Chapter 2 Solutions Manual

5/14/2018 Chapter 2 Solutions Manual

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Chapter 2: Heat, Work, Internal Energy, Enthalpy, and the First Law ofThermodynamics

Problem numbers in italics indicate that the solution is included in the Student's SolutionsManual.Questions on ConceptsQ2.1) Electrical current is passed through a resistor immersed in a liquid in an adiabatic container. Thetemperature of the liquid is varied by 1C. The system consists solely of the liquid. Does heat or workflow 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 theboundary between the surroundings and the system because of the temperature difference between them.Q2.2) Two ideal gas systems undergo reversible expansion under different conditions starting from thesame P and V. At the end of the expansion, the two systems have the same volume. The pressure in thesystem that has undergone adiabatic expansion is lower than in the system that has undergone isothermalexpansion. Explain this result without using equations.In the system undergoing adiabatic expansion, all the work done must come through the lowering of I1U,and therefore of the temperature. By contrast, some of the work done in the isothermal expansion cancome at the expense of the heat that has flowed across the boundary between the system andsurroundings.Q2.3) You have a liquid and its gaseous form in equilibrium in a piston and cylinder assembly in aconstant temperature bath. Give an example of a reversible process.Water drips onto the piston, increasing the conversion of gas to liquid. This process is essentiallyreversible if the rate of evaporation of the water is nearly equal to the rate of water deposition.Q2.4) Describe how reversible and irreversible expansions differ by discussing the degree to whichequilibrium is maintained between the system and the surroundings.In a reversible expansion, the system and surroundings are always in equilibrium with one another. Inan irreversible expansion, they are not in equilibrium with one another.Q2.5) For a constant pressure process, I1H = ql'. Does it follow that qr is a state function? Explain.Heat is never a state function, even if the heat transferred in a process has a unique value because asystem at equilibrium does not possess heat.

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Chapter 2 1 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynami

Q2.6) A cup of water at 278 K (the system) is placed in a microwave oven and the oven is turned onI minute during which it begins to boil. Which of q,w, and I1 U are positive, negative, or zero?The heat q is positive because heat flows across the system-surrounding boundary into the system. Thwork w is negative because the vaporizing water does work on the surroundings. I1 U is positive becauthe temperature increases and some of the liquid is vaporized.Q2.7) In the experiment shown in Figure 2.4a, I1 U u < 0, but I 1 T u > O.Explain how this is possible.This is only possible if a process, such as a chemical reaction in this case, occurs to changecomposition of the system.Q2.8) What is wrong with the following statement?: Burns caused by steam at 1000e can be msevere than those caused by water at 1000e because steam contains more heat than water. Rewritesentence 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 toliquid phase. In doing so, energy is released that is absorbed by the skin. Hot water does not releasemuch energy in the same situation, because no phase change occurs.Q2.9) 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, astate at equilibrium is not associated with values of heat or work.Q2.10) You have a liquid and its gaseous form in equilibrium in a piston and cylinder assemblyconstant temperature bath. Give an example of an irreversible process.A large mass is placed on the piston, leading to complete conversion of gaseous to liquid water.Q2.11) What is wrong with the following statement?: Because the well-insulated house stored a loheat, the temperature didn't fall much when the furnace failed. Rewrite the sentence to convey the sainformation in a correct way.Heat can't be stored because it exists only as a transient. A possible rephrasing follows. Becausehouse was well insulated, the walls were nearly adiabatic. Therefore, the temperature of the housenot fall as rapidly when in contact with the surroundings at a lower temperature as would have beencase if the walls were diathermal.Q2.12) Explain how a mass of water in the surroundings can be used to determine q for a proceCalculate q if the temperature of a I.OO-kg water bath in the surroundings increases by I.2SoC. Assuthat the surroundings are at a constant pressure.

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Chapter 2 1 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

If heat flows across the boundary between the system and the surroundings, it will lead to a temperaturechange in the surroundings given by I'1T = _ 5 f _ . For the case of interest,C I'q =-Q,"rrolndll1gs =-m Cpl'1T = -1.00 x I03 gx4.19 J g-IK-1x 1.25 K =-5.24x1 03 J .Q2.13) A chemical reaction occurs in a constant volume enclosure separated from the surroundings bydiathermal walls. Can you say whether the temperature of the surroundings increases, decreases, orremains the same in this process? Explain.No, because any of these possibilities could occur, depending on the potential energy stored in the bondsof the reactants and those of the products.Q2.14) Explain the relationship between the terms exact differential and state function.In order for a functionj(x,y) to be a state function, it must be possible to write the total differential dfinthe form df = (af) d x + (afJ dy. If the form dfas written exists, it is an exact differential.aXy ayxQ2.15) In the experiment shown in Figure 2.4b, the weight drops in a very short time. How will thetemperature of the water change with time?Initially, there will not be a unique value of the temperature for the whole system, because diffusion istoo slow to ensure equilibrium. At a location far from the filament, the temperature will be observed torise and asympotically approach its equilibrium value.

