Which of these molecules exhibit hydrogen bonding h5 methane chloro h5 methanolWhich of these molecules exhibit hydrogen bonding h5 methane chloro h5 methanol
Instructor’s Solutions Manual for Physical Chemistry Thomas Engel University of Washington Philip Reid University of Washington San Francisco Boston New York Cape Town Hong Kong London Madrid Mexico City Montreal Munich Paris Singapore Sydney Tokyo Toronto ISBN 0-8053-3854-3 Copyright ©2006 Pearson Education, Inc., publishing as Benjamin Cummings, 1301 Sansome St., San Francisco, CA 94111. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E. Lake Ave., Glenview, IL 60025. For information regarding permissions, call (847) 486-2635. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. www.aw-bc.com Preface This Instructor’s Solutions Manual has a twofold purpose. First, and most obvious, is to provide worked solutions for the use of instructors. Second, but equally important, is to provide examples of good problem-solving techniques and strategies that will benefit your students if you share these solutions with them. Most education researchers believe that it is more beneficial for students to study a smaller number of carefully chosen problems in detail, including variations, than to race through a larger number of poorly understood calculations. The solutions presented here are intended to provide a basis for this practice. Please note that Benjamin Cummings has copyrighted the Instructor's Solutions Manual and permits posting these solutions to password-protected sites only. Posting of these solutions to open-access sites is forbidden in order to prevent dissemination of the solutions via the Internet. Should you have any questions concerning this policy, please direct them to the Chemistry Editor at Benjamin Cummings. We have made every effort to be accurate and correct in these solutions. However, if you do find errors or ambiguities, we would be very grateful to hear from you. Please contact us at: [email protected] Thomas Engel University of Washington Philip Reid University of Washington Chapter 1: Fundamental Concepts of Thermodynamics Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q1.1) The location of the boundary between the system and the surroundings is a choice that must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight room. Is the system open or closed if you place the boundary just outside the liquid water? Is the system open or closed if you place the boundary just inside the walls of the room? If the system boundaries are just outside of the liquid water, the system is open because water can escape from the top surface. The system is closed if the boundary is just inside the walls, because the room is airtight. Q1.2) Real walls are never totally adiabatic. Order the following walls in increasing order with respect to being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cmthick copper, 1-cm-thick cork. 1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of interest a necessary condition for equilibrium between two systems? Equilibrium is a dynamic process in which the rates of two opposing processes are equal. However, if the rate in each direction is zero, no exchange is possible, and therefore the system can not reach equilibrium. Q1.4) At sufficiently high temperatures, the van der Waals equation has the form RT . Note that the attractive part of the potential has no influence in this Vm − b expression. Justify this behavior using the potential energy diagram in Figure 1.6. P≈ At high temperatures, the energy of the molecule is large as indicated by the colored rectangular area in the figure below. 1-1 Chapter 1/Fundamental Concepts of Thermodynamics In this case, the well depth is a small fraction of the total energy. Therefore, the particle is unaffected by the attractive part of the potential. Q1.5) The parameter a in the van der Waals equation is greater for H2O than for He. What does this say about the form of the potential function in Figure 1.6 for the two gases? It says that the depth of the attractive potential is greater for H2O than for He. Problems P1.1) A sealed flask with a capacity of 1.00 dm3 contains 5.00 g of ethane. The flask is so weak that it will burst if the pressure exceeds 1.00 × 106 Pa. At what temperature will the pressure of the gas exceed the bursting temperature? 1.00 × 106 Pa × 10−3 m3 PV = = 723 K T= 5.00 × g nR −1 −1 × 8.314J mol K 30.07 g mol−1 P1.2) Consider a gas mixture in a 2.00-dm3 flask at 27.0ºC. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent. a) 1.00 g H2 and 1.00 g O2 b) 1.00 g N2 and 1.00 g O2 c) 1.00 g CH4 and 1.00 g NH3 1-2 Chapter 1/Fundamental Concepts of Thermodynamics a) nH 2 RT 1.00 2.016 mol × 8.314 J mol−1 K −1 × 300 K = 6.24 × 105 Pa −3 3 V 2.00 × 10 m nO2 RT 1.00 32.00 mol × 8.314 J mol−1 K −1 × 300 K PO2 = = = 3.90 × 104 Pa −3 3 2.00 × 10 m V 5 Ptotal = 6.57 × 10 Pa mol H 2 1.00 2.016 mol % H 2 = 100 × = 100 × = 94.1% mol H 2 + mol O2 1.00 2.016 + 1.00 32.00 PH 2 = = mol % O2 = 100 × mol O2 1.00 32.00 = 100 × = 5.9% mol H 2 + mol O 2 1.00 2.016 + 1.00 32.00 b) nN2 RT 1.00 28.02 mol × 8.314 J mol−1 K −1 × 300 K = 4.45 × 104 Pa −3 3 2.00 × 10 m V nO RT 1.00 32.00 mol × 8.314 J mol−1 K −1 × 300 K = = 3.90 × 104 Pa PO2 = 2 2.00 × 10−3 m3 V Ptotal = 8.35 × 104 Pa mol N 2 1.00 28.02 mol % N 2 = 100 × = 100 × = 53.3% mol N 2 + mol O2 1.00 28.02 + 1.00 32.00 PN2 = = mol % O2 = 100 × mol O2 1.00 32.00 = 100 × = 46.7% mol N 2 + mol O 2 1.00 28.02 + 1.00 32.00 c) nNH3 RT 1.00 17.03 mol × 8.314 J mol−1 K −1 × 300 K PNH3 = = = 7.32 × 104 Pa −3 3 V 2.00 × 10 m nCH 4 RT 1.00 16.04 mol × 8.314 J mol−1 K −1 × 300 K PCH 4 = = = 7.77 × 104 Pa V 2.00 × 10−3 m3 Ptotal = 1.51 × 105 Pa mol NH 3 1.00 17.03 mol % NH 3 = 100 × = 100 × = 48.5% mol NH 3 + mol CH 4 1.00 17.03 + 1.00 16.04 mol % O2 = 100 × mol CH 4 1.00 16.04 = 100 × = 51.5% mol NH 3 + mol CH 4 1.00 17.03 + 1.00 16.04 P1.3) Suppose that you measured the product PV of one mole of a dilute gas and found that PV = 22.98 L atm at 0ºC and 31.18 L atm at 100ºC. Assume that the ideal gas law is valid, with T = t(ºC) + a, and that the value of R is not known. Determine R and a from the measurements provided. Expressing the ideal gas law in the form PV = R(t + a) = m(t + a), 1-3 Chapter 1/Fundamental Concepts of Thermodynamics PV ( 31.18-22.98) l atm mol = = 0.08200 atm mol−1 o C−1 = R o t (100-0) C −1 m= PV 31.18l atm mol−1 −t = − 100D C = 280.2D C a= −1 o −1 R .08200 atm mol C P1.4) A compressed cylinder of gas contains 1.50 × 103 g of N2 gas at a pressure of 2.00 × 107 Pa and a temperature of 17.1ºC. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 1.80 × 105 Pa? Assume ideal behavior and that the gas temperature is unchanged. Let ni and nf be the initial and final number of mols of N2 in the cylinder. ni RT n f RT = Pi Pf n f = ni Pf Pi = 1.50×103g 1.80 × 105 Pa × = 0.482 mol 28.01g mol−1 2.00 × 107 Pa 1.50×103g = 53.55 mol ni = 28.01g mol−1 The volume of gas released into the atmosphere is given by n f − ni RT ( 53.55 − 0.482) mol×8.2057×10−2 L atm mol−1K −1 ×290.2 K V= = 1atm P ( ) = 1.26 × 103 L P1.5) A balloon filled with 10.50 L of Ar at 18.0ºC and 1 atm rises to a height in the atmosphere where the pressure is 248 Torr and the temperature is –30.5ºC. What is the final volume of the balloon? 760 Torr × ( 273.15 − 30.5) K P Tf Vf = i Vi = ×10.50 L = 26.8 L Pf Ti 248 Torr × ( 273.15 + 18.0 ) K P1.6) Consider a 20.0-L sample of moist air at 60ºC and 1 atm in which the partial pressure of water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mole percent N2, 21.0 mole percent O2, and 1.00 mole percent Ar. a) What are the mole percentages of each of the gases in the sample? PH O b) The percent relative humidity is defined as % RH = *2 where PH 2O is the partial PH 2O pressure of water in the sample and PH∗2O = 0.197 atm is the equilibrium vapor pressure of water at 60ºC. The gas is compressed at 60ºC until the relative humidity is 100%. What volume does the mixture now occupy? 