∆ 1 50 10 0 250 0 7505. . . )) = − ×7 50 104. J

(c) The work done on the gas is the negative of the work the gas does on the environment because of Newton’s third law . When the environment exerts a force on the gas, the gas exerts an equal magnitude force in the opposite direction on the environment.

12.2 (a) W P V Vab a b a= −( )

= ×( )⎡⎣ ⎤⎦ −3 1 013 10 3 0 1 05. . .Pa L LLm

L

J

3

( )⎛⎝⎜

⎞⎠⎟

=

−10

1

610

3

(b) W P V Vbc c b= −( ) = 0

(c) W P V Vcd c d c= −( )= ×( )⎡⎣ ⎤⎦ −2 1 013 10 1 0 3 05. . .Pa L LL

m

L

J

3

( )⎛⎝⎜

⎞⎠⎟

= −

−10

1

410

3

(d) W P V Vda a d= −( ) = 0

(e) W W W W Wab bc cd danet J J= + + + = + + − + = +610 0 410 0 200 JJ

12.3 The constant pressure is P = ( ) ×( ) = ×1 5 1 013 10 1 52 105 5. . . atm Pa atm Pa and the work done on the gas is W P V= − ( )∆ .

(a) ∆V = + 4 0. m3 and

W P V= − ( ) = − ×( ) +( ) = − ×∆ 1 52 10 4 0 6 1 105 5. . . Pa m J3

(b) ∆V = − 3 0. m3, so

W P V= − ( ) = − ×( ) −( ) = + ×∆ 1 52 10 3 0 4 6 105 5. . . Pa m J3

P (atm)

V (L)

3a b

cd

2

1

03210

P (atm)

V (L)

3a b

cd

2

1

03210



12.4 (a) The work done by the gas on the projectile is given by the area under the curve in the PV diagram. This is

Wby gas triangular area rectangular area= ( ) + ( ))

= −( ) −( ) + −( ) = +( ) −1

2

1

20 0 0 0P P V V P V V P P V Vf f f f f f 00

51

211 1 0 10 40 0 8 0

( )

= +( ) ×⎡⎣ ⎤⎦ −( )⎡⎣ ⎤. . . Pa cm3⎦⎦

⎛⎝⎜

⎞⎠⎟

=1

10196

m

cm J

3

3

From the work-energy theorem, W KE m= = −∆ 12

2 0v where W is the work done on the projectile by the gas. Thus, the speed of the emerging projectile is

v = = ( )

×=−

2 2 19

40 0 10313

W

m

J

kgm s

.

(b) The air in front of the projectile would exert a retarding force of

F P Ar = = ×( ) ( )( )air2 2 2Pa cm m cm1 0 10 1 0 1 105 4. .⎡⎡⎣ ⎤⎦ = 10 N

on the projectile as it moves down the launch tube. The energy spent overcoming this retarding force would be

W F srspent N m J= ⋅ = ( )( ) =10 0 32 3 2. .

and the needed fraction is

W

Wspent J

19 J= =3 2

0 17.

.

12.5 In each case, the work done on the gas is given by the negative of the area under the path on the PV diagram. Along those parts of the path where volume is constant, no work is done. Note that 1 1 013 10 105 3 atm Pa and 1 Liter m3= × = −. .

(a) W W W P V VIAF IA AF I A I= + = − −( ) +

= − ×

0

4 00 1 013 105. . Pa(( )⎡⎣ ⎤⎦ −( ) ×⎡⎣ ⎤⎦ = −−4 00 2 00 10 810. . 3 3 m J

(b) WIF = − ( ) − ( )

= −

triangular area rectangular area

11

2

1

2P P V V P V V P P V VI B F B B F B I B F B−( ) −( ) − −( ) = − +( ) −(( )

= − +( ) ×( )⎡⎣ ⎤⎦ −1

24 00 1 00 1 013 10 4 00 25. . . . . Pa 000 10

507

3( ) ×

= −

− m

J

3

(c) W W W P V VIBF IB BF B F I= + = − −( )

= − ×(

0

1 00 1 013 105. . Pa))⎡⎣ ⎤⎦ −( ) ×⎡⎣ ⎤⎦ = −−4 00 2 00 10 203. . 3 3 m J


602 Chapter 12

12.6 The sketches for (a) and (b) are shown below:

(a) (b)

P

V V

P

P2

P1

V1 V2 V1 V2

P2

P1

(c) As seen from the areas under the paths in the PV diagrams above, the higher pressure during the expansion phase of the process results in more work done by the gas in (a) than in (b).

12.7 With P P Pf i= = , the ideal gas law gives P V PV P V nR Tf f i i− = ( ) = ( )∆ ∆ , so the work done by the gas is

W P V nR Tm

MR T Tf ienv

He

= + ( ) = ( ) =⎛⎝⎜

⎞⎠⎟

−( )∆ ∆

If Wenv J= 20 0. , the mass of helium in the gas sample is

mW M

R T Tf i

=( )

−( ) =( )( )env He J g mol20 0 4 00

8 31

. .

. J mol K K K g mg

⋅( ) −( ) = =373 273

0 0963 96 3. .

12.8 (a) The initial absolute temperature is Ti = × + =1 50 10 273 15 4232. . K, and the ideal gas law with P Pf i= = 2 00. atm yields

P V

P V

nRT

nRT

f f

i i

f

i

= or TV

VTf

f

ii=

⎛⎝⎜

⎞⎠⎟

= ( ) =2

3423 282 k K

(b) The work done on the gas is

W P V PV P nRT

P

nRTi i i= − ( ) = − −⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

=∆3 3 3

or

W =( ) ⋅( )( )

= × =1 8 31 423

1 17 10 13 mol J mol K K

3 J

.. ..17 kJ

12.9 (a) From the ideal gas law, nR PV T PV Tf f i i= = . With pressure constant this gives

T TV

Vf if

i

=⎛⎝⎜

⎞⎠⎟

= ( )( ) = ×273 4 1 09 103 K K.

(b) The work done on the gas is

W P V PV PV nR T Tf i f i= − ( ) = − −( ) = − −( )

= − ( )

∆

1 00 8. mol ..

. .

31 1 092 273

6 81 10 63

J mol K J K

J

⋅( ) −( )

= − × = − 881 kJ



12.10 (a) The work done on the fl uid is the negative since V Vf i>( ) of the area under the curve on the PV diagram. Thus,

Wif = − ×( ) −( ){ 6 00 10 2 00 1 006. . . Pa m

3

Pa+ −( ) ×⎡⎣ ⎤⎦ −( )1

26 00 2 00 10 2 00 1 006. . . . m

3

+ ×2 00. 110 4 00 2 006 Pa m3( ) −( ) }. .

