1 Te Laws of Termodynamics CLICKER QUESTIONS Question J.01 Description: Relating termodynamic processes to PV curves: isobar. Question A quantity of ideal gas undergoes a termodynamic process. Wic curve represents an isobaric pat? 1) T ) P ) V P 4) V V 5) P T T V Commentary Purpose: To develop your ability to interrelate te basic termodynamic quantities of pressure, temperature, and volume. Discussion: Isobaric means constant pressure. None of te depicted graps wit pressure on one axis sow a constant pressure. However, grap (4) displays a linear relationsip between volume and temperature; according to te ideal gas law, if volume and temperature are proportional to eac oter, te pressure must be constant. Key Points: Isobaric means constant pressure. Be able to relate te beavior of pressure, volume, and temperature via te ideal gas law. If pressure is constant for a process, volume, and temperature will be directly proportional to eac oter. For Instructors Only Tis question provokes students to reason qualitatively wit grapical representations and fundamental relationsips, wic is valuable for developing conceptual understanding and analysis skills. 557
558 Capter 1 Question J.0 Description: Interpreting PV process diagrams. Question A vertical cylinder wit a movable cap is cooled. Te process corresponding to tis is: M 0 B P(Pa) 10 A 1 V (m ) C 1. CB. AB. AC 4. CA 5. Not sown Commentary Purpose: To link te grapical representation of an isobaric process to te description of suc a process. Discussion: Forces on te moveable cap must always balance, wic means tat te internal pressure of te gas must always be enoug larger tan atmosperic pressure to counteract te weigt of te cap. Tus, te process described is isobaric (constant pressure). A cooled gas contracts, so te volume must be decreasing. Only one segment of te PV grap, C to A, depicts decreasing volume at constant pressure; terefore, te correct answer must be (4).
Te Laws of Termodynamics 559 Key Points: Interpreting process diagrams and relating tem to pysical situations is a vital skill to develop. A cylinder wit a freely moving cap, supported only by te pressure of te contained gas, is a mecanism for insuring constant pressure. For Instructors Only Some follow-up discussion questions you may wis to use: Is work done during tis process? (By or on te gas?) How does te temperature at A compare to tat at C? How muc eat was extracted during te process? Question J.0 Description: Linking grapical representations of termodynamic processes to quantities: work. Question An ideal gas is taken around te process sown. Te net work done on te gas is most nearly: 0 P(Pa) 10 1 V (m ) 1. 0 J. 0 J. 15 J 4. 10 J 5. None of te above 6. Cannot be determined Commentary Purpose: To one te concept of work for a termodynamic cycle, and connect it to a grapical representation of a cycle. Discussion: On a pressure vs. volume grap, te area under a curved segment is te work done by te system during te process represented by tat segment: te integral of pressure over a cange in volume. (For a segment moving rigt to left, te work done by te system is te negative of tat area.) For a cycle a process tat ends in te same state it began te net work done by te system must be equal to te area inside te closed PV pat if te cycle is traversed clockwise. Te work done on te gas is te negative of te work done by te gas. So, for tis situation, te work done on te gas is te negative of te area of te triangle: answer (4).
560 Capter 1 Key Points: Te area under a pressure vs. volume curve is te work done by te system. Te area inside a closed pat on a PV diagram is te work done by te system during a cycle. Te work on a system is te negative of te work by a system. Pay attention to signs: work on vs. work by, pat traveled left-to-rigt vs. rigt-to-left, cycle traveled clockwise vs. counterclockwise, etc. For Instructors Only Students selecting a positive answer (1) or () sould be sensitized to te difference between work on and work by a gas. Question J4.01a Description: Relating internal energy to and reasoning wit a PV cycle. Question Two moles of a perfect gas are taken along te cycle sown below. D C Pressure, P P 1 A B 0 V 1 Volume, V Wic of te following is true about te internal energy U? 1. U 0 = U A. U D < U A. U B > U C 4. U D = U B 5. U C = U A 6. None of tese is true.
Te Laws of Termodynamics 561 Commentary Purpose: To connect te idea of internal energy to a PV cycle and reason wit te information contained in a PV diagram. Discussion: Te internal energy of a perfect gas depends only on its temperature and te number of moles. Te number of moles is not canging, so, by applying te perfect gas law (PV = nrt), we can find out about ow te temperature is canging. Te product PV is smallest in state A, so T is also smallest in state A. In states B and D, te product PV is twice as large as it is in state A, so te temperature in states B and D are te same and twice as large as te temperature in state A. In state C, PV is four times as large as it is in state A, so te temperature is four times as large. Matematically, since n = moles: T A = P 1 V 1 R T B = P 1 (V 1 ) R = P 1 V 1 R = T A T C = (P 1 )(V 1 ) R = P 1 V 1 R = 4T A T D = (P 1 )V 1 R = P 1 V 1 R = T A Looking at te given expressions involving U, we see tat #4 is true. Since te temperature of state B is equal to te temperature of state D, te internal energy must be te same too. Statement (1) is not necessarily true, because we do not need to begin te cycle at state A. (U 0 indicates te initial internal energy.) For Instructors Only Many students will tink tat statement (1) is true and, not needing to analyze te situation any furter, will stop. Tis distracter answer is a trap for students wo assume tat all cycles begin wit state A. We often begin a cycle at state A, but tis is not a requirement; is not even a common convention. (If students ask wat te notation U 0 means, you sould tell tem it means te initial internal energy. ) Students need to make two connections to successfully analyze tis situation. First, tey must recognize tat internal energy depends on temperature. Second, tey must recognize tat tey can compare temperatures by looking at te specific values of P and V, even toug tey can t calculate numerical values. We intend students to use te grap to scale. From A to B, te volume doubles at constant pressure. From B to C, te pressure doubles at constant volume, etc.