ProblemsP2.1) 4.25 moles of an ideal gas with CV,m=3/2R initially at a temperature T,= 325 K and Pi = 1.00 baris enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 575 kgmass on the piston of diameter 20.0 ern. Calculate the work done in this process and the distance that thepiston travels. Assume that the mass of the piston is negligible.We first calculate the external pressure and the initial volume.~xlerna/ = F =] 05 Pa + m~ =1.00 x 105 Pa + 575 kgx9.81 m2s-

2=2.80x1 05 PaA 1Cr nx (0. 100 m )

V = nR T = 4.25 molx8.314 Jmor1K-1 x325 K =0.115 m ' =115 LI ~ 105 Pa

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Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynami

Following Example Problem 2.6,C, + RP.cxlernalr,m p

I = 298 Kxc, + R~xlernal~,m PI

= 558 Kv = nR T = 4.25 molx8.314 JmorlK-1 x558 K = 7.06xlO-2 m'j PI 2.80x105 Pa

w=-~xlernal(VI-V,)=-2.80Xl05 Pax(7.05XIO-2 m3-11.5x10-2 m3)=12.4xl03 J

P2.2) The temperature of2.50 moles of an ideal gas increases from 13.5C to 55.l C as the gas iscompressed adiabatically. Calculate q, w, I"l.U, and I"l.H for this process assuming that CV,m = 3 /2 R .q =0 because the process is adiabatic.W = I"l.U = nCv,ml" l.T = 2.50 molx 3x8.314 ~mor1K-1 x( 55.1C -13.5C) = 1.30xl03 JW=I"l.U + I" l.(PV) =I"l.U + nRI"l.T = 1.30x 103 J + 2.50 molx8.314 JmorlK-1 x( 55.lC -I3.5C)W= 2.16xl03 JP2.3) 1.65 moles of an ideal gas, for which CV im = 3 /2 R , is subjected to two successive changes in sta(1) From 39.0C and 100. x 103 Pa, the gas is expanded isothermally against a constant pressure of 16x 103 Pa to twice the initial volume. (2) At the end of the previous process, the gas is cooled at constavolume from 39.0C to -25.0C. Calculate q, w, I"l.U, and I"l.H for each of the stages. Also calculatew, I"l.U, and I"l.H for the complete process.a) V = n R T =1.65 molx8.314 Jmol-1K-1 x312 K=4.28xIO-2 m'

I P , 100.x103 PaVI = 2V, = 8.56x 10-2 rrr'W = -~Xl (VI - v , ) = -16.5x 103 Pa x( 8.56x 10-2 m' -4.28x 10-2 m3) = -707 JI"l.U and I"l.H = 0 because I"l.T = 0q=-w=707 J

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b) I1U =n C v , T I I ( T t - T , ) =1.65 molx 1.5x 8.314 J m o r ' K - ' x (248 K - 312 K)=-1.32x103 J

W = 0 because I1V = 0q = I1U = -1.32 x103 JI1H = n C r , T I I ( T t - T , ) = n ( C V ,m + R )( T t - T , )

=1.65 molx2.5x8.314 J mor'K-' x(248 K-312 K)= -2.19 x103 J

I1Ulolal =0-1.32x103 J= -1.32x103 JWlolal = 0 - 707 J = - 707 JQl"Ia/ =707 J-1.32x103 J=-610.JI1Hloial = 0 - 2.19 x 103J = - 2.19 xl 03 J

P2.4).A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packingemergency rations which if completely metabolized will release 30. kJ of heat per gram of rationsconsumed. How much rations must the hiker consume to avoid a reduction in body temperature of 4.0 Kas a result of heat loss? Assume the heat capacity of the body equals that of water and that the hikerweighs 55 kg. State any additional assumptions.

QmratlOn5 =--

qrations

55x103 gx75.3 J K-' mol-' x4.0 K =919 kJ18.02 g mor '

919kJ =3130. kJ g-I g

P2.5) Count Rumford observed that using cannon boring machinery a single horse could heat 11.6 kg ofice water (T= 273 K) to T= 355 Kin 2.5 hours. Assuming the same rate of work, how high could ahorse raise a 150. kg weight in one minute? Assume the heat capacity of water is 4.18 J K-1 g-l.

Cpm .I1T 4.18JK-' glx(355-273)K ,Rate = walel = = 442 J s"time, 2.5 hr x 3600 s hr'h= Rate x time2 = 442 J s-' x60 s =18 m

mwelghlg 150. kg x9 .81 m S2

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P2.6) 2.25 moles of an ideal gas at 35.6C expands isothermally from an initial volume of26.0 drrr' tfinal volume of70.0 dm3. Calculate w for this process (a) for expansion against a constant externalpressure of 1.00 x l 05 Pa and (b) for a reversible expansion.a) w = -P",{erna/~V = -1.00x1 05 Pax(70.0 - 26.0 )x10-3 m' = -4.40 x103 J

V. 70.0 drrr'W"v'''ihle =-nRTln-f =-2.25 mol=B.Ll-l JmorIK-1x(273.15+35.6) Kxln 3b) e e. v : 26.0 dm= -5.72x103 J

P2.7) Calculate q, w, ~u,and ~H if 1.65 mol of an ideal gas with CV,m = 3/2R undergoes a reversibadiabatic expansion from an initial volume Vi = 7.75 rrr ' to a final volume Vj= 20.5 rrr'. The initialtemperature is 300. K.

q = 0 because the process is adiabatic.