Wif = − × = −1 20 10 12 07. . J MJ

(b) When the system follows the process curve in the reverse direction with V Vf i<( ), the work done on the fl uid equals the area under the process curve, which is the negative of that computed in (a). Thus,

W Wfi if= − = + 12 0. MJ

12.11 From kinetic theory, the average kinetic energy per molecule is

KE k T

R

NTmolecule B

A

= =⎛⎝⎜

⎞⎠⎟

3

2

3

2

For a monatomic ideal gas containing N molecules, the total energy associated with random molecular motions is

U N KE

N

NRT nRT= ⋅ =

⎛⎝⎜

⎞⎠⎟

=molecule

A

3

2

3

2

Since PV nRT= for an ideal gas, the internal energy of a monatomic ideal gas is found to be given by U PV= ( )3 2 .

12.12 (a) The initial absolute temperature is Ti = + =20 0 273 15 293. .° K, so the initial pressure is

PnRT

Vii

i

= =( ) ⋅( )( )10 0 8 31 293. . mol J mol K K

0.3000 m K3 = ×8 12 104.

(b) For a monatomic ideal gas, the internal energy is U nRT= 3 2. Thus,

U nRTi i= = ( ) ⋅( )( ) =3

2

3

210 0 8 31 293. . mol J mol K K 33 65 104. × J

(c) The work done on the gas in this isobaric expansion is

W P V= − ( ) = − ×( ) −( ) =∆ 8 12 10 1 000 0 3004. . . Pa m m3 3 −− ×5 68 104. J

(d) TP V

nRff f= =

×( )( )( )

8 12 10 1 00

10 0 8

4. .

.

Pa m

mol

3

..31977

J mol K K

⋅( ) =

(e) U nRTf f= = ( ) ⋅( )( ) =3

2

3

210 0 8 31 977. . mol J mol K K 11 22 105. × J

continued on next page


604 Chapter 12

(f ) ∆U U Uf i= − = × − × = ×1 22 10 3 65 10 105 4. . J J 8.55 J4

(g) ∆U W− = × − − ×( ) = + ×8.55 J J J410 5 68 10 1 42 104 5. .

(h) Since ∆U W− > 0, the increase in the internal energy of the gas exceeds the energy

transferred to the gas by work. Thus, additional energy must be transferred to thee gas by heat.

(i) The additional energy that must be transferred to the gas by heat is

∆U W− = + ×1 42 105. J [See part (g) above]

( j) The suggested relationship is Q U W= −∆ , which is a statement of the fi rst law of thermodynamics.

12.13 (a) Along the direct path IF, the work done on the gas is

W area under curve= − ( )

= − ( ) −

atm L1 00 4 00 2 0. . . 001

24 00 1 00 4 00 2 00 L atm atm L L( ) + −( ) −( ). . . .⎡⎡

⎣⎢⎤⎦⎥

= − ⋅( ) ×⎛⎝⎜

⎞W 5.00 atm L

1.013 10 Pa

1 atm

5

⎠⎠⎟⎛⎝⎜

⎞⎠⎟

= −−10

506 53 m

1 L J

3

.

Thus, ∆U Q W= + = − = −418 506 5 88 5J J J. . .

(b) Along path IAF, the work done on the gas is

W = − ( ) −( ) ×4 4 00 2 00.00 atm L L

1.013 10 Pa

1 a

5

. .ttm

m

1 L J

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= −−10

8103

From the fi rst law, Q U W= − = − − −( ) =∆ 88 5 810 722. J J J .

12.14 From the fi rst law,

Q U W= − = − − = −∆ 500 220 720 J J J

The negative sign in the result means that energy is transferred from the system by heat.

12.15 (a) W P V= − ( ) = − ( ) −( ) ×∆ 0 800 7 00. . atm L1.013 10 Pa

1

5

atm

m

1 L J

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=−10

5673

(b) ∆U Q W= + = − + =400 167J 567 J J



12.16 The work done on the gas is the negative since V Vf i>( ) of the area under the curve on the PV diagram, or

W P V V P P V V P= − −( ) + −( ) −( )⎡

⎣⎢⎤⎦⎥

= −0 0 0 0 0 0 021

22 2

3

2 00 0V

From the result of problem 11,

∆U P V PV P V P V P Vf f i i= − = ( )( ) − =3

2

3

2

3

22 2

3

2

9

20 0 0 0 0 0

Thus, from the fi rst law,

Q U W P V P V P V= − = − −⎛⎝⎜

⎞⎠⎟ =∆ 9

2

3

260 0 0 0 0 0

12.17 (a) The change in the volume occupied by the gas is

∆V V V A L Lf i f i= − = −( ) = ( ) −( ) = −0 150 0 200 3 0. . . m m2 00 10 2× − m3

and the work done by the gas is

W P Vby gas

3 Pa m= + ( ) = ( ) − ×( ) = −−∆ 6 000 3 00 10 182. 00 J

(b) The fi rst law of thermodynamics is ∆U Q W Q W= + = − −input on gas output by gas. Thus, if ∆U = −8 00. J, the energy transferred out of the system by heat is

Q U Woutput by gas J J= − − = − −( ) − −( ) = +∆ 8 00 180 18. 88 J

12.18 The work done on the gas is the negative since V Vf i>( ) of the area under the curve on the PV diagram, so

W P V V P P V V P= − −( ) + −( ) −( )⎡⎣⎢

⎤⎦⎥

= +0 0 0 0 0 0 021

22 2

3

2 00 0V , or W > 0

From the result of problem 11,

∆U P V PV P V P Vf f i i= − = ( )( ) − ( )( ) =3

2

3

2

3

22

3

220 0 0 0 0

Then, from the fi rst law, Q U W P V P V= − = − = −∆ 0 32 0 0

32 0 0, or Q < 0 .

12.19 (a) The change in the internal energy of an ideal gas is ∆ ∆U nC Tv= ( ). Thus, for this isothermal process, ∆U = 0 .

(b) From the fi rst law of thermodynamics, the energy transfer by heat is Q U W= −∆ . Therefore, if the work done on the gas in this isothermal process is W = 75 J, we have

Q U W= − = − = −∆ 0 75 75 J J


606 Chapter 12

12.20 (a) Since the process is adiabatic, the energy transfer by heat is Q = 0 .

(b) If the work done on the gas is W = +125 J, the fi rst law of thermodynamics gives

∆U Q W= + = + =0 125 125 J J

12.21 (a) In an isothermal process involving an ideal gas, the work done on the gas is W W nRT V Vf i= − = −env ln ( ). But, when temperature is constant, the ideal gas law gives

PV P V nRTi i f f= = and we may write the work done on the gas as

W PV

V

Vi if

i

= −⎛⎝⎜

⎞⎠⎟

= − ×( )(ln . .1 00 10 0 5005 Pa m3 )) ⎛⎝⎜

⎞⎠⎟

= − ×ln.

..

1 25

0 5004 58 104m

mJ

3

3

(b) The change in the internal energy of an ideal gas is ∆ ∆U nC T= ( )v , and for an isothermal process, we have ∆U = 0. Thus, from the fi rst law of thermodynamics, the energy transfer by heat in this isothermal expansion is

Q U W= − = − − ×( ) = + ×∆ 0 4 58 10 4 58 104 4. .J J

(c) ∆U = 0 [See part (b) above.]