1-4 Chapter 1/Fundamental Concepts of Thermodynamics c) What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 200 atm? a) mol % N 2 = 100 × mol % O2 = 100 × mol % Ar = 100 × PN 2 Ptotal PO2 Ptotal = 100 × 0.78 × 0.88 atm = 68.6% 1 atm = 100 × 0.21 × 0.88 atm = 18.5% 1 atm PAr 0.01 × 0.88 atm = 100 × = 0.9% Ptotal 1 atm mol % H 2 O = 100 × PH 2O Ptotal = 100 × 0.12 atm = 12.0% 1 atm b) PH 2OV = nH 2O RT V PH′2OV ′ = PH 2OV where the primed quantities refer to 100% RH PH 2OV V′ = PH′ 2O = 0.12 atm × 20.0 L = 12.2 L 0.197 atm c) If all the water remained in the gas phase, the partial pressure of water at a total pressure of 200 atm would be PH 2O = Ptotal × mol fraction H 2 O = 200 atm × 0.12 = 24.0 atm However, the partial pressure of water cannot be greater than 0.197 atm, and the excess will condense. The fraction that condenses is given by 24.0 atm − 0.197 atm fraction condensed = = 0.992 24.0 atm P1.7) A mixture of 2.50 × 10–3 g of O2, 3.51 × 10–3 mol of N2, and 4.67 × 1020 molecules of CO are placed into a vessel of volume 3.50 L at 5.20ºC. a) Calculate the total pressure in the vessel. b) Calculate the mole fractions and partial pressures of each gas. a) 2.50×10−3g 4.67 × 1020 molecules −5 = ; n 7.81×10 mol = = 7.75×10−4 mol CO 32.0 g mol−1 6.022 × 1023 molecules mol−1 = nO2 + nN2 + nCO = 7.81×10−5 mol + 3.51×10−3 mol + 7.75×10−4 mol = 4.36×10−3 mol nO2 = ntotal Ptotal nRT 4.36×10−3mol×8.314×10−2 L bar mol−1K −1 × 278.3 K = = = 2.88×10−2 bar V 3.50 L 1-5 Chapter 1/Fundamental Concepts of Thermodynamics b) 7.81×10−5 mol 3.51×10−3 mol xO2 = = 0.0179 ; xN2 = = 0.803 ; 4.36×10−3 mol 4.36×10−3 mol 7.75×10−4 mol = 0.178 4.36×10−3 mol PO2 = xO2 Ptotal = 0.0179×2.88×10−2 bar = 5.16×10−4 bar xCO = PN2 = xN2 Ptotal = 0.803×2.88×10−2 bar = 2.31×10−2 bar PCO = xCO Ptotal = 0.177×2.88×10−2 bar = 5.10×10−3bar P1.8) Liquid N2 has a density of 875.4 kg m–3 at its normal boiling point. What volume does a balloon occupy at 18.5ºC and a pressure of 1.00 atm if 2.00 × 10–3 L of liquid N2 is injected into it? ρ N Vliq nN 2 = 2 M N2 VN2 = nN2 RT P = ρ N Vliq RT 2 M N2 P 875.4 g L−1 ×2.00×10−3 L×8.2057×10−2 L atm K −1mol−1 × ( 273.15 + 18.5) K = 28.01 g mol−1 ×1 atm = 1.50 L P1.9) A rigid vessel of volume 0.500 m3 containing H2 at 20.5ºC and a pressure of 611 × 103 Pa is connected to a second rigid vessel of volume 0.750 m3 containing Ar at 31.2ºC at a pressure of 433 × 103 Pa. A valve separating the two vessels is opened and both are cooled to a temperature of 14.5ºC. What is the final pressure in the vessels? PV 611 × 103 Pa × 0.500m3 nH 2 = = = 125 mol RT 8.314 J mol−1K −1 × ( 273.15 + 20.5) K nAr = P= PV 433×103 Pa ×0.750m3 = = 128 mol RT 8.314J mol−1K −1× ( 273.15 + 31.2) K −1 −1 nRT (125 + 128 ) mol×8.314J mol K × ( 273.15 + 14.5 ) K = = 4.84×105 Pa 3 V (0.500 + 0.750) m P1.10) A sample of propane (C3H8) placed in a closed vessel together with an amount of O2 that is 3.00 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation assuming that the H2O is present as a gas. The reaction is C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g). If m mol of propene are present initially, there must be 15m mol of O2. After the reaction is complete, there are 3m mol of CO2, 4m mol of H2O, and 10m mol of O2. 1-6 Chapter 1/Fundamental Concepts of Thermodynamics xCO2 = 3m 4m 10m = 0.176 ; xH 2O = = 0.235 ; xO2 = = 0.588 17m 17m 17m P1.11) A glass bulb of volume 0.136 L contains 0.7031 g of gas at 759.0 Torr and 99.5ºC. What is the molar mass of the gas? m PV RT = ;M = m n= M RT PV 8.2057×10−2 L atm mol−1K −1 × ( 273.15 + 99.5 ) K = 158 amu M = 0.7031g × 759 atm×0.136 L 760 P1.12) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.400 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole percent? 2H2(g) + O2(g) → 2H2O(l) nOD 2 0 initial moles nHD 2 at equilibrium nHD 2 − 2α nOD 2 − α 2α If the O2 is completely consumed, nOD 2 − α = 0 or α = nOD 2 . The number of moles of H2 remaining is nHD 2 − 2α = nHD 2 − 2nOD 2 . Let P1 be the initial total pressure and P2 be the total pressure after all the O2 is consumed. RT RT and P2 = nHD 2 − 2nOD 2 P1 = nHD 2 + nOD 2 V V Dividing the second equation by the first ( ) ( ) nHD 2 nOD 2 P2 = −2 D = xH 2 − 2 xO2 = 1 − xO2 − 2 xO2 = 1 − 3xO2 P1 nHD 2 + nOD 2 nH 2 + nOD 2 P 1 xO2 = 1 − 2 = P1 3 1 0.400 atm 1 − = 0.20; 3 1.00 atm1 xH 2 = 0.80 P1.13) A gas sample is known to be a mixture of ethane and butane. A bulb of 200.0- cm3 capacity is filled with the gas to a pressure of 100.0 × 103 Pa at 20.0ºC. If the weight of the gas in the bulb is 0.3846 g, what is the mole percent of butane in the mixture? 1-7 Chapter 1/Fundamental Concepts of Thermodynamics n1 = moles of ethane n2 = moles of butane PV 100.0 × 103 Pa × 0.200 × 10−3 m3 = = 8.21 × 10−3 mol −1 −1 RT 8.314 J mol K × 293 K The total mass is n1M 1 + n2 M 2 = 0.3846 g n1 + n2 = Dividing this equation by n1 + n2 n1M 1 nM 0.3846 g + 2 2 = = 46.87 g mol−1 n1 + n2 n1 + n2 8.21 × 10−3 mol x1M 1 + x2 M 2 = (1 − x2 ) M 1 + x2 M 2 = 46.87 g mol−1 x2 = 46.87 g mol−1 − M 1 46.87 g mol−1 − 30.069 g mol−1 = = 0.599 M 2 − M1 58.123 g mol−1 − 30.069 g mol−1 mole % = 59.9% P1.14) When Julius Caesar expired, his last exhalation had a volume of 500 cm3 and contained 1.00 mole percent argon. Assume that T = 300 K and P = 1.00 atm at the location of his demise. Assume further that T and P currently have the same values throughout the Earth's atmosphere. If all of his exhaled CO2 molecules are now uniformly distributed throughout the atmosphere (which for our calculation is taken to have a thickness of 1.00 km), how many inhalations of 500 cm3 must we make to inhale one of the Ar molecules exhaled in Caesar's last breath? Assume the radius of the Earth to be 6.37 × 106 m. [Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of the Earth’s surface and a height equal to the thickness of the atmosphere. See Problem P1.15 for the dependence of the barometric pressure on the height above the Earth’s surface.] ∞ D D is the The total number of Ar atoms in the atmosphere is N Ar = ∫ N Ar Adz where N Ar 0 number of Ar atoms per m at the surface of the earth. N is given by 3 D Ar N P 6.023 × 1023 × 0.0100 × 1 × 105 Pa D N Ar = A Ar = = 2.41 × 1023 m −3 −1 −1 RT 8.314 J K mol × 300 K The total number of Ar atoms in the atmosphere is ∞ ∞ M gz − Ar RT D D N Ar = ∫ N Ar Adz = N Ar ∫ e RT Adz = N Ar A M Ar g 0 0 = ( ) 2 2.41 × 1023 m −3 × 4π × 6.37 × 106 m × 8.314 J mol−1 K −1 × 300 K 39.9 × 10−3 kg × 9.81 m s −2 = 7.85 × 1041 The fraction of these atoms that came from Caesar’s last breath, f, is given by 1-8 Chapter 1/Fundamental Concepts of Thermodynamics D N Ar V 2.41 × 1023 m −3 × 0.500 × 10−3 m3 f = = = 1.53 × 10−22 41 N Ar 7.85 × 10 The number of Ar atoms that we inhale with each breath is PV 10−2 × 1 × 105 Pa × 0.500 × 10−3 m3 23 N = NA = 6.023 × 10 × = 1.21 × 1020 −1 −1 RT 8.314 J mol K × 300 K The number of these that came from Caesar’s last breath is fN fN = 1.53 × 10−22 × 1.21 × 1020 = 1.85 × 10−2 The reciprocal of this result, or 54, is the number of breaths needed to inhale one Ar atom that Caesar exhaled in his last breath. P1.15) The barometric pressure falls off with height above sea level in the Earth's − M i gz atmosphere as Pi = Pi 0 e RT where Pi is the partial pressure at the height z, Pi 0 is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, and T is the absolute temperature. Consider an atmosphere that has the composition xN 2 = 0.600 and xCO2 = 0.400 and that T = 300 K. Near sea level, the total pressure is 1.00 bar. Calculate the mole fractions of the two components at a height of 50.0 km. Why is the composition different from its value at sea level? M gz − i 28.04 × 10−3 kg × 9.81 m s −2 × 50 × 103 m 0 5 RT PN2 = PN2 e = 0.600 ×1.0125 ×10 Pa exp − 8.314 J mol−1 K −1 × 300 K = 242 Pa 0 CO2 PCO2 = P e xCO2 − M i gz RT 44.04 × 10−3 kg × 9.81 m s −2 × 50 × 103 m = 0.400 ×1.0125 × 10 Pa exp − 8.314 J mol−1 K −1 × 300 K 5 = 6.93 Pa PCO2 6.93 = = = 0.028 PCO2 + PN2 6.93 + 242 xN2 = 1 − xCO2 = 0.972 The mole fraction of CO2 at the high altitude is much reduced relative to its value at sea level because it has a larger molecular mass than N2. P1.16) Assume that air has a mean molar mass of 28.9 g mol–1 and that the atmosphere has a uniform temperature of 25.0ºC. Calculate the barometric pressure at Denver, for which z = 1600 m. Use the information contained in Problem P1.15. M gz − i 28.9 × 10−3 kg × 9.81 m s −2 × 1600 m 4 P = P 0e RT = 105 Pa exp − = 8.34 × 10 Pa −1 −1 8.314 J mol K × 300 K P1.17) Calculate the pressure exerted by Ar for a molar volume 1.42 L at 300 K using the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1-9 Chapter 1/Fundamental Concepts of Thermodynamics 1.355 bar dm6 mol–2 and 0.0320 dm3 mol–1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions? RT a P= − 2 Vm − b Vm = 8.314×10−2 bar dm3 mol−1K −1 ×300K 1.355 bar dm 6 mol−2 − = 17.3 bar 2 1.42 dm3 mol−1 − 0.0321dm3 mol−1 1.42 dm3 mol−1 ( Pideal = −2 −1 ) −1 RT 8.3145 × 10 ×L bar mol K ×300 K = = 17.6 bar 1.42 L V Because P < Pideal, the attractive part of the potential dominates. P1.18) Calculate the pressure exerted by benzene for a molar volume 1.42 L at 790 K using the Redlich- Kwong equation of state: RT a 1 nRT n2a 1 P= − = − . The Redlich-Kwong Vm − b T Vm (Vm + b ) V − nb T V (V + nb ) parameters a and b for benzene are 452.0 bar dm6 mol–2 K1/2 and 0.08271 dm3 mol–1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions? RT a 1 − P= Vm − b T Vm (Vm + b ) = 8.314×10−2 bar dm3 mol−1K −1 ×790 K 1.42 dm3 mol−1 − 0.08271dm3 mol−1 6 − −2 1 2 452.0 bar dm mol K 1 × −1 3 3 1.42 dm mol × 1.42 dm mol−1 + 0.08271dm3 mol−1 790 K ( P = 41.6 bar RT 8.3145×10−2 ×L bar mol−1K −1 ×790 K = = 46.3 bar Pideal = V 1.42 L Because P < Pideal, the attractive part of the potential dominates. P1.19) Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas constant is 1.00 J G–1 mol–1. Let T and T ′ represent the Kelvin and G scales, and R and R ′ represent the gas constant in each of these scales. Then PV = nRT = nR ′T ′ T′ = R T = 8.314T R′ The temperature on the G scale is the value in K multiplied by 8.314. 