Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap06

a After heating to 320 K at constant pressure, the system is still gaseous.. b Isothermal compression at 320 K to 100 atm pressure brings the sample into the supercritical region.. c Aft

pure substances

Solutions to exercises

Discussion questions

E6.1(b) Refer to Fig 6.8 The white lines represent the regions of superheating and supercooling The chemical

potentials along these lines are higher than the chemical potentials of the stable phases represented by the colored lines Though thermodynamically unstable, these so-called metastable phases may persist for a long time if the system remains undisturbed, but will eventually transform into the thermo-dynamically stable phase having the lower chemical potential Transformation to the condensed phases usually requires nucleation centers In the absence of such centers, the metastable regions are said to be kinetically stable

E6.2(b) At 298 K and 1.0 atm, the sample of carbon dioxide is a gas (a) After heating to 320 K at constant

pressure, the system is still gaseous (b) Isothermal compression at 320 K to 100 atm pressure brings the sample into the supercritical region The sample is now not much different in appearance from ordinary carbon dioxide, but some of its properties are (see Box 6.1) (c) After cooling the sample to

210 K at constant pressure, the carbon dioxide sample solidifies (d) Upon reducing the pressure to 1.0 atm at 210 K, the sample vapourizes (sublimes); and finally (e) upon heating to 298 K at 1.0 atm, the system has resumed its initial conditions in the gaseous state Note the lack of a sharp gas to liquid transition in steps (b) and (c) This process illustrates the continuity of the gaseous and liquid states

E6.3(b) First-order phase transitions show discontinuities in the first derivative of the Gibbs energy with

respect to temperature They are recognized by finite discontinuities in plots ofH , U, S, and V

against temperature and by an infinite discontinuity in C p Second-order phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous The second-order transitions are recognized by kinks in plots ofH, U, S,

andV against temperature, but most easily by a finite discontinuity in a plot of C pagainst temperature

Aλ-transition shows characteristics of both first and second-order transitions and, hence, is difficult

to classify by the Ehrenfest scheme It resembles a first-order transition in a plot ofC pagainstT , but

appears to be a higher-order transition with respect to other properties See the book by H E Stanley listed under Further reading for more details

Numerical exercises

E6.4(b) Assume vapour is a perfect gas and vapH is independent of temperature

lnp∗

p = +

vapH R

1

T −

1

T∗



1

T =

1

T∗ + R

vapH ln

p∗ p

293.2 K+

8.314 J K−1mol−1

32.7 × 103J mol−1 × ln

58.0

66.0



= 3.378 × 10−3K−1

3.378 × 10−3K−1 = 296 K = 23◦C

E6.5(b) dp

dT =

Sm

Vm

fusS = Vm

dp

dT



≈ Vm p T

assuming fusS and Vm independent of temperature

fusS = (152.6 cm3mol−1− 142.0 cm3mol−1) × (1.2 × 106Pa) − (1.01 × 105Pa)

429.26 K − 427.15 K

= (10.6 cm3mol−1) ×



1 m3

106cm3



× (5.21 × 105Pa K−1)

= 5.52 Pa m3K−1mol−1= 5.5 J K−1mol−1

fusH = Tf S = (427.15 K) × (5.52 J K−1mol−1)

= 2.4 kJ mol−1

E6.6(b) Use



d lnp =  vapH

RT2 dT

lnp = constant − vapH

RT

Terms with 1

T dependence must be equal, so

−3036.8 K

T /K = −

vapH RT vapH = (3036.8 K)R = (8.314 J K−1mol−1) × (3036.8 K)

= 25.25 kJ mol−1

E6.7(b) (a) logp = constant − vapH

RT (2.303)

Thus

vapH = (1625 K) × (8.314 J K−1mol−1) × (2.303)

= 31.11 kJ mol−1

(b) Normal boiling point corresponds top = 1.000 atm = 760 Torr

log(760) = 8.750 −1625T /K

1625

T /K = 8.750 − log(760)