12.22 (a) From the ideal gas law, PV nRTi i i= and P V nRTf f f= . Thus, if P P Pi f= = , subtracting

these two expressions gives PV PV nRT nRTf i f i− = − , or

P V nR T∆ ∆( ) = ( )

(b) For a monatomic, ideal gas containing N gas atoms, the internal energy is U N m nN k T nRTA B= = =( ) ( ) ( )1

22 3

232v . Thus, the change in internal energy of this gas in a

thermodynamic process is ∆ ∆U nR T= 32 ( ). But, using the result of part (a) above, we have,

for an isobaric process involving an monatomic ideal gas,

∆ ∆ ∆U nR T P V W= ( ) = ( ) =3

2

3

2

3

2 env

(c) We recall that the work done on the gas is W W= − env, and use the fi rst law of thermodynamics to fi nd that the energy transferred to the gas by heat to be

Q U W U W W W= − = + = +∆ ∆ env env env

3

2 or Q W= 5

2 env

(d) In an isobaric expansion ⇒ >( ) ∆V 0 , the work done on the environment is W P Venv = ( ) >∆ 0. Thus, from the result of part (c) above, the energy transfer as heat is Q > 0, meaning that the energy fl ow is into the gas. Therefore, it is

impossible for the gas to exhaust thermal ennergy in an isobaric expansion



12.23 (a) W P V= − ( )

= − ×( ) −( )∆

1 013 10 1 09 1 005. . . Pa cm cm3 3 11

109 12 106

3 m

cm J

3

3

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = − × −.

(b) To freeze the water, the required energy transfer by heat is

Q mL f= − = − ×( ) ×( ) = −−1 00 10 3 33 10 3333 5. . kg J kg J

The fi rst law then gives

∆U Q W= + = − − × = −−333 333J 9.12 10 J J3

12.24 Volume is constant in process BC, so WBC = 0 . Given that QBC < 0 , the fi rst law shows that

∆U Q W QBC BC BC BC= + = + 0. Thus, ∆UBC < 0 .

For process CA, ∆V V VCA A C= − < 0, so W P V= − ( )∆ shows that WCA > 0 . Then, given that

∆UCA < 0 , the fi rst law gives Q U WCA CA CA= −∆ and QCA < 0 .

In process AB, the work done on the system is W AB= − ( )area under curve where

area under curve AB P V V P P VA B A B A B( ) = −( ) + −( )1

2−−( ) >VA 0

Hence, WAB < 0 . For the cyclic process, ∆ ∆ ∆ ∆U U U UAB BC CA= + + = 0, so,

∆ ∆ ∆U U UAB BC CA= − +( ). This gives ∆UAB > 0 , since both ∆ ∆U UBC CAand are negative.

Finally, from the fi rst law, Q U W= −∆ shows that QAB > 0 since both ∆UAB ABand − W are positive.

12.25 (a) The original volume of the aluminum is

Vm

035 0

101 85 10= =

×= × −

ρ.

. kg

2.70 kg m m3 3

3

and the change in volume is ∆ ∆ ∆V V T V T= ( ) = ( ) ( )β α0 03 , or

∆V = × ( )⎡⎣ ⎤⎦ ×( )( ) =− − −3 24 10 1 85 10 70 9 36 1 3°C °C. . ×× −10 6 m3

The work done by the aluminum is then

W P Vby system3 Pa m= + ( ) = ×( ) × −∆ 1 013 10 9 3 105 6. .(( ) = 0 95. J

(b) The energy transferred by heat to the aluminum is

Q mc T= ( ) = ( ) ⋅( )( ) = ×Al kg J kg °C °C∆ 5 0 900 70 3 2 1. . 005 J

(c) The work done on the aluminum is W W= − = −by system J0 95. , so the fi rst law gives

∆U Q W= + = × − = ×3 2 10 0 95 3 2 105 5. . .J J J


608 Chapter 12

12.26 (a) The work done on the gas in each process is the negative of the area under the process curve on the PV diagram.

For path IAF, W W W WIAF IA AF AF= + = +0 , or

WIAF = − ( ) ×⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥

1 50 1 013 10 05. . .atmPa

atm5500

10

1

76 0

3

Lm

L

J

3

( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= −

−

.

For path IBF, W W W WIBF IB BF IB= + = + 0, or

WIBF = − ( ) ×⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

2 00 1 013 105. . atm Pa

atm00 500

10

101

3

. L m

1 L

J

3

( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= −

−

For path IF, W W triangular areaIF AF= − ( ), or

WIF = − − ( ) ×⎛

⎝⎜⎞76 0

1

20 500 10. . J m 1.013

Pa

atm3 5

⎠⎠⎟⎡⎣⎢

⎤⎦⎥

( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= −

−

0 50010

1

8

3

. L m

L

3

88 7. J

(b) Using the fi rst law, with ∆U U UF A= − = −( ) =180 91 0 89 0. .J J, for each process gives

Q U WIAF IAF= − = − −( ) =∆ 89 0 76 0 165. .J J J

Q U WIBF IBF= − = − −( ) =∆ 89 0 101 190. J J J

Q U WIF IF= − = − −( ) =∆ 89 0 88 7 178. . J J J



12.27 (a) For adiabatic processes in ideal gases, P V PVf f i iγ γ= = constant. From the ideal gas law,

P nRT V= , so the above expression becomes

nRT

VV

nRT

VVf

ff

i

ii

⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

γ γ or T V T Vf f i iγ γ− −=1 1

which can be summarized as TV Cγ − =1 , where C is a constant.

(b) The current radius of the universe is assumed to be rf = ×1 4 1026. m and the temperature is Tf = 2 7. K. Since γ = =1 67 5 3. for monatomic ideal gases, the temperature Ti of the universe when its radius was ri = = × −2 0 2 0 10 2. .cm m must have been

T T

V

VT

V

VTi f

f

if

f

if=

⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

=−

−

−γ

γ

γ π1

1

1 43 rr

rT

r

rf

if

f

i

3

43

3

1 3 3

π

γ γ⎛

⎝⎜

⎞

⎠⎟ =

⎛⎝⎜

⎞⎠⎟

− −

or

Ti = ( ) ××

⎛⎝⎜

⎞⎠⎟−

( )−

2 71 4 10

2 0 10

26

2

3 5 3 3

..

.K

m

m== ( ) ×( ) = ×2 7 7 0 10 1 3 1027 2 0 56. . .

.K K

12.28 (a) The number of atoms per mole in any monatomic gas is Avogadro’s number NA = ×6 02 1023. atoms mol. Thus, if the density of gas in the Universe is 1 hydrogen atom per cubic meter, the number of moles per unit volume is

n

V NA

= ==×

=1 1

6 02 10123

atom m atom m

atoms mol

3 3

...66 10 24× − mol m3

(b) With the density of gas found in part (a) and an absolute temperature of T = 2 7. K, the ideal gas law gives the pressure of the Universe as

Pn

VRT= ⎛

⎝⎞⎠ = ×( ) ⋅( )−1 66 10 8 31 224. . .mol m J mol K3 77 3 7 10 23K Pa( ) = × −.

