Chapter 18: The Laws of ThermodynamicsAnswers to Even-Numbered Conceptual Questions2. (a) Yes. Heat can flow into the system if at the same time the system expands, as in an isothermal expansionof a gas. (b) Yes. Heat can flow out of the system if at the same time the system is compressed, as in anisothermal compression of a gas.4. No. The heat might be added to a gas undergoing an isothermal expansion. In this case, there is no change inthe temperature.6. Yes. In an isothermal expansion, all the heat added to the system to keep its temperature constant appears aswork done by the system.8. The final temperature of an ideal gas in this situation is T; that is, there is no change in temperature. Thereason is that as the gas expands into the vacuum, it does no work because it has nothing to push against.The gas is also insulated, so no heat can flow into or out of the system. It follows that the internal energy ofthe gas is unchanged, which means that its temperature is unchanged as well.10. This would be a violation of the second law of thermodynamics, which states that heat always flows from ahigh-temperature object to a low-temperature object. If heat were to flow spontaneously between objects ofequal temperature, the result would be objects at different temperatures. These objects could then be used torun a heat engine until they were again at the same temperature, after which the process could be repeatedindefinitely.12. Yes. In fact, the heat delivered to a room is typically 3 to 4 times the work done by the heat pump.14. The law of thermodynamics most pertinent to this situation is the second law, which states that physicalprocesses move in the direction of increasing disorder. To decrease the disorder in one region of spacerequires work to be done, and a larger increase in disorder in another region of space.Solutions to Problems and Conceptual Exercises1. Picture the Problem: Thermodynamic systems change their internal energy when heat flows and when work is done.Strategy: Use equation 18-3 to find the change in internal energy for each system.Solution: 1. (a) Calculate U: ΔU =Q −W =50 J −50 J = 0 J2. (b) Calculate U: ΔU =Q −W =−50 J − −50 J( )= 0 J3. (c) Calculate U: ΔU =Q −W =−50 J − 50 J( )= −100 JInsight: In part (c) 50 J of heat energy flows out of the system, and the system does work on the external world, removing an additional 50 J of energy from the system. The net effect is 100 J of energy is removed from the system.2. Picture the Problem: A gas expands, doing 100 J of work while its internal energy increases by 200 J.Strategy: Use equation 18-3 to find the heat flow into the system.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.18 – 1Chapter 18: The Laws of Thermodynamics James S. Walker, Physics, 4th EditionSolution: Solve equation 18-3 for Q: Q =W +ΔU =100 J +200 J = 300 JInsight: The work is positive because the system is doing work on the external world. The heat flow into the system provides the energy to do this work plus an additional 200 J for the system to store as internal energy.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.18 – 2Chapter 18: The Laws of Thermodynamics James S. Walker, Physics, 4th Edition3. Picture the Problem: A swimmer does work and gives off heat during a workout.Strategy: The heat and work are given in the problem. Use equation 18-3 to find the change in internal energy.Solution: 1. Write down the heat: Q = −4.1×105J2. Write down the work: W = 6.7×105J3. Find the change in internal energy: ΔU =Q −W =−4.1×105J −6.7×105J = −10.8×105JInsight: The heat is negative because the system (the swimmer) loses heat. Q is positive when the system gains heat.4. Picture the Problem: The temperature of one mole of an ideal gas increases as heat is added.Strategy: Use the first law of thermodynamics to determine the work, recognizing that the change in internal energy of a monatomic gas is given by ΔU =32nRΔT .Solution: Solve equation 18-3 for W: W =Q −ΔU =Q −32nRΔT=1210 J −321mol( )8.31J/ mol ⋅K( )⎡⎣⎤⎦276K −272K( )=1160 JInsight: Of the 1210 J of heat flow into the gas, only 50 J was converted to internal energy. The rest was energy for thegas to do work on the external world.5. Picture the Problem: Three different processes act on a system, resulting in different final states.Strategy: Use the first law of thermodynamics (equation 18-3) to solve for the unknown quantity in each process.Solution: 1. (a) Apply equation 18-3 directly: ΔU =Q −W =77 J −(−42 J ) = 119 J2. (b) Apply equation 18-3 directly: ΔU =Q −W =77 J −42 J = 35 J3. (c) Solve equation 18-3 for Q: Q =ΔU +W =−120 J +120 J = 0Insight: Conservation of energy requires that the net energy flow (Q and −W) into the system must equal the change in internal energy, irregardless of the process by which the energy exchange occurs.6. Picture the Problem: A cycle of four processes is shown on the pressure–volume diagram at right.Strategy: Set the sum of the changes in internal energy equal to zero. Then solve for ΔUCD.Solution: 1. Sum the changes in internal energy: ΔUAB+ΔUBC+ΔUCD+ΔUDA=02. Solve for ΔUCD: ΔUCD=−ΔUAB−ΔUBC−ΔUDA3. Insert the numeric values: ΔUCD=−82 J −15 J − −56 J( )= −41JInsight: In a complete cycle the system returns to its original state, which means that the internal energy must return to its initial value. Therefore the net change in internal energy must be zero.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.18 – 3Chapter 18: The Laws of Thermodynamics James S. Walker, Physics, 4th Edition7. Picture the Problem: In a basketball game, the player does work and gives off heat in the form of perspiration.Strategy: Use the first law of thermodynamics (equation 18-3) to
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