1-10 ) Chapter 1/Fundamental Concepts of Thermodynamics P1.20) A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the reaction CuO + H2 → Cu + H2O and oxygen reoxidizes the copper formed according to Cu + 1/2 O2 → CuO. At 25ºC and 750 Torr, 100.0 cm3 of the mixture yields 84.5 cm3 of dry oxygen measured at 25ºC and 750 Torr after passage over CuO and the drying agent. What is the original composition of the mixture? Two equilibria must be considered: CuO(s) + H2(g) → H2O(l) + Cu at equilibrium α−β nOD 2 − α Cu (s) + ½ O2(g) → CuO 1 at equilibrium α−β nOD 2 − β 2 In the final state, only O2 is present. Therefore α = nOD 2 . In an excess of O2, all the 1 copper is oxidized. Therefore α−β = 0 or β = nOD 2 . We conclude that nO2 = nOD 2 − nHD 2 . 2 Let V1 and V2 be the initial and final volumes. RT 1 RT V1 = nHD 2 + nOD 2 V2 = nOD 2 − nHD 2 P 2 P ( ) Dividing the second equation by the first yields D nOD V2 1 nH 2 1 1 3 = D 2 D − = xOD 2 − xHD 2 = 1 − xHD 2 − xHD 2 = 1 − xHD 2 D D V1 nH 2 + nO2 2 nH 2 + nO2 2 2 2 xHD 2 = 2 V2 2 84.5 cm 3 1 1 − = − = 0.103; 3 V1 3 100.0 cm3 1-11 xOD 2 = 1 − xHD 2 = 0.897 Questions on Concepts Q2.1) Electrical current is passed through a resistor immersed in a liquid in an adiabatic container. The temperature of the liquid is raised by 1ºC. The system consists solely of the liquid. Does heat or work flow across the boundary between the system and surroundings? Justify your answer. Although work is done on the resistor, this work is done in the surroundings. Heat flows across the boundary between the surroundings and the system because of the temperature difference between them. Q2.2) Explain how a mass of water in the surroundings can be used to determine q for a process. Calculate q if the temperature of 1.00 kg of water in the surroundings increases by 1.25ºC. Assume that the surroundings are at a constant pressure. If heat flows across the boundary between the system and the surroundings, it will lead to q a temperature change in the surroundings given by T = . For the case of interest, CP q = − qsurroundings = − mCP T = −1000 g × 4.19 J g -1 K -1 × 1.25 K = −5.24 × 103 J . Q2.3) Explain the relationship between the terms exact differential and state function. In order for a function f(x,y) to be a state function, it must be possible to write the total ∂f ∂f differential df in the form df = dx + dy . If the form df as written exists, it is ∂x y ∂y x an exact differential. Q2.4) Why is it incorrect to speak of the heat or work associated with a system? Heat and work are transients that exist only in the transition between equilibrium states. Therefore, a state at equilibrium is not associated with values of heat or work. Q2.5) Two ideal gas systems undergo reversible expansion starting from the same P and V. At the end of the expansion, the two systems have the same volume. The pressure in the system that has undergone adiabatic expansion is lower that in the system that has undergone isothermal expansion. Explain this result without using equations. In the system undergoing adiabatic expansion, all the work done must come through the lowering of U, and therefore of the temperature. By contrast, some of the work done in the isothermal expansion can come at the expense of the heat that has flowed across the boundary between the system and surroundings. 13 Q2.6) A cup of water at 278 K (the system) is placed in a microwave oven and the the oven is turned on for one minute, during which it begins to boil. Which of q, w, and U are positive, negative or zero? The heat q is positive because heat flows across the system-surrounding boundary into the system. The work w is negative because the vaporizing water does work on the surroundings. U is positive because the temperature increases and some of the liquid is vaporized. Q2.7) What is wrong with the following statement?: Because the well insulated house stored a lot of heat, the temperature didn't fall much when the furnace failed. Rewrite the sentence to convey the same information in a correct way. Heat can’t be stored because it exists only as a transient. A possible rephrasing follows. Because the house was well insulated, the walls were nearly adiabatic. Therefore, the temperature of the house did not fall as rapidly when in contact with the surroundings at a lower temperature as would have been the case if the walls were diathermal. Q2.8) What is wrong with the following statement?: Burns caused by steam at 100ºC can be more severe than those caused by water at 100ºC because steam contains more heat than water. Rewrite the sentence to convey the same information in a correct way. Heat is not a substance that can be stored. When steam is in contact with your skin, it condenses to the liquid phase. In doing so, energy is released that is absorbed by the skin. Hot water does not release as much energy in the same situation, because no phase change occurs. Q2.9) Describe how reversible and irreversible expansions differ by discussing the degree to which equilibrium is maintained between the system and the surroundings. In a reversible expansion, the system and surroundings are always in equilibrium with one another. In an irreversible expansion, they are not in equilibrium with one another. Q2.10) A chemical reaction occurs in a constant volume enclosure separated from the surroundings by diathermal walls. Can you say whether the temperature of the surroundings increase, decrease, or remain the same in this process? Explain. No. The temperature will increase if the reaction is exothermic, decrease if the reaction is endothermic, and not change if no energy is evolved in the reaction. Problems 14 P2.1) 3.00 moles of an ideal gas at 27.0ºC expands isothermally from an initial volume of 20.0 dm3 to a final volume of 60.0 dm3. Calculate w for this process a) for expansion against a constant external pressure of 1.00 x 105 Pa, and b) for a reversible expansion. a) w = − Pexternal V = −1.00×105 Pa × ( 60.0-20.0 ) ×10-3 m 3 = −4.00×103J b) wreversible 60.0 dm3 = − nRT ln = −3.00mol×8.314 J mol K ×300 K ×ln = −8.22×103J 3 20.0 dm Vi Vf -1 -1 P2.2) 3.00 moles of a gas are compressed isothermally from 60.0 L to 20.0 L using a constant external pressure of 5.00 atm. Calculate q, w, U, and H. w = − Pexternal V = −5 × 1.013 × 105 Pa × ( 60 × 10−3 L − 20 × 10−3 L ) = −2.03 × 104 J U = 0 and H = 0 because T = 0 q = − w = 2.03 × 104 J P2.3) A system consisting of 57.5 g of liquid water at 298 K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 1.50 A passes through the 10.0 Ohm resistor for 150 s, what is the final temperature of the water? The heat capacity for water can be found in the Data Tables in Appendix A. q = I 2 Rt = nCP ,m ( T f − Ti ) ; T f = (1.50A ) 2 I 2 Rt + nCP ,mTi ×10.0 Ohm × 150 s + Tf = nCP ,m 57.5g ×75.291 J mol-1 K -1 ×298 K 18.02 g mol -1 57.5 g ×75.291 J mol-1 K -1 18.02 g mol-1 = 312 K P2.4) For one mole of an ideal gas, Pexternal = P = 200 x 103 Pa. The temperature is changed from 100ºC to 25.0ºC at constant pressure. CV,m = 3/2 R. Calculate q, w, U, and H. 3 × 8.314 J mol-1 K -1 × ( 298 K − 373 K ) = −935 J 2 H = nCP ,m T = n ( CV ,m + R ) T U = nCV ,m T = 5 × 8.314 J mol -1K -1 × ( 298 K − 373 K ) 2 = −1.56 × 103 J = = qP 15 w = U − qP = −935 J + 1.56 × 103 J = 624 J P2.5) Consider the isothermal expansion of 5.25 moles of an ideal gas at 450 K from an initial pressure of 15.0 bar to a final pressure of 3.50 bar. Describe the process that will result in the greatest amount of work being done by the system with Pexternal ≥ 3.50 bar and calculate w. Describe the process that will result in the least amount of work being done by the system Pexternal ≥ 3.50 bar and calculate w. What is the least amount of work done without restrictions on the external pressure? The greatest amount of work is done in a reversible expansion. The work is given by wreversible = −nRT ln Vf Vi = − nRT ln Pi 15.0 bar = −5.25 mol×8.314 J mol-1 K -1 ×450 K ×ln Pf 3.50 bar = −28.6×103 J The least amount of work is done in a single stage expansion at constant pressure with the external pressure equal to the final pressure. The work is given by 1 1 w = − Pexternal (V f − Vi ) = − nRTPexternal − Pf Pi 1 1 3 = −5.25 mol×8.314 J mol -1 K -1 ×450 K ×3.50 bar × − = −15.1×10 J 3.50 bar 15.0 bar The least amount of work done without restrictions on the pressure is zero, which occurs when Pexternal = 0. P2.6) Calculate H and U for the transformation of one mole of an ideal gas from 27.0ºC and 1.00 atm to 327ºC and 17.0 atm if T CP ,m = 20.9 + 0.042 in units of J K -1mol-1 . K Tf H = n ∫ CP ,m dT Ti T′ 20.9 + 0.42 dT ′ K 300 K 600 K = ∫ = 20.9 × ( 600 K − 300 K ) J + 0.21T 2 600 K 300K = 6.27 × 103 J + 56.7 × 103 J = 63.0 × 103 J 16 J U = H − ( PV ) = H − nRT = 63.0 × 103 J − 8.314 J K -1mol −1 × 300 K = 60.5 × 103 J P2.7) Calculate w for the adiabatic expansion of one mole of an ideal gas at an initial pressure of 2.00 bar from an initial temperature of 450 K to a final temperature of 300 K. Write an expression for the work done in the isothermal reversible expansion of the gas at 300 K from an initial pressure of 2.00 bar. What value of the final pressure would give the same value of w as the first part of this problem? Assume that CP,m = 5/2 R. 3 wad = U = n ( CP , m − R ) T = − ×8.314 J mol-1K -1 ×150 K = −1.87×103 J 2 wreversible = −nRT ln ln − wreversible Pi P ; ln i = Pf Pf nRT Pi nRT 1.87×103 J = = = 0.7497 Pf wreversible 1 mol×8.314 J mol -1 K -1 ×300 K Pf = 0.472 Pi = 0.944 bar P2.8) In the adiabatic expansion of one mole of an ideal gas from an initial temperature of 25ºC, the work done on the surroundings is 1200 J. If CV,m = 3/2R, calculate q, w, U, and H. q = 0 because the process is adiabatic U = w = −1200 J U = nCV ,m (T f − Ti ) Tf = U + nCV ,mTi nCV ,m −1200 J + 7.5 × 8.314 J mol-1 K −1 × 298 K 1.5 × 8.314 J mol -1 K -1 = 202 K H = nCP ,m (T f − Ti ) = n ( CV ,m + R ) (T f − Ti ) = = 2.5 × 8.314 J mol-1 K −1 ( 202 K − 298 K ) = −2.00 × 103 J P2.9) An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to a final state described by Pf, Vf, T in a) a process at the constant external pressure Pf, and b) in a reversible process. Derive expressions for the largest mass that can be lifted through a height h in the surroundings in these processes. 