(c) For an adiabatic expansion, PV P Vi i f fγ γ= with γ = =1 67 5 3. for monatomic ideal gases,

so the initial pressure of the Universe is estimated to be

P PV

VP

V

VP

ri f

f

if

f

if

f=⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

=γ

γ

γ ππ

43

3

43 rr

Pr

rff

f

i3

3⎛

⎝⎜

⎞

⎠⎟ =

⎛⎝⎜

⎞⎠⎟

γ γ

or

Pi = ×( ) ××

⎛⎝⎜

⎞⎠⎟

−−3 7 10

1 4 10

2 0 1023

26

2..

.Pa

m

m

33 5 3

23 27 5 03 7 10 7 0 10

( )−= ×( ) ×( ). .

.Pa

giving

Pi = ×( )( )⎡⎣ ⎤⎦( ) = ×−3 7 10 7 0 10 3 723 5 0 27 5 0. . .. .

Pa 110 1 7 10 1023 4 135−( ) ×( )⎡⎣ ⎤⎦( )Pa .

and

Pi = ×⎡⎣ ⎤⎦( ) = ×−6 3 10 10 6 3 1019 135 116. .Pa Pa


610 Chapter 12

12.29 The net work done by a heat engine operating on the cyclic process shown in Figure P12.29 equals the triangular area enclosed by this process curve. Thus,

Wnet3 3atm atm m m= −( ) −( )1

26 00 2 00 3 00 1 00. . . .

== ⋅ ×⎛⎝⎜

⎞⎠⎟

= ×4 001 013 10

4 05 15

..

.atm mPa

1 atm3 00

405 10 405

5

3

J

J kJ= × =

12.30 The net work done by a heat engine operating on the cyclic process shown in Figure P12.30 equals the triangular area enclosed by this process curve. This is,

W base altitudenet

3m

= ( )( )

= −( )⎡⎣ ⎤

1

21

24 00 1 00. . ⎦⎦ −( ) ×⎡⎣ ⎤⎦

= ( ) ×

6 00 2 00 10

1

23 00 4 00 10

5. .

. .

Pa

m3 55 56 00 10Pa J( ) = ×.

12.31 The maximum possible effi ciency for a heat engine operating between reservoirs with absolute temperatures of Tc = ° + =25 273 298 K and Th = ° + =375 273 648 K is the Carnot effi ciency

e

T

TCc

h

= − = − =1 1298

0 540K

648 Kor 54.0%.

12.32 (a) The absolute temperature of the cold reservoir is Tc = ° + =20 273 293 K. If the Carnot effi ciency is to be eC = 0 65. , it is necessary that

1 0 65− =T

Tc

h

. or T

Tc

h

= 0 35. and TT

hc=

0 35.

Thus,

Th = =293

0 35837

KK

. or Th = − =837 564273 °C

(b) No. Any real heat engine will have an effi ciency less that the Carnot effi ciency because it operates in an irreversible manner.

12.33 (a) The effi ciency of a heat engine is e W Qh= env , where Wenv is the work done by the engine and Qh is the energy absorbed from the higher temperature reservoir. Thus, if W Qhenv = 4, the effi ciency is e = =1 4 0 25. or 25% .

(b) From conservation of energy, the energy exhausted to the lower temperature reservoir is Q Q Wc h= − env. Therefore, if W Qhenv = 4, we have Q Qc h= 3 4 or Q Qc h = 3 4 .

V (m3)

6.00

P (atm)

4.00

2.00

1.00 2.00 3.00V (m3)

6.00

P (atm)

4.00

2.00

1.00 2.00 3.00

V (m3)

6.00

P (105 Pa)

4.00

2.00

1.00 2.00 3.00 4.00V (m3)

6.00

P (105 Pa)

4.00

2.00

1.00 2.00 3.00 4.00



12.34 (a) From e W Q Q Qh c h≡ = −eng 1 , the energy intake each cycle is

QQ

ehc=

−=

−= =

1

8 00010 667 10 7

J

1 0.250J kJ.

(b) From P = =W t e Q tceng , the time for one cycle is

t

e Qc= =( )( )

×=

P0 250 10 667

100 533

..

J

5.00 Ws3

12.35 (a) The maximum effi ciency possible is that of a Carnot engine operating between the specifi ed reservoirs.

e

T T

T

T

TCh c

h

c

h

= − = − = − =1 17

0 67203 K

2 143 Kor 6. 77.2%( )

(b) From eW

Qh

= eng , we fi nd W e Qheng J J= = ×( ) = ×0 420 1 40 10 5 88 105 4. . . , so

P = = × = × =W

teng J

1.00 sW kW

5 88 105 88 10 58 8

44.

. .

12.36 The work done by the engine equals the change in the kinetic energy of the bullet, or

W mb feng kg m s= − = ×( )( ) =−1

20

1

22 40 10 320 122 3 2

v . 33 J

Since the effi ciency of an engine may be written as

eW

Q

W

W Qh c

= =+

eng eng

eng

where Qc is the exhaust energy from the engine, we fi nd that

Q W

ec = −⎛⎝⎜

⎞⎠⎟ = ( ) −⎛

⎝⎜⎞⎠⎟ =eng J

11 123

1

0 1101 1 1

.. 00 104× J

This exhaust energy is absorbed by the 1.80-kg iron body of the gun, so the rise in temperature is

∆TQ

m cc= = ×

( ) ⋅gun iron

J

1.80 kg J kg °

1 10 10

448

4.

CC°C( ) = 13 7.

12.37 (a) eW

Q

Q

Qh

c

h

≡ = − = − =eng J

1 700 Jor 29.41 1

1 2000 294. %%( )

(b) W Q Qh ceng J J J= − = − =1 700 1 200 500

(c) P = = = × =W

teng J

0.300 sW kW

5001 67 10 1 673. .


612 Chapter 12

12.38 (a) The coeffi cient of performance of a heat pump is COP = Q Wh , where Qh is the thermal energy delivered to the warm space and W is the work input required to operate the heat pump. Therefore,

Q W th = ⋅ = ⋅( ) ⋅ = ×⎛⎝⎜

⎞⎠⎟

COP COPJ

shP ∆ 7 03 10 8 003. .(( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = ×3 600

3 80 7 69 108s

1 hJ. .

(b) The energy extracted from the cold space (outside air) is

Q Q W Q

QQc h h

hh= − = − = −⎛

⎝⎞⎠COP COP

11

or

Qc = ×( ) −⎛⎝

⎞⎠ = ×7 69 10 1

15 67 108 8. .J

3.80J

12.39 (a) W t= ⋅ =⎛

⎝⎜⎞

⎠⎟×⎛

⎝⎜⎞⎠⎟

P ∆ 4573 60 106kWh

y

J

1 kWh

. 11 y

365.242 dd J

⎛

⎝⎜⎞

⎠⎟⋅ ( ) = ×1 4 50 106.

(b) From the defi nition of the coeffi cient of performance for a refrigerator, COP R( ) = Q Wc , the thermal energy removed from the cold space each day is

Q Wc = ( ) ⋅ = ×( ) = ×COP J JR 6 30 4 50 10 2 84 106 7. . .