56 Capter 1 Question J4.01b Description: Interrelating energy and ideal gas ideas for a PV cycle. Question An ideal monatomic gas is taken around te cycle ABCDA as sown below. D C Pressure, P P 1 A B 0 V 1 Volume, V Wat is te amount of energy removed by eat during one cycle? 1. 0. P 1 V 1. P 1 V 1 4. P 1 V 1 5. 4P 1 V 1 6. 5P 1 V 1 7. Impossible to determine Commentary Purpose: To interrelate energy and ideal gas ideas witin te context of a PV cycle. Discussion: Te internal energy of a perfect gas is a state function, wic means tat any time te gas is in a particular state (a point on te PV diagram), it must ave te same internal energy. If te gas goes around a cycle and returns to a state, it must return to te same internal energy. Tis means te cange in internal energy U is equal to zero for tis cycle. Since te internal energy is not canging, te work done on te gas must be equal to te amount of energy removed from te gas by eat, according to te first law of termodynamics. Wen a gas expands, it does work on its environment, and wen it is compressed, te environment does work on it. At constant pressure, te work done by a gas is P V. From A to B, te gas is expanding, so te work done by te gas is P 1 V 1, and te work done on te gas is P 1 V 1. (Since te grap is drawn to scale, V B = V A = V 1.) From B to C, te volume is constant, so no work is done on te gas, even toug te temperature is increasing. From C to D, work is done on te gas to compress it at constant pressure. Te work done on te gas is P 1 V 1. From D to A, te work done is zero.
Te Laws of Termodynamics 56 Te total amount of work done on te gas is P 1 V 1. Te amount of energy removed by eat from te gas to te environment is terefore also P 1 V 1 : answer (). Note tat te area of te enclosed pat in te PV diagram is te work done by te gas wen te cycle is counterclockwise (as above), and te area is te work done on te gas wen te cycle is clockwise. Note also tat te amount of energy removed by eat is a net amount. During two legs of te cycle, energy is added to te gas by eat, and during te oter two, energy is removed by eat. Te net amount is te difference between tese two amounts. Key Points: Internal energy is a state function. If a gas returns to a state (a point on te PV diagram) tat is as been in previously, it returns to te same internal energy. Wen a gas travels around a closed cycle, net work is done on it and an equal amount of net eat is expelled to te environment. (Te net work may be positive or negative, depending on te direction of te cycle.) No work is done on a gas during a constant-volume process. During a constant-pressure process, te work done on a gas is P V. For Instructors Only Tis question requires students to put many ideas togeter. Some students migt not realize tat te cange in internal energy is zero for a complete cycle. Tey migt not connect work wit te amount of energy transferred by eat. Or, tey migt not realize tat tey can find te work done using te grap of P vs. V. A possible misconception is to tink tat since te system returns to its initial state, te energy transferred by eat is zero. In oter words, tey migt inadvertently tink tat Q is a state variable. Students migt ave difficulty understanding ow net eat can be expelled wit no net cange in internal energy. Students selecting answer (4) migt be considering only te magnitudes of te work done during te steps of te cycle, and neglecting te signs. Students selecting answer (7), Impossible to determine, migt tink tat te prase amount of energy removed by eat is ambiguous: does it refer to te net amount, or does it refer only to tose legs during wic energy is actually removed (as opposed to added)? Alternatively, tey may simply be overwelmed by te complexity of te question. Students migt benefit from a concrete description of ow eac leg can be accomplised. For instance, from A to B, we would put a certain weigt on te piston to maintain constant pressure, ten put te gas in termal contact wit a temperature bat at T A until equilibrium is reaced, tus transferring energy to te gas by eat. From D to A, te piston is eld fixed to maintain constant volume, ten energy is removed by eat by placing te gas (at T D = T A ) in termal contact wit a temperature bat at T A, tus removing energy from te gas by eat. It turns out tat during te first two legs, 4P 1 V 1 of energy is added to te gas by eat, and during te last two, 5P 1 V 1 of energy is removed.
564 Capter 1 Question J5.01a Description: Understanding free expansion of a gas: cange in pressure. Question Two moles of an ideal gas fill volume V = 10 liters at pressure P =.4 atm. Te gas is termally insulated from te surroundings. A membrane is broken wic allows te gas to expand into te new volume wic is times as large as te old volume. Te new pressure is: 1. 0.4 atm. 0.8 atm. 1. atm 4. 1.6 atm 5..4 atm 6. None of te above 7. Cannot be determined Commentary Purpose: To explore te application of te ideal gas law to te free expansion of a gas. Discussion: Te temperature of a gas undergoing free expansion does not cange. For discussion of tat fact, see question b of tis question set. For a constant-temperature process, te ideal gas law may be used to relate te cange pressure to te cange in volume: P f V f = (constant) = P i V i. So, if te volume triples, te pressure must drop to 1 its original value, or 0.8 atm. We don t actually need to know te temperature of te gas or te number of moles. Te subtlety about tis question is recognizing tat te temperature is in fact a constant. If te gas ad expanded via some different mecanism for example, if it ad pused a piston to te far wall wile termally insulated te answer would be different. Key Points: In a constant-temperature process, te ideal gas law allows us to relate canges in volume to canges in pressure. In termodynamics processes, ow a system gets from one state to anoter is frequently important.