17 a) w = mgh = − Pf (V f − Vi ) ; m = − b) w = mgh = −nRT ln Vf Vi ; m= − Pf (V f − Vi ) gh nRT V f ln gh Vi P2.10) An automobile tire contains air at 320 × 103 Pa at 20ºC. The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of 100 × 103 Pa until P = Pexternal. For air, CV,m = 5/2 R. Calculate the final temperature of the gas in the tire. Assume ideal gas behavior. because q = 0, U = w nCV ,m ( T f − Ti ) = − Pext (V f − Vi ) nRT f nRTi nCV ,m ( T f − Ti ) = − Pext − P Pi f The factor n cancels out. Rearranging the equation RPext CV ,m + Pf Tf Ti CV ,m = CV ,m RPext T f = CV ,m + Ti P i RP + ext Pi RP + ext Pf 8.314 J mol -1 K −1 × 105 Pa 3.20 × 105 Pa = 8.314 J mol-1 K −1 × 105 Pa 2.5 × 8.314 J mol-1 K −1 + 105 Pa T f = 0.804Ti T f = 235 K 2.5 × 8.314 J mol -1 K -1 + P2.11) 3.50 moles of an ideal gas is expanded from 450 K and an initial pressure of 5.00 bar to a final pressure of 1.00 bar. CP,m = 5/2R. Calculate w for the following two cases. a) The expansion is isothermal and reversible. b) The expansion is adiabatic and reversible. Without resorting to equations, explain why the result to part b) is greater than (or less than) the result to part a). 18 a) w = −nRT ln Vf Vi = − nRT ln Pi Pf 5.00 bar = −21.1 × 103J 1.00 bar b) Because q = 0, w = U. In order to calculate U, we first calculate Tf. = −3.50 mol×8.314J mol -1 K -1 ×450 K ×ln 1−γ Vf = Ti Vi Tf 1−γ Tf = Ti Pi Pf 1−γ 1−γ γ P Tf ; = i P Ti f P = i ; Ti Pf Tf 1−γ γ 5 3 5 3 1− 5.00 bar = = 0.525 Ti 1.00 bar T f = 0.525×450 K = 236 K Tf 3 × 8.314 J mol-1 K -1 × ( 236 K − 450 K ) = −9.34×103J 2 Less work is done on the surroundings in part b) because in the adiabatic expansion, the temperature falls and therefore the final volume is less than that in part a). w = U = nCV ,m T = 3.50 mol× P2.12) An ideal gas described by Ti = 300 K, Pi = 1.00 bar and Vi = 10.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed cycle process in a P-V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction? 19 PV 1.00 bar ×10.0 L i i = = 0.401 mol RTi 8.3145×10-2 L bar mol-1 K -1 ×300 K The process can be described by step 1: Pi,Vi,Ti → P1 = 10.0 bar,Vi, T1 step 2: P1,Vi, T1 → Pi,V2 T1 step 3: Pi, V2, T1 → Pi,Vi,Ti. n= In step 1, Pi,Vi,Ti → P1,Vi, T1, w = 0 because V is constant. In step 2, P1,Vi, T1 → Pi, V2, T1 Before calculating the work in step 2, we first calculate T1. P 10.0 bar T1 = Ti 1 = 300 K × = 3000 K 1.00 bar Pi Vf P w = −nRT1 ln = − nRT1 ln i Vi Pf = −0.401 mol× 8.314 J mol-1 K -1 × 3000 K ×ln 10.0 bar = −23.0×103J 1.00 bar In step 3, PV 1 i = PV i 2 ; V2 = PV 1 i = 10Vi = 100 L Pi 105 Pa 10-3 m3 × (10 L − 100 L ) × = 9.00×103 J bar L 3 3 3 = 0 − 23.0 × 10 J + 9.00 × 10 J = −14.0 × 10 J w = − Pexternal V = −1.00 bar × wcycle If the cycle were traversed in the opposite direction, the magnitude of each work term would be unchanged, but all signs would change. P2.13) 3.00 moles of an ideal gas with CV,m = 3/2 R, initially at a temperature Ti = 298 K and Pi = 1.00 bar, is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 625 kg mass on the piston of diameter 20.0 cm. Calculate the work done in this process and the distance that the piston travels. Assume that the mass of the piston is negligible. F mg 625 kg×9.81m s-2 Pexternal = = 2 = = 1.95×105 Pa 2 A πr π × ( 0.100 m ) nRT 3.00 mol×8.314 J mol-1 K -1 ×298 K = = 7.43 × 10-2 m 3 = 74.3 L Pi 105 Pa Following Example Problem 2.6, Vi = 20 -1 -1 5 RPexternal -1 -1 8.314 J mol K ×1.95 × 10 Pa + C 12.47 J mol K + V ,m Pi 1.00 × 105 Pa = 298 K × T f = Ti -1 -1 5 C + RPexternal -1 -1 8.314 J mol K ×1.95 × 10 Pa V ,m 12.47 J mol K + Pf 1.95 × 105 Pa = 411 K nRT 3.00 mol×8.314 J mol-1 K -1 ×411 K Vf = = = 5.25 × 10-2 m3 5 Pf 1.95×10 Pa w = − Pexternal (V f − Vi ) = −1.95×105 Pa× ( 5.25×10-2 m3 − 7.43 × 10-2 m3 ) = 4.25 × 103J h= w 4.25×103J = = 0.69 m mg 625 kg×9.81 m s-2 P2.14) A bottle at 21.0ºC contains an ideal gas at a pressure of 126.4 x 103 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against Pexternal = 101.9 x 103 Pa, and some gas is expelled from the bottle in the process. When P = Pexternal, the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 21.0ºC. What is the final pressure in the bottle for a monatomic gas, for which CV,m = 3/2 R and a diatomic gas, for which CV,m = 5/2 R? In this adiabatic expansion, U = w nCV ,m (T f − Ti ) = − Pext (V f − Vi ) nRT nRT nCV ,m (T f − Ti ) = − Pext − V Vi f RPext CV ,m + Pf Tf Ti CV ,m = CV ,m RPext T f = CV ,m + Ti Pi RP + ext Pi RP + ext Pf 8.314 J mol-1 K −1 × 101.9 × 103 Pa 126.4 × 103 Pa = 8.314 J mol -1 K −1 × 101.9 × 103 Pa -1 −1 1.5 × 8.314 J mol K + 101.9 × 103 Pa 1.5 × 8.314 J mol-1 K -1 + Tf Ti = 0.923 , T f = 271 K Once the stopper is put in place, the gas makes a transformation from 21 Ti = 214 K, Pi = 101.9 × 103 Pa to T f = 294 K and Pf PV PV i i = f f , but Vi = V f Ti Tf Pf = Tf Ti Pi = 294 K × 101.9 × 103 Pa = 110.5 × 103 Pa 271 K The same calculation carried out for CV ,m = Tf Ti 5 R gives 2 = 0.945, T f = 278 K Pf = 107.8 × 103 Pa P2.15) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a pressure P = 1.00 bar and temperature T = 298 K. What is the reaction that occurs? Calculate w for the process. Zn(s) + H2SO4(aq) → Zn2+(aq) + SO42-(aq) +H2(g) The volume of H2 produced is given by V = 10.0 g 65.39 g ( mol Zn ) × -1 1mol H 2 8.314 J mol -1 K -1 × 298 K = 3.79 × 10-3 m 3 × 5 1mol Zn 1×10 Pa w = − Pexternal V V ≈ volume of H 2 produced. w = −1×105 Pa ×3.79×10-3 m 3 = −379 J P2.16) One mole of an ideal gas for which CV,m = 20.8 J K-1 mol-1 is heated from an initial temperature of 0ºC to a final temperature of 275ºC at constant volume. Calculate q, w, U and H for this process. w = 0 because V = 0. U = q = CV T = 20.8 J mol -1 K -1 ×275 K = 5.72 × 103J H = U + ( PV ) = U + RT = 5.72×103 J + 8.314 J mol -1 K -1 × 275 K = 8.01 × 103 J P2.17) One mole of an ideal gas, for which CV,m = 3/2 R , initially at 20.0ºC and 1.00 x 106 Pa undergoes a two stage transformation. For each of the stages described below, calculate the final pressure, as well as q, w, U and H. Also calculate q, w, U and H for the complete process. a) The gas is expanded isothermally and reversibly until the volume doubles. 22 b) Beginning at the end of the first stage, the temperature is raised to 80.0ºC at constant volume. a) P2 = PV P 1 1 = 1 = 0.500 × 106 Pa V2 2 w = −nRT ln V2 = −8.314 J mol-1 K -1 × ln 2 = −1.69 × 103 J V1 U = 0 and H = 0 because T = 0 q = − w = 1.69 × 103 J b) T1 T2 T P 353 K × 0.500 × 106 Pa = ; P2 = 2 1 = = 6.02 × 105 Pa 293 K P1 P2 T1 U = nCV ,m T = 1.5 × 8.314 J mol-1 K −1 × ( 353 K − 293 K ) = 748 J w = 0 because V = 0 q = U = 748 J H = nCP ,m T = n ( CV ,m + R ) T = 3 × 8.314 J mol-1 K −1 × ( 353 K − 293 K ) 2 = 1.25 × 103 J For the overall process, q = 1.69 × 103 J + 748 J = 2.44 × 103 J w = −1.69 × 103 J + 0 = −1.69 × 103 J U = 0 + 748 J = 748 J H = 0 + 1.25 × 103 J = 1.25 × 103 J P2.18) One mole of an ideal gas with CV,m = 3/2R initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 1.00 × 106 Pa. Calculate the final temperature of the gas. Calculate q, w, U and H for this process. 