(c) The water must be cooled 20.0°C before it will start to freeze, so the thermal energy that must be removed from mass m of water to freeze it is Q mc T mLc w f= +∆ . The mass of water that can be frozen each day is then

mQ

c T Lc

w f

=+

= ×⋅( )( )∆

2 84 10

4 186 20 0

7.

.

J

J kg °C °C ++ ×=

3 33 1068 25.

.J kg

kg

12.40 (a) The coeffi cient of performance of a heat pump is defi ned as

COP hp( ) = =−

=−

Q

W

Q

Q Q Q Qh h

h c c h

1

1

But when a device operates on the Carnot cycle, Q Q T Tc h c h= . Thus, the coeffi cient of performance for a Carnot heat pump would be

COP hp,C( ) =−

=−

1

1 T T

T

T Tc h

h

h c

(b) From the result of part (a) above, we observe that the COP of a Carnot heat pump would increase if the temperature difference T Th c− became smaller.

(c) If T Tc h= ° + = = ° + =50 273 323 70 273 343K and K, the COP of a Carnot heat pump would be

COPK

343 K Khp,C( ) =−

=−

=T

T Th

h c

343

32317 2.



12.41 The actual effi ciency of the engine is

eQ

Qc

h

= − = − =1 13

0 40000 J

500 J.

If this is 60.0% of the Carnot effi ciency, then

ee

C = = =0 600

0 400

0 600

2

3.

.

.

Thus, from e T TC c h= −1 , we fi nd

T

Tec

hC= − = − =1 1

2

3

1

3

12.42 (a) The Carnot effi ciency represents the maximum possible effi ciency. With T Th c= = = =20 0 293 5 00 278. .°C K and °C K, this effi ciency is given by

eT

TCc

h

= − = − = ( )1 1278

2930 0512

K

K or 5.12%.

(b) The effi ciency of an engine is e W Qh= eng , so the minimum energy input by heat each hour is

QW

e

t

eh minmax max

.= = ⋅ =

×( )eng J s sP ∆ 75 0 10 3 6006 (( )= ×

0 05125 27 1012

.. J

(c) As fossil-fuel prices rise, this could be an attractive way to use solar energy. However, the potential environmental impact of such an engine would require serious study. The energy output, Q Q W Q ec h h= − = −( )eng 1 , to the low temperature reservoir (cool water deep in the ocean) could raise the temperature of over a million cubic meters of water by 1°C every hour.

12.43 From P = =W t e Q theng , the energy input by heat in time t is Qt

eh = ⋅P

Thus, from e Q Q Qh c h= − , the energy expelled in time t is

Q Q et

ee t

ec h= −( ) = ⋅⎛⎝⎜

⎞⎠⎟ −( ) = ⋅ −⎛

⎝⎜⎞⎠⎟1 1

11

P P

In time t, the mass of cooling water used is m t= ×( ) ⋅1 0 106. kg s , and its rise in temperature is

∆TQ

mc

t

t c ec= = ⋅

×( ) ⋅ ⋅−⎛

⎝⎜⎞⎠⎟

=

P1 0 10

11

1 000

6. kg s

××( )×( ) ⋅( ) −

10

1 0 10 4 186

1

0 331

6

6

J s

kg s J kg °C. .⎛⎛⎝⎜

⎞⎠⎟

or

∆T = 0 49. °C


614 Chapter 12

12.44 (a) The Carnot effi ciency is

eT

TCc

h

= − = − =1 1353

0 433K

623 K.

so the maximum power output is

Pmax

eng max kJ

1.00 s=

( )= =

( ) =W

t

e Q

tC h 0 433 21 0. .

99 10. kW

(b) From e Q Qc h= −1 , the energy expelled by heat each cycle is

Q Q ec h= −( ) = ( ) −( ) =1 21 0 1 0 433 11 9. . .kJ kJ

12.45 The thermal energy transferred to the room by the water as the water cools from 1 00 102. × °C to 20.0°C is

Q mc Tw= = ( ) ⋅( )( ) = ×∆ 0 120 4 186 80 4 0 1. .kg J kg °C °C 004 J

If the room has a constant absolute temperature of T = ° + =20 0 273 293. K, the increase in the entropy of the room is

∆S

Q

T= = × =4 0 10

1404. J

293 KJ K

12.46 The total momentum before collision is zero, so the combined mass must be at rest after the collision. The energy dissipated by heat equals the total initial kinetic energy,

Q m= ⎛

⎝⎞⎠ = ( )( ) = × =2

1

22 000 20 8 0 10 802 2 5v kg m s J. 00 kJ

With the environment at an absolute temperature of T = + =23 273 296° K, the change in entropy is

∆ ∆S

Q

Tr= = =800

2 7kJ

296 KkJ K.

12.47 The energy transferred from the water by heat, and absorbed by the freezer, is

Q mL V Lf f= = ( ) = ( ) ×( )⎡⎣ ⎤⎦−ρ 10 1 0 10 3 33 3kg m m3 3. . 33 10 3 3 105 5×

⎛⎝⎜

⎞⎠⎟

= ×J

kgJ.

Thus, the change in entropy of the water is

(a) ∆∆

SQ

Tr

waterwater J

273 K=

( )= − × = − ×3 3 10

1 2 105.

. 33 1 2J

KkJ K= − .

and that of the freezer is

(b) ∆∆

SQ

Tr

freezerfreezer J

273 K=

( )= + × = +3 3 10

15.

.22 kJ K



12.48 The energy added to the water by heat is

∆Q mLr = = ( ) ×( ) = ×v 1 00 2 26 10 2 26 106 6. . .kg J kg J

so the change in entropy is

∆ ∆S

Q

Tr= = × = × =2 26 10

6 06 10 6 066

3.. .

J

373 K

J

KkJ KK

12.49 The potential energy lost by the log is transferred away by heat, so

Q mgh= = ( )( )( ) = ×70 9 80 25 1 7 104kg m s m J2. .

and the change in entropy is ∆ ∆S

Q

Tr= = × =1 7 10

574. J

300 KJ K

12.50 (a) In a game of dice, there is only one way you can roll a 12. You must have a 6 on each die.

(b) There are six ways to obtain a 7 with a pair of dice. The combinations that yield a 7 are1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1. Note that 1 + 6 and 6 + 1 are different combina-tions in that the 6 occurs on different members of the pair of dice in the two combinations.

12.51 A quantity of energy, of magnitude Q, is transferred from the Sun and added to Earth. Thus,

∆SQ

TSunSun

= − and ∆S

Q

TEarthEarth

= +

so the total change in entropy is

∆ ∆ ∆S S SQ

T

Q

Ttotal Earth SunEarth Sun

J

= + = −

= (1 000 )) −⎛⎝⎜

⎞⎠⎟

=1

290

1

5 7003 27

K KJ K.

12.52 The change in entropy of a reservoir is ∆S Q Tr= , where Qr is the energy absorbed Qr >( )0 or

expelled Qr <( )0 by the reservoir, and T is the absolute temperature of the reservoir.

(a) For the hot reservoir:

∆Sh = − × = −2 50 103 45

3..

