Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap04

Energy in the form of heat enthalpy supplied to a liquid manifests itself as turbulent motion kinetic energy of the molecules.. The enthalpy of vaporization is the heat energy enthalpy r

Solutions to exercises

Discussion questions

E4.1(b) Trouton’s rule is that the ratio of the enthalpy of vaporization of a liquid to its boiling point is a

constant Energy in the form of heat (enthalpy) supplied to a liquid manifests itself as turbulent motion (kinetic energy) of the molecules When the kinetic energy of the molecules is sufficient

to overcome the attractive energy that holds them together the liquid vaporizes The enthalpy of vaporization is the heat energy (enthalpy) required to accomplish this at constant pressure It seems reasonable that the greater the enthalpy of vaporization, the greater the kinetic energy required, and the greater the temperature needed to achieve this kinetic energy Hence, we expect that
vapH ∝ Tb, which implies that their ratio is a constant

E4.2(b) The device proposed uses geothermal heat (energy) and appears to be similar to devices currently in

existence for heating and lighting homes As long as the amount of heat extracted from the hot source (the ground) is not less than the sum of the amount of heat discarded to the surroundings (by heating the home and operating the steam engine) and of the amount of work done by the engine to operate the heat pump, this device is possible; at least, it does not violate the first law of thermodynamics However, the feasability of the device needs to be tested from the point of view of the second law as well There are various equivalent versions of the second law, some are more directly useful in this case than others Upon first analysis, it might seem that the net result of the operation of this device

is the complete conversion of heat into the work done by the heat pump This work is the difference between the heat absorbed from the surroundings and the heat discharged to the surroundings, and all

of that difference has been converted to work We might, then, conclude that this device violates the second law in the form stated in the introduction to Chapter 4; and therefore, that it cannot operate

as described However, we must carefully examine the exact wording of the second law The key words are “sole result.” Another slightly different, though equivalent, wording of Kelvin’s statement

is the following: “It is impossible by a cyclic process to take heat from a reservoir and convert it into work without at the same time transferring heat from a hot to a cold reservoir.” So as long as some heat is discharged to surroundings colder than the geothermal source during its operation, there is no reason why this device should not work A detailed analysis of the entropy changes associated with this device follows

Pump

“ground” water at Th

Flow

Flow

Environment at Tc

Figure 4.1 C V andC pare the temperature dependent heat capacities of water

Three things must be considered in an analysis of the geothermal heat pump: Is it forbidden by the first law? Is it forbidden by the second law? Is it efficient?

Etot =
Ewater+
Eground+
Eenvironment

Ewater= 0

Eground = −C V (Th){Th − Tc}

Eenvironment = −C V (Th){Th − Tc}

adding terms, we find that
Etot = 0 which means that the first law is satisfied for any value of Th andTc

Stot =
Swater+
Sground+
Senvironment

Swater= 0

Sground=qground Th =−C p (Th){Th Th − Tc}

Senvironment= qenvironment

C p (Tc){Th − Tc}

Tc

adding terms and estimating thatC p (Th)  C p (Tc) = C p, we find that

Stot = C p {Th− Tc}

1

Tc −

1

Th



This expression satisfies the second law(
Stot > 0) only when Th > Tc We can conclude that, if the proposal involves collecting heat from environmentally cool ground water and using the energy

to heat a home or to perform work, the proposal cannot succeed no matter what level of sophisticated technology is applied Should the “ground” water be collected from deep within the Earth so that

Th > Tc, the resultant geothermal pump is feasible However, the efficiency, given by eqn 4.11, must

be high to compete with fossil fuels because high installation costs must be recovered during the lifetime of the apparatus

Erev= 1 − Tc Th

withTc ∼ 273 K and Th = 373 K (the highest value possible at 1 bar), Erev= 0.268 At most, about

27% of the extracted heat is available to do work, including driving the heat pump The concept works especially well in Iceland where geothermal springs bring boiling water to the surface

E4.3(b) See the solution to exercises 4.3 (a)

Numerical exercises



dqrev

q T

(a)
S = 50× 103J

273 K = 1.8 × 102J K−1

(b)
S = 50× 103J

(70 + 273) K = 1.5 × 102J K−1

E4.5(b) At 250 K, the entropy is equal to its entropy at 298 K plus
S where

S =



dqrev

 C

V,mdT

T = C V,mln

Tf Ti

soS = 154.84 J K−1mol−1+ [(20.786 − 8.3145) J K−1mol−1]× ln250 K

298 K

S = 152.65 J K−1mol−1



dqrev

 C p,mdT

T = C p,mln

Tf Ti

S = (1.00 mol) ×

 5

2 + 1



× (8.3145 J K−1mol−1) × ln (100 + 273) K

273 K = 9.08 J K−1

E4.7(b) However the change occurred,
S has the same value as if the change happened by reversible heating

at constant pressure (step 1) followed by reversible isothermal compression (step 2)

