Exerc Cios Resolvidos - Cap. 02-Atkins (a)

8/22/2019 Exerc Cios Resolvidos - Cap. 02-Atkins (a)

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TRAPP: CHAP02 2006/3/8 17:05 PAGE 23 #1

2 The First Law

Answers to discussion questions

D2.1 Work is a precisely defined mechanical concept. It is produced from the application of a force through adistance. The technical definition is based on the realization that both force and displacement are vector

quantities and it is the component of the force acting in the direction of the displacement that is used in

the calculation of the amount of work, that is, work is the scalar product of the two vectors. In vector

notation w = f d = fdcos , where is the angle between the force and the displacement. The

negative sign is inserted to conform to the standard thermodynamic convention.

Heat is associated with a non-adiabatic process and is defined as the difference between the adiabatic

work and the non-adiabatic work associated with the same change in state of the system. This is the

formal (and best) definition of heat and is based on the definition of work. A less precise definition of

heat is the statement that heat is the form of energy that is transferred between bodies in thermal contact

with each other by virtue of a difference in temperature.

At the molecular level, work is a transfer of energy that results in orderly motion of the atoms andmolecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular

interpretation 2.1 for a more detailed discussion.

D2.3 The difference results from the definition H = U + PV; hence H = U + (PV). As (PV) is

not usually zero, except for isothermal processes in a perfect gas, the difference between H and U

is a non-zero quantity. As shown in Sections 2.4 and 2.5 of the text, H can be interpreted as the heat

associated with a process at constant pressure, and U as the heat at constant volume.

D2.5 In the Joule experiment, the change in internal energy of a gas at low pressures (a perfect gas) is zero.

Hence in the calculation of energy changes for processes in a perfect gas one can ignore any effect due

to a change in volume. This greatly simplifies the calculations involved because one can drop the first

term of eqn 2.40 and need work only with dU = CV dT. In a more sensitive apparatus, Joule would

have observed a small temperature change upon expansion of the real gas. Joules result holds exactly

only in the limit of zero pressure where all gases can be considered perfect.

The solution to Problem 2.33 shows that the JouleThomson coefficient can be expressed in terms of

the parameters representing the attractive and repulsive interactions in a real gas. If the attractive forces

predominate, then expanding the gas will reduce its energy and hence its temperature. This reduction

in temperature could continue until the temperature of the gas falls below its condensation point. This

is the principle underlying the liquefaction of gases with the Linde refrigerator, which utilizes the

JouleThomson effect. See Section 2.12 for a more complete discussion.


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24 SOLUTIONS MANUAL

D2.7 The vertical axis of a thermogram represents Cp, and the baselines represent the heat capacity associated

with simple heating in the absence of structural transformations or similar transitions. In the exampleshown in Fig. 2.16, the sample undergoes a structural change between T1 and T2, so there is no reason to

expect Cp after the transition to return to its value before the transition. Just as diamond and graphite have

different heat capacities because of their different structures, the structural changes that occur during

the measurement of a thermogram can also give rise to a change in heat capacity.

Solutions to exercises

E2.1(a) The physical definition of work is dw = Fdz [2.4].

In a gravitational field the force is the weight of the object, which is F = mg.

Ifg is constant over the distance the mass moves, d w may be integrated to give the total work,

w =

zfzi

Fdz =

zfzi

mgdz = mg(zf zi) = mgh where h = (zf zi).

On earth: w = (65kg) (9.81 m s2) (4.0m) = 2.6 103 J = 2.6 103 J needed .

On the moon: w = (65kg) (1.60m s2) (4.0m) = 4.2 102 J = 4.2 102 J needed .

E2.2(a) This is an expansion against a constant external pressure; hence w = pexV [2.8].

pex = (1.0 atm) (1.013 105 Pa atm1) = 1.01 105 Pa.

The change in volume is the cross-sectional area times the linear displacement:

V = (100 cm2) (10cm)

1 m

100 cm

3= 1.0 103 m3,

so w = (1.01 105 Pa) (1.0 103 m3) = 1.0 102 J as 1 Pa m3 = 1 J .

E2.3(a) For all cases U = 0, since the internal energy of a perfect gas depends only on temperature. (See

Molecular interpretation 2.2 and Section 2.11(b) for a more complete discussion.) From the definition

of enthalpy, H = U+pV, so H = U+ (pV) = U+ (nRT) (perfect gas). Hence, H = 0 as

well, at constant temperature for all processes in a perfect gas.

(a) U = H = 0 .

w = nRT ln

VfVi

[2.11]

= (1.00 mol) (8.314 J K1 mol1) (273K) ln

44.8 dm3

22.4 dm3

= 1.57 103 J = 1.57 kJ .

q = U w [First Law] = 0 + 1.57 kJ = +1.57 kJ .

(b) U = H = 0 .

w = pexV[2.8], V = (44.8 22.4) dm3 = 22.4 dm3.


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THE FI RST LAW 25

pex can be computed from the perfect gas law,

pV = nRT,

so pex = pf =nRT

Vf=

(1.00 mol) (0.08206 dm3 atm K1 mol1) (273K)

44.8 dm3= 0.500 atm

w = (0.500 atm)

1.013 105 Pa

1atm

(22.4 dm3)

1 m3

103 dm3

= 1.13 103 Pa m3 = 1.13 103 J = 1.13 kJ .

q = U w = 0 + 1.13 kJ = +1.13 kJ .

(c) U = H = 0Free expansion is expansion against no force, so w = 0 and q = U w = 0 0 = 0 .

COMMENT. An isothermal free expansion of a perfect gas is also adiabatic.

E2.4(a) For a perfect gas at constant volume

p

T=

nR

V= constant; hence,

p1

T1=

p2

T2.

p2 =

T2

T1

p1 =

400K

300K

(1.00 atm) = 1.33 atm .

U = nCV,mT[2.16b] = (n) (3

2

R) (400K 300K)

= (1.00 mol) (3

2) (8.314 J K1 mol1) (100K)

= 1.25 103 J = +1.25 kJ .

w = 0 [constant volume] q = U w [First Law] = 1.25 kJ 0 = +1.25 kJ .