J

725 KJ K

(b) For the cold reservoir:

∆Sc = + × = +2 50 108 06

3..

J

310 KJ K

(c) For the Universe:

∆ ∆ ∆S S SU h c= + = − + = +3 45 8 06 4 61. . .J K J K J K

(d) The magnitudes of the thermal energy transfers, appearing in the numerators, are the same for the two reservoirs, but the cold reservoir necessarily has a smaller denominator. Hence, its positive change dominates.


616 Chapter 12

12.53 (a) The table is shown below. On the basis of the table, the most probable result of a toss is

2 H and 2 T .

End Result Possible TossesTotal Number of

Same Result

All H HHHH 1

1T, 3H HHHT, HHTH, HTHH, THHH 4

2T, 2H HHTT, HTHT, THHT, HTTH, THTH, TTHH 6

3T, 1H TTTH, TTHT, THTT, HTTT 4

All T TTTT 1

(b) The most ordered state is the least likely. This is seen to be all H or all T .

(c) The least ordered state is the most likely. This is seen to be 2 H and 2 T .

12.54 The change in entropy of a reservoir is ∆S Q Tr= , where Qr is the energy absorbed Qr >( )0 or expelled Qr <( )0 by the reservoir, and T is the absolute temperature of the reservoir.

(a) For the hot reservoir, Q Qr h= − , and

∆SQ

Thh

h

=−

(b) For the cold reservoir, Q Qr h= + , and

∆SQ

Tch

c

=+

(c) For the Universe,

∆ ∆ ∆S S SQ

T

Q

TU h ch

h

h

c

= + = − +

12.55 By substituting 1 hour of study for 1 hour of sleep each day for a year, energy consumed while sleeping would change by ∆ ∆E tsleep W= − ( ) ⋅80 , and the energy consumed while studying would change by ∆ ∆E tstudy W= + ( ) ⋅230 . The net change in energy consumed would be

∆ ∆ ∆ ∆ ∆E E E t tnet sleep study W W= + = − ( ) ⋅ + ( ) ⋅80 230 == + ×( ) ⋅1 5 102. W ∆t

The total time during which this change occurs is 1 hour per day for a year or a total of 365 hours. Thus,

∆Enet

J

sh

s

1 h= + ×⎛

⎝⎜⎞⎠⎟ ( )⎛

⎝⎜⎞

1 5 10 3653 6002.

⎠⎠⎟= + ×2 0 108. J

The additional fat burned to yield this increased consumption will be

Fat BurnedE

Energy Equivalentnet= = + ×∆ 2 0 108. JJ

1.7 J lblb7×

=10

12



12.56 (a) At the sleeping rate of 80 W, the time required for the body to use the 450 Cal of energy supplied by the bagel is

∆ ∆

tU= = ⎛

⎝⎜⎞⎠⎟P

450 4 186 1 Cal

80 J s

J

1 Cal

h

3 600 s h

⎛⎝⎜

⎞⎠⎟

= 6 5.

(b) The increased metabolic rate while working out is P = + = ×80 650 7 3 102W W W. , and the time to use the energy from the bagel at this rate is

∆ ∆

tU= =

×⎛⎝⎜

⎞⎠⎟P

450

10

4 186 1 Cal

7.3 J s

J

1 Cal

2

hh

3 600 s h

⎛⎝⎜

⎞⎠⎟

= 0 72.

(c) W F x mghperlift

2kg m s m= ⋅ = = ( )( )( ) =∆ 120 9 8 2 0. . 22 4 103. × J

(d) The additional energy consumed in 1 minute while working out (instead of sleeping) is ∆U = ( ) = ( )( ) = ×Pincrease min J s s1 650 60 3 9 104. JJ. The number of barbell lifts this should allow in 1 minute is

NU

W= = ×

×=∆

perlift

3

J

2.4 J

3 9 10

1016

4.

(e) No. The body is only about 25% effi cient in converting chemical energy to mechanical energy.

12.57 The maximum rate at which the body can dissipate waste heat by sweating is

∆∆

∆∆

Q

t

m

tL=

⎛⎝⎜

⎞⎠⎟

=⎛

⎝⎜⎞

⎠⎟×v 1 5 2 430 103.

kg

h

J

kg

⎛⎛

⎝⎜⎞

⎠⎟⎛⎝⎜

⎞⎠⎟

= ×11 0 103h

3 600 sW.

If this represents 80% of the maximum sustainable metabolic rate i.e., ∆ ∆ ∆ ∆Q t U t=[ ]0 80. ( )max , then that maximum rate is

∆∆

∆ ∆U

t

Q t⎛⎝⎜

⎞⎠⎟

= = × = ×max

.

.

..

0 80

1 0 10

0 801 3 10

3 W 33 W

12.58 Operating between reservoirs having temperatures of Th = =100 373°C K and Tc = =20 293°C K, the theoretical effi ciency of a Carnot engine is

e

T

TCc

h

= − = − =1 1293

0 21K

373 K.

If the temperature of the hotter reservoir is changed to ′ = =Th 550 823°C K, the theoretical effi ciency of the Carnot engine increases to

′ = −

′= − =e

T

TCc

h

1 1293

0 64K

823 K.

The factor by which the effi ciency has increased is

′ = =e

eC

C

0 64

0 213 0

.

..


618 Chapter 12

12.59 The work output from the engine in an interval of one second is Weng kJ= 1 500 . Since the effi ciency of an engine may be expressed as

e

W

Q

W

W Qh c

= =+

eng eng

eng

the exhaust energy each second is Q Wec = −⎛

⎝⎞⎠ = ( ) −⎛

⎝⎞⎠ = ×eng kJ

11 1 500

1

0 251 4 5 10

.. 33 kJ

The mass of water fl owing through the cooling coils each second is

m V= = ( )( )( ) =−ρ 10 60 10 603 3kg m L m 1 L kg3 3

so the rise in the temperature of the water is

∆TQ

mcc= = ×

( ) ⋅( ) =water

J

60 kg J kg °C

4 5 10

4 186

6.118°C

12.60 The energy exhausted from a heat engine is

Q Q WW

eW W

ec h= − = − = −⎛⎝

⎞⎠eng

engeng eng

11

where Qh is the energy input from the high temperature reservoir, Weng is the useful work done, and e W Qh= eng is the effi ciency of the engine.

For a Carnot engine, the effi ciency is e T T T T TC c h h c h= − = −( )1 , so we now have

Q WT

T TW

T

T Tch

h c

c

h c

=−

−⎛⎝⎜

⎞⎠⎟

=−

⎛⎝⎜

⎞⎠⎟eng eng1

Thus, if T Th c= = = =100 373 20 293°C K and °C K, the energy exhausted when the engine has done 5 0 104. × J of work is

Qc = ×( ) −⎛⎝

⎞⎠ = ×5 0 10

293

373 2931 83 104 5. .J

K

K KJ

The mass of ice (at 0°C) this exhaust energy could melt is

mQ

Lc

f

= = ××

=,water

J

3.33 J kgkg

1 83 10

100 55

5

5

..