1−γ Vf = Ti Vi Tf 1−γ Tf = Ti Pi Pf 1−γ 1−γ γ P Tf ;= = i P Ti f 5 3 5 3 1− 1.00×105 Pa −0.4 = = ( 0.100 ) = 2.51 6 Ti 1.00×10 Pa T f = 2.51×298 K = 749 K q = 0 because the process is adiabatic. Tf 23 P = i ; Ti Pf Tf 1−γ γ 3×8.314 J mol -1 K -1 × ( 749 K − 298 K ) = 5.62 × 103 J 2 H = U + ( PV ) = U + RT = 5.62 × 103 J + 8.314 J mol-1 K -1 × ( 749 K − 298 K ) w = U = nCV ,m T = 1 mol× H = 9.37 × 103 J P2.19) One mole of an ideal gas, for which CV,m = 3/2R, is subjected to two successive changes in state: a) From 25.0ºC and 100 × 103 Pa, the gas is expanded isothermally against a constant pressure of 20.0 × 103 Pa to twice the initial volume. b) At the end of the previous process, the gas is cooled at constant volume from 25.0ºC to –25.0ºC. Calculate q, w, U and H for each of the stages. Also calculate q, w, U and H for the complete process. a) Vi = nRT 8.314 J mol-1K -1 × 298 K = = 2.48 × 10−2 m3 3 Pi 100 R × 10 Pa V f = 2Vi = 4.96 × 10−2 m3 w = − Pext (V f − Vi ) = −20.0 × 103 Pa × ( 4.96 × 10−2 m3 − 2.48 × 10−2 m3 ) = −496 J U and H = 0 because T = 0 q = − w = 496 J b) U = nCV ,m (T f − Ti ) = 1.5 × 8.314 J mol-1K -1 × ( 248 K − 298 K ) = −623 J w = 0 because V = 0 q = U = −623 J H = nCP ,m (T f − Ti ) = n ( CV ,m + R ) (T f − Ti ) = 2.5 × 8.314 J mol -1K −1 × ( 248 K − 298 K ) = −1.04 × 103J U total = 0 − 623 J = 623 J wtotal = 0 − 496 J = − 496 J qtotal = 496 J − 623 J = − 127 J H total = 0 − 1.04 × 103J = − 1.04 × 103 J P2.20) The temperature of one mole of an ideal gas increases from 18.0ºC to 55.1ºC as the gas is compressed adiabatically. Calculate q, w, U and H for this process assuming that CV,m = 3/2 R. q = 0 because the process is adiabatic. 24 3×8.314 J mol -1 K -1 × ( 55.1o C − 18.0o C ) = 463 J 2 H = U + ( PV ) = U + RT = 463 J + 8.314 J mol -1 K -1 × ( 55.1o C − 18.0o C ) w = U = nCV ,m T = 1 mol× H = 771 J P2.21) A one mole sample of an ideal gas, for which CV,m = 3/2R, undergoes the following two-step process. a) From an initial state of the gas described by T = 28.0ºC and P = 2.00 × 104 Pa, the gas undergoes an isothermal expansion against a constant external pressure of 1.00 × 104 Pa until the volume has doubled. b) Subsequently, the gas is cooled at constant volume. The temperature falls to –40.5ºC. Calculate q, w, U and H for each step and for the overall process. a) For the first step, U = H = 0 because the process is isothermal. -1 -1 nRTi 1 mol×8.314 J mol K × ( 273.15 + 28.0 ) K Vi = = = 1.25 × 10-2 m 3 4 2.00×10 Pa Pi w = − q = − Pexternal V = −1.00×104 Pa ×0.125×10-2 m3 = −1.25×103 J b) For the second step, w = 0 because V = 0. 3×8.314 J mol-1 K -1 q = U = CV T = 1 mol× × ( 28.0o C + 40.5o C ) = 854 J 2 H = U + ( PV ) = U + RT = 854 J +8.314 J mol-1 K -1 × ( 28.0o C + 40.5o C ) H = 1.42 × 103J For the overall process, w = – 1.25×103 J, q = 854 + 1.25×103 J = 2.02×103 J, U = 854 J, and H = 1.42 × 103 J. P2.22) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with CV,m = 3/2R. Initially, Ti = 298 K and Pi = 1.00 bar in each part. Heat is slowly introduced into the left part using an electrical heater until the piston has moved sufficiently to the right to result in a final pressure Pf = 7.50 bar in the right part. Consider the compression of the gas in the right part to be a reversible process. a) Calculate the work done on the right part in this process, and the final temperature in the right part. b) Calculate the final temperature in the left part and the amount of heat that flowed into this part. The number of moles of gas in each part is given by PV 1.00 bar ×50.0 L n= i i = = 2.02 mol RTi 8.3145×10-2 L bar mol-1 K -1 ×298 K 25 a) We first calculate the final temperature in the right side. 1−γ 1−γ Vf = Ti Vi Tf Tf = Ti Pi Pf 1−γ γ P Tf ;= = i P Ti f 1−γ ; P = i Ti Pf Tf 1−γ γ 5 3 5 3 1− 1.00 bar = = 2.24 Ti 7.50 bar T f = 2.24×298 K = 667 K Tf 3×8.314 J mol-1 K -1 × ( 667 K − 298 K ) = 9.30 × 103 J 2 b) We first calculate the final volume of the right part. nRTrf 2.02 mol×8.314×10-2 L bar mol-1K -1 ×667 K Vrf = = = 14.9 L Prf 7.50 bar Therefore, Vlf = 100.0 L – 14.9 L = 85.1 L. PV 7.50 bar×85.1 L = 3.80 ×103 K Tlf = lf lf = -2 -1 -1 2.02 mol×8.314×10 L bar mol K nR U = w = nCV ,m T = 2.02 mol× U = nCV ,m T = 2.02 mol×3 2 × 8.314 J mol-1K -1 × ( 3.80 ×103 K − 298 K ) = 88.2 ×103 J From part a), w = –9.30 × 103J q = U – w = 88.2 × 103 J + 9.30 × 103 J = 97.5 × 103 J P2.23) A vessel containing one mole of an ideal gas with Pi = 1.00 bar and CP,m = 5/2R is in thermal contact with a water bath. Treat the vessel, gas and water bath as being in thermal equilibrium, initially at 298 K, and as separated by adiabatic walls from the rest of the universe. The vessel, gas and water bath have an average heat capacity of CP = 7500 J/K. The gas is compressed reversibly to Pf = 10.5 bar. What is the temperature of the system after thermal equilibrium has been established? Assume initially that the temperature rise is so small that the reversible compression can be thought of as an isothermal reversible process. If the answer substantiates this assumption, it is valid. w = −nRT1 ln Vf Vi = − nRT1 ln Pi Pf = −1 mol× 8.314 J mol-1 K -1 × 298 K ×ln 1.00 bar = 5.83 × 103 J 10.5 bar U combined system = C P T T = U combined system CP = 5.83×103J = 0.777 K 7500 J K -1 T f ≈ 299 K 26 The result justifies the assumption. P2.24) The heat capacity of solid lead oxide is given by T in units of J K -1mol -1 . Calculate the change in enthalpy of 1 K mol of PbO(s) if it is cooled from 500 K to 300 K at constant pressure. CP ,m = 44.35 + 1.47 × 10 −3 Tf H = n ∫ C p ,m dT Ti 300 = ∫ 44.35 + 1.47 ×
10 −3 500 T T d K K = 44.35 × ( 300 K − 500 K ) 300 K 1.47 × 10−3 T 2 + 2 K 500 K = −8870 J − 117 J = −8.99 × 103 J P2.25) Consider the adiabatic expansion of 0.500 moles of an ideal monatomic gas with CV,m = 3/2R. The initial state is described by P = 3.25 bar and T = 300 K. a) Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a final pressure P = 1.00 bar. b) Calculate the final temperature if the same gas undergoes an adiabatic expansion against an external pressure of to P = 1.00 bar to a final pressure P = 1.00 bar. Explain the difference in your results for parts a) and b). a) 1−γ Vf = Ti Vi Tf 1−γ Tf = Ti Pi Pf 1−γ 1−γ γ P Tf ;= = i P Ti f 5 3 5 3 1− 3.25 bar = = 0.626 Ti 1.00 bar T f = 0.626 × 300 K = 188 K Tf b) U = nCV ,m (T f − Ti ) = − Pexternal (V f − Vi ) 27 P = i ; Ti Pf Tf 1−γ γ T f Ti − nCV ,m (T f − Ti ) = − nRPexternal P Pi f nRPexternal T f nCV ,m + Pf nRPexternal = Ti nCV ,m + Pi RPexternal CV ,m + Pi T f = Ti C + RPexternal V ,m Pf T f = 217 K -1 -1 -1 -1 8.314 J mol K ×1.00 bar 1.5×8.314 J mol K + 3.25 bar = 300 K × -1 -1 -1 -1 8.314 J mol K ×1.00 bar 1.5×8.314 J mol K + 1.00 bar More work is done on the surroundings in the reversible expansion, and therefore U and the temperature decrease more than for the irreversible expansion. P2.26) An ideal gas undergoes a single stage expansion against a constant external pressure Pexternal from T, Pi, Vi, to T, Pf, Vf. a) What is the largest mass, m, that can be lifted through the height h in this expansion? b) The system is restored to its initial state in a single state compression. What is the smallest mass, m' that must fall through the height h to restore the system to its initial state? c) If h = 10.0 cm, Pi = 1.00 × 106 Pa, Pf = 0.500 × 106 Pa , T = 300 K, and n = 1.00 mole, calculate the values of the masses in parts a and b. Consider the expansion a) w = mgh = − Pext (V f − Vi ) m= − Pext (V f − Vi ) gh for the final volume to be V f , the external pressure can be no bigger than Pf mmax = − Pf (V f − Vi ) gh b) Consider the compression m′ = − Pext (Vi − V f ) gh for the final volume to be Vi , the pressure can be no smaller than Pi 28 ' mmin = − Pi (Vi − V f ) gh c) Vi = nRT 1.00 mol× 8.314 J mol -1 K -1 × 300 K = = 2.49 × 10−3 m 3 6 1.00×10 Pa Pi Pf V f = PV i i PV 1.00 × 106 Pa × 2.49 × 10-3 m3 i i = = 4.98 × 10−3 m3 Vf = 6 0.500 × 10 Pa Pf mmax = ′ = mmin −0.500 × 106 Pa × ( 4.98 × 10-3 m 3 − 2.49 × 10−3 m3 ) 9.81 m s × 0.100 m -2 −1.00 × 106 Pa × ( 2.49 × 10−3 m3 − 4.98 × 103 m 3 ) −2 9.81 m s × 0.100 m = 1.27 × 103 kg = 2.54 × 103 kg P2.27) Calculate q, w, U and H if 1.00 mole of an ideal gas with CV,m = 3/2R undergoes a reversible adiabatic expansion from an initial volume Vi = 5.25 m3 to a final volume Vf = 25.5 m3. The initial temperature is 300 K. q = 0 because the process is adiabatic. 1−γ Tf V f = Ti Vi 1− 5 25.5 L 3 = = 0.349 Ti 5.25L T f = 0.349×300 K = 105 K Tf 3 × 8.314 J mol -1 K -1 × (105 K − 300 K ) = −2.43 × 103J 2 3 H = U + nRT = −2.43×10 J + 1.00 mol×8.314 J mol-1 K -1 × (105 K − 300 K ) U = w = nCV ,n T = 1.00 mol× H = −4.05 × 103 J P2.28) A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf = 1.00 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire. Assume that CV,m = 5/2 R. 29 1−γ Vf = Ti Vi Tf 1−γ Tf = Ti Pi Pf 1−γ 1−γ γ Tf P ;= = i Ti Pf P ; = i Ti Pf Tf 1−γ γ 7 5 7 5 1− −0.285 1.00 bar = = 1.58 = ( 0.200 ) Ti 5.00 bar T f = 1.58×298 K = 472 K Tf P2.29) One mole of an ideal gas with CV,m = 3/2R is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure are Ti = 300 K and Pi = 25.0 bar. The final pressure is Pf = 1.00 bar. Calculate q, w, U and H for the process. U = nCV ,m (T f − Ti ) = − Pexternal (V f − Vi ) = w q = 0 because the process is adiabatic. T f Ti nCV ,m (T f − Ti ) = − nRPexternal − P f Pi nRPexternal T f nCV ,m + Pf nRPexternal = Ti nCV ,m + Pi RPexternal CV ,m + Pi T f = Ti C + RPexternal V ,m Pf T f = 185 K -1 -1 -1 -1 8.314 J mol K ×1.00 bar 1.5×8.314 J mol K + 25.0 bar = 300 K × -1 -1 -1 -1 8.314 J mol K ×1.00 bar 1.5×8.314 J mol K + 1.00 bar 3×8.314 J mol-1 K -1 × (185 K − 300 K ) = −1.43 × 103 J 2 H = U + nRT = −1.43×103 J + 1.00mol×8.314J mol-1 K -1 × (185 K − 300 K ) U = w = nCV ,n T = 1.00 mol× H = −2.39×103J P2.30) One mole of N2 in a state defined by Ti = 300 K and Vi = 2.50 L undergoes an isothermal reversible expansion