Te Laws of Termodynamics 565 For Instructors Only Tis problem is meant to be used alongside Question J5.01b. We recommend presenting te question, collecting answers, and allowing students to voice arguments in favor of teir selections, but not resolving any of te disagreements or revealing anyting about te correct logic beyond wat students articulate temselves. Instead, proceed on to 18b, wic addresses te question of temperature ead-on. Bot questions can ten be discussed and resolved togeter. (If te constancy of te temperature arises as a focal issue, you can say Let s look at tis next question, wic addresses tat directly.) One reason for tis approac is to catc students wo may apply Boyle s law or some oter form of te ideal gas law blindly, witout pausing to consider weter te temperature is in fact constant. We ope tat te second question will cause tem to stop and ponder weter tey approaced te first correctly. One can invert te order of te two questions, wic makes tem more straigtforward for students, but tat ordering as less potential for causing students to wake up and tink ard. Question J5.01b Description: Understanding free expansion of a gas: cange in temperature. Question Two moles of an ideal gas fill a volume of 10 liters wit a pressure of.4 atm. Te gas is termally insulated from te surroundings. A membrane is broken wic allows te gas to expand into te new volume wic is times as large as te old volume. Wat is te new temperature? 1. Te same as before. Lower tan before. Higer tan before 4. Cannot be determined Commentary Purpose: To explore ow te free expansion of a gas affects its temperature. Discussion: Wen te membrane is broken, te gas expands into a vacuum. It does no work, since it is not pusing on anyting: it exerts zero force troug some distance. (If it were doing work, were would te energy be going?)
566 Capter 1 Also, no eat is absorbed or expelled by te gas, since te container is insulating. Even if it were not insulating, te expansion appens essentially instantaneously, so tere is no time for any significant eat transfer to occur. According to te first law of termodynamics, if a gas does no work and absorbs no eat, its internal energy does not cange. If internal energy does not cange, temperature does not cange. (Consider te equation of state of a gas, relating internal energy to temperature.) Te answer is tus (1). One way to tink about tis is to realize tat te temperature of a gas describes te average speed of te molecules bouncing around witin te gas. Wen te membrane disappears, te molecules spread out into te new space. Wen tey ave finised spreading out into te larger volume, tey ave te same velocity distribution as before; tere is noting to slow down te molecules. (Wile te spreading out is occurring, te gas is not in equilibrium and does not ave a well-defined temperature. Temperature is defined only for a system in equilibrium.) Many students ave learned tat a gas cools as it expands, and may terefore coose answer (). A gas does cool as it expands, if it does work during te process of expansion for example, by pusing back a piston. In tis case, te energy lost by te gas is gained by te piston or by someting connected to te piston. In te molecular picture, gas molecules are rebounding from te moving wall of te piston; bouncing from a surface tat is receding causes a loss of speed. Key Points: Te temperature of a gas does not cange wen te gas undergoes free expansion. Temperature is defined only for a system in equilibrium. For Instructors Only Tis question is meant to follow Question J5.01a, bringing to te fore te issue (implicit in 18a) of weter te temperature of te gas canges. See te commentary for tat question for suggestions about using te questions togeter. Tis situation provides a good context for discussion wat temperature means, and wen a system as a well-defined temperature. It may be elpful to draw a PV diagram, and indicate tat during te gas s nonequilibrium expansion te system is not represented by any point on te diagram; it as disappeared from te plot at one equilibrium point and reappeared at te oter.
Te Laws of Termodynamics 567 Question J5.0a Description: Developing understanding of adiabatic (termally isolated) processes. Question A gas is placed in a termally insulated container as sown below. F As te piston is slowly lowered, te temperature: 1. Increases. Decreases. Stays te same 4. Te cange in te temperature cannot be determined. Commentary Purpose: To explore wat te prase termally insulated means. Discussion: Te prase termally insulated means tat te gas is termally isolated from its environment. No eat can flow between te gas and te environment, even if tey are at different temperatures. Tat is, Q = 0 for any process. However, te force F can do work on te gas, wic means its internal energy can cange. In tis case, as te piston is lowered, positive work is done on te gas, so te internal energy increases. If te internal energy increases te temperature must rise. Key Points: Termally insulated means no eat flows into or out of te system. Te temperature of a termally insulated system can cange if work is done on it.
568 Capter 1 For Instructors Only Students coosing answer () migt tink tat termally insulated means tat te temperature is constant. Some migt know tat termally insulated means Q = 0, but interpret tat to mean te temperature is constant. Students coosing answer () migt assume tat te force F is constant, wic means te pressure is constant and te temperature must decrease to keep te pressure constant at a smaller volume of te gas. (Te next question will tease out tis facet even more, so you do not need to deal wit te idea explicitly. It is not really relevant, since even a constant force would do positive work on te gas.) Students coosing answer (1), te correct answer, migt be using te ideal gas law (PV = nrt ), assuming tat P is constant, and concluding (correctly) tat since V is getting smaller, T must be getting larger. Since te pressure is not constant, tis reasoning is flawed even toug te conclusion is rigt. Terefore, it is critical tat you find out wy students cose te answers tey did. Even if everyone answers tis question correctly, you need to discuss te reasoning. Question J5.0b Description: Developing understanding of adiabatic (termally isolated) processes. Question A gas is placed in a termally insulated container as sown below. F As te piston is slowly lowered, te applied force F: 1. Increases. Decreases. Stays te same 4. Te cange in te applied force cannot be determined.
