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
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www.aw-bc.com Preface This Instructor’s Solutions Manual has a
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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
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