E2.5(a) (a) w = pexV [2.8].

pex = (200 Torr) (133.3 Pa Torr1) = 2.666 104 Pa.

V = 3.3 dm3 = 3.3 103 m3.

Therefore, w = (2.666 104 Pa) (3.3 103 m3) = 88 J .

(b) w = nRT ln

Vf

Vi

[2.11].

n =4.50 g

16.04 g mol1= 0.2805 mol, RT = 2.577 kJ mol1, Vi = 12.7 dm

3, Vf = 16.0 dm3.

w = (0.2805 mol) (2.577 kJ mol1) ln

16.0 dm3

12.7 dm3

= 167 J .


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26 SOLUTIONS MANUAL

E2.6(a) H = condH = vapH = (1mol) (40.656kJ mol1) = 40.656 kJ .

Since the condensation is done isothermally and reversibly, the external pressure is constant at 1.00 atm.

Hence,

q = qp = H = 40.656 kJ .

w = pexV[2.8] where V = Vliq Vvap Vvap because Vliq


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THE FI RST LAW 27

q = H = 2.83 104 J = +28.3 kJ .

w = pexV[2.8] where pex = p.

w = pV = (pV)[constant pressure] = (nRT) [perfect gas] = nRT

= (1.00 mol) (8.314JK1 mol1) (473 K 298K)

= 1.45 103 J = 1.45 kJ .

U = q + w = (28.3 kJ) (1.45 kJ) = +26.8 kJ .

(b) The energy and enthalpy of a perfect gas depend on temperature alone (Molecular interpretation

2.2 and Exercise 2.3); hence it does not matter whether the temperature change is brought about at

constant volume or constant pressure; H and U are the same.

H = +28.3 kJ , U = 26.8 kJ .

Under constant volume, w = 0 .

q = U w = +26.8 kJ .

E2.9(a) For reversible adiabatic expansion

Tf = Ti

Vi

Vf

1/c[2.28a],

where c =CV,m

R=

Cp,m R

R=

(20.786 8.3145) J K1 mol1

8.3145J K1 mol1= 1.500,

so the final temperature is

Tf = (273.15K )

1.0 dm3

3.0 dm3

1/1.500= 131K .

E2.10(a) Reversible adiabatic work is

w = CVT[2.27] = n(Cp,m R) (Tf Ti)

where the temperatures are related by

Tf = Ti

Vi

Vf

1/c[2.28a] where c =

CV,m

R=

Cp,m R

R= 3.463.

So Tf = [(27.0 + 273.15) K]

500 103 dm3

3.00 dm3

1/3.463= 179 K

and w =

2.45g

44.0 g mol1

[(37.11 8.3145) J K1mol1] (179 300) K = 194 J .


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28 SOLUTIONS MANUAL

E2.11(a) For reversible adiabatic expansion

pfV

f= piV

i[2.29]sopf = pi

Vi

Vf

= (57.4 kPa)

1.0 dm3

2.0 dm3

1.4= 22 kPa

E2.12(a) Cp =qp

T[2.24] =

229J

2.55K= 89.8JK1

so Cp,m = (89.8JK1)/(3.0mol) = 3 0 J K1 mol1 .

For a perfect gas Cp,m CV,m = R[2.26].

CV,m = Cp,m R = (30 8.3) J K1 mol1 = 22 J K1mol1 .

E2.13(a) qp = CpT [2.24] = nCp,mT = (3.0 mol) (29.4 J K1 mol1) (25 K) = +2.2 kJ .

H = qp [2.23b] = +2.2 kJ .

U = H (pV) [from H U + pV] = H (nRT) [perfect gas] = H nRT

= (2.2 kJ) (3.0 mol) (8.314JK1 mol1) (25 K) = (2.2 kJ) (0.62 kJ) = +1.6 kJ .

E2.14(a) In an adiabatic process, q = 0 . Work against a constant external pressure is

w = pex

V = (600 Torr) 1.013 105 Pa760 Torr

(40 103 m3) = 3.2 kJU = q + w = 3.2 kJ .

One can also relate adiabatic work to T (eqn 2.27):

w = CVT = n(Cp,m R)T so T =w

n(Cp,m R),

T =3.2 103 J

(4.0 mol) (29.355 8.3145) J K1 mol1= 38 K .

H = U + (pV) = U + nRT

= (3.2 kJ) + (4.0 mol) (8.3145J K1 mol1) (38 K) = 4.5 kJ .

Question. Calculate the final pressure of the gas.

E2.15(a) In an adiabatic process, the initial and final pressures are related by (eqn 2.29)

pfV

f= piV

i

where =Cp

CV=

CV + nR

CV=

20.8 J K1 + (1.0 mol)(8.31J K1 mol1)

20.8 J K1= 1.40.


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THE FI RST LAW 29

Find Vi from the perfect gas law:

Vi =nRTi

pi=

(1.0 mol)(8.31 J K1 mol1)(310K)

3.25 atm

1atm

1.013 105 Pa

Vi = 7.83 103 m3.

so Vf = Vi

pi

pf

1/= (7.83 103 m3)

3.25 atm

2.50 atm

1/1.40= 0.0113 m3 .

Find the final temperature from the perfect gas law:

Tf =pfVf

nR=

(2.50 atm) (0.0113 m3)

(1.0 mol)(8.31 J K1 mol1)

1.013 105 Pa

1atm

Tf = 344 K .

Adiabatic work is (eqn 2.27)

w = CVT = (20.8 J K1)(344 310) K = 7.1 102 J .

E2.16(a) At constant pressure

q = H = nvapHTf = (0.50 mol) (26.0 kJ mol

1) = 13.0 kJ

and w = pV pVvap = nRT = (0.50mol) (8.3145 J K1 mol1) (250 K),

w = 1.0 103 J = 1.0 kJ .

U = w + q = 13.0 1.0 kJ = 12.0 kJ .

COMMENT. Because the vapor is here treated as a perfect gas, the specific value of the external pressure

provided in the statement of the exercise does not affect the numerical value of the answer.

E2.17(a) The reaction is

C6H5C2H5(l) +212

O2(g) 8CO2(g) + 5H2O(1)

cH = 8fH

(CO2, g) + 5fH(H2O, l) fH

(C6H5C2H5, l)

= [(8) (393.51) + (5) (285.83) (12.5)] kJ mol1

= 4564.7 kJ mol1 .