Te Laws of Termodynamics 569 Commentary Purpose: To reason wit te ideal gas law in an adiabatic (termally insulated) process. Discussion: Te prase termally insulated means tat Q = 0 for tis process. Terefore, according to te First Law of Termodynamics, te work W done on te gas is equal to te cange in internal energy E. Since te work done on te gas is positive, te internal energy increases. If te internal energy increases, te temperature increases. As te piston is lowered, te temperature is increasing and te volume is decreasing. Terefore, by te Ideal Gas Law (PV = nrt ) te pressure must be increasing. Tus, te force F must increase also in order to keep te piston in equilibrium. Key Points: Termally insulated means no eat flows into or out of te system. Te ideal gas law relates te pressure, volume, and temperature of a gas. For Instructors Only Students coosing answer (4) migt not tink of using te ideal gas law, peraps because te question asks about te force but te ideal gas law refers to pressure. Or, tey migt be confused by te fact tat te piston s mass is not specified (toug tis is in fact irrelevant). Question J5.0c Description: Developing understanding of adiabatic (termally isolated) processes. Question A gas is placed in a termally insulated container as sown below. F As te piston is slowly raised, te work done on te gas is: 1. Positive. Negative. Zero 4. Te sign of te work done on te gas cannot be determined.
570 Capter 1 Commentary Purpose: To ceck your understanding of te termodynamic definition of work. Discussion: Te termal insulation of te container is irrelevant. All tat matters is te direction of te displacement of te force as compared to te direction of te force. As te piston is raised, te force F points down, but te displacement of te piston is up, so te work done by F is negative. Tus, te work done on te gas (by te piston) is negative. (Note tat te work done by te gas (on te piston) is positive, because te gas is pusing up on te piston as it moves in te same direction.) Anoter way of tinking about tis is in terms of pressure. Te container olding a gas exerts a pressure (a force per unit area) on te gas on all sides, directed inward. If te gas expands, it is doing positive work against tis pressure, so te 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: Te work done by a force is positive if te object acted upon moves in te direction te force puses, and negative if it moves in te opposite direction. Mecanics ideas, definitions, and principles apply to termodynamic systems as well. An expanding gas as negative work done on it by its environment. For Instructors Only Some students migt still ave trouble understanding ow te work done can be negative. Depending on ow long it as been since students ave used work and energy ideas, it migt be a good idea to review tem in more familiar contexts. Tis question provides a good opportunity to link and integrate mecanics wit termodynamics. Students often get confused about te work done on te gas (by an external agent) vs. te work done by te gas (on te environment). Students migt be unduly distracted by te fact tat te container is insulated, and try to apply termodynamic principles to deduce te sign of te work done on te gas rater tan simply applying te definition of work.
Te Laws of Termodynamics 571 Question J5.0d Description: Developing understanding of adiabatic (termally isolated) processes. Question A gas is placed in a termally insulated container as sown below. F As te piston is slowly raised: 1. Te gas gives off eat to te environment (Q < 0). Te gas absorbs eat from te environment (Q > 0). Te gas does not excange any eat wit te environment (Q = 0) 4. Te eat excanged wit te environment cannot be determined. Commentary Purpose: To revisit te concept of termally insulated. Discussion: Termally insulated means tat no eat can be excanged wit te environment. Te only way te environment can interact wit te gas is troug te piston and te applied force F. Tis does not mean te temperature of te gas must remain constant. As work is done on te gas by te piston, te energy of te gas canges according to te first law of termodynamics. If te internal energy canges, te temperature canges. Key Points: Termally insulated means no eat flows into or out of te system. Te temperature of a termally insulated system can cange if work is done on it. For Instructors Only Toug tis question may seem redundant wit te first in tis set, many students will remember only tat te temperature can cange (a surprising result!) and incorrectly conclude tat eat must be excanged wit te environment.
57 Capter 1 Question J5.0a Description: Understanding adiabatic processes: pressure. Question An ideal gas is allowed to expand slowly. Te system is termally isolated. Wic statement regarding te final pressure is true? M F M P, T P, T Before After 1. P < P. P = P. P > P 4. Not enoug information Commentary Purpose: To understand pressure cange in an adiabatic process. Discussion: Te system is termally isolated, so te expansion process is adiabatic (no eat flows in or out). For adiabatic processes, PV γ is constant. Tus, if te volume increases, te pressure decreases. Anoter way to determine tis is to realize tat for an adiabatic process, te first law of termodynamics (energy conservation) requires tat if te gas does work on its environment, its internal energy and tus its temperature must drop (since eat transfer is zero). If te temperature drops and te volume increases, te ideal gas law indicates tat te pressure must decrease. A tird way to reason to te answer is to realize tat te gas won t expand unless te pressure inside te container is greater tan te pressure outside plus te pressure caused by te weigt of piston ead (te moveable mass). Te gas will expand until te forces on te piston ead are balanced, so te final internal pressure must be lower tan te initial internal pressure. Key Points: For adiabatic processes, PV γ is constant. For adiabatic processes, te work done by te gas equals te internal energy lost by te gas. Tere often more tan one way to answer a question, and some ways are simpler tan oters.