S =
S1 +
S2

For the first step

S1 =



dqrev

 C p,mdT

T = C p,mln

Tf Ti

S1 = (2.00 mol) ×

 7 2



× (8.3145 J K−1mol−1) × ln (135 + 273) K

(25 + 273) K = 18.3 J K−1

and for the second

S2=



dqrev

qrev T

whereqrev = −w =



p dV = nRT ln Vf Vi = nRT ln pi pf

so
S2 = nR ln pi

pf = (2.00 mol) × (8.3145 J K−1mol−1) × ln

1.50 atm

7.00 atm = −25.6 J K−1

S = (18.3 − 25.6) J K−1= −7.3 J K−1

The heat lost in step 2 was more than the heat gained in step 1, resulting in a net loss of entropy Or the ordering represented by confining the sample to a smaller volume in step 2 overcame the disordering represented by the temperature rise in step 1 A negative entropy change is allowed for a system as long as an increase in entropy elsewhere results in
Stotal > 0.

E4.8(b) q = qrev= 0 (adiabatic reversible process)

S =

 f i

dqrev

T = 0

U = nC V,m
T = (2.00 mol) × (27.5 J K−1mol−1) × (300 − 250) K

= 2750 J = +2.75 kJ

w =
U − q = 2.75 kJ − 0 = 2.75 kJ

H = nC p,m
T

C p,m = C V,m + R = (27.5 J K−1mol−1+ 8.314 J K−1mol−1) = 35.814 J K−1mol−1

So
H = (2.00 mol) × (35.814 J K−1mol−1) × (+50 K)

= 3581.4 J = 3.58 kJ

E4.9(b) However the change occurred,
S has the same value as if the change happened by reversible heating

at constant volume (step 1) followed by reversible isothermal expansion (step 2)

S =
S1 +
S2

For the first step

S1 =



dqrev

 C

V,mdT

T = C V,mln

Tf Ti



C V,m = C p,m − R

= (3.50 mol) ×

 3 2



× (8.3145 J K−1mol−1) × ln700 K

250 K = 44.9 J K−1

and for the second

S2=



dqrev

qrev T

whereqrev = −w =



p dV = nRT ln Vf Vi,

so
S2 = nR ln pf pi = (3.50 mol) × (8.3145 J K−1mol−1) × ln60.0 L

20.0 L = 32.0 J K−1

S = 44.9 + 32.0 J K−1= 76.9 J K−1

E4.10(b)
S = qrev

T If reversibleq = qrev qrev = T
S = (5.51 J K−1) × (350 K)

= 1928.5 J

rev rev; therefore the process is not reversible

E4.11(b) (a) The heat flow is

q = C p
T = nC p,m
T

=



2.75 kg

63.54 × 10−3kg mol−1



× (24.44 J K−1mol−1) × (275 − 330) K

= −58.2 × 103J

(b)
S =



dqrev

 C pdT

T = nC p,mln

Tf Ti

=



2.75 kg

63.54 × 10−3kg mol−1



× (24.44 J K−1mol−1) × ln275 K

330 K = −193 J K−1

E4.12(b)
S =



dqrev

qrev

T whereqrev = −w = nRT ln

Vf

Vi = nRT ln

pi pf

so
S = nR ln pi pf =



35 g

28.013 g mol−1



× (8.3145 J K−1mol−1) × ln21.1 atm

4.3 atm = 17 J K−1

E4.13(b)
S =



dqrev

qrev

T whereqrev = −w = nRT ln

Vf Vi

so
S = nR ln Vf Vi andVf = Viexp


S

nR



We need to compute the amount of gas from the perfect gas law

pV = nRT so n = pV RT = (1.20 atm) × (11.0 L)

(0.08206 L atm K−1mol−1) × (270 K) = 0.596 mol

SoVf = (11.0 L) exp



−3.0 J K−1 (0.596 mol) × (8.3145 J K−1mol−1)

= 6.00 L

E4.14(b) Find the final temperature by equating the heat lost by the hot sample to the heat gained by the cold

sample

−n1Cp,m (Tf − Ti1) = n2Cp,m (Tf − Ti2)