E2.18(a) First fH[(CH2)3, g] is calculated, and then that result is used to calculate rH

for the isomerization

(CH2)3(g) +9

2O2(g) 3 CO2(g) + 3H2O(l), cH

= 2091kJ mol1.

fH[(CH2)3, g] = cH

+ 3fH(CO2, g) + 3fH

(H2O,g)

= [+2091 + (3) (393.51) + (3) (285.83)] kJ mol1

= +53 kJ mol1 .


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30 SOLUTIONS MANUAL

(CH2)3(g) C3H6(g), rH =?

rH = fH(C3H6, g) fH[(CH2)3, g]

= (20.42 53) kJmol1 = 33 kJ mol1 .

E2.19(a) For naphthalene the reaction is C10H8(s) + 12O2(g) 10CO2(g) + 4H2O(l).

A bomb calorimeter gives qV = ncU rather than qp = ncH

; thus we need

cU = cH

ngRT [2.21], ng = 2 mol.

cH = 5157 kJmol1 [Table 2.6].

cU = (5157kJ mol1) (2) (8.3 103 kJ K1 mol1) (298K)

= 5152 kJ mol1 .

|q| = |qV| = |ncU| =

120 103 g

128.18 g mol1

(5152 kJmol1) = 4.823 kJ

C =|q|

T=

4.823 kJ

3.05 K= 1.58 kJ k1

When phenol is used the reaction is

C6H5OH(s) +152

O2(g) 6CO2(g) + 3H2O(l).

cH = 3054 kJmol1 [Table2.6].

cU = cH

ngRT, ng = 3

2

= (3054 kJ mol1) + ( 32

) (8.314 103 kJ K1 mol1) (298K)

= 3050 kJ mol1.

|q| =10 103 g

94.12 g mol1 (3050 kJmol1) = 0.3241 kJ.

T =|q|

C=

0.3241 kJ

1.58kJ K1= +0.205 K .

COMMENT. Inthis case cU and cH

differed by about 0.1 per cent. Thus, to within 3 significant figures,

it would not have mattered if we had used cH instead ofcU

, but for very precise work it would.

E2.20(a) The reaction is AgCl(s) Ag+(aq) + Cl(aq).

solH = fH

(Ag+, aq) + fH(Cl, aq) fH

(AgCl, s)

= [(105.58) + (167.16) (127.07)] kJ mol1 = +65.49 kJ mol1 .


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THE FI RST LAW 31

E2.21(a)NH3SO2(s) NH3(g) + SO2(g), rH

= +40kJmol1.

rH = fH

(NH3, g) + fH(SO2, g) fH

(NH3SO2, s).

Solving for fH(NH3SO2, s) yields

fH(NH3SO2, s) = fH

(NH3, g) + fH(SO2, g) rH

= (46.11 296.83 40) kJ mol1 = 383 kJ mol1 .

E2.22(a) (a) Reaction (3) = (2) reaction (1) + reaction (2) and ng = 1.

The enthalpies of reactions are combined in the same manner as the equations (Hesss law).

rH(3) = (2) rH

(1) + rH(2)

= [(2) (184.62) + (483.64)] kJ mol1

= 114.40 kJ mol1 .

rU = rH

ngRT [2.21] = (114.40 kJ mol1) (1) (2.48kJ mol1)

= 111.92 kJ mol1 .

(b) fH refers to the formation of one mole of the compound; hence

fH(J) =

rH(J)

J

.

fH(HCl,g) =

184.62

2kJ mol1 = 92.31 kJ mol1 .

fH(H2O, g) =

483.64

2kJmol1 = 241.82 kJ mol1 .

E2.23(a) rH = rU

+ ngRT [2.21]; ng = +2

= (1373 kJ mol1) + 2 (2.48 kJ mol1) = 1368 kJ mol1 .

COMMENT. As a number of these exercises have shown, the use ofrH as an approximation for rU

is often valid.E2.24(a) In each case, the strategy is to combine reactions in such a way that the combination corresponds to the

formation reaction desired. The enthalpies of the reactions are then combined in the same manner as the

equations to yield the enthalpies of formation.

(a) rH/(kJ mol1)

K(s) + 12

Cl2(g) KCl(s) 436.75

KCl(s) + 32

O2(g) KClO3(s)12

(89.4)

K(s) + 12

Cl2(g) +32

O2(g) KClO3(s) 392.1

Hence, fH(KClO3, s) = 392.1 kJ mol

1 .


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32 SOLUTIONS MANUAL

(b) rH/(kJmol1)

Na(s) +12 O2(g) +

12 H2(g) NaOH(s) 425.61

NaOH(s) + CO2(g) NaHCO3(s) 127.5

C(s) + O2(g) CO2(g) 393.51

Na(s) + C(s) + 12

H2(g) +32

O2(g) NaHCO3(s) 946.6

Hence, fH(NaHCO3, s) = 946.6 kJ mol

1 .

E2.25(a) When the heat capacities of all substances participating in a chemical reaction are assumed to be constant

over the range of temperatures involved Kirchoffs law [2.36] integrates to

rH(T2) = rH

(T1) + rC

p (T2 T1) [Example 2.6].

rC

p =

Products

C

p,m

Reactants

C

p,m [2.37],

rC

p = C

p (N2O4, g) 2 C

p (NO2, g) = (77.28) (2) (37.20 J K1 mol1)

= +2.88J K1 mol1,

rH(373K) = rH

(298K) + rC

p T

= (57.20 kJ mol1) + (2.88 J K1) (75 K)

= [(57.20) + (0.22)] kJ mol1

= 56.98 kJ mol1 .

E2.26(a) (a) rH =

Products

fH

Reactants

fH [2.34].

rH(298K) = [(110.53) (241.82)] kJ mol1 = +131.29 kJ mol1 .

rU(298K) = rH

(298K) ngRT [2.21]

= (131.29 kJ mol1) (1) (2.48 kJ mol1) = +128.81 kJ mol1 .