( JI-ri . = ~

5T =(20.5 L)I-3 xT =157 Kf 7.75 L I

~u= W= nCv,n~T = 1.65 mol> 3x8.314 ~morlK-1 x(157 K -300. K) = -2.95x1 03 J~H = si:+nR~T = -2.95x103 J + 1.65 molx8.314 J morlK-1 x(157 K -300. K)W= -4.91x103 J

P2.8) Calculate Wfor the adiabatic expansion of 1 mol of an ideal gas at an initial pressure of 2.25 bafrom an initial temperature of 475 K to a final temperature of322 K. Write an expression for the workdone in the isothermal reversible expansion of the gas at 322 K from an initial pressure of2.25 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 = 5/2R.

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Chapter 2/ Heat, Work, Intemal Energy, Enthalpy, and the First Law of Thermodynamics

w . ::::-nRTln!i_'ln!i_:::: -Wrel'erslhlereversible PI' r, nRT

In!i_= 1.91x103J =-0.7127P I 1 molx8.314 Jmol-1K-1x322 K

PI = 0.472 P ; = 1.10 barP2.9) At 298 K and 1 bar pressure, the density of water is 0.9970 g ern", the coefficient of thermalexpansion is 2.07 x 10-4 K-', and CP,m = 75.30 J K' mor'. If the temperature of250. g of water isincreased by 38.0 K, calculate w, q, Mi,and i1U.v , = m = 0.250 kg = 2.51x 10--4rrr'p 997 kg m-3i1V = V,Pi1T = 2.51 x 10--4m' x 2.07 x 10--4K-1 x 38 K=1.97x10-6 m'W=-~xli1V =-105 Pax1.97x10-6m3 =-0.197 Ji1H = q = nCp IIi1T = 250. g 1x 75.3 J K-1 mor ' x 38 K

,1 18.02 g mol-=39.7x103 Ji1U=q+w=39.7x103 J

P2.10) A muscle fiber contracts by 2.0 em and in doing so lifts a weight. Calculate the work performedby the fiber and the weight lifted. Assume the muscle tiber obeys Hooke's law F = -k x with a forceconstant, k, of 800. N m-I .

P2.11) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionlessadiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with C";m = 3/2R. Initially, T,=325 K and Pi = 2.50x 105 Pa in each part. Heat is slowly introduced into the left part using an electricalheater until the piston has moved sufficiently to the right to result in a final pressure Pj= 7.50 bar in theright 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.

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The number of moles of gas in each part is given byn = ~ v , = 2.50x105 Pax50.0x10-3 m3 = 4.63 mol

RT, 8.314Pam3mor1K-1x325 Ka) We first calculate the final temperature in the right side.

1_23T=(2.50x105paJ~ x325K=504Kf 7.50x 105Pa

~U = w = nCv,m~T = 4.63 mol> 3x8.314 ~morlK-1 x(504 K -325 K)= 10.3x103 Jb) We first calculate the final volume of the right part.V rt = _ n _ R _ ~ _ f= 4.63 molx8.314x10-

2 LbarmorlK-1 x504 K = 25.9 L. P , t 7.50 bar

Therefore, Vif= 100.0 L - 25.9 L = 74.1 L.T , . = ~tv,t = 7.50 barx74.1 L = 1.45x103K'j n R 4.63 molx8.314x10-2 LbarmolK"

~U = nCv,II1~T = 4.63 molx3/2 x8.314 Jmol-1K-1X(1.45x103K -325 K) = 64.7x103 JFrom part a), W = 10.3 x 103.q = I1U - W = 64.7 x 103 J + 10.3 X 103 J =75.0 x 103 JP2.12) In the adiabatic expansion of 2.25 mol of an ideal gas from an initial temperature of 32.0C, twork done on the surroundings is 1450. J. If CV,m = 3/2R, calculate q, W, ~U, and ~H .q = 0 because the process is adiabatic~U = W = -1450. J~U = nC v ,1I1 (Tf - T, )

~U +nC v /liT,T= 'j Cn V,1Il-1450. J +2.25x1.5x8.314 J mor1K-1 x305 K=----------------------~~-------2.25x1.5x8.314 J mor1K-1

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Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

M! = n C I ' , 1 1 1 ( T j - T ; ) = n ( C v , m + R ) ( T j - T ; )= 2.25 x 2.5 x8.314 J mol-1K 1(253 K -305 K)=-2.42xI03 J

P2.13) A system consisting of73.2 g ofliquid water at 289 K is heated using an immersion heater at aconstant pressure of 1.00 bar. If a current of 2.25 A passes through the 10.0-0Iml resistor for 125 s, whatis the final temperature of the water?