Te Laws of Termodynamics 57 For Instructors Only Tis is te first of two related questions. It is a simple question, but good for introducing or cecking students understanding of adiabatic processes and te PV γ rule. It is also valuable for demonstrating tat a question may be answered troug multiple arguments. Te tird argument, in particular, is elegant and simple, and carries te moral tat making pysical sense of a situation is as important as remembering and applying abstract matematical laws. Question J5.0b Description: Understanding adiabatic processes: temperature. Question An ideal gas is allowed to expand slowly. Te system is termally isolated. Wic statement regarding te final temperature is true? M F M P, T P, T Before After 1. T < T. T = T. T > T 4. Not enoug information Commentary Purpose: To understand temperature cange in an adiabatic process. Discussion: Te system is termally isolated, so te expansion process is adiabatic (no eat flows in or out). For adiabatic processes, T V γ 1 ; tis can be derived from te ideal gas law and te fact tat PV γ is constant. Tus, if te volume increases, te temperature decreases. Anoter way to determine tis is to realize tat for an adiabatic process, te first law of termodynamics (energy conservation) requires tat if te gas does work on its environment, its internal energy must drop (since eat transfer is zero). For an ideal gas, internal energy depends only on temperature, so if te internal energy drops, so also must te temperature.
574 Capter 1 Key Points: For adiabatic processes, TV γ 1 is constant. For adiabatic processes, te work done by te gas equals te internal energy lost by te gas. Tere often more tan one way to answer a question, and some ways are simpler tan oters. For Instructors Only Tis is te second of two related questions. Te answer to tis question is likely to be implicitly addressed wile covering te previous one, making tis one good for cecking weter students really got it. As wit 74a, it is wort making sure students appreciate te multiple ways of determining te answer. Question J6.01 Description: Sensitizing to assumptions wen working wit te ideal gas law. Question Consider te two systems below, labeled A and B. F F A B Wic gas as te iger pressure? 1. A. B. Neiter; te pressures are te same. 4. Impossible to determine
Te Laws of Termodynamics 575 Commentary Purpose: To develop your awareness of te role different assumptions play in answering an ideal gas law question. Discussion: Te ice-water mixture keeps te gas in te container at 0 C. Tus, we know te temperatures of te two systems are te same, and te volume of system A is larger tan tat of system B. If bot systems contain te same amount (number of moles) of gas, ten te ideal gas tells us tat system B must ave te iger pressure. However, we don t know anyting about te amounts of gas, so answer () is unjustified. If te applied force F is te same for bot systems, and if bot pistons ave te same size, sape, and mass, ten te pressure applied to te gas must be te same in bot cases. Te question statement doesn t say tat tese conditions are true, but we migt assume tat te label F is a variable tat represents te same value for bot cases, and te picture seems to sow two identical pistons. So answer () is defensible, but requires some assumptions. Te only ting we can say for sure is tat te volumes are different and te temperatures are te same, wic means tat eiter te pressure or te amount of gas must differ. Te most conservative answer is terefore (4). Key Points: Te ideal gas law relates te pressure, volume, temperature, and amount of a gas. Don t overlook te amount (number of moles) as a variable. Be aware of te assumptions you make, and coose tem carefully. For Instructors Only Answer (), tat system B as a iger pressure, tends to be common; tis is generally because students implicitly assume bot systems contain te same amount of gas. As always, we recommend focusing on te consequences and reasonableness of various possible assumptions rater tan on te correctness of answers. Additional Questions: 1. Wic system as te iger temperature?. Wic system as te iger number of moles of gas?. Assume now tat a termodynamic process takes system A and turns it into system B. How would you accomplis tis? Description: Understanding gas pressure. Question Question J6.0a An unknown amount of gas is placed in a container wit a moveable piston of negligible mass. Te crosssectional area of te container is A. Wen a mass M is placed on te piston, te eigt of te piston above te bottom of te container is.
576 Capter 1 M Wat is te pressure of te gas in te container? 1. Mg. MgA. MA 4. MA 5. Mg 6. M A 7. Mg A 8. M A 9. None of te above 10. Impossible to determine from te given information Commentary Purpose: To develop your understanding of te pressure of a gas. Discussion: Tis is a perfectly static situation. We are not yet discussing any processes; we simply want to make sure we understand te connection between te mass on te piston and te pressure of te gas inside te container. Focusing on te piston and block as a single system is particularly useful. It is at rest, so its acceleration is zero, and tus te net force on it must be zero. Pressure is defined as force per unit area (P = F A), so te gas puses up on te piston wit a force of PA, were A is te cross-sectional area of te piston. Tis makes intuitive sense: te larger te area or te larger te pressure, te larger te force sould be. Gravitation is pulling down on te block and piston wit a force of Mg. Since te piston and block are at rest, te total force up must balance te total force down. So, ignoring te atmosperic pressure pusing down on te piston and block, PA = Mg, or P = Mg A. Te outside air puses down on te top of te block and piston wit one atmospere of pressure (P atm ), for an additional downward force of P atm A. If we include tis, we find tat te pressure of te gas in te cylinder must be P = P atm + Mg A. Note tat te pressure does not depend on te eigt. Key Points: Pressure is force exerted per unit surface area. For a flat surface tat exerts a force on a gas, P = F A. A gas contained by a floating (movable) piston will expand or contract until its pressure as te rigt value to make te net force on te piston zero.