(b) rU(378K) = rH

(298K) (T2 T1)rC

p [Example 2.6].

rC

p = C

p,m(CO,g) + C

p,m(H2,g) C

p,m(C,gr) C

p,m(H2O,g)

= (29.14 + 28.82 8.53 33.58) 103 kJ K1 mol1

= 15.85 103 kJ K1 mol1.

rH(378K) = (131.29 kJ mol1) + (15.85 103 kJ K1 mol1) (80 K)

= (131.29 + 1.27) kJmol1 = +132.56 kJ mol1 .

rU(378K) = rH

(378K) (1) (8.31 103 kJ K1 mol1) (378 K)

= (132.56 3.14) kJ mol1 = +129.42 kJ mol1 .

COMMENT. The differences in both rH and rU

between the two temperatures are small and justify

the use of the approximation that rCp is a constant.


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34 SOLUTIONS MANUAL

E2.30(a) The internal energy is a function of temperature and volume, Um = Um(T, Vm), so

dUm =

Um

T

Vm

dT +

Um

Vm

T

dVm [T = (Um/V)T].

For an isothermal expansion dT = 0; hence

dUm =

Um

Vm

T

dVm = T dVm =a

V2mdVm,

Um =

Vm,2Vm,1

dUm =

Vm,1Vm,2

a

V2mdVm = a

24.8dm3 mol11.00dm3 mol1

dVm

V2m=

a

Vm

24.8dm3 mol1

1.00dm3 mol1

= a

24.8 dm3 mol1+

a

1.00 dm3 mol1=

23.8a

24.8 dm3 mol1= 0.9597 a mol dm3.

From Table 1.6, a = 1.352 dm6 atm mol1.

Um = (0.9597mol dm3) (1.352 dm6 atm mol2)

= (1.30 dm3atm mol1)

1m

10dm

3

1.013 105 Pa

atm

= +131 J mol1 .

w =

p dVm where p =

RT

Vm b

a

V2mfor a van der Waals gas. Hence,

w =

RT

Vm b

dVm +

a

V2mdVm = q + Um.

Therefore,

q =

24.8dm3 mol11.00dm3 mol1

RT

Vm b

dVm = RT ln(Vm b)|

24.8dm3 mol1

1.00dm3 mol1

= (8.314 J K1 mol1) (298K) ln

24.8 3.9 102

1.00 3.9 102

= +8.05 103 J mol1 .

and w = q + Um = (8.05 103 Jmol1) + (131 J mol1) = 7.92 103 J mol1 .

E2.31(a) = 1

VV

T

p

[2.43]; 320 = 1

V320

V

T

p, 320

.

V

T

p

= V300(3.9 104/K + 2.96 106T/K2),

V

T

p, 320

= V300(3.9 104/K + 2.96 106 320/K) = 1.34 103K1V300.

V320 = V300{(0.75) + (3.9 104) (320) + (1.48 106) (320)2} = (V300) (1.026).


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THE FI RST LAW 35

so 320 = 1V320

VT

p, 320

= 11.026V300

(1.34 103 K1 V300),

320 =1.34 103 K1

1.026= 1.31 103 K1 .

COMMENT. Knowledge of the density at 300 K is not required to solve this exercise, but it would be required

to obtain numerical values of the volumes at the two temperatures.

E2.32(a) The isothermal compressibility is

T =

1

V

V

p

T

[2.44] so

V

p

T

= TV.

At constant temperature

dV =

V

p

T

dp so dV = TV dp ordV

V= T dp.

Substituting V =m

and dV =

m

2d yields

dV

V=

d

= Tdp.

Therefore,

Tp.

For

= 0.08 102 = 8 104, p

8 104

T=

8 104

7.35 107 atm1= 1.1 103 atm .

E2.33(a) The isothermal JouleThomson coefficient isHm

p

T

= Cp,m = (0.25K atm1) (2 9 J K1 mol1) = 7.2 J atm1 mol1 .

dH = n

Hm

p

T

dp = nCp,m dp,

H =

p2p1

(nCp,m)dp = nCp,m(p2 p1) [andCpare constant],

H = (15mol) (+7.2 J atm1 mol1) (75 atm) = +8.1 kJ

so q(supplied) = +H = +8.1 kJ .

Solutions to problems

Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are

for 298 K.


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36 SOLUTIONS MANUAL

Solutions to numerical problems

P2.1 The temperatures are readily obtained from the perfect gas equation, T =pV

nR,

T1 =(1.00 atm) (22.4 dm3)

(1.00 mol) (0.0821dm3 atm mol1 K1)= 273 K = T3[isotherm].

Similarly, T2 = 546 K .

In the solutions that follow all steps in the cycle are considered to be reversible.

Step 1 2

w = pexV = pV = nRT [(pV) = (nRT)],

w = (1.00 mol) (8.314 J K1 mol1) (546 273) K = 2.27 103 J .

U = nCV,mT = (1.00 mol) 3

2 (8.314 J K1 mol1) (273K) = +3.40 103 J.

q = U w = +3.40 103J (2.27 103 J) = +5.67 103 J .

H = qp = +5.67 103 J .

If this step is not reversible, then w, q, and H would be indeterminate.

Step 2 3

w = 0 [constant volume].

qV = U = nCV,mT = (1.00 mol) (3

2) (8.314 J K1 mol1) (273K)

= 3.40 103 J .

From H U + pV

H = U + (pV) = U + (nRT) = U + nRT

= (3.40 103 J) + (1.00 mol) (8.314JK1 mol1) (273 K) = 5.67 103 J .

Step 3 1

U and H are zero for an isothermal process in a perfect gas; hence for the reversible compression

q = w = nRT lnV1

V3= (1.00 mol) (8.314JK1 mol1) (273 K) ln

22.4 dm3

44.8 dm3

= +1.57 103 J , q = 1.57 103 J .

If this step is not reversible, then q and w would have different values which would be determined by

the details of theprocess.


15/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 37 #15

THE FI RST LAW 37

Total cycle

State p/atm V/dm3 T/K

1 1.00 22.44 273

2 1.00 44.8 546

3 0.50 44.8 273

Thermodynamic quantities calculated for reversible steps

Step Process q/kJ w/kJ U/kJ H/kJ

1 2 p constant = pex +5.67 2.27 +3.40 +5.67

2 3 V constant 3.40 0 3.40 5.67

3 1 Isothermal, reversible 1.57 +1.57 0 0Cycle +0.70 0.70 0 0

COMMENT. All values can be determined unambiguously for the reversible cycle. The net result of the overall

process is that 700 J of heat has been converted to work.