until Vf = 23.0 L. Calculate w assuming a) that the gas is described by the ideal gas law, and b) that the gas is described by the van der Waals equation of state. What is the percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for N2 are tabulated in Appendix A. a) for the ideal gas 30 wreversible = − nRT ln Vf = −1mol×8.314 J mol -1 K -1 ×300 K ×ln Vi 23.0 L = −5.54 × 103 J 2.50 L for the van der Waals gas Vf Vf RT a − 2 dV w = − ∫ Pexternal dV = − ∫ V − b Vm Vi Vi m Vf V f RT a = −∫ + dV 2 dV ∫ V −b V Vi m Vi m The first integral is solved by making the substitution y = Vm – b. V y f f RT RT dV −∫ = − y ∫ V −b Vi m yi dy = − RT ln (V f + b ) − ln (Vi + b ) Therefore, the work is given by w = −nRT ln (V − b) 1 1 +a − (Vi − b ) Vi V f f 23.0 L − 0.0380 L 2.50 L − 0.0380 L 5 -6 6 10 Pa 10 m 1 1 + 1.366 L2 bar × − × 2 -3 3 -3 3 bar L 2.50×10 m 23.0×10 m = −1 mol×8.314 J mol -1 K -1 ×300 K ×ln w = −5.56 × 103 J + 48.7 J = −5.52 × 103 J Percent error = 100 × −5.52×103J + 5.54×103J = −0.4% −5.52 × 103 J 31 Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q3.1) Why is CP,m a function of temperature for ethane, but not for argon? Argon has only translational degrees of freedom, which are fully excited at very low temperatures because Etranslational 0 and w < 0. P3.10) Starting with the van der Waals equation of state, find an expression for the total differential dP in terms of dV and dT. By calculating the mixed partial derivatives ∂ ∂P ∂ ∂P and , determine if dP is an exact differential. ∂T ∂V T V ∂V ∂T V T ∂ ∂V RT a 2a RT V − b − V 2 = V 3 − 2 (Vm − b ) m T m m ∂ ∂T RT a R − V − b V 2 = V − b ( m ) m V m 2a RT R dP = 3 − dT dV + Vm (V − b )2 (Vm − b ) m 3-6 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂ ∂P ∂ = ∂T ∂V T V ∂T 2a RT R =− 3− 2 2 (Vm − b ) Vm (Vm − b ) V ∂ R ∂ ∂P R = − = 2 (Vm − b ) ∂V ∂T V T ∂V (Vm − b ) T Therefore, dP is an exact differential. 1 ∂V P3.11) Obtain an expression for the isothermal compressibility κ = − for a V ∂P T van der Waals gas. 1 ∂V κ =− m =− Vm ∂P T κ =− 1 =− ∂P ∂ Vm Vm ∂Vm ∂Vm T 1 RT a − V − b V 2 m T m 1 2a RT Vm 3 − 2 Vm (Vm − b ) P3.12) Regard the enthalpy as a function of T and P. Use the cyclic rule to obtain the ∂H ∂P T . expression CP = − ∂T ∂P H ∂H ∂P ∂T = −1 ∂P T ∂T H ∂H P ∂H ∂H ∂H ∂P ∂P T CP = = − = − ∂T ∂T P ∂P T ∂T H ∂P H P3.13) Equation (3.38), CP = CV + TV β2 , links CP and CV with β and κ. Use this κ equation to evaluate CP – CV for an ideal gas. 3-7 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy β= 1 ∂V 1 nR 1 ∂V nRT 1 ; κ =− = = = V ∂T P V P V ∂P T VP 2 P 2 β2 n2 R2 1 nR CP − CV = TV = TV P = TV = nR κ V 2P V P ∂U βT − κ P ∂U P3.14) Use Equation (3.58), = to calculate for an ideal gas. ∂V T κ ∂V T 1 ∂V 1 nR 1 ∂V nRT 1 ; κ =− = = = V ∂T P V P V ∂P T VP 2 P 1 nRT −1 βT − κ P V P ∂U = = = P (1 − 1) = 0 1 κ ∂V T P P3.15) An 80.0-g piece of gold at 650 K is dropped into 100.0 g of H2O(l) at 298 K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP,m for Au and H2O is constant at their values for 298 K throughout the temperature range of interest. β= nAu CPAu,m (T f − Ti Au ) + nH 2O CPH,2mO (T f − Ti H 2O ) = 0 Tf = nAu CPAu, mTi Au + nH 2O CPH,2mOTi H 2O nAu CPAu, m + nH 2O CPH,2mO 80.0 g Au 100.0 g H 2 O × 25.42 J K −1mol−1 × 650 K + × 75.291 J K −1mol−1 × 298 K −1 −1 196.97 g Au mol 18.02 g H 2 O mol = 80.0 g Au 100.0 g H 2 O × 25.42 J K −1mol−1 + × 75.291 J K −1mol−1 −1 −1 196.97 g Au mol 18.02 g H 2 O mol T f = 306 K P3.16) A mass of 35.0 g of H2O(s) at 273 K is dropped into 180.0 g of H2O(l) at 325 K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP,m for H2O is constant at its values for 298 K throughout the temperature range of interest. 3-8 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy H 2O ice nice H ice ) + nH2OCPH,2mO (T f − Ti H2O ) = 0 fusion + nice C P , m ( T f − Ti Tf = nAu CPAu, mTi Au + nH 2O CPH,2mOTi H 2O − nice H ice fusion nAu CPAu,m + nH 2O CPH,2mO = 35.0 g ice 180.0 g H 2O × 75.291 J K −1mol−1 × 273 K + × 75.291 J K −1mol−1 × 325 K −1 −1 18.02 g ice mol 18.02 g H 2 O mol 35.0 g ice × 6008 J mol−1 18.02 g ice mol−1 35.0 g ice 100.0 g H 2 O × 75.291 J K −1mol−1 + × 75.291 J K −1mol−1 −1 18.02 g ice mol 18.02 g H 2 O mol−1 − T f = 303 K P3.17) A mass of 20.0 g of H2O(g) at 373 K is flowed into 250 g of H2O(l) at 300 K and 1 atm. Calculate the final temperature of the system once equilibrium has been reached. Assume that CP,m for H2O is constant at its values for 298 K throughout the temperature range of interest. water − nsteam H vaporization + nsteamCPH,2mO (T f − Ti ice ) + nH 2O CPH,2mO (T f − Ti H 2O ) = 0 Tf = water nsteamCPH,2mOTi steam + nH 2O CPH,2mOTi H 2O + nsteam H vaporization nsteamCPH,2mO + nH 2O CPH,2mO = 250.0 g H 2 O 20.0 g steam × 75.291 J K −1mol−1 × 373 K + × 75.291 J K −1mol−1 × 300 K −1 −1 18.02 g H 2 O mol 18.02 g steam mol 20.0 g steam × 40656 J mol−1 18.02 g steam mol−1 20.0 g steam 250.0 g H 2 O × 75.291 J K −1mol−1 + × 75.291 J K −1mol−1 −1 18.02 g H 2 O mol−1 18.02 g steam mol + T f = 345 K P3.18) Calculate w, q, H, and U for the process in which 1 mol of water undergoes the transition H2O(l, 373 K) → H2O(g, 460 K) at 1 bar pressure. The volume of liquid water at 373 K is 1.89 × 10–5 m3 mol–1 and the volume of steam at 373 and 460 K is 3.03 and 3.74 × 10–2 m3 mol–1, respectively. For steam, CP,m can be considered constant over the temperature interval of interest at 33.58 J mol–1 K–1. 3-9 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy q = H = nH vaporization + nCPsteam , m T = 40656 J + 1 mol × 33.58 J mol−1K −1 × ( 460 K − 373 K ) = 4.35 ×104 J w = − Pexternal V = −105 Pa × ( 3.03 ×10−2 m3 − 1.89 ×10−5 m3 ) − 105 Pa × ( 3.74 ×10−2 m3 − 3.03 ×10−2 m3 ) = −3028 J − 710 J = −3.74 ×103 J U = w + q = 4.35 ×104 J − 3738 J = 3.98 ×104 J ∂H P3.19) Because = −CP µ J −T , the change in enthalpy of a gas expanded at ∂P T constant temperature can be calculated. In order to do so, the functional dependence of µ J −T on P must be known. Treating Ar as a van der Waals gas, calculate H when 1 mol of Ar is expanded from 400 bar to 1.00 bar at 300 K. Assume that µ J −T is independent of 2a pressure and is given by µ J −T = − b /CP,m = 5/2R for Ar. What value would H RT have if the gas exhibited ideal gas behavior? Pf H m = − ∫ CP ,m µ J −T dP ≈ −CP ,m µ J −T ( Pf − Pi ) Pi = −CP ,m x 1 2 × 0.1355 m 6 Pa mol−2 − 0.03201×10−3 m3 mol−1 × (1.00 ×105 Pa − 400 ×105 Pa ) −1 −1 CP , m 8.314 J mol K × 300 K = 3.06 ×103 J For an ideal gas, Hm = 0 because µ J −T is zero for an ideal gas. ∂P β P3.20) Using the result of Equation (3.8), = , express β as a function of κ and ∂T V κ Vm for an ideal gas, and β as a function of b, κ, and Vm for a van der Waals gas. For the ideal gas, R β κR ∂P = ; β= = Vm ∂T P Vm κ For the van der Waals gas, ∂ ∂T RT κR a R V − b − V 2 = V − b ; β = V − b ( m ) m V m m 3-10 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂T ∂P 1 P3.21) The Joule coefficient is defined as = P − T . Calculate the ∂V U CV Joule coefficient for an ideal gas and for a van der Waals gas. ∂T V For an ideal gas 1 1 nRT ∂T ∂ nRT P −T P− =0 = = V ∂V U CV ,m ∂T V V CV ,m For a van der Waals gas ∂ 1 ∂T P − T = ∂V U CV ,m ∂T RT a 1 RT 1 a V − b − V 2 = C P − V − b = − C V 2 ( m ) m V V ,m V m m ∂U ∂P P3.22) Use the relation = T − P and the cyclic rule to obtain an ∂V T ∂T V ∂U expression for the internal pressure , in terms of P, β, T, and κ. ∂V T ∂V β Vβ ∂U ∂P ∂T P −P=T −P =T −P =T − P = −T ∂V κ Vκ ∂V T ∂T V ∂P T ∂U 3a P3.23) Derive the following relation, , for the internal = ∂Vm T 2 TVm (Vm + b ) pressure of a gas that obeys the Redlich-Kwong equation of state, RT a 1 . P= − Vm − b T Vm (Vm + b ) The internal pressure of a gas is given by ∂V ∂P =T −P ∂T T ∂T V Using the Redlich-Kwong equation of state, 3-11 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy R 1 a ∂P + = 3/ 2 ∂T V Vm − b 2 T Vm (Vm + b ) RT RT 1 a a ∂U ∂P = T − P = + − − Vm − b 2 T Vm (Vm + b ) Vm − b TVm (Vm + b ) ∂V T ∂T V = 1 a a 3a + = 2 T Vm (Vm + b ) TVm (Vm + b ) 2 TVm (Vm + b ) P3.24) Derive an expression for the internal pressure of a gas that obeys the Bethelot equation of state, P = RT a − . Vm − b TVm2 The internal pressure of a gas is given by ∂V ∂P =T −P ∂T T ∂T V Using the Bethelot equation of state R a ∂P + 2 2 = ∂T V Vm − b T Vm RT a ∂U + = 2 ∂V T Vm − b TVm RT a 2a − = 2 − V b TV TVm2 m m P3.25) For a gas that obeys the equation of state Vm = dB (T ) ∂H . = B (T ) − T dT ∂P T ∂H ∂V =V −T ∂P T ∂T P RT + B (T ) P R dB ∂V = + ∂T P P dT For Vm = 3-12 RT + B (T ) , derive the result P Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy dB ( T ) RT RT ∂H R dB RT + B (T ) − T + + B (T ) − −T = = P P P dT ∂P T P dT dB (T ) = B (T ) − T dT P3.26) Derive the following expression for calculating the isothermal change in the ∂2 P ∂C constant volume heat capacity: V = T 2 . ∂V T ∂T V ∂ ∂U ∂ ∂U ∂CV ∂V = ∂V ∂T = ∂T ∂V V T T T T The order of differentiation can be reversed because U is a state function. ∂U ∂P Using the equation =T −P ∂V T ∂T V ∂ ∂P ∂CV T = − P ∂V T ∂T ∂T V V ∂ 2 P ∂P ∂2P ∂P = + T − = T ∂T 2 ∂T ∂T 2 ∂T V V V V ∂C P3.27) Use the result of the Problem P3.26 to show that V for the van der Waals ∂V T gas is zero. We use the relationship 3-13 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂2 P ∂CV T = T2 ∂V T ∂ V RT a P= − 2 Vm − b Vm R ∂P = ∂T V Vm − b R ∂ ∂ P Vm − b =0 = T2 ∂ V ∂T V 2 ∂2 P ∂C therefore V = T 2 = T × 0 = 0 ∂V T ∂T V ∂C P3.28) Use the result of Problem P3.26 to derive a formula for V for a gas that ∂V T RT a 1 obeys the Redlich-Kwong equation of state, P = . − Vm − b T Vm (Vm + b ) We use the relationship ∂2 P ∂CV T = T2 ∂V T ∂ V RT −a P= Vm − b TVm (Vm + b ) R a ∂P + 3/ 2 = ∂T V Vm − b 2T Vm (Vm + b ) ∂2 P −3a T 2 = 4T 5/ 2V V b ∂ V m ( m + ) 3a ∂CV ∂V = − 4T 3/ 2V V + b ) T m ( m P3.29) For the equation of state Vm = ( RT P ) + B (T ) , show that d 2 B (T ) ∂CP ,m ∂P = −T dT 2 .