2 () /2 Rt + n C I ' m T ;q = 1 Rt = n C 1 ' , m Tj - T ; ; '0 = "'-c.,(2.25 A ) 2 x10.0 ohmx125 s + 73.2 g Ix75.291 Jmor1K-1x289 KT = 18.02 gmor

j 73.2 g Ix75.291 JmorIK-'18.02 gmol=310. KP2.14) 2.75 moles of an ideal gas is expanded from 375 K and an initial pressure of 4.75 bar to a finalpressure of 1.00 bar, and 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 topart (a).a)

V j ~w = -nRTln- = -nRTln-V, P I4.75 bar 3= -2.75 molx8.314 J mol-1K-1x450 Kxln = -13.3x10 J1. 00 bar

b) Because q = 0, w = I.!u. In order to calculate I.!U,we first calculate T ; ;

I~3T=Tx(4.75bar)-f =201K

j I l.00 bar

w=I.!U=nC, I.!T=2.75 molx3x8.314 JmorlK-1 x(201 K-375 K)=-5.96x103 J~~ 2

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Less work is done on the surroundings in part b) because in the adiabatic expansion, the temperaturefalls and therefore the final volume is less than that in part a).P2.1S) A bottle at 285 K contains an ideal gas at a pressure of 155.7 x 103 Pa. The rubber stopperclosing the bottle is removed. The gas expands adiabatically against Pexternal = 111.4 x 103 Pa, and somgas is expelled from the bottle in the process. When P = Pexternal, the stopper is quickly replaced. Theremaining in the bottle slowly warms up to 285 K.What is the final pressure in the bottle for amonatomic gas, for which CV,m = 3 / 2 R , and a diatomic gas, for which CV,m = 5 1 2 R ?In this adiabatic expansion, I1U = wn C v ,1Il ( 7 ; - T ; ) = -~XI ( V r - v , )C (r - T) =-P [ n R 7 ; - n R . T ; JV,m I I ext R P

1 I

(c, + R~x ' J r = ( C + R~ .x , ~TV .m R 1 V,m P I1 I

=253 KOnce the stopper is put in place, the gas makes a transformation from7 ; = 253 K, ~ = 111.4xl03 Pa to Tf = 285 K and r,

T 285 KP, = _I P , = x 111.4 x 103 Pa = 125.7 x 103 Pa. T ; 253 KThe same calculation carried out for C; =2 . R gives.m 2

T r =262 KP r = 121.3 x 103 Pa

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P2.16) One mole of an ideal gas with CVm = 3/2R initially at 325 K and I.50x 105 Pa undergoes areversible adiabatic compression. At the end of the process, the pressure is 2.50 x 106 Pa. Calculate thefinal temperature of the gas. Calculate q, w, f...U, and f...H for this process.

q = 0 because the process is adiabatic.I - C " , , , , / C , , , , , 1-5/3

( P ~ c p m / c " ' " (I50XIOSpa]5f3T = T -' ' , =325 Kx . . =1.00xl03 K/ t PI 2.50xIOsPa

w=f...U=nCv,mf...T=1 mOlx3x8.314~mol-lK-1 x(1000 K-325 K)=8.44x103 Jf...H = /'."U + f . . . ( PV) = f...U + Rf...T = 5.62xl03J +8.314 Jmol-1K-1 x(1000 K -325 K)f...H = 14.1x103 JP2.17) A vessel containing 2.25 mol of an ideal gas with Pi = 1.00 bar and CP, m = 5/2R is in thermalcontact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initiallyat 312 K, and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and waterbath have an average heat capacity of Cp = 6250. J K-1 The gas is compressed reversibly to Pj= 10.5bar. 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 ofas an isothermal reversible process. If the answer substantiates this assumption, it is valid.

V f P ,w = -nRT In- = -nRT.ln-1 V ; I PI

= -2.25 mol> 8.314 Jmor1K-1 x 312 Kxln 1.00 bar = 13.7x103 J10.5 barf...U. -C f...Tcombined system - PWe use C; in the above equation because the heat capacity is dominated by the waterbath, and for a liquid C; ~C;sr = / l " Uc ombmed s v s t em 13.7x103 J = 0.976 Kc, 6250 JK-1Tf ~ 313 KThe result justifies the assumption.

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P2.18) An ideal gas undergoes an expansion from the initial state described by Pi, ~, T to a final statdescribed by PI, VI, Tin (a) a process at the constant external pressure Pr and (b) in a reversible procesDerive expressions for the largest mass that can be lifted through a height h in the surroundings in theprocesses.

VI nRT Vb) w=mgh=-nRTln-" m= ---In-V' h VI g IP2.19) An ideal gas described by T, = 300. K, Pi = 1.10 bar, and Vi = 10.0 L is heated at constantvolume until P = 9.80 bar. Itthen undergoes a reversible isothermal expansion until P = 1.10 bar. It isthen restored to its original state by the extraction of heat at constant pressure. Depict this closed-cycleprocess in a P-V diagram. Calculate w for each step and for the total process. What values for w woulyou calculate if the cycle were traversed in the opposite direction?

12

8P(bar)6

42

20 40 60V e l )

80 100 120

n= P ; ~ = 1.10barxl0.0L =0.481 molRT, 8.3145xl0-2L bar molK" x275 KThe process can be described bystep 1: Pi,vi,Ti - - - - -+ PI = 9.80 bar.F, TI

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Before calculating the work in step 2, we first calculate T,.