P2.3 Since the volume is fixed, w = 0 .

Since U = q at constant volume, U = +2.35 kJ .

H = U + (pV) = U + Vp [V = 0].

From the van der Waals equation [Table 1.6]

p =RT

Vm b

a

V2mso p =

RT

Vm b[Vm = 0 at constant volume].

Therefore, H = U +RVT

Vm b.

From the data,

Vm =15.0 dm3

2.0 mol= 7.5 dm3 mol1, T = (341 300) K = 41 K.

Vm b = (7.5 4.3 102) dm3 mol1 = 7.46 dm3 mol1.

RVT

Vm b=

(8.314 J K1 mol1) (15.0 dm3) (41 K)

7.46 dm3 mol1= 0.68 kJ.

Therefore,H = (2.35 kJ) + (0.68 kJ) = +3.03 kJ .


16/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 38 #16

38 SOLUTIONS MANUAL

P2.5 This cycle is represented in Figure 2.2. Assume that the initial temperature is 298 K .

2

(b)

V/dm3

(c)

3

(a)

10050 150

2.0

1.0

p/atm

1

Figure 2.2

(a) First, note that w = 0 (constant volume). Then calculate U since T is known (T = 298K)

and then calculate q from the First Law.

U = nCV,mT [2.16b]; CV,m = Cp,m R =7

2R R =

5

2R,

U = (1.00 mol)

5

2

(8.314 J K1 mol1) (298K) = 6.19 103 J = +6.19 kJ .

q = qV = U w = 6.19 kJ 0 = +6.19 kJ .

H = U + (pV) = U + (nRT) = U + nRT

= (6.19 kJ) + (1.00 mol) (8.31 103 kJ mol1) (298K) = +8.67 kJ .

(b) q = 0 (adiabatic).

Because the energy and enthalpy of a perfect gas depend on temperature alone,

U(b) = U(a) = 6.19 kJ , since T(b) = T(a).

Likewise H(b) = H (a) = 8.67 kJ .

w = U = 6.19 kJ [First Law with q = 0].

(c) U = H = 0 [isothermal process in perfect gas].

q = w [First Law with U = 0]; w = nRT1 lnV1

V3[2.11].

V2 = V1 =nRT1

p1=

(1.00 mol) (0.08206 dm3 atm K1 mol1) (298 K)

1.00 atm= 24.45 dm3.


17/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 39 #17

THE FI RST LAW 39

V2Tc2 = V3T

c3 [2.28b]; hence V3 = V2 T2

T3

c

where c =CV,m

R

=5

2

so V3 = (24.45 dm3)

(2) (298K)

298K

5/2= 138.3 dm3.

w = (1.00 mol) (8.314 J K1 mol1) (298 K) ln

22.45 dm3

138.3 dm3

= 4.29 103 J = +4.29 kJ .

q = 4.29 kJ .

P2.7 The formation reaction is

2C(s) + 3H2(g) 2C6H6(g), fH(298K) = 84.68kJ mol1.

In order to determine fH(350K) we employ Kirchhoffs law [2.36] with T2 = 350 K, T1 = 298 K,

fH(T2) = fH

(T1) +

T2T1

rCp dT

where rCp =J

JCp,m(J) = Cp,m(C6H6) 2Cp,m(C) 3Cp,m(H2).

From Table 2.2

Cp,m(C6H6)/(J K1 mol1) = 14.73 + 0.1272

KT,

Cp,m(C, s)/(J K1 mol1) = 16.86 +

4.77 103

K

T

8.54 105 K2

T2

,

Cp,m(H2, g)/(J K1 mol1) = 27.28 +

3.26 103

K

T

0.50 105 K2

T2

,

rCp/(J K1 mol1) = 100.83 +

0.1079T

K

1.56 106 K2

T2

.

T2T1

rCp dT

J K1 mol1= 100.83 (T2 T1) +

12

0.1079K1

(T22 T21 )

(1.56 106 K2)

1

T2

1

T1

= 100.83 (52 K) +

1

2

(0.1079)(3502 2982) K

(1.56 106)

1

350

1

298

K

= 2.65 103 K.


18/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 40 #18

40 SOLUTIONS MANUAL

Multiplying by the units J K1mol1, we obtain

T2T1

rCpdT = (2.65 103 K) (J K1mol1) = 2.65 103 J mol1

= 2.65 kJ mol1.

Hence fH(350 K) = fH

(298 K) 2.65 kJ mol1

= 84.68 kJ mol1 2.65 kJ mol1 = 87.33 kJ mol1 .

P2.9 Cr(C6H6)2(s) Cr(s) + 2C6H6(g), ng = +2 mol.

rH = rU

+ 2RT, from [2.21]

= (8.0 kJ mol1) + (2) (8.314 J K1 mol1) (583 K) = +17.7 kJ mol1 .

In terms of enthalpies of formation

rH = (2) fH

(benzene, 583 K) fH(metallocene, 583 K)

or rH(metallocene, 583 K) = 2fH

(benzene, 583 K) 17.7 kJ mol1.

The enthalpy of formation of benzene gas at 583 K is related to its value at 298 K by

fH(benzene, 583 K) = fH

(benzene, 298 K)

+ (Tb 298 K)Cp,m(l) + vapH

+ (583 K Tb)Cp,m(g) 6 (583 K 298 K)Cp,m(gr) 3 (583 K 298 K)Cp,m(H2, g)

where Tb is theboiling temperature of benzene (353 K).We shall assumethat theheat capacities of graph-

ite and hydrogen are approximately constant in the range of interest and use their values from Table 2.15.

fH(benzene, 583 K) = (49.0 kJ mol1) + (353 298) K (136.1 J K1 mol1)

+ (30.8 kJ mol1) + (583 353) K (81.67 J K1 mol1)

(6) (583 298) K (8.53 J K1 mol1)

(3) (583 298) K (28.82 J K1 mol1)

= {(49.0) + (7.49) + (18.78) + (30.8) (14.59) (24.64)} kJ mol1

= +66.8 kJ mol1.