[Hint: Use Equation (3.44) and the property of states functions T with respect to the order of differentiation in mixed second derivatives.] Using Equation (3.44), 3-14 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂ ∂H ∂ ∂H ∂CP ,m T P = P T = P ∂ ∂ T P ∂ ∂ P T ∂ T Equation 3.44 states that ∂H ∂V =V −T ∂P T ∂T P Therefore, ∂ ∂CP ,m ∂V ∂P = ∂T V − T ∂T P P T ∂ = ∂T ∂ ( RT P ) + B (T ) ( RT P ) + B (T ) − T T ∂ P P ∂ = ∂T R dB ( T ) ( RT P ) + B (T ) − T + dT P P P R dB (T ) R dB ( T ) d 2 B (T ) = + − − +T dT P dT dT 2 P P =T d 2 B (T ) dT 2 ∂Vm ∂P , the cyclic rule, and the van der ∂T P ∂T P Waals equation of state to derive an equation for CP ,m − CV ,m in terms of Vm and the gas P3.30) Use the relation CP ,m − CV ,m = T constants R, a, and b. ∂V We use the cyclic rule to evaluate m . ∂T P 3-15 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂P = −1 ∂Vm T ∂Vm ∂P ∂Vm ∂T = − ∂T ∂P Vm P T ∂Vm ∂T ∂T ∂P Vm P ∂V ∂P CP ,m − CV ,m = T m ∂T P ∂T Vm P= ∂P 2 ∂P ∂Vm ∂T Vm = −T = −T ∂P ∂T Vm ∂P T ∂Vm T 2 RT a − 2 Vm − b Vm R ∂P = ∂T Vm Vm − b 3 ∂P − RT 2a − RTVm + 2a (Vm − b ) + 3 = = 2 2 Vm3 (Vm − b ) ∂Vm T (Vm − b ) Vm 2 CP ,m − CV ,m R Vm − b R R = −T = −T = 2 2 − RT 2a 2a (Vm − b ) 2a (Vm − b ) + 2 3 −T + 1− (Vm − b ) Vm RVm3 RTVm3 In the ideal gas limit, a = 0, and CP,m – CV,m = R. ∂U βT − κ P P3.31) Show that = . ∂V T κ ∂U ∂P =T −P ∂V T ∂T V Using the cyclic rule 3-16 Chapter 3/ The Importance of State Functions: Internal Energy and Enthalpy ∂P ∂T ∂V = −1 ∂T V ∂V P ∂P T ∂V ∂T P β ∂P = = − ∂V κ ∂T V ∂P T Tβ Tβ −κP ∂V −P= = κ R ∂V T ∂U ∂P P3.32) Show that the expression = T − P can be written in the form ∂V T ∂T V ∂ [P T ] ∂U 2 ∂ [P T ] . = − =T ∂V T ∂T V ∂ [1 T ] V ∂U ∂P =T −P ∂V T ∂T V ∂ [P / T ] ∂ [1/ T ] 1 ∂P = P + ∂T V ∂T V T ∂T V P 1 ∂P =− 2 + T T ∂T V ∂ [P / T ] P ∂P + 2 = T ∂T V ∂T V T P ∂U 2 ∂ [P / T ] T = + −P ∂T T 2 ∂V T V ∂ [P / T ] 2 ∂ [P / T ] = T2 +P−P =T ∂T V ∂T V We now change the differentiation to the variable 1/T. ∂ [ P / T ] ∂ [1/ T ] ∂ [P / T ] 1 ∂ [P / T ] = =− 2 T ∂ [1/ T ] V ∂T V ∂ [1/ T ] V ∂T V 1 ∂ [P / T ] ∂ [P / T ] ∂U 2 ∂ [P / T ] 2 = − = T − 2 =T ∂V T ∂T V T ∂ [1/ T ] V ∂ [1/ T ] V 3-17 Questions on Concepts Q4.1) Under what conditions are H and U for a reaction involving gases and/or liquids or solids identical? H = U + (PV). H ≈ U for reactions involving only liquids or solids because, to a good approximation, the volume does not change as a result of the reaction. If gases are iunvolved, (PV)= nRT. Therfore, H ≈ U if the number of moles of reactants and products that are gases are the same. Q4.2) If H of for the chemical compounds involved in a reaction are available at a given temperature, how can the reaction enthalpy be calculated at another temperature? The reaction enthalpy can be calculated to high accuracy at another temperature can be calculated only if the heat capacities of all reactants and products are known using T H = H o T o 298.15 K + ∫ CP ( T ′) dT ′ . If the heat capacities of reactants and products 298.15 K are similar, the reaction enthalpy will not vary greatly over a limited temperature range. Q4.3) Does the enthalpy of formation of compounds containing a certain element change if the enthalpy of formation of the element under standard state conditions is set equal to 100 kJ mol-1 rather than to zero? If it changes, how will it change for the compound AnBm if the formation enthalpy of element A is set equal to 100 kJ mol-1? Yes, because part of the enthalpy change of the reaction will be attributed to the element. For the reaction nA +mB → AnBm, o H reaction = H of ( A n Bm ) − nH of ( A ) − mH of ( B ) o H of ( A n Bm ) = H reaction + nH of ( A ) + mH of ( B) Therefore H of ( A n Bm ) for the compound will increase by 100n kJ mol-1. Q4.4) Is the enthalpy for breaking the first C-H bond in methane equal to the average CH bond enthalpy in this molecule? Explain your answer. No. The average bond enthalpy is the average of the enthalpies of the four steps leading to complete dissociation. The enthalpy of each successive dissociation step will increase. Q4.5) Why is it valid to add the enthalpies of any sequence of reactions to obtain the enthalpy of the reaction that is the sum of the individual reactions? Because H is a state function, any path between the reactants and products, regardless of which intermediate products are involved, has the same value for H. 48 Q4.6) The reactants in the reaction 2NO(g) + O2(g) → 2NO2(g) are initially at 298 K. Why is the reaction enthalpy the same if the reaction is (a) constantly kept at 298 K or (b) if the reaction temperature is not controlled and the heat flow to the surroundings is measured only after the temperature of the products is returned to 298 K? o Q4.7) In calculating H reaction at 285.15 K, only the H of of the compounds that take part in the reactions listed in Tables 4.1 and 4.2 (Appendix A, Data Tables) are needed. Is this o statement also true if you want to calculate H reaction at 500 K? No. At any temperature other than 298.15 K, the heat capacities of all elements and compounds that appear in the overall reaction enter into the calculation. Q4.8) What is the point of having an outer water bath in a bomb calorimeter (see Figure 4.3), especially if its temperature is always equal to that of the inner water bath? This water bath effectively isolates the calorimeter and the inner water bath from the rest of the universe. Because the temperatures of the water baths are the same, there is no heat follow between them. Because the container of the inner water bath has rigid walls, no work is done on the composite system consisting of the calorimeter and the inner water bath. Therefore, the alorimeter and inner water bath form an isolated composite system. Q4.9) What is the advantage of a differential scanning calorimeter over a bomb calorimeter in determining the enthalpy of fusion of a series of samples? All the samples can be measured in parallel, rather than sequentially. This reduces both the measurement time, and the possibility of errors. Q4.10) You wish to measure the heat of solution of NaCl in water. Would the calorimetric technique of choice be at constant pressure or constant volume? Why? Constant pressure calorimetry is the technique of choice because none of the reactants or products is gaseous, and it is therefore not necessary to contain the reaction. Constant pressure calorimetry is much easier to carry out than constant volume calorimetry. Problems o o P4.1) Calculate H reaction and U reaction at 298.15 K for the following reactions. a) 4NH3(g) + 6NO(g) → 5 N2(g) + 6H2O(g) b) 2NO(g) + O2(g) → 2NO2(g) c) TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(g) d) 2 NaOH(aq) + H2SO4(aq) → Na2SO4(s) + 2H2O(l) e) CH4(g) + H2O(g) → CO(g) + 3H2(g) 49 f) CH3OH(g) + CO(g) → CH3COOH(l) a) 4NH3(g) + 6NO(g) → 5 N2(g) + 6H2O(g) o H reaction = 5H of ( N 2 , g ) + 6H of ( H 2 O, g ) − 4H of ( NH 3 , g ) − 6H of ( NO, g ) = 0 − 6 × 241.8 kJ mol-1 + 4 × 45.9 kJ mol-1 − 6 × 91.3kJ mol-1 = −1815.0 kJ mol-1 o o U reaction = H reaction − nRT = −1815.0 kJ mol-1 − 8.314 J K -1mol-1 × 298.15 K = −1817.5 kJ mol-1 b) 2NO(g) + O2(g) → 2NO2(g) o H reaction = 2H of ( NO 2 , g ) − H of ( O 2 , g ) − 2H of ( NO, g ) = 2 × 33.2 kJ mol-1 − 0 − 2 × 91.3kJ mol-1 = −116.2 kJ mol-1 o o U reaction = H reaction − nRT = −116.2 kJ mol-1 + 8.314J K -1mol-1 × 298.15K = −113.7 kJ mol-1 c) TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(g) o H reaction = H of ( TiO 2 , s ) + 4H of ( HCl, g ) − H of ( TiCl 4 , l ) − 2H of ( H 2 O, l ) = −944 − 4 × 92.3 kJ mol -1 + 804.2 kJ mol-1 + 2 × 285.8 kJ mol -1 = 62.6 kJ mol -1 o o U reaction = H reaction − nRT = 62.6 kJ mol -1 − 4×8.314 J K -1mol -1 × 298.15 K = 52.7 kJ mol-1 d) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) +2H2O(l) Assume that NaOH(aq) and H2SO4(aq) are completely dissociated. The overall reaction is 2OH - (aq) + 2H + (aq) → 2H 2 O(l ) o H reaction = 2H of ( H 2 O, l ) − 2H of ( H + , aq ) − 2H of ( OH - , aq ) = −2 × 285.8 kJ mol-1 − 0 + 2 × 230.0 kJ mol-1 = −111.6 kJ mol-1 o o U reaction = H reaction − nRT = −111.6 kJ mol-1 − 0 = −111.6 kJ mol-1 e) CH4(g) + H2O(g) → CO(g) + 3H2(g) 50 o H reaction = H of ( CO, g ) + 3H of ( H 2 , g ) − H of ( CH 4 , g ) − H of ( H 2 O, g ) = −110.5 kJ mol-1 + 0 + 74.6 kJ mol-1 + 241.8 kJ mol-1 = 205.9 kJ mol-1 o o U reaction = H reaction − nRT = 205.9 kJ mol-1 − 2 × 8.3145 J K -1mol-1 × 298.15 K = 200.9 kJ mol-1 f) CH3OH(g) + CO(g) → CH3COOH(l) o H reaction = H of ( CH 3COOH, l ) − H of ( CH 3OH, g ) − H of ( CO, g ) = − 484.3 kJ mol-1 + 201.0 kJ mol-1 + 110.5 kJ mol -1 = −172.8 kJ mol -1 o o U reaction = H reaction − nRT = −172.8 kJ mol -1 + 2 × 8.3145 J K -1mol-1 × 298.15 K = −167.8 kJ mol-1 o o P4.2) Calculate H reaction and U reaction for the total oxidation of benzene. Also calculate o o H reaction − U reaction . o H reaction 15/2 O2(g) + C6H6(l) → 3H2O(l) + 6CO2(g) From the data tables, o H combustion = 3H of ( H 2 O, l ) + 6H of ( CO 2 , g ) − H of ( C6 H 6 , l ) = − 3 × 285.8 kJ mol-1 − 6 × 393.5 kJ mol -1 − 49.1kJ mol -1 = 3268 kJ mol -1 o o U reaction = H reaction − nRT = −3268 kJ mol-1 + 1.5 × 8.314J K -1mol-1 × 298.15 K = −3264 kJ mol -1 o o H reaction − U reaction −3268 kJ mol-1 + 3264 kJ mol -1 = = 0.0122 o H reaction −3268 kJ mol-1 P4.3) Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of formation of CO2(g) and H2O(l) to determine H of for benzene. 