Chapter 17: Phases and Phase ChangesAnswers to Even-Numbered Conceptual Questions2. If the temperature of the air in a house is increased, and the amount of air in the house remains constant, itfollows from the ideal gas law that the pressure will increase as well.4. Yes. If the pressure and volume are changed in such a way that their product remains the same, it followsfrom the ideal gas law that the temperature of the gas will remain the same. If the temperature of the gas isthe same, the average kinetic energy of its molecules will not change.6. No. The temperature at which water boils on a mountaintop is less than its boiling temperature at sea leveldue to the low atmospheric pressure on the mountain. Therefore, if the stove is barely able to boil water onthe mountain, it will not be able to boil it at sea level, where the required temperature is greater.8. If we look at the phase diagram in Figure 17-16, we can see that in order to move upward in the graph fromthe sublimation curve to the fusion curve, the pressure acting on the system must be increased.10. No. Water is at 0˚C whenever it is in equilibrium with ice. The ice cube thrown into the pool will soon melt,however, showing that the ice cube–pool system is not in equilibrium. As a result, there is no reason toexpect that the water in the pool is at 0˚C.Solutions to Problems and Conceptual Exercises1. Picture the Problem: We are to compare a mole of N2 and a mole of O2.Strategy: Use the concept of a mole to answer the conceptual questions. Solution: 1. (a) By the definition of a mole, the number of molecules in one mole of N2 is equal to the number of molecules in one mole of O2.2. (b) The mass of one mole of N2 (28 g) is less than the mass of one mole of O2 (32 g) because each molecule of O2 is more massive than an N2 molecule. Insight: A mole is similar to a dozen; the words always refer to the same number, 6.02×1023 and 12, respectively.2. Picture the Problem: We are to compare a mole of He and a mole of O2.Strategy: Use the concept of a mole to answer the conceptual question. Solution: By the definition of a mole, the number of molecules in one mole of He is equal to the number of molecules in one mole of O2. However, each molecule of O2 has two atoms, but the He molecules are monatomic, so the number of atoms in one mole of helium is less than the number of atoms in one mole of oxygen.Insight: A mole is similar to a dozen; the words always refer to the same number, 6.02×1023 and 12, respectively.3. Picture the Problem: A helium-filled balloon at room temperature is placed in the refrigerator.Strategy: Use the ideal gas model to answer the conceptual question. Solution: 1. (a) Treating the helium as an ideal gas, it satisfies the relation PV = nRT. When we put the balloon in the refrigerator, the pressure and the number of moles remain constant. Under these conditions, a decrease in temperature implies a decrease in volume. Therefore, if you put a helium-filled balloon in the refrigerator its volume will decrease.2. (b) The best explanation is I. Lowering the temperature of an ideal gas at constant pressure results in a reduced volume. Statement II neglects the effect of the temperature change, and statement III is false.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.17 – 1Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th EditionInsight: Likewise, either placing the balloon at constant pressure inside a hot automobile or placing it at constant temperature in a low pressure region will result in an increase in the balloon’s volume.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.17 – 2Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition4. Picture the Problem: Two containers hold ideal gases at the same temperature. Container A has twice the volume and half the number of molecules as container B.Strategy: Use the ideal gas model to answer the conceptual question. Solution: From the given information we know that VA = 2VB and nA = nB/2. The pressure of an ideal gas is given by P = nRT/V. If PB = nBRT/VB it follows that PA = (nB/2)RT/(2VB) = PB/4, or PAPB=1 4.Insight: A mole is similar to a dozen; the words always refer to the same number, 6.02×1023 and 12, respectively.5. Picture the Problem: An ideal gas at standard temperature and pressure occupies a fixed volume.Strategy: Solve the ideal gas law for the volume at standard temperature and pressure. Solution: Solve equation 17-5 for the volume: V =nRTP=1 mol 8.31J/ mol⋅K( )⎡⎣⎤⎦273.15 K1.013×105Pa= 0.0224 m3Insight: This is slightly smaller than a cubic foot.6. Picture the Problem: A person inhales and holds her breath. The volume of the air in her lungs expands as the temperature increases.Strategy: Use equation 17-8 to solve for the final volume and then subtract the initial volume to find the increase in volume.Solution: 1. Use equation 17-8 to find the volume ratio: ViTi=VfTf ⇒ VfVi=TfTi=273.15+37( ) K273.15 K=1.143. Subtract the initial volume: ΔV =Vf−Vi=1.14Vi−Vi=0.14 4.1 L( )= 0.57 LInsight: Another approach would be to calculate the final volume (4.7 L) and then subtract the initial volume (4.1 L) butthe 0.6 L result has only one significant digit due to the rules of subtraction. 7. Picture the Problem: Air contained within a car’s tires has a fixed volume and mass. The pressure inside the tires increases as the temperature increases.Strategy: Use the ideal gas law to develop a relationship between temperature and pressure. Solve this expression for the final temperature.Solution: 1. Arrange equation 17-5 to relate the initial and final conditions of the gas, and eliminate V and n: PiVnTi=PfVnTf ⇒ PiTi=PfTf2. Solve for the final temperature: Tf=PfTiPi=554 kPa( )286 K( )501 kPa= 316 KInsight: This final temperature is 107F, a very warm afternoon.8. Picture the Problem: An automobile tire has a fixed volume and temperature. Adding air to the tire increases the tire pressure.Strategy: Solve the ideal gas law, equation