Therefore fH (metallocene, 583 K) = (2 66.8 17.7) kJ mol1 = + 116.0 kJ mol1 .

P2.11 (a) and (b). Thetable displays computed enthalpies of formation(semi-empirical, PM3level, PC Spartan

ProTM), enthalpies of combustion based on them (and on experimental enthalpies of formation of H2O(l)

and CO2(g), 285.83 and 393.51 kJ mol1 respectively), experimental enthalpies of combustion

(Table 2.5), and the relative error in enthalpy of combustion.


19/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 41 #19

THE FI RST LAW 41

Compound fH/kJ mol1 cH

/kJ mol1(calc.) cH/kJ mol1(expt.) %error

CH4(g) 54.45 910.72 890 2.33

C2H6(g) 75.88 1568.63 1560 0.55

C3H8(g) 98.84 2225.01 2220 0.23

C4H10(g) 121.60 2881.59 2878 0.12

C5H12(g) 142.11 3540.42 3537 0.10

The combustion reactions can be expressed as:

CnH2n+2(g) +

3n + 1

2

O2(g) n CO2(g) + (n + 1) H2O(1).

The enthalpy of combustion, in terms of enthalpies of reaction, is

cH = nfH(CO2) + (n + 1)fH(H2O) fH(CnH2n+2),

where we have left out fH(O2) = 0. The % error is defined as:

%error =cH

(calc.) cH(expt.)

cH(expt.) 100%.

The agreement is quite good.

(c) If the enthalpy of combustion is related to the molar mass by

cH = k[M/(g mol1)]n

then one can take the natural log of both sides to obtain:

lncH = ln |k| + n ln M/(g mol1).

Thus, if one plots lncH vs. ln [M/(g mol1)], one ought to obtain a straight line with slope n and

y-intercept ln |k|. Draw up the following table.

Compound M/(g mol1) cH/kJ mol1 ln M/(g mol1) ln

cH/ kJ mol1CH4(g) 16.04 910.72 2.775 6.814

C2H6(g) 30.07 1568.63 3.404 7.358

C3H8(g) 44.10 2225.01 3.786 7.708

C4H10(g) 58.12 2881.59 4.063 7.966

C5H12(g) 72.15 3540.42 4.279 8.172

The plot is shown in Fig 2.3.

The linear least-squares fit equation is:

ln |cH/kJ mol1| = 4.30 + 0.903 lnM/(g mol1) R2 = 1.00

These compounds support the proposed relationships, with

n = 0.903 and k = e4.30 kJ mol1 = 73.7 kJ mol1 .


20/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 42 #20

42 SOLUTIONS MANUAL

ln M/(g mol1)

ln|

cH

/kJmol1|

26

7

8

9

3 4 5

Figure 2.3

The agreement of these theoretical values ofk and n with the experimental values obtained in P2.10 is

rather good.

P2.13 The reaction is

C60(s) + 60O2(g) 60CO2(g).

Because the reaction does not change the number of moles of gas, cH = cU [2.21]. Therefore

cH = (36.0334 kJ g1) (60 12.011g mol1) = 25968 kJ mol1 .

Now relate the enthalpy of combustion to enthalpies of formation and solve for that of C60.

cH

= 60fH

(CO2) 60fH

(O2) fH

(C60),

fH(C60) = 60fH

(CO2) 60fH(O2) cH

= [60(393.51) 60(0) (25968)] kJ mol1 = 2357 kJ mol1 .

P2.15 (a) rH = fH

(SiH2) + fH(H2) fH

(SiH4)

= (274 + 0 34.3) kJ mol1 = 240 kJ mol1 .

(b) rH = fH

(SiH2) + fH(SiH4) fH

(Si2H6)

= (274 + 34.3 80.3) kJ mol1

= 228 kJ mol1

.

P2.17 The temperatures and volumes in reversible adiabatic expansion are related by eqn 2.28a:

Tf = Ti

Vf

Vi

1/cwhere c =

CV,m

R.

From eqn 2.29, we can relate the pressures and volumes:

pf = pi

Vf

Vi

where =

Cp,m

CV,m.


21/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 43 #21

THE FI RST LAW 43

We are looking for Cp,m, which can be related to c and .

c =

CV,m

R

Cp,m

CV,m

=

Cp,m

R.

Solving both relationships for the ratio of volumes, we have

pf

pi

1/=

Vf

Vi=

Tf

Ti

cso

pf

pi=

Tf

Ti

c.

Therefore

Cp,m = R

ln

pf

pi

ln

TfTi

= (8.314 J K1 mol1)

ln

202.94 kPa

81.840 kPa

ln

298.15K248.44K

= 41.40 J K1 mol1 .

P2.19 Hm = Hm(T,p).

dHm =

Hm

T

p

dT +

Hm

p

T

dp.

Since dT = 0,

dHm =

Hm

p

T

dp where

Hm

p

T

= Cp,m[2.53] =

2a

RT b

.

Hm =

pf

pi

dHm =

pf

pi

2a

RT b

dp =

2a

RT b

(pf pi)

= (2) (1.352 dm6 atm mol2)

(0.08206 dm3 atm K1 mol1) (300 K) (0.0387 dm3 mol1)

(1.00 atm 500 atm)

= (35.5 atm)

1 m

10 dm

3

1.013 105 Pa

1 atm

= 3.60 103 J = +3.60 kJ .

COMMENT. Note that it is not necessary to know the value ofCp,m.

Solutions to theoretical problems

P2.21 (a) dz =

z

x

y

dx+

z

y

x

dy [definition of total differential].

z

x

y

= (2x 2y + 2),

z

y

x

= (4y 2x 4),

dz = (2x 2y + 2) dx+(4y2x4)dy .


22/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 44 #22

44 SOLUTIONS MANUAL

(b)

y z

x =

y(2x 2y + 2) = 2,

xz

y =

x(4y 2x 4) = 2.

(c)

z

x

y

=

y +

1

x

,

z

y

x

= (x 1),

dz =

y +

1

x

dx + (x - 1) dy .