T. = T ~ = 275 K x 9.80 bar = 2450 KI I P ; 1.10 barVI ~W = -nRT. In- -nRT. In-

I v : I PI9.80 bar j= -0.481 mol> 8.314 J morlK-1 x 2450 Kxln = -21.4xlO- J1.10 bar

In step 3,

v = ~ v : = 9.80 V =89.1 L2 P ; 1.10 I105 Pa 10-3m'w=-~x'erna,i1.V=-1.00barx bar x(9.80L-89.l L)x L =8.70xl03 J

W,ycle =0-21.4x103 J +8.70xl03 J =-12.7xl03 JIf the cycle were traversed in the opposite direction, the magnitude of each work term would beunchanged, but all signs would change.P2.20) In an adiabatic compression of 1 mol of an ideal gas with CV,m = 5/2 R, the temperature risesfrom 293 K to 325 K.Calculate q, w , i1.H, and st:q = 0 because the process is adiabatici1.U = nC v,mi1.T= 1mol x 2.5 x 8.314 J K-I mol' x 32.0 Ki1.U = w = 665 Ji1.H = n ( Cv,m + R)I1T = 932 JP2.21) The heat capacity of solid lead oxide is given by

TC". = 44.35 + 1.47 x 10-3- in units of J K-Imor ',111 K

Calculate the change in enthalpy of3.25 mol ofPbO(s) ifit is cooled from 750. to 300. K at constantpressure.

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ij/l.H = n f Cp,mdT

300( T } ( T )3.25x f 44.35+1.47xl0-3_ _7~ K K

[

44.35X(300K-750K) 1~ 3.25x + [ 1.47;102 (~)2[= -64.9 x103 J - 1.13x 103 J=-66.0xl03 J

P2.22) One mole of carbon dioxide, for which CP,m = 37.1 J K-' mor' at 298 K, is expanded reversiblyand adiabatically from a volume of2.85 L and temperature of300. K to a final volume of 16.5 L.calculate the final temperature, q, W, / l.H, and / l.U. Assume that CP,m is constant over the temperatureinterval.

q = 0 because the process is adiabatic

(V J 1 - C P / C v ( J 1-[371/(371-8.314)JTl=~ u. =300.Kx 16.5L =181K. V, 2.85 r . ,

/l.U = C; (Tl - ~) = (37.1-8.314) J K-1 mol " x (181 K -300. K)/ l .U=w=-3.44xl03 Jmorl/l.H = (C v + R )( Tl - ~) = -4.43 X103 J mol-IP2.23) One mole of an ideal gas for which P = 1.00 bar and T = 300. K is expanded adiabatically agaan external pressure of 0.100 bar until the final pressure is 0.100 bar. Calculate the final temperature,w, / I .H, and / l.Ufor a) CV,m = 3/2 R, and b) CV,m = 5/2 R.a) for Cv; = 3/2 R

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' ( ) [ T l T , 1-T =-nRP _'--n e v , m j / external PI p ,T [ n e , + n R P . : x l e m a . l =T [ n e ' + n R p , x , e r n a ' )j f ,m P / I,m p

j /

[e v + R P . : x l e , r I I " I1 [ 1.5x 8.314 J mol' K,I+ 8.314 J mol' K,I x0.1 00 bar 1T =T ,m P , = 300. K x 1.00 bar

j / C. + B P ' x , e r n a , 1.5X8.314JmorIK,]+8.314Jmo!,IK, lxI.00bart .m P j 1.00 barT l =192 K

I1U=w=nCv,mI1T= 1 molxl.5x8.314JKl mor-1(192-300.)K=-1.35xI03 JMI = nC p ,,,,I1T = 1 mol x (CV,tI/+ R )I1T = -2.24 x 103Jb) for CV,m = 5/2 R

[e v , m + R P . : x , e r n a ' 1 [ 2.5x 8.314] mo],IK,1 + 8.314 J molK ,IX0.100 bar 1

T = T P , = 300. K x 1.00 barf / c. + R P , x l e r n a i 25x8314J !,1K,1 8.314 Jmor1K']xI.00bar

f m P . . mo +, f 1.00barTf =222 KI1U=w=nCvI1T= 1 molx2.5x8.314JK-l mor1(222-300.)K= -1. 60 x 103 JM/=n(Cv +R)I1T=-2.24x10 3 JP2.24) 2.25 moles of N, in a state defined by T, = 300. K and Vi = 1.75 L undergoes an isothermalreversible expansion until Vj= 20.5 L. Calculate w assuming (a) that the gas is described by the ideal gaslaw and (b) that the gas is described by the van der Waals equation of state. What is the percent error inusing the ideal gas law instead of the van der Waals equation? The van der Waals parameters for N2 arelisted in Table 7.4.a) for the ideal gasWrcl'crsible =-nRTln VI =-2.25 molx8.314 JmorlK,l x300 K xln 20.5 LV , 1.00 L= -17.0 x l 03 J

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Chapter 21Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynam

for the van der Waals gaso 0 ( }RT aw-- P -- ----- f externa,dV - f V -b V2 V~ . ~ t / / m m

v > ( RT} VI ( a }=-f- V+f-2 VV Vm-b V V " ,,The first integral is solved by making the substitutiony = Vm b.