Te Laws of Termodynamics 577 For Instructors Only Tis set of four questions introduces constant-pressure processes, beginning wit te concept of gas pressure. Students migt not fully appreciate te balance tat occurs to keep te piston and block at rest, especially if teir understanding of Newtonian mecanics is rusty. Tis question provides an opportunity to review elementary statics. Students migt tink tat te eigt sould be part of te answer. Tey migt tink tat te larger te pressure, te smaller te eigt. (Te next question in tis set sould elp tem sort out tis feature.) Students can use proportional reasoning. A larger mass would create a larger pressure, so te pressure sould depend on M in te numerator. A larger area would require a smaller pressure to produce te same force. And so on... Question J6.0b Description: Understanding gas pressure. Question Consider te two systems below, labeled A and B. M M A B Wic gas as te iger pressure? 1. A. B. Neiter; te pressures are te same. 4. Impossible to determine
578 Capter 1 Commentary Purpose: To ceck your understanding of gas pressure. Discussion: Assuming te piston is free to move, te force pusing up due to pressure in te gas must exactly balance te forces pusing down on te piston. Te forces down are te same for te two situations: te weigt of te piston and block and atmosperic pressure from above. Terefore, te pressures inside te two cylinders must be te same. How can te pressures be te same wile te volumes are clearly different? According to te ideal gas law (PV = nrt ), if pressure is constant, a larger V means a larger nrt. So te larger volume in situation A could be due to a iger temperature in te gas, or due to a larger quantity of gas, or bot. We are not told enoug to determine wic. Key Points: For a floating piston supported by a gas under pressure, te upward force due to te gas pressure must balance all oter forces acting on te piston. You cannot necessarily infer anyting about a gas pressure from information about its volume, since temperature and number of moles are also relevant variables. For Instructors Only Students coosing answer (A) migt be tinking tat A must ave a greater pressure because we can see it as pused te piston iger. Students coosing answer (B) migt be tinking tat B must ave a greater pressure because it as been compressed more. Bot groups are imagining a dynamic process, rater tan considering a static balance of forces. Even if students recognize at some point tat te pressures inside te containers are te same, tey will likely tink tat te temperature is iger for A tan B, because tey will implicitly assume tat te amounts of gas in te containers are te same. Tey sould be made aware of tis assumption. Some students migt tink tat te answer is impossible to determine, peraps because tey do not know if atmosperic pressure sould be included or not or because tey do not know its value. Tis means only tat a value for te pressure cannot be computed; it does not mean tat tey cannot compare te pressures. Discussion Questions 1. Wic system as te iger temperature?. Wic system as te iger number of moles of gas?. Assume now tat a termodynamic process takes system A and turns it into system B. How would you accomplis tis? Question J6.0c Description: Understanding isobaric (constant pressure) processes. Question An unknown amount of gas is placed in a container wit a moveable piston as sown below. Te crosssectional area of te container is A. Wen a mass M is placed on te piston, te eigt of te piston above te bottom of te container is.
Te Laws of Termodynamics 579 M Wat appens wen an amount of energy Q is transferred by eat to te gas? (Q > 0.) 1. T increases.. increases.. T and increase. 4. T and P increase. 5. and P increase. 6. T,, and P increase. 7. T decreases. 8. increases; T decreases. 9. None of te above 10. Impossible to determine Commentary Purpose: To understand constant pressure processes. Discussion: Q is positive, so termal energy is being added to te gas. According to te first law of termodynamics, te added eat must eiter increase te internal energy and tus te temperature of te gas, cause te gas to expand and do work on te piston, or bot. Te energy as to go somewere. Te pressure must remain constant to balance te downward forces on te piston and mass. According to te ideal gas law, if te temperature rises wit constant pressure, te volume must increase, and vice-versa. So, te added eat must raise te gas temperature and cause it to expand. Te best answer is terefore (). Key Points: Wen eat is added to a gas at constant pressure, te gas temperature and volume increase. Te first law of termodynamics elps you keep track of energy tat gets added to or removed from a termodynamic system. Te ideal gas law relates te pressure, volume, temperature, and quantity of a gas. For Instructors Only Some students migt intuit tat te gas will expand, but not be sure wat appens to te temperature. Since te expanding gas does work on te piston, tey migt not be sure if tis uses up all of te added eat: weter te work is larger or smaller tan Q, and terefore weter te temperature falls or rises. Te first law of termodynamics is not enoug to resolve tis; te ideal gas law is required as well.