A differential is exact if it satisfies the condition

x

z

y

=

y

z

x

,

y

z

x

=

y

y +

1

x

= 1,

x

z

y

=

x(x 1) = 1.

COMMENT. The total differential of a function is necessarily exact.

P2.23 U = U(T, V) so dU =

U

T

V

dT +

U

V

T

dV = CVdT +

U

V

T

dV.

For U = constant, dU = 0, and

CVdT =

U

V

T

dV or CV =

U

V

T

dV

dT

U

=

U

V

T

V

T

U

.

This relationship is essentially Eulers chain relation [Further information 2.2].

H = H(T,p) so dH =

H

T

p

dT +

H

p

T

dp = CpdT +

H

p

T

dp.

According to Eulers chain relationH

p

T

p

T

H

dT

dH

p

= 1

so, using the reciprocal identity [Further information 2.2],

H

p T

= T

p H

dH

dTp

= - Cp .

P2.25 (a) H = U + pV so

H

U

p

= 1 + p

V

U

p

= 1 +p

(U/V)p.

(b)

H

U

p

=(H/V)p

(U/V)p=

((U + pV)

V

p

(U/V)p=

(U/V)p + p

(U/V)p

so

H

U

p

= 1 +p

(U/V)p= 1 + p

V

Up.


23/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 45 #23

THE FI RST LAW 45

P2.27 w =

V2

V1

p dV.

InsertingV

n= Vm into the virial equation for p we obtain

p = nRT

1

V+

nB

V2+

n2C

V3+ ...

.

Therefore, w = nRT

V2V1

1

V+

nB

V2+

n2C

V3+ ...

dV,

w = nRT lnV2

V1+ n2RTB 1

V2

1

V1+ 1

2n3RTC 1

V22

1

V22 + ....

For n = 1 mol: nRT = (1.0 mol) (8.314 J K1 mol1) (273 K) = 2.27 kJ.

From Table 1.4, B = 21.7 cm3 mol1 and C = 1200 cm6 mol2, so

n2BRT = (1.0 mol) (21.7 cm3 mol1) (2.27 kJ) = 49.3 kJ cm3,

1

2n3CRT =

1

2(1.0 mol)2 (1200cm6 mol2) (2.27 kJ) = +1362 kJ cm6.

Therefore,

(a)w = 2.27 kJ ln 2 (49.3 kJ)

1

1000

1

500+ (1362 kJ) 1

10002

1

5002= (1.57) + (0.049) (4.1 103) kJ = 1.52 kJ = 1.5 kJ .

(b) A perfect gas corresponds to the first term of the expansion ofp, so

w = 1.57 kJ = 1.6 kJ .

P2.29 =

T

p

H

= 1

Cp

H

p

T

[2.51 and 2.53],

=1

Cp

T

V

T

p

V

[See problem 2.34 for this result].

But V =nRT

p+ nb or

V

T

p

=nR

p.

Therefore,

=1

Cp

nRT

p V

=

1

Cp

nRT

p

nRT

p nb

=

nb

Cp.

Since b > 0 and Cp > 0, we conclude that for this gas < 0 or

T

p

H

< 0. This says that when the

pressure drops during a JouleThomson expansion the temperature must increase .


24/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 46 #24

46 SOLUTIONS MANUAL

P2.31p =

nRT

V nb

n2a

V2[Table 1.7].

Hence T = p

nR

(V nb) +

naRV2

(V nb) ,

T

p

V

=V nb

nR=

Vm b

R=

1p

T

V

.

For Eulers chain relation, we need to show thatT

p

V

pV

T

VT

p= 1.

Hence, in addition to

T

p

V

we need

p

V

T

and

V

T

p

=1

T

V

p

.

p

V

T

=nRT

(V nb)2+

2n2a

V3

which can be found from

TV

p= p

nR

+ na

RV2

2na

RV3

(V nb),

T

V

p

=

T

V nb

2na

RV3

(V nb).

Therefore,

T

p

V

p

V

T

V

T

p

=

T

p

V

p

V

T

T

V

p

=

V nb

nR

nRT(V nb)2

+ 2n2a

V3

T

V nb

2na

RV3

(V nb)

=

T

V nb

+

2na

RV3

(V nb)

T

V nb

2na

RV3

(V nb)

= 1.


25/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 47 #25

THE FI RST LAW 47

P2.33 Cp = TVT

p

V =T

TV

p

V [reciprocal identity, Further information 2.2]

T

V

p

=T

V nb

2na

RV3(V nb) [Problem 2.31]

Introduction of this expression followed by rearrangement leads to

Cp =(2na) (V nb)2 nbRTV2

RTV3 2na(V nb)2 V.

Then, introducing =RTV3

2na(V nb)2to simplify the appearance of the expression

Cp =

1

nb

V

1

V =

1 b

Vm

1

V.

For xenon, Vm = 24.6 dm3 mol1, T = 298 K, a = 4.137 dm6 atmmol2, b = 5.16 102 dm3 mol1,

nb

V =

b

Vm =

5.16 101 dm3 mol1

24.6 dm3 mol1 = 2.09 103

,

=(8.206 102 dm3 atm K1 mol1) (298 K) (24.6 dm3 mol1)3

(2) (4.137 dm6 atmmol2) (24.6 dm3 mol1 5.16 102 dm3 mol1)2= 73.0.

Therefore, Cp =1 (73.0) (2.09 103)

72.0 (24.6 dm3 mol1) = 0.290 dm3 mol1.

Cp = 20.79 J K1 mol1[Table 2.6], so

=0.290 dm3 mol1

20.79 J K1 mol1=

0.290 103 m3 mol1

20.79 J K1 mol1

= 1.393 105 K m3 J1 = 1.393 105 K Pa1

= (1.393 105) (1.013 105 K atm1) = 1.41 K atm1 .

The value of changes at T = TI and when the sign of the numerator 1 nb

Vchanges sign ( 1 is

positive). Hence

b

Vm= 1 at T = TI or

RTbV3

2na(V nb)2Vm= 1 implying that TI =

2a(Vm b)2

RbV2m,


26/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 48 #26

48 SOLUTIONS MANUAL

that is, TI = 2a

Rb

1

b

Vm2

=

27

4 Tc

1

b

Vm2

.