Tf = n1Ti1 + n2Ti2

n1 + n2 = M1(m1Ti11 + m2Ti2)

M (m1 + m2)

= m1Ti1 m1 + m + m2Ti2

2

= (25 g) × (323 K) + (70 g) × (293 K)

25 g+ 70 g = 300.9 K

S =
S1 +
S2= n1Cp,mln



Tf Ti1



+ n2Cp,mln



Tf Ti2



46.07 g mol−1

 ln



300.9

323

+



70 g

46.07 g mol−1



×ln



300.9

293

C p,m

=−3.846 × 10−2+ 4.043 × 10−2 C p,m

= (0.196 × 10−2mol) × (111.5 J K−1mol−1)

= 0.2 J K−1

E4.15(b)
Htotal= 0 in an isolated container

Since the masses are equal and the heat capacity is assumed constant, the final temperature will be the average of the two initial temperatures

Tf = 1

2(200◦C+ 25◦C) = 112.¯5◦C

nC m = mCs whereCsis the specific heat capacity

S = mCsln

Tf

Ti



200◦C= 473.2 K; 25◦C= 298.2 K; 112.¯5◦C= 385.7 K

S1 = (1.00 × 103g) × (0.449 J K−1g−1) × ln



385.7

298.2



= 115.5 J K−1

S2 = (1.00 × 103g) × (0.449 J K−1g−1) × ln



385.7

473.2



= −91.802 J K−1

Stotal =
S1+
S2= 24 J K−1

E4.16(b) (a) q = 0 [adiabatic]

(b) w = −pex
V = −(1.5 atm) ×



1.01 × 105Pa atm

× (100.0 cm2) × (15 cm) ×



1 m3

106cm3

= −227.2 J = −230 J

(c)
U = q + w = 0 − 230 J = −230 J

(d)
U = nC V,m
T

T =
U

nC V,m =

−227.2 J (1.5 mol) × (28.8 J K−1mol−1)

= −5.3 K

(e)
S = nC V,mln



Tf Ti



+ nR ln



Vf Vi



Tf = 288.15 K − 5.26 K = 282.9 K

Vi=nRT pi = (1.5 mol) × (8.206 × 10−2L atm K−1mol−1) × (288.¯2 K)

9.0 atm

= 3.942 L

Vf = 3.942 L + (100 cm2) × (15 cm) ×



1 L

1000 cm3



= 3.942 L + 1.5 L = 5.44 L

S = (1.5 mol) × (28.8 J K−1mol−1) × ln



282.9

288.2

+ (8.314 J K−1mol−1) × ln



5.4¯4

3.942

= 1.5 mol(−0.5346 J K−1mol−1+ 2.678 J K−1mol−1) = 3.2 J K−1

E4.17(b) (a)
vapS−−=
vapH−−

35.27 × 103J mol−1

(64.1 + 273.15) K = + 104.58 J K−1= 104.6 J K−1

(b) If vaporization occurs reversibly, as is generally assumed

Ssys +
Ssur= 0 so
Ssur= −104.6 J K−1

E4.18(b) (a)
r S−− = Sm−−(Zn2 +, aq) + Sm−−(Cu, s) − Sm−−(Zn, s) − Sm−−(Cu2 +, aq)

= [−112.1 + 33.15 − 41.63 + 99.6] J K−1mol−1= −21.0 J K−1mol−1

(b)
r S−− = 12Sm−−(CO2, g) + 11Sm−−(H2O, l) − Sm−−(C12H22O11, s) − 12Sm−−(O2, g)

= [(12 × 213.74) + (11 × 69.91) − 360.2 − (12 × 205.14)] J K−1mol−1

= + 512.0 J K−1mol−1

E4.19(b) (a)
r H−−=
fH−−(Zn2 +, aq) −
f H−−(Cu2 +, aq)

= −153.89 − 64.77 kJ mol−1 = −218.66 kJ mol−1

r G−− = −218.66 kJ mol−1− (298.15 K) × (−21.0 J K−1mol−1) = −212.40 kJ mol−1

(b)
rH−− =
cH−− = −5645 kJ mol−1

rG−−= −5645 kJ mol−1− (298.15 K) × (512.0 J K−1mol−1) = −5798 kJ mol−1

E4.20(b) (a)
rG−−=
fG−−(Zn2 +, aq) −
f G−−(Cu2 +, aq)

= −147.06 − 65.49 kJ mol−1 = −212.55 kJ mol−1

(b)
rG−−= 12
fG−−(CO2, g) + 11
f G−−(H2O, l) −
f G−−(C12H22O11, s)