3H2O(l) + 6CO2(g) → 15/2 O2(g) + C6H6(l) o −H combustion ( C6 H 6 , l ) 6C(s) + 6O2(g) → 6CO2(g) 6H of ( CO 2 , g ) 51 3H of ( H 2 O, l ) 3H2(g) + 3/2O2(g) → 3H2O(l) _________________________________________________________ o 3H2(g) + 6C(s) → C6H6(l) −H combustion ( C6 H 6 , l ) + 6H of ( CO2 , g ) + 3H of ( H 2 O, l ) H of ( C6 H 6 , l ) = 3268 kJ mol -1 − 6 × 393.5 kJ mol-1 − 3 × 285.8 kJ mol-1 = 49.6 kJ mol -1 P4.4) Calculate H for the process in which N2(g) initially at 298.15 K at 1 bar is heated to 650 K at 1 bar. Use the temperature dependent heat capacities in the data tables. How large is the relative error if the molar heat capacity is assumed to be constant at its value of 298.15 K over the temperature interval? H o ( N 2 , g , 298.15 K → N 2 , g , 650 K ) = H of ( N 2 , g , 298.15 K ) + 650 T T CP ,m d K K 298.15 ∫ 650 T T2 T3 T = ∫ 30.81 − 0.01187 + 2.3968 × 10−5 2 − 1.0176 × 10−8 3 d J K -1mol-1 K K K K 298.15 = (10841 − 1980 + 1982 − 434.0 ) J mol-1 = 10.41 kJ mol-1 If is is assumed that the heat capacity is constant at its value at 298 K, 650 T H o ≈ ∫ ( 29.13) d J K -1mol-1 = 10.25 kJ mol-1 K 298.15 10.25 kJ mol-1 − 10.41 kJ mol-1 Error = 100 × = −1.54% 10.41 kJ mol-1 P4.5) The standard reaction enthalpies at 25ºC for the following reactions are given below. CaC2(s) + 2H2O(l) → Ca(OH)2(s) + C2H2(g) Ca(s) + ½ O2(g) → CaO(s) CaO(s) + H2O(l) → Ca(OH)2(s) o H reaction (kJ mol-1) –127.9 –635.1 –65.2 The standard enthalpy of combustion of graphite and C2H2(g) are –393.51 kJ mol-1 and –1299.58, kJ mol-1 respectively. Calculate the standard enthalpy of formation of CaC2(s) at 25ºC. Ca(OH)2(s) + C2H2(g) → CaC2(s) + 2H2O(l) CaO(s) + H2O(l) → Ca(OH)2(s) 2CO2(g) + H2O(l) → C2H2(s) + 5/2 O2(s) 2C(s) + 2O2(g) → 2CO2(g) Ca(s) + 1/2O2(g) → CaO(s) 52 o (kJ mol-1) H reaction +127.9 –65.2 1299.58 2 × (–393.51) –635.1 H of = –59.8 kJ mol-1 2C(s) + Ca(s) → CaC2(s) P4.6) From the following data at 25C o H reaction (kJ mol-1) 492.6 155.8 –393.51 –282.98 Fe2O3(s) + 3C(graphite) → 2Fe(s) + 3CO(g) FeO(s) + C(graphite) → Fe(s) + CO(g) C(graphite) + O2(g) → CO2(g) CO(g) + ½ O2(g) → CO2(g) Calculate the standard enthalpy of formation of FeO(s) and of Fe2O3(s). o H reaction (kJ mol-1) Fe(s) + CO(g) → FeO(s) + C(graphite) –155.8 282.98 CO2(s) → CO(g) + 1/2O2(g) –393.51 C(graphite) + O2(g) → CO2(g) __________________________________________________________________ H of = –266.3 kJ mol-1 Fe(g) + 1/2O2(g) → FeO(s) o H reaction (kJ mol-1) –492.6 2Fe(g) + 3CO(g) → Fe2O3(s) + 3C(graphite) –3×393.51 3C(graphite) + 3O2(g) → 3CO2(g) 3×282.98 3CO2(g) → 3CO(g) + 3/2O2(g) __________________________________________________________________ H of = –824.2 kJ mol-1 2Fe(g) + 3/2O2(g) → Fe2O3(s) P4.7) Calculate H of for NO(g) at 840 K assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at 298.15 K. 650 H o f T T C P d K K 298.15 ( NO, g , 840 K ) = H ( NO, g , 298.15K ) + ∫ o f 1 1 CP = CP ,m ( NO, g ) − CP ,m ( N 2 , g ) − CP ,m ( O 2 , g ) 2 2 = ( 29.86 − 0.5 × 29.13 − 0.5 × 29.38 ) J K -1mol-1 = 0.605 J K -1mol -1 650 T H of ( NO, g , 840 K ) = H of ( NO, g , 298.15 K ) + ∫ 0.605 d J mol-1 K 298.15 = H of ( NO, g , 298.15 K ) + 0.328 kJ mol -1 = 91.3 kJ mol -1 + 0.328 kJ mol-1 = 91.6 kJ mol-1 53 o P4.8) Calculate H reaction at 650 K for the reaction 4NH3(g) +6NO(g) → 5N2(g) + 6H2O(g) using the temperature dependent heat capacities in the data tables. H o reaction ( 650 K ) = H 650 o reaction T T C P d K K 298.15 ( 298.15K ) + ∫ CP = 5CP ,m ( N 2 , g ) + 6CP ,m ( H 2 O, g ) − 4CP , m ( NH 3 , g ) − 6CP ,m ( NO, g ) ( 5 × 30.81 + 6 × 33.80 − 4 × 29.29 − 6 × 33.58 ) − ( 5 × 0.01187 + 6 × 0.00795 + 4 × 0.01103 − 6 × 0.02593) T K 2 J K -1mol -1 = T − 5 + ( 5 × 2.3968 + 6 × 2.8228 − 4 × 4.2446 − 6 × 5.3326 ) ×10 K2 3 − ( 5 ×1.0176 + 6 ×1.3115 − 4 × 2.7706 − 6 × 2.7744 ) ×10−8 T K 3 2 3 T T T −4 T −8 T CP d = 38.21 + 0.00441 − 2.0053 ×10 + 1.4772 ×10 J K -1mol-1 2 3 ∫ K K K K K 298.15 650 650 T T2 T3 T = ∫ 38.21 + 0.00441 − 2.0053×10−4 2 + 1.4772 ×10−8 3 d J mol-1 K K K K 298.15 = (13.444 − 0.736 − 16.585 + 0.630 ) kJ mol-1 = −1.775 J mol-1 o H reaction ( 298.15 K ) = 5H of ( N 2 , g ) + 6H of ( H 2O, g ) − 4H of ( NH3 , g ) − 6H of ( NO, g ) o H reaction ( 298.15 K ) = −6 × 241.8 kJ mol-1 + 4 × 45.9 kJ mol-1 − 6 × 91.3 kJ mol-1 = −1814 kJ mol-1 o H reaction ( 650 K ) = −1814 kJ mol-1 − 1.775 kJ mol-1 = −1812 kJ mol-1 P4.9) From the following data at 298.15 K as well as the data tables, o H reaction (kJ mol-1) –137.0 –562.0 Fe(s) + 2H2S(g) → FeS2(s) + 2H2(g) H2S(g) + 3/2 O2(g) → H2O(l) + SO2(g) calculate the standard enthalpy of formation of H2S(g) and of FeS2 (s). o (kJ mol-1) H reaction 562.0 H2O(l) + SO2(g) → H2S(g) + 3/2O2(g) S(s) + O2(g) → SO2(g) –296.8 H2(g) + 1/2O2(g) → H2O(l) –285.8 __________________________________________________________________ H2(g) + S(s) → H2S(g) H of = –20.6 kJ mol-1 o (kJ mol-1) H reaction –137.0 Fe(s) + 2H2S(g) → FeS2(s) + 2H2(g) 54 –2 × 20.6 2H2(g) + 2S(s) → 2H2S(g) _________________________________________________________________ Fe(s) + 2S(s) → FeS2(s) H of = –178.2 kJ mol-1 P4.10) Calculate the average C-H bond enthalpy in methane using the data tables. Calculate the percent error in equating C-H bond energy in Table 4.3 with the bond enthalpy. CH4(g) → C(g) + 4H(g) o H reaction = 4H of ( H, g ) + H of ( C, g ) − H of ( CH 4 , g ) = 4 × 218.0 kJ mol-1 + 716.7 kJ mol-1 + 74.6 kJ mol-1 = 1663 kJ mol-1 1663 kJ mol -1 = 415.8 kJ mol-1 4 415.8 kJ mol -1 − 411 kJ mol -1 Relative Error = 100 × = 1.2% 415.8 kJ mol -1 Average Bond Enthalpy = o P4.11) Use the average bond energies in Table 4.3 to estimate U reaction for the reaction o C2H4(g) + H2(g) → C2H6(g). Also calculate U reaction from the tabulated values of H of for reactants and products (Appendix A, Data Tables). Calculate the percent error o in estimating U reaction from the average bond energies for this reaction. Ureaction = –(C–C bond energy + 6 C–H bond energy – H–H bond energy – C=C bond energy –4 C–H bond energy) Ureaction = – (346 kJ mol-1 + 6 × 411 kJ mol-1 – 432 kJ mol-1 – 602 kJ mol-1 – 4 × 411 kJ mol-1)= –134 kJ mol-1. Using the data tables, o H reaction ( 298.15 K ) = H of ( C2 H 6 , g ) − H of ( C2 H 4 , g ) − H of ( H 2 , g ) o H reaction ( 298.15 K ) = −84.0 kJ mol-1 − 52.4 kJ mol-1 = −136.4 kJ mol-1 o o U reaction ( 298.15 K ) = H reaction ( 298.15 K ) − nRT = −136.4 kJ mol -1 + 8.314 J mol-1 K -1 × 298.15 K= − 133.9 kJ mol-1 +134 kJ mol -1 − 133.9 kJ mol-1 Relative Error = 100 × ≈ 0% −133.9 kJ mol-1 P4.12) Calculate the standard enthalpy of formation of FeS2(s) at 300ºC from the data below at 25ºC and from the information that for the reaction o =–1655 kJ mol-1. 2FeS2(s) + 11/2O2(g) → Fe2O3(s) + 4 SO2(g), H reaction Assume that the heat capacities are independent of temperature. 55 Substance Fe(s) FeS2(s) Fe2O3(s) S(rhombic) SO2(g) -1 o –824.2 –296.81 H f (kJ mol ) CP ,m /R 3.02 7.48 2.72 o H reaction = H of ( Fe 2 O3 , s ) + 4 H of (SO 2 , g ) –2 H of ( FeS2 , s ) =–824.2 kJ mol-1–1655 kJ mol-1– 1655 kJ mol-1–1655 kJ mol-1 o = –1655 kJ mol-1 2FeS2(s) + 11/2O2(g) → Fe2O3(s) + 4SO2(g) H reaction −1655 kJ mol −1 = H of ( Fe 2 O3 ,s ) + 4H of (SO 2 , g ) − 2H of ( Fe 2S2 , s ) H o f ( Fe2S2 , s,298 K ) = 1655 kJ mol -1 + H of ( Fe 2 O3 , s ) + 4H of ( SO2 , g ) 2 1655 − 824.2 − 4 × 296.81 kJ mol -1 = 2 -1 = −178.2 kJ mol The enthalpy of formation at 300° C is given by H of ( FeS2 ( s ) ,573 K ) − H of ( FeS2 ( s ) , 298 K ) + 573 K ∫ C p (T ) dT 298 K Because the heat capacities are assumed to be independent of T, H of ( FeS2 ( s ) , 573 K ) = H of ( FeS2 ( s ) , 298 K ) + CP ,m ( FeS2 , s ) − CP ,m ( Fe, s ) − 2CP ,mS ( s ) [573 K − 298 K ] = −178.2 kJ mol -1 + 8.314 JK -1 mol-1 × ( 7.48 − 3.02 − 2 × 2.70 ) × [573 K − 298 K ] = −180.0 kJ mol-1 o P4.13) At 1000 K, H reaction = –123.77 kJ mol-1 for the reaction N2(g) + 3H2(g) → 2NH3(g). CP ,m = 3.502R, 3.466R, and 4.217R for N2(g), H2(g), and NH3(g), respectively. Calculate H of of NH3(g) at 300 K from thi