For xenon,2a

Rb=

(2) (4.137 dm6 atmmol2)

(8.206 102 dm3 atmK1 mol1) (5.16 102 dm3 mol1)= 1954K

and so TI = (1954 K)

1

5.16 102

24.6

2= 1946 K .

Question. An approximate relationship for of a van der Waals gas was obtained in Problem 2.30. Use

it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result

above.

P2.35 Cp,m CV,m = 2

TVnT

[2.49] = TVn

pT

V

[2.57].

p

T

V

=nR

V nb[Problem 2.31].

V =

V

T

p

=1

T

V

p

.

Substituting,

Cp,m CV,m =

Tp

T

V

n

T

V

p

.

Also

T

V

p

=T

V nb

2na

RV3(V nb) [Problem 2.31].

Substituting,

Cp,m CV,m =

RT

(V nb)

T

(V nb)

2na

RV3 (V nb)

= R

with =1

1 2na

(RTV3) (V nb)2

or1

= 1

2a(Vm b)3

RTV3m.

Now introduce the reduced variables and use Tc =8a

27Rb, Vc = 3b.

After rearrangement,

1

= 1

(3Vr 1)2

4TrV3r.


27/30

TRAPP: CHAP02 2006/3/8 17:05 PAGE 49 #27

THE FI RST LAW 49

For xenon, Vc = 118.1 cm3 mol1, Tc = 289.8 K. The perfect gas value for Vm may be used as any error

introduced by this approximation occurs only in the correction term for 1

.

Hence, Vm 2.45 dm3, Vc =118.8 cm

3 mol1, Tc = 289.8 K, and Vr = 20.6 and Tr =1.03; therefore

1

= 1

(61.8 1)2

(4) (1.03) (20.6)3= 0.90, giving 1.1

and

Cp,m CV,m 1.1R = 9.2JK1 mol1 .

P2.37 (a) =

1

Cp

H

p

T

=1

Cp

T

Vm

T

p

Vm

[2.53 and Problem 2.34].

Vm =RT

p+ aT2 so

Vm

T

p

=R

p+ 2aT.

=1

Cp

RT

p+ 2aT2

RT

p aT2

= =

aT2

Cp.

(b)CV = Cp TVm

p

T

V

= Cp T

Vm

T

p

p

T

V

.

But, p =RT

Vm aT2

.

p

T

V

=R

Vm aT2

RT(2aT)

(Vm aT2)2

=R

(RT/p)+

2aRT2

(RT/p)2=

p

T+

2ap2

R.

Therefore

CV = Cp T

R

p+ 2aT

p

T+

2ap2

R

= Cp RT

p

1 +

2apT

R

1 +

2apT

R

pT

,

CV = Cp R

1 +

2apT

R

2.


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50 SOLUTIONS MANUAL

Solutions to applications

P2.39 Taking the specific enthalpy of digestible carbohydrates to be 17 kJ g1 (Impact I.2.2), the serving of

pasta yields

q = (40 g) (17kJg1) = 680kJ.

Converting to Calories (kcal) gives:

q = (680kJ) 1Cal

4.184 kJ= 162Cal.

As a percentage of a 2200-Calorie diet, this serving is

162Cal2200 Cal

100% = 7.4% .

P2.41 (a) q = ncH =

1.5 g

342.3 g mol1 (5645kJ mol1) = 25kJ .

(b) Effective work available is 25kJ 0.25 = 6.2kJ.

Because w = mgh, and m 65kg

h 6.2 103 J

65kg 9.81 m s2= 9.7 m .

(c) The energy released as heat is

q = rH = ncH =

2.5g

180gmol1

(2808 kJ mol1) = 39 kJ .

(d) If one-quarter of this energy were available as work a 65 kg person could climb to a height h

given by

1

4q = w = mgh so h =

q

4mg=

39 103 J

4(65kJ) (9.8 m s2)= 15 m .

P2.43 First, with the pure sample, record a thermogram over a temperature range within which P undergoes a

structural change, as canbe inferred from a peak in thethermogram. Thearea under thethermogram is the

enthalpy change associated with the structural change for the given quantity of P. Then, with an identical

mass of the suspected sample, record a thermogram over the same temperature range. Assuming that

the impurities in P undergo no structural change over the temperature rangea reasonable assumption

if the impurities are monomers or oligomers and if the temperature range is sufficiently narrowthen

the peak in the test sample thermogram is attributable only to P. The ratio of areas under the curve in the

test sample to the pure sample is a measure of the purity of the test sample.

P2.45 The coefficient of thermal expansion is

=1

V

V

T

p

V

VTso V VT.


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THE FI RST LAW 51

This change in volume is equal to the change in height (sea level rise, h) times the area of the ocean

(assuming that area remains constant). We will use of pure water, although the oceans are complexsolutions. For a 2C rise in temperature

V = (2.1 104 K1) (1.37 109 km3) (2.0K) = 5.8 105 km3

so h =V

A= 1.6 103 km = 1.6m .

Since the rise in sea level is directly proportional to the rise in temperature, T = 1C would lead to

h = 0.80 m and T = 3.5C would lead to h = 2.8 m .

COMMENT. More detailed models of climate change predict somewhat smaller rises, but the same order

of magnitude.

P2.47 We compute from

= 1

Cp

H

p

T

and we estimate

H

p

T

from the enthalpy and pressure data. We are given both enthalpy and heat

capacity data on a mass basis rather than a molar basis; however, the masses will cancel, so we need not

convert to a molar basis.

(a)

(b)

Figure 2.4


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52 SOLUTIONS MANUAL

(a) At 300 K.

The regression analysis gives the slope as 18.0Jg1 MPa1

Hp

T

,

so = 18.0 kJkg1 MPa1

0.7649 kJ kg1 K1 = 23.5 K MPa1= 23.5 K MPa1 .

(b) At 350 K.

The regression analysis gives the slope as 14.5Jg1 MPa1

H

p

T

,

so = 14.5kJ kg1 MPa1

1.0392 kJkg1 K1= 14.0K MPa1 .

Exerc Cios Resolvidos - Cap. 02-Atkins (a)

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