V ( JRT Yf RT-f. -_- dV=- f(-JdY=-RT[ln(v/+b)-ln(v,+b)]V Vm b y, y,

Therefore, the work is given byw=-nRTln (VI-b) +a(J_ __ l . ](v, -b) V , VI=-2.25 molx8.314 1morlK-1 x300 K xln 20.5 L-0.0387 L1.00 L - 0.0387 L

+1.366 L2 barx10SPa)0-6m6 (1 1 Jbar L2 1.00x10-3m3 20.5x10-3m3w=-17.2xl03 1+657 1=-16.5xI031

-16 5x103 1 + ] 7 0 x103 1Percent error = 100x' 3' = -2.7% (before rounding oft)-16.5x10 1P2.25) A major league pitcher throws a baseball with a speed of 150. kilometers per hour. If the basebweighs 220. grams and its heat capacity is 2.0 1 g-l K-1, calculate the temperature rise of the ball whenis stopped by the catcher's mitt. Assume no heat is transferred to the catcher's mitt and that the catchearm does not recoil when he/she catches the ball.v =150.x 103 m hr-I x hr =41.7m S-I3600 sq p = CI'!J.T= limv'

h m v 2 0.5xO.220 kgx( 41.7 m S-Ir!J.T= = = 0.43 KC 1 , m 20001 g-I K-IP2.26) A 1.65 mol sample of an ideal gas for which C't:m=3/2R undergoes the following two-sprocess: (1) From an initial state of the gas described by T = 14.5C and P = 2.00 x 104 Pa, theundergoes an isothermal expansion against a constant external pressure of 1.00 x 104 Pa untilvolume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature fall-35.6C. Calculate q, w, !J.U, and !J.H for each step and for the overall process.

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a) For the first step, I1U = f...H = 0 because the process is isothermal.v = nRT,

I pI

1.65 molxS.314 J mol-1K-1x(273.15 + 14.5)K 1= = 0.197m-2.00xl04Paw = -q = = -~xlernaII1V = = -1.00xl04 Pax2 x 0.197 m '= -1.97Xl 03 Jb) For the second step, W = 0 because 1 1V = O .q = = I1U = = nC vl1T = = 1.65 mol> 3xS.314 J mor1K-

1 x( -35.6C - 14.5C) = -1.03 x103 J2I1H = = i1U + 1 1 ( PV) = I1U + nRI1T = -1.03x 103 J

+1.65 molxS.314 Jmor1K-1 x( -35.6C - 14.5C)f...H = -1.72xl03 JFortheoverallprocess,w=-1.97xl03 J,q=-1.03 x 103 J+ 1.97xIQ3 J= 941 J, I1U= -1.03 X 103 J, and I1H= - 1.72 x 103 J.P2.27) 1.75 moles of an ideal gas, for which Cv. = 3 /2R , initially at 32.0C and 2.50 x 106 Paundergoes a two-stage transformation. For each of the stages described in the following list, calculatethe final pressure, as well as q, w, I1U, and I1H. Also calculate q, w, I1U, and I1H for the completeprocess.a. The gas is expanded isothermally and reversibly until the volume doubles.b. Beginning at the end of the first stage, the temperature is raised to 92.0C at constant volume.a) P = P o V ; = F a = l.25 xl 06 Pa

I V 22w=-nRT InV 2 =-1.75 molxS.314Jmor1K-1x305 Kxln2=-3.0Sxl03 J

V ;I1U =0 and f...H = 0 because I1T = 0q=-w=3.0Sxl03 J

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b) ~ = 1;; ~ = T2~ = 365 Kx1.25xl06 Pa =15.0xI05 Pa~ ~ t; 305 Ksi: = nCv,mb.T = 1.75 molx1.5x8.314 J mor1K-1 x(365 K-305 K)

=1.31xl03 JW = 0 because b .V = 0q = b .U = 1.31x 103 J

3b .H=nC1, ,b .T= n(C" +R )b .T= 1.75 molx-x8.314Jmor1K-1~.~ 2x(365 K-305 K)=2.18xl03 J

For the overall process,q = 3.08x 103 J + 1.31X 103 J = 4.39x 103 Jw=-3.08xl03 J+0=-3.08xl03 Jb .U = 0 + 1.31x 103 J = 1.31x 103 Jb .H = 0 + 2.18 x103 J = 2.18 x 103 JP2.28) 1.75 mole of an ideal gas with CV,m = 3/2R is expanded adiabatically against a constant externpressure of 1.00 bar. The initial temperature and pressure are T, = 290. K and Pi = 19.5 bar. The finalpressure is Pr= 1.00 bar. Calculate q, w, b .U, and b .H for the process.b .U =nCv,m (Tr - r ; ) = -Pexternal (Vr - V ; ) =wq = 0 because the process is adiabatic.

C' R~xterna'u +----="-',m PI

C + R~xterna'V,m Pf

r, =180. Kb .U = w = nCv,nb.T = 1.75 mol> 3x8.314 ~mor1K-1 x(180 K - 290 K) = -2.40xl03 JM l = b .U + nRb .T = -2.40xl03 J + 1.75molx8.314 Jmol-1K-1x(180 K - 290 K)b .H = -4.00x103 J

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P2.29) A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximatethis process as a reversible adiabatic compression. Assume the initial pressure and temperature of the airbefore it is put in the tire to be Pi = 1.00 bar and T,= 305 K. The final volume of the air in the tire is Vj=1.00 L and the final pressure is Pj= 5.75 bar. Calculate the final temperature of the air in the tire.Assume that Cv~m=5/2R.

1 _ 25

T = ( 1.00bar ) f x305 K = 503 Kf 5.75 bar

P2.30) For 2.25 mol of an ideal gas, Pexternal = P = 200. x 103 Pa. The temperature is changed from122C to 28.5C, and C V,f / l = 3/2R. Calculate q, w, I 1U , and I 1 H .