12.61 (a) The work done by the system in process AB equals the area under this curve on the PV diagram. Thus,

Wby system triangular area rectangular ar= ( ) + eea( ) or

Wbysystem

atm L

= ( )( )⎡⎣ 12 4 00 40 0. .

atm L+ ( )( )⎤⎦1 00 40 0 1 013. . . ××⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= × =

−

1010

1 22 10

53

4

Pa

atm

m

L

J

3

. 112 2. kJ

Note that the work done on the system is W WAB = − = −bysystem

kJ12 2. for this process.

(b) The work done on the system (that is, the work input) for process BC is the negative of the area under the curve on the PV diagram, or

WBC = − ( ) −( )[ ] ×1 00 10 0 50 0 1 013 105. . . .atm L LPaa

atm

m

L

kJ

3⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

=

−10

1

4 05

3

.

(c) The change in internal energy is zero for any full cycle, so the fi rst law gives

Q U W W W WAB BC CAcycle cycle cycle= − = − + +( )

= − −

∆ 0

0 12.. . .2 4 05 0 8 15kJ kJ kJ+ +( ) =

12.62 (a) From the fi rst law, ∆U Q W1 3 123 123 418 167 251→ = + = + + −( ) =J J J .

(b) The difference in internal energy between states 1 and 3 is independent of the path used to get from state 1 to state 3.

Thus, ∆U Q W1 3 143 143 251→ = + = J, and

Q W143 143251 251 63 0 314= − = − −( ) =J J J J.

(c) W W W W W12341 123 341 123 143 167 63 0= + = + −( ) = − − −J . JJ J( ) = −104

or 104 J of work is done the gasby in the cyclic process 12341.

(d) W W W W W14321 143 321 143 123 63 0 167= + = + −( ) = − − −. J JJ J( ) = +104

or 104 J of work is done the gason in the cyclic process 14321.

(e) The change in internal energy is zero for both parts (c) and (d) since both are cyclic processes.


620 Chapter 12

12.63 (a) The change in length, due to linear expansion, of the rod is

∆ ∆L L T= ( ) = × ( )⎡⎣ ⎤⎦( ) −− −α 06 111 10 2 0 40 20°C m °C. °°C m( ) = × −4 4 10 4.

The load exerts a force F mg= = ( )( ) = ×6 000 9 80 5 88 104kg m s N2. . on the end of the rod in the direction of movement of that end. Thus, the work done on the rod is

W F L= ⋅ = ×( ) ×( ) =−∆ 5 88 10 4 4 10 264 4. .N m J

(b) The energy added by heat is

Q mc T= ( ) = ( )⋅

⎛⎝⎜

⎞⎠⎟

( ) =∆ 100 448 20 9 0kgJ

kg °C°C . ×× 105 J

(c) From the fi rst law, ∆U Q W= + = × + = ×9 0 10 9 0 105 5. .J 26 J J .

12.64 (a) The work done by the gas during each full cycle equals the area enclosed by the cycle on the PV diagram. Thus

W P P V V P Vby gas = −( ) −( ) =3 3 40 0 0 0 0 0

(b) Since the work done on the gas is W W P V= − = −by gas 4 0 0 and ∆U = 0 for any cyclic process, the fi rst law gives

Q U W P V P V= − = − −( ) =∆ 0 4 40 0 0 0

(c) From the ideal gas law, P V nRT0 0 0= , so the work done by the gas each cycle is

W nRTby gas molJ

mol K= = ( )

⋅⎛⎝

⎞⎠4 4 1 00 8 31 2730 . . K

J kJ

( )

= × =9 07 10 9 073. .

12.65 (a) The energy transferred to the gas by heat is

Q mc T= ( ) = ( )

⋅⎛⎝⎜

⎞⎠⎟ (∆ 1 00 20 79 120. . mol

J

mol K K))

= × =2 49 10 2 493. . J kJ

(b) Treating the neon as an ideal gas, the result of problem 11 gives the change in internal energy as

∆ ∆U P V PV nRT nRT nR Tf f i i f i= −( ) = −( ) = ( )3

2

3

2

3

2

or

∆U = ( )⋅

⎛⎝

⎞⎠ ( ) = ×3

21 00 8 31 120 1 50. . .mol

J

mol KK 110 1 503 J kJ= .

(c) From the fi rst law, the work done on the gas is

W U Q= − = × − × = −∆ 1 50 10 2 49 10 9903 3. .J J J



12.66 Assuming the gravitational potential energy given up by the falling water is transformed into thermal energy when the water hits the bottom of the falls, the rate of thermal energy production is

∆∆

∆∆

∆∆

Q

t

m

tgh

V

tghw=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

ρ

Then, if the absolute temperature of the environment is TK ° K= + =20 0 273 293. , the rate of entropy production is

∆∆

∆ ∆ ∆ ∆S

t

Q t

T

V t gh

Tw= =

( )K K

ρ

or

∆∆

S

t=

⎛⎝⎜

⎞⎠⎟

×⎛

⎝⎜⎞

⎠⎟1

10 10 9 83

293 K

kg

m5

m

s3

33

.mm

sm

kg m s m

K s2

2⎛⎝

⎞⎠ ( ) = ×

⋅( ) ⋅⋅

50 0 8 106.

and

∆∆

S

t= × ⋅( )

⋅= × ⋅8 10 8 106 6N m

K sJ K s

12.67 (a) TP V

nRAA A=

=×( )( )

( )10 0 10 1 00

10 0 8

3. .

.

Pa m

mol

3

..31 J mol K⋅( )

= ×1 20 102. K

TP V

nRBB B=

=×( )( )

( )10 0 10 6 00

10 0 8

3. .

.

Pa m

mol

3

..31 J mol K⋅( ) = 722 K

(b) As it goes from A to B, the gas is expanding and hence, doing work on the environment. The magnitude of the work done equals the area under the process curve from A to B. We subdivide this area into 2 rectangular and 2 triangular parts:

Wenv

3Pa m= ×( ) −( )⎡⎣ ⎤⎦ + −10 0 10 6 00 1 00 40 0 13. . . . 00 0 10 1 003. .( ) ×⎡⎣ ⎤⎦( )Pa m3

2 m3+ ( )1

21 00 40 0. . −−( ) ×⎡

⎣⎢⎤⎦⎥

= ×10 0 10 1 10 103 5. .Pa J

(c) The change in the internal energy of a monatomic, ideal gas is ∆ ∆U nR T= ( )32 , so

∆U nR T TA B B A→ = −( ) = ( ) ⋅( )3

2

3

210 0 8 31. .mol J mol K 7722 120 7 50 104K K J−( ) = ×.

(d) From the fi rst law of thermodynamics, Q U W= −∆ , where W is the work done on the gas. In this case, W W= − = − ×env J1 10 105. , and

Q U W= − = × − − ×( ) = ×∆ 7 50 10 1 10 10 1 85 104 5 5. . .J J J

A B

P (kPa)

V (m3)

40.0

30.0

20.0

10.0

1.00 2.00 3.00 4.00 5.00 6.00

A B

P (kPa)

V (m3)

40.0

30.0

20.0

10.0

1.00 2.00 3.00 4.00 5.00 6.00


622 Chapter 12

12.68 (a) The constant volume occupied by the gases is V r= = ( ) =43

3 34 0 500 3 0 524π π . .m m3

and the initial absolute temperature is Ti = + =20 0 273 293. ° K.