= [12 × (−394.36) + 11 × (−237.13) − (−1543)] kJ mol−1= −5798 kJ mol−1

Comment In each case these values of
rG−−agree closely with the calculated values in Exercise 4.19(b)

E4.21(b) CO(g) + CH3OH(l) → CH3COOH(l)

rH−− = νJ
f H−−(J)

= −484.5 kJ mol−1− (−238.66 kJ mol−1) − (−110.53 kJ mol−1)

= −135.31 kJ mol−1

rS−− = νJS−−(J)

= 159.8 J K−1mol−1− 126.8 J K−1mol−1− 197.67 J K−1mol−1

= −164.67 J K−1mol−1

rG−− =
rH−−− T
rS−−

= −135.31 kJ mol−1− (298 K) × (−164.67 J K−1mol−1)

= −135.31 kJ mol−1+ 49.072 kJ mol−1 = −86.2 kJ mol−1

E4.22(b) The formation reaction of urea is

C(gr) +1

2O2(g) + N2(g) + 2H2(g) → CO(NH2)2(s) The combustion reaction is

CO(NH2)2(s) +3

2O2(g) → CO2(g) + 2H2O(l) + N2(g)

cH =
f H−−(CO2, g) + 2
fH−−(H2O, l) −
fH−−(CO(NH2)2, s)

f H−−(CO(NH2)2, s) =
f H−−(CO2, g) + 2
f H−−(H2O, l) −
cH (CO(NH2)2, s)

= −393.51 kJ mol−1+ (2) × (−285.83 kJ mol−1) − (−632 kJ mol−1)

= −333.17 kJ mol−1

f S−−= Sm−−(CO(NH2)2, s) − Sm−−(C, gr) −1

2Sm−−(O2, g) − Sm−−(N2, g) − 2Sm−−(H2, g)

= 104.60 J K−1mol−1− 5.740 J K−1mol−1−1

2(205.138 J K−1mol−1)

− 191.61 J K−1mol−1− 2(130.684 J K−1mol−1)

= −456.687 J K−1mol−1

f G−−=
fH−−− T
fS−−

= −333.17 kJ mol−1− (298 K) × (−456.687 J K−1mol−1)

= −333.17 kJ mol−1+ 136.093 kJ mol−1

= −197 kJ mol−1

E4.23(b) (a)
S(gas) = nR ln



Vf Vi



=



21 g

39.95 g mol−1



× (8.314 J K−1mol−1) ln 2

= 3.029 J K−1= 3.0 J K−1

S(surroundings) = −
S(gas) = −3.0 J K−1 [reversible]

S(total) = 0

(b) (Free expansion)

S(gas) = +3.0 J K−1 [S is a state function]

S(surroundings) = 0 [no change in surroundings]

S(total) = +3.0 J K−1

(c) qrev = 0 so
S(gas) = 0

S(surroundings) = 0 [No heat is transfered to the surroundings]

S(total) = 0

E4.24(b) Because entropy is a state function, we can choose any convenient path between the initial and final

states

Choose isothermal compression followed by constant-volume heating

S = nR ln

Vf

Vi



+ nC V,mln

Tf

Ti



= −nR ln 3 + nC V,mln 3

= n(C V,m − R) ln 3 C V,m =5

2R for a diatomic perfect gas

S = 3

2nR ln 3

E4.25(b) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

rG−−= 3
fG−−(CO2, g) + 4
f G−−(H2O, l) −
f G−−(C3H8, g) − 0

= 3(−394.36 kJ mol−1) + 4(−237.13 kJ mol−1) − 1(−23.49 kJ mol−1)

= −2108.11 kJ mol−1

The maximum non-expansion work is 2108.11 kJ mol−1 since|we| = |
rG|

E4.26(b) (a) ε = 1 − Th Tc

= 1 − 500 K

1000 K = 0.500

(b) Maximum work= ε|qh| = (0.500) × (1.0 kJ) = 0.50 kJ

(c) εmax = εrev and |wmax| = |qh| − |qc,min|

|qc,min | = |qh| − |wmax|

= 1.0 kJ − 0.50 kJ

= 0.5 kJ

Solutions to problems

Assume that all gases are perfect and that data refer to 298 K unless otherwise stated

Solutions to numerical problems

P4.1 (a) Because entropy is a state function
trsS(l → s, −5◦C) may be determined indirectly from the

following cycle

H2O(1, 0◦C) −−−−−−−−−→ H
trsS(1→s,0◦C) 2O(s, 0◦C)