3I1 U = .c; I1 T = 2.25 molx-x8.314 J morlK1 x(302 K-395 K) = -2.63xl03 J,m 2I1 H = n C p , m l 1 T = n ( C V,f / l + R ) I1 T

5= 2.25 molx-x8.314 J mol'K" x(302 K -395 K)2= -4.37 X 103 J= s,

w=I1U-ql'=-2.63x103 J+4.37x103 J=1.75x103 JP2.31) Suppose an adult is encased in a thermally insulating barrier so that all the heat evolved bymetabolism of foodstuffs is retained by the body. What temperature does her body reach after 3.0 hours?Assume the heat capacity of the body is 4.18 J g-'K-' and that the heat produced by metabolism is 10. kJk -Ih-Igr.

P2.32) Consider the isothermal expansion of 4.50 mol of an ideal gas at 450. K from an initial pressureof 12.0 bar to a final pressure of2.75 bar. Describe the process that will result in the greatest amount ofwork being done by the system with ~xlema/ ;::: 2.75 bar and calculate w. Describe the process that willresult in the least amount of work being done by the system with ~Xlet11a/ ;::: 3.50 bar and calculate w.What is the least amount of work done without restrictions on the external pressure?

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The greatest amount of work is done in a reversible expansion. The work is given byVI P 12.0 bar

wreverstble = -nRTln-'- = -nRTln-' = -4.50 molxS.314 J mol-1K-1 x450 K xln---V , PI 2.75 bar= -24.Sxl03 J

The least amount of work is done in a single stage expansion at constant pressure with the externalpressure equal to the final pressure. The work is given by

) ( 1 1 J--P .- . -- T ---- external (V I V , - nR ~xtertlGl PI P ;= -4.50 molxS.314 J molK.' x450 Kx2.75 barx( 1 _ 1 )2.75 bar 12.0 bar

=-13.0xl03 JThe least amount of work done without restrictions on the pressure is zero, which occurs whenPexterna/ = O .P2.33) An automobile tire contains air at 275. x 103 Pa at 27.5C. The stem valve is removed and theair is allowed to expand adiabatically against the constant external pressure of one bar until P =PexternFor air, CV,m = 5 1 2 R . Calculate the final temperature. Assume ideal gas behavior.because q = 0, !1U = wnC V,m ( T j - T;) = - r;(VI - V , )

( ) (nRTI nRT;I


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P2.34) One mole of an ideal gas is subjected to the changes below. Calculate the change in temperaturefor each case if Cv,m =3/2 R.

a) q = -500. J, W = 150. Jb) q = 425. J, W = -425 Jc) q = 0, W = 175 J

a) I J , . U = q +W = -350. JCv,m = 1.5R = 12.47 J K-1 mol "I J . . T = I J , . U = -350. J mol-I = -28.1 K

Cv,m 12.47 J K-1 mol"

b) I J ,. U = q + w = OCv,m = 1.5R = 12.47 J K-1 mol'I J , . T = I 1 U = 0 = 0

CV,m 12.47 J K-1 mol'c ) I J . . U = q + w = 1 7 5 JCv,m = 1.5R = 12.47 J K-1 mol"I J . . T = I 1 U = 175 J mol-' = 14.0 K

Cv,m 12.47 J K-' mol-IP2.35) Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with CII,m = 3/2R. Theinitial state is described by P = 5.50 bar and T = 285. K.a. Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a final

pressure of P = 1.00 bar.b. Calculate the final temperature if the same gas undergoes an adiabatic expansion against an external

pressure of P = 1.00 bar to a final pressure P = 1.00 bar.Explain the difference in your results for parts (a) and (b).a)

T f = ( V f J I - r = ( T ~ J I - r [ l l J I - r / T j J r [ ~ I I - r . r, [ ~ J I ~ rt, V ; 7; PI ~ t: = PI) 'r; = PI1 - 2 .3

T =(S.50bar) f x285 K=144 Kf 1.00 bar

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Chapter 2 1 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamib)I 1 U = n C v , m ( T t - T ; ) = - P .x le r n a l ( V t - v , )n C v , m ( T t - T ; ) = - n R P . x ' e r n a ' [ T , t - T ; JPt P ;T ( n c , + n R P . x l e m a ' J = T ( n c , + n R P . x l e r n a ' Jj " , 1 1 1 P I I v,m p ;T =Tj I

c + R P . x l e r n a 'V,III P

I = 285 Kx 1.5x8.314 ] morlK-1+ 8.314] mo[-IK-1x1.00bar1.00bar

1.5x8.314J morIK-1+ 8.314] mo[-IK-1xI.00bar5.50 bar

c + R P . x l e r n a lV,m PI

TI = 192 KMore work is done on the surroundings in the reversible expansion, and therefore I 1 U and thetemperature decrease more than for the irreversible expansion.P2.36) A pellet of Zn of mass 14.5 g is dropped into a flask containing dilute H2S04 at a pressure of1.00 bar and temperature of T= 325K. What is the reaction that occurs? Calculate w for the process.Zn(s) + H2S04(aq) -+ Zn2+(aq) + SO/-(aq) +H2(g)The volume of H2 produced is given byV= 14.5 g 1 mol H7 8.314 JmorlK-1 x325 K 599 10-3 3----.;:c_--c-x - X 5 = . x m