580 Capter 1 Discussion Questions 1. Wat is te relationsip between T and V for tis process?. Wat is te relationsip between T and for tis process?. Wic is larger, Q or W? 4. If te same amount of energy Q is transferred to te gas, but tis time te piston is eld fixed, ow would te cange in temperature compare to te cange in tis situation? Question J6.0d Description: Understanding isobaric (constant pressure) processes. Question An unknown amount of gas is placed in a container wit a moveable piston as sown below. Te crosssectional area of te container is A. Wen a mass M is placed on te piston, te eigt of te piston above te bottom of te container is. M An amount of energy Q is transferred by eat to te gas. (Q > 0.) Put te magnitudes of te energy Q, work W, and cange in internal energy E in order from smallest to largest. 1. Q < W < E. Q < E < W. 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 te above 10. Impossible to determine Commentary Purpose: To understand constant pressure processes. Discussion: Since Q is positive, some of te energy raises te temperature of te gas and some of te energy does work to raise te piston. We can see tis directly from te first law, E = Q + W. Te temperature increases, so E is positive. Te work W done on te gas by te piston is negative. Te only way all
Te Laws of Termodynamics 581 of tese statements can be true is if bot E and W are smaller tan Q. Tus, we know ow E and W compare to Q, but we do not know yet ow tey compare to eac oter. To compare E and W, we need to dig a little deeper. Te internal energy E of a gas is proportional to its temperature. For a monatomic gas, te relationsip is E = nrt. For all oter gases, te proportionality constant is even larger. Tus, te cange in internal energy may be written E nr T. For a constant pressure process suc as tis one, te work done on te gas is P V. (A positive cange in volume means te work done on te gas is negative.) Assuming te gas in te container is ideal, it obeys te ideal gas law, PV = nrt, so P V = nr T. Tis means te magnitude of te work done on te gas is W = nr T. Te cange in internal energy is always larger tan tis, so te work done is smallest. Tus, W < E < Q. For Instructors Only Many students will likely be able to reason tat Q is larger tan W and E, but unable to sort out wy W is smaller tan E. Some students coosing te correct answer, (4), may be doing so erroneously: even toug te question asks for a comparison between magnitudes, tey may tink tat W is smallest since it is negative. Some students migt tink tat te cange in internal energy is always te largest of te tree, peraps because E = Q + W. In oter words, tey are not tinking tat eiter Q or W could be negative. If students assume te gas is monatomic, callenge tem to answer te question and justify teir answer witout tat assumption. Additional Questions: An amount of energy Q is added by eat suc tat te volume exactly doubles. In terms of known quantities (M, A,, etc.), wat are (a) Q, (b) W, and (c) E for tis process? Question J7.01a Description: Introducing isotermal processes, linking to latent eat and pase cange. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F
58 Capter 1 As te piston is slowly lowered, te temperature: 1. Increases. Decreases. Stays te same 4. Te cange in te temperature cannot be determined. Commentary Purpose: To introduce te idea of an isotermal process. Discussion: A mixture of ice and water remains at 0 C until all te ice is melted or until all te water as turned into ice. If you add energy (eat) to te mixture, some of te ice melts, but te temperature remains constant. Te energy added literary goes into melting te ice rater tan canging te temperature of te mixture. Wen all te ice as melted, ten adding more energy would increase te temperature of te water. Likewise, removing energy from te mixture causes some water to turn into ice, and again, te temperature remains constant. Wen all of te water as turned into ice, ten removing more energy would lower te temperature of te ice. Tus, an ice-water mixture is a mecanism for creating a constant temperature environment. Te gas camber is not insulated, so we will assume tat eat can flow freely between te gas and te ice-water mixture. Since te piston is lowered slowly, te gas must stay at 0 C: if te temperature of te gas rises (or drops) infinitesimally, eat will flow to (or from) te mixture, bringing it back into termal equilibrium. We call suc a process isotermal (meaning constant temperature ). If te piston were pused down rapidly, te temperature of te gas migt rise suddenly, and ten slowly fall back to 0 C as eat flows troug te cylinder wall and te system returns to termal equilibrium. Key Points: An isotermal process as a constant temperature. An ice-water bat is a mecanism for maintaining a system at approximately constant temperature. For Instructors Only Depending on students understanding of latent eat and pase canges, tey migt not understand ow te temperature of an ice-water mixture can remain constant wen energy is added or removed. A demonstration migt be elpful. Tis is a good question for linking students understanding of termodynamic processes wit teir understanding of latent eat and pase cange. Students migt ave trouble understanding ow ice is formed by removing energy from te water; adding energy and melting ice seem muc more easily accomplised. An apparent counterexample tat may confuse students is a freezer, wic requires electrical energy to form ice. A refrigerator or freezer is in fact a more complicated system. Heat only flows spontaneously from warm to cold; to pump eat from a cold object to a warmer one, additional energy is required. A discussion of tis can be used as a forward reference to entropy and te second law of termodynamics. Students migt select answer (1) because tey do not recognize tat eat will flow out of te gas into te ice-water bat, or because tey tink eat cannot flow rapidly enoug out of te gas to maintain constant temperature.
Te Laws of Termodynamics 58 Students selecting answer () migt be misinterpreting te situation, not realizing tat te gas as already equilibrated to 0 C wen te piston is lowered. One possible reason students migt give for claiming te temperature cange cannot be determined (answer 4) is tat tey do not know weter te mixture contains enoug ice. Question J7.01b Description: Developing understanding of te ideal gas law in an isotermal context. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F As te piston is slowly lowered, te applied force F: 1. Increases. Decreases. Stays te same 4. Te cange in te applied force cannot be determined. Commentary Purpose: To understand te ideal gas law for an isotermal process. Discussion: As te piston is lowered, te temperature of te gas remains constant at 0 C, but te volume decreases. According to te ideal gas law (PV = nrt ), if te temperature is constant but te volume is decreasing, te pressure must increase. Terefore, as te piston is lowered, te force must increase to keep it in equilibrium. Key Points: For an ideal gas at constant temperature, pressure and volume ave an inverse relationsip. Tat is, if one decreases, te oter must increase.