S1 
S

s

H2O(1, −5◦C) −−−−−−−−−−−→ H
trsS(1→s,−5◦C) 2O(s, −5◦C)

trsS(l → s, −5◦C) =
Sl +
trsS(l → s, 0◦C) +
Ss

Sl = C p,m (l) ln Tf

T [θf= 0◦C, θ = −5◦C]

Ss = C p,m (s) ln T

Tf

Sl +
Ss= −
C plnT

Tf with
C p = C p,m (l) − C p,m (s) = +37.3 J K−1mol−1

trsS(l → s, Tf ) = −
TffusH

Thus,
trsS(l → s, T ) = −
fusH

Tf −
C pln

T Tf

=−6.01 × 103J mol−1

273 K − (37.3 J K−1mol−1) × ln268

273

= −21.3 J K−1mol−1

Ssur=
fusH(T ) T

fusH(T ) = −
Hl +
fusH (Tf) −
Hs

Hl +
Hs= C p,m (l)(Tf − T ) + C p,m (s)(T − Tf ) =
Cp(Tf − T )

fusH(T ) =
fusH(Tf ) −
C p (Tf − T )

Thus,
Ssur=
fusH(T ) T =
fusH (Tf T ) +
C p (T − Tf) T

= 6.01 kJ mol−1

268 K + (37.3 J K−1mol−1) ×



268− 273 268



= +21.7 J K−1mol−1

Stotal = (21.7 − 21.3) J K−1mol−1= +0.4 J K−1mol−1 Since
Stotal > 0, the transition l → s is spontaneous at −5◦C

(b) A similar cycle and analysis can be set up for the transition liquid→ vapour at 95◦C However, since the transformation here is to the high temperature state (vapour) from the low temperature

state (liquid), which is the opposite of part (a), we can expect that the analogous equations will

occur with a change of sign

trsS(l → g, T ) =
trsS(l → g, Tb) +
C pln T

Tb

=
vapH

Tb +
C pln

T

Tb ,
Cp = −41.9 J K−1mol−1

trsS(l → g, T ) = 40.7 kJ mol−1

373K − (41.9 J K−1mol−1) × ln368

373

= +109.7 J K−1mol−1

Ssur= −
vapH(T )

vapH (Tb)

C p (T − Tb) T

=



−40.7 kJ mol−1

368 K

− (−41.9 J K−1mol−1) ×



368− 373 368



= −111.2 J K−1mol−1

Stotal = (109.7 − 111.2) J K−1mol−1= −1.5 J K−1mol−1 Since
Stotal < 0, the reverse transition, g → l, is spontaneous at 95◦C

 T2

T1

C p,mdT

T [19]=

 T2

T1

a + bT

T



dT = a ln

T2

T1



+ b(T2− T1)

a = 91.47 J K−1mol−1, b = 7.5 × 10−2J K−2mol−1

Sm = (91.47 J K−1mol−1) × ln



300 K

273 K



+ (0.075 J K−2mol−1) × (27 K)

= 10.7 J K−1mol−1 Therefore, for 1.00 mol,
S = +11 J K−1

P4.8

Process (a) +5.8 J K−1 −5.8 J K−1 0 0 −1.7 kJ −1.7 kJ

Process (b) +5.8 J K−1 −1.7 J K−1 0 0 −1.7 kJ −1.7 kJ

Process (c) +3.9 J K−1 0 −8.4 × 102J −41 K ? ?

Process (a)

H =
T = 0 [isothermal process in a perfect gas]

Stot = 0 =
S +
Ssurr

S = nR ln



Vf Vi

 [4.17] = (1.00 mol) × (8.314 J K−1mol−1) × ln



20 L

10 L



= +5.8 J K−1

Ssurr = −
S = −5.8 J K−1

A =
U − T
S [36]
U = 0 [isothermal process in perfect gas]

A = 0 − (298 K) × (5.76 J K−1) = −1.7 × 103

J

G =
H − T
S = 0 − T
S = −1.7 × 103J

... class="page_container" data-page="9">

Solutions to problems

Assume that all gases are perfect and that data refer to 298 K unless otherwise stated

Solutions to numerical problems...

S(total) = +3.0 J K−1

(c) qrev = so
S(gas) = 0

S(surroundings) = 0 [No heat is transfered to the surroundings]

S(total) =... K−1

E4.15(b)
Htotal= in an isolated container

Since the masses are equal and the heat capacity is assumed constant, the final temperature will be the