To determine the initial pressure, we treat each component of the mixture as an ideal gas and compute the pressure it would exert if it occupied the entire volume of the container. The total pressure exerted by the mixture is then the sum of the partial pressures exerted by the components of the mixture. This gives

Pn RT

VHH

2

2= Pn RT

VOO

2

2=

and

P P Pn RT

V

n RT

V

n n RT

Vi = + = + =+( )

H OH O H O

2 2

2 2 2 2

or

Pi =+( ) ⋅( )( )14 4 7 2 8 31 293

0

. . .

.

mol mol J mol K K

55241 00 105

mPa3 = ×.

(b) Treating both the hydrogen and oxygen as ideal gases, each with internal energy U nC T= v where Cv is the molar specifi c heat at constant volume, we use Table 12.1 and fi nd the initial internal energy of the mixture as

U U U n C n C Ti i i i= + = +( )H O H H O O2 2 2 2 2 2, , , ,v v

or

Ui = ( ) ⋅( ) + ( )14 4 20 4 7 2 21 1. . . .mol J mol K mol J mool K K J⋅( )⎡⎣ ⎤⎦( ) = ×293 1 31 105.

(c) During combustion this mixture produces 14.4 moles of water (1 mole of water for each mole of hydrogen used) with a conversion of 241.8 kJ of chemical potential energy per mole. Since the volume is constant, no work is done and the additional internal energy generated in the combustion is

∆U = ( )( ) = × =14 4 241 8 3 48 10 3 483. . . .mol kJ mol kJ ×× 106 J

continued on next page



(d) After combustion, the internal energy of the system is

U U Uf i= + = × + × = ×∆ 1 31 10 3 48 10 3 61 105 6. . .J J J6

Treating the steam as an ideal gas, so U nC T= v K, and obtaining the molar heat capacity for water vapor (a polyatomic gas) from Table 12.1, we fi nd

TU

n Cff= = ×

( )water water

J

14.4 molv,

.

.

3 61 10

27

6

009 28 103

J mol KK

⋅( ) = ×.

and the fi nal pressure is

P

n RT

Vff= =

( ) ⋅( ) ×water mol J mol K14 4 8 31 9 28 1. . . 00

0 5242 12 10

36

K

mPa3

( )= ×

..

(e) The total mass of steam present after combustion is

m n Msteam steam water mol kg m= = ( ) × −14 4 18 0 10 3. . ool kg( ) = 0 259.

and its density is

ρs

m

V= = =steam

33kg

0.524 mkg m

0 2590 494

..

(f ) Assuming the steam is essentially at rest within the container v1 0≈( ), P2 0= (since the steam spews into a vacuum), and y y2 1= , we use the pressure from part (d) above and

Bernoulli’s equation P gy P gys s s s212 2

22 1

12 1

2+ + = + +( )ρ ρ ρ ρv v ` to fi nd the exhaust speed as

v21

632 2 2 12 10

0 4942 93 10= =

×( )= ×P

sρ.

..

Pa

kg m3 mm s

12.69 The work that you have done is

W mg heng lb

N

1 lb= ( ) = ( )⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

∆ 1504 448

90.

.00 30 0 8 002step

minmin

in

step⎛⎝

⎞⎠ ( )⎛

⎝⎜⎞⎠⎟

. ..554 10 2×⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

− m

1 in

or Weng J= ×3 66 105. .

If the energy input by heat was Qh = ( )( ) = ×600 2 51 106kcal 4 186 J 1 kcal J. , your effi ciency has been

eW

Qh

= = ××

=eng6

J

2.51 10 Jor 14.6

3 66 100 146

5.. %%

If the actual effi ciency was e = 0 180. or 18.0%, the actual energy input was

QW

eh actual

eng

actual

J

0.180= = × = ×3 66 10

2 03 15.

. 00 1 4866 J kcal 4 186 J kcal( ) =


624 Chapter 12

12.70 (a) The energy transferred from the water by heat as it cools is

Q mc T V c Th = = ( )

= ⎛⎝⎜

⎞⎠⎟ ( )

∆ ∆ρ

1 0 1 0103

. . g

cm L

cm3

33

L

cal

g °C

J

1 11 0

4 186⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ ⋅

⎛⎝⎜

⎞⎠⎟

..

ccal°C °C⎛

⎝⎜⎞⎠⎟ −( )570 4 0.

or

Qh = ×2 4 106. J

(b) The maximum effi ciency of a heat engine is the Carnot effi ciency. Thus,

eT

TCc

h

= − = − +( )+( ) = −1 1

4 0 2731

277. K

570 273 K

K

8443 K= 0 67.

The maximum useful work output is then

W e QC heng J J( ) = = ( ) ×( ) = ×max

. . .0 67 2 4 10 1 6 106 6

(c) The energy available from oxidation of the hydrogen sulfi de in 1.0 L of this water is

U n= ( ) = ×⎛⎝

⎞⎠ ( )⎡

⎣⎢−310 0 90 10 1 03kJ mol

mol

LL. . ⎤⎤

⎦⎥×⎛

⎝⎞⎠ = ×310 10 2 8 103 2J

molJ.

12.71 (a) With an overall effi ciency of e = 0 15. and a power output of Pout MW= 150 , the required power input (from burning coal) is

P P

inout W

0.15 J s= = × = ×

e

150 101 0 10

69.

The coal used each day is

∆∆m

t heat of combustion= =

×( )Pin J s1 0 10 8 649. . ××( )

×⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

10

1010

4

3

s d

7.8 calg

g1 kg

3 44 1861

2 64 106

..

J cal

kg

d⎛⎝⎜

⎞⎠⎟

= ×

or

∆∆m

t= ×⎛

⎝⎞⎠

⎛⎝⎜

⎞2 64 10

1

106.

kg

d

metric ton

kg3 ⎠⎠⎟= ×2 6 103. metric ton d

(b) The annual fuel cost is: cost coal used yearly rate= ( ) ⋅ ( ), or

cost = ×( )( )( ) = ×2 64 10 365 8 0 7 73. $ . $ .ton d d y ton 1106 y

(c) The rate of energy transfer to the river by heat is

P P Pexhaust in out W W= − = × − × = ×1 0 10 150 10 8 5 19 6. . 008 W

Thus, the fl ow required if the maximum rise in temperature is 5.0°C is

flow ratem

t c T water exhaust

water

= = ( ) = ×∆∆ ∆

P 8 5. 110

4 186 5 04 1 10

8 J s

J kg °C °C kg s4

⋅( )( ) = ×.

.


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The Laws of Thermodynamics - Croom Physics€¦ · The Laws of Thermodynamics CLICKER QUESTIONS Question J3.01 ... 558 Chapter 12 Question J3.02 ... Question J3.03