584 Capter 1 For Instructors Only One possible source of confusion ere is tat students migt not appreciate te connection between te pressure of te gas and te force applied to keep te piston from moving. Constructing a free-body diagram for te piston may elp. Students occasionally assume tat te letter F indicates a constant value for te force, and coose answer (). Question J7.01c Description: Exploring termodynamic work in an isotermal context. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F As te piston is slowly raised, te work done on te gas is: 1. Positive. Negative. Zero 4. Te sign of te work done on te gas cannot be determined. Commentary Purpose: To develop your understanding of termodynamic work for an isotermal process. Discussion: As te piston is raised, te work done by te gas on te piston is positive: te force is up, and te displacement of te piston is up also. It does not matter weter te force is constant or canging. However, te work done on te gas is negative, because te displacement is again up, but te force F is down. Key Points: Te work done by a force is positive if te object acted upon moves in te same direction as te force, and negative if it moves in te opposite direction. Te work done on a gas and te work done by te gas ave opposite signs. Be careful not to confuse tem. If a gas expands, it does positive work on its environment.
Te Laws of Termodynamics 585 For Instructors Only Students often get confused about te work done on te gas (by an external agent) vs. te work done by te gas (on te environment). Some students migt still ave trouble understanding ow te work done can be negative. Depending on ow long it as been since students ave used work and energy ideas, it migt be a good idea to review tem in more familiar contexts. Students migt be unduly distracted by te isotermal process, not able to ignore it even toug it is irrelevant. Tey migt try to apply termodynamic principles to deduce te sign of te work done on te gas, rater tan simply apply te definition of work. For instance, tey migt conclude tat te work done must be zero, since bot te temperature and internal energy are constant. Question J7.01d Description: Developing understanding of eat transfer during an isotermal process. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F As te piston is slowly raised: 1. Te gas gives off eat to te environment (Q < 0). Te gas absorbs eat from te environment (Q > 0). Te gas does not excange any eat wit te environment (Q = 0) 4. Te eat excanged wit te environment cannot be determined. Commentary Purpose: To understand eat excange during an isotermal process. Discussion: An ice-water mixture maintains a constant temperature of 0 C unless all te ice melts or all te water freezes. A gas in good termal contact wit suc a mixture remains at 0 C; if energy is added to (or removed from) te gas, eat flows out of (or into) te gas so its temperature stays constant. If temperature is constant, internal energy E is also ( E = 0). As te piston is raised, te work W done on te gas is negative. According to te first law of termodynamics, te eat Q absorbed by te gas must be positive (Q + W = 0). Tus, te gas absorbs eat from te environment.
586 Capter 1 Key Points: An isotermal system as constant internal energy ( E = 0). According to te first law of termodynamics, any work done on (or by) an isotermal system must cause an equal amount of eat to be expelled (or absorbed). For Instructors Only Students migt not realize tat wen a gas is at a constant temperature, te internal energy is constant too. Students migt not fully appreciate tat work and eat cause canges in internal energy. Students migt not believe tat te work done on a gas can be negative. You need to create sufficient conflict to elp tem callenge tis belief. For instance, wen te gas expands it does work on te environment. Were does it get te energy to do tis work? Its internal energy is constant. Te only way to get energy is to absorb energy from te ice-water mixture, so Q > 0. If Q > 0 and E = 0, ten W < 0. Tat is, te work done on te gas is negative. Students migt not be sure exactly wat is meant by te environment. In tis case, it is te ice-water mixture, since tat is wat it is in termal contact wit. In oter words, tere are two environments. Te gas is in termal contact wit an environment wit wic it excanges eat (microscopic) energy, and it is in mecanical contact wit an environment wit wic it excanges mecanical (macroscopic) energy. In tis case, te ice-water mixture is te termal environment, and te piston and applied force F is te mecanical environment. Question J7.01e Description: Honing te concept of a termodynamic system s environment. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F As te piston is slowly raised, is it possible for te temperature of te gas to be different from te temperature of te environment? 1. Yes. No
Te Laws of Termodynamics 587 Commentary Purpose: To develop your understanding of wat te environment of a termodynamic system refers to. Discussion: In pysics, we typically separate te world into a system and its environment. Sometimes te environment is poorly defined. Generally, te environment refers to tat part of te world te system is in contact wit or tat affects its beavior. In tis case, te gas is te system, and it is in termal contact wit te ice-water mixture and mecanical contact wit te piston. We usually assume tat te piston is insulating, so tat we do not need to worry about eat excange wit it. Since tis question asks about temperatures, we can interpret environment as referring to wat te gas is in termal contact wit: te ice-water mixture. Terefore, te best answer is No, te gas cannot ave a different temperature tan te environment (as long as te piston is raised sufficiently slowly). Key Points: Te environment of a system is usually tat part of te rest of te world tat contacts or oterwise directly affects te system. How we interpret environment depends on te specific situation and question asked. For Instructors Only We tend to use a rater narrow definition of environment in tese situations, toug students are not always sensitized to its narrow use. It is reasonable for tem to tink tat te temperature of te rest of te world is different tan 0 C, so students wo answer Yes are not necessarily confused about te pysics involved. Tey migt understand completely tat te gas is always at 0 C. Tey simply are not using te same definition of environment as you are. We recommended letting students debate tis question, explaining teir reasoning and assumptions and also callenging eac oter s answers. After a few minutes, you can provide te narrow definition of environment typically used in termodynamics. Question J7.01f Description: Developing understanding of adiabatic (termally isolated) processes. Question A gas is placed in a container, wic is immersed in an ice-water mixture as sown below. F