65.39 gj rnol Znt 1mol Zn i-to Paw = - P . x l e r n a I I 1 VI 1 V ; : : : : :volume of H, produced.w=-1.00xl05 Pax5.99xlO-3 rrr' =-599JP2.37) Calculate I 1 H and I 1 U for the transformation of 1 mol of an ideal gas from 35.0C andatm to 422C and 17.0 atm if

TC, = 20.9 + 0.042-in units of J K'mol " ..m K

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T,MI=nfC1 , dT.m

1 ;

= 69T(20.9+0.042 T'lAT'308K K r

=20.9 x (695 K - 308 K) J + [ 0.021T2 J : : : : J= 8.088 x 103J + 8.153 x l03 J= 1.62x 104 J

1lU =Mf - i1(P V) =i1H - nR i1T= 1.62 X 104 J - 8.314 J K-1morl x (695 -308) K= 13.0x 103J

P2.38) 2.50 moles of an ideal gas for which CV,m = 20.8 J K-1 mol-1 is heated from an initial temperatureof 10.5C to a final temperature of 305C at constant volume. Calculate q, W, llU, and i1H for thisprocess.W = 0 because llV= O.i1U = q = nC\.i1T = 2.50 molx 20.8 J mor1K-1 x294.5 K = 15.3xl03 Ji1H = !1U + i1(PV) = i1U + nRi1T = 15.3xl03 J + 2.50 molx 8.314 JmollK" x275 K=21.4xl03 JP2.39) An ideal gas undergoes a single-stage expansion against a constant external pressure Pexternal atconstant temperature from T, Pi, Vi, to T, r; V ; :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 ern, Pi = 2.50 x 106 Pa, P]= 0.750 x 106 Pa, T =300. K, and n = 1.50 mol, calculate the

values of the masses in parts (a) and (b).Consider the expansiona) W = mgh = - ~ x , (VI - V , )

-~x, (VI - 1 1 , )m= ghfor the final volume to be VI' the external pressure can be no bigger than Pf

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b) Consider the compressionm' = -~x, ( V ; - V I )

ghfor the final volume to be V ; , the pressure can be no smaller than ~

c)v = nRT = 1.50 molx 8.314 J morlK-lx 300. K =1.50x10-3 m3I t; 2.50x 106 PaPjVI = ~ V ;V = ~ V ; = 2.50 x 106Pa x 1.50 x 10-3 m3 = 4.99 x 10-3 m'f PI 0.750 x 106 Pa

-0.750x106 Pax( 4.99x10-3 m' -1.50x10-3 m3)m - = 2.67 x 103 kgmax - 9.81 m S- 2 x 0.100 m

-2.50x106 Pax(1.50xlO-3 m3 -4.99x103 m3)m~in = 2 = 8.90x103 kg9.81 m s x 0.100 m

P2.40) The formalism of the Young's modulus is sometimes used to calculate the reversible workinvolved in extending or compressing an elastic material. Assume a force F is applied to an elastic roof cross sectional area A a and length L a . As a result of this force the rod changes in length by I ! . . L . TheYoung's modulus E is defined asE = tensile stress = A , = F L o

tensile strain ~" A a I ! . . La) Relate k in Hooke's Law to the Young's modulus expression given above.

b) Using your result in part a show that the magnitude of the reversible work involved in changing thelength L a of an elastic cylinder of cross sectional area A a by M is w = _ . ! _ ( I ! . . L J 2 E A o L o .2 t,

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W ~ , . " ~ J J it t< ~ IILdAL ~ ~ k'M' ~ ~ EAcI. (~~ JP2.41) The Young's modulus (see P 2 A 1 ) of muscle fiber is approximately 3.12xl07 Pa. A muscle fiber2.00 cm in length and 0.100 em in diameter is suspended with a mass Mhanging at its end. Calculate themass required to extend the length of the fiber by 10%.

7 (0.0010 m)23.12x10 Pax0.10xO.0200mxJ[x ~~-EAoM 2m = = J = 0.25 kgLog 0.0200 mx9.81 m s~-P2.42) DNA can be modeled as an elastic rod which can be twisted or bent. Suppose a DNA moleculeof length L is bent such that it lies on the arc of a circle of radius R e. The reversible work involved inbending DNA without twisting is Wh . I = BL2 where B is the bending force constant. The DNA in ae n c 2 R cnucleosome particle is about 680. A in length. Nucleosomal DNA is bent around a protein complexcalled the histone octamer into a circle of radius 55 A . Calculate the reversible work involved in bendingthe DNA around the histone octamer if the force constant B = 2.00x 10-28 J m-I

= BL =2.00x10-28Jmx680.x10~IO m=22x10~19 JWbend 2 2'2 R c 2x(55x10-'O m ]P2.43) 3.00 moles of an ideal gas are compressed isothermally from 56.0 to 24.0 L using a constantexternal pressure of3.35 atm. Calculate q, W, I1U, and I1H.

= -3.35x l.013 X105 Pa x (24.0xlO~3 L -56.0x 10-3 L ) = l.09 X104 JI1U = 0 and MI = 0 because I1T = 0q = -w= -1.09xl04 J

44

Chapter 2 Solutions Manual

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