8.750 − log(760) = 276.87

Tb= 276.9 K

E6.8(b) T = fusV

fusS × p =

Tf fusV fusH × p =

Tf pM fusH ×

1

ρ



[Tf = −3.65 + 273.15 = 269.50 K]

T = (269.50 K) × (99.9 MPa)M

8.68 kJ mol−1 ×

1

0.789 g cm−3 − 1

0.801 g cm−3



= (3.1017 × 106K Pa J−1mol) × (M) × (+ 01899 cm3/g) ×



m3

106cm3



= (+ 5.889 × 10−2K Pa m3J−1g−1mol)M = (+ 5.889 × 10−2K g−1mol)M

T = (46.07 g mol−1) × (+ 5.889 × 10−2K g−1mol)

= + 2.71 K

Tf = 269.50 K + 2.71 K = 272 K

dt =

dn

dt × MH 2 O wheren = q

vapH

dn

dt =

dq/dt vapH =

(0.87 × 103W m−2) × (104m2)

44.0 × 103J mol−1

= 197.7 J s−1J−1mol

= 200 mol s−1

dm

dt = (197.7 mol s−1) × (18.02 g mol−1)

= 3.6 kg s−1

E6.10(b) The vapour pressure of ice at −5◦C is 3.9 × 10−3atm, or 3 Torr Therefore, the frost will sublime

A partial pressure of 3 Torr or more will ensure that the frost remains

E6.11(b) (a) According to Trouton’s rule (Section 4.3, eqn 4.16)

vapH = (85 J K−1mol−1) × Tb= (85 J K−1mol−1) × (342.2 K) = 29.1 kJ mol−1

Start Gas

Temperature

a d

Critical point

Figure 6.1

(b) Use the Clausius–Clapeyron equation [Exercise 6.11(a)]

ln

p2

p1



= vapH

1

T1− 1

T2



AtT2= 342.2 K, p2= 1.000 atm; thus at 25◦C

lnp1= −



29.1 × 103J mol−1

8.314 J K−1mol−1



×

1

298.2 K−

1

342.2 K



= −1.509

p1= 0.22 atm = 168 Torr

At 60◦C,

lnp1= −



29.1 × 103J mol−1

8.314 J K−1mol−1



×

1

333.2 K−

1

342.2 K



= −0.276

p1= 0.76 atm = 576 Torr

E6.12(b) T = Tf(10 MPa) − Tf(0.1 MPa) = Tf pM

fusH

1

ρ



fusH = 6.01 kJ mol−1

T =



(273.15 K) × (9.9 × 106Pa) × (18 × 10−3kg mol−1)

6.01 × 103J mol−1



×



1

9.98 × 102kg m−3 − 1

9.15 × 102kg m−3

= −0.74 K

Tf(10 MPa) = 273.15 K − 0.74 K = 272.41 K

E6.13(b) vapH = vapU + vap(pV )

vapH = 43.5 kJ mol−1

vap(pV ) = p vapV = p(Vgas− Vliq) = pVgas= RT [per mole, perfect gas]

vap(pV ) = (8.314 J K−1mol−1) × (352 K) = 2927 J mol−1

Fraction= vap(pV )

vapH =

2.927 kJ mol−1

43.5 kJ mol−1

= 6.73 × 10−2 = 6.73 per cent

E6.14(b) Vm= M

ρ =

18.02 g mol−1

999.4 × 103g m−3 = 1.803 × 10−5m3mol−1

2γ Vm

rRT =

2(7.275 × 10−2N m−1) × (1.803 × 10−5m3mol−1) (20.0 × 10−9m) × (8.314 J K−1mol−1) × (308.2 K)

= 5.119 × 10−2

p = (5.623 kPa)e0.05119 = 5.92 kPa

E6.15(b) γ = 1

2ρghr = 1

2(0.9956 g cm−3) × (9.807 m s−2) × (9.11 × 10−2m)

× (0.16 × 10−3m) ×



1000 kg m−3

g cm−3



= 7.12 × 10−2N m−1

E6.16(b) pin− pout =2γ

r =

2(22.39 × 10−3N m−1) (220 × 10−9m)

= 2.04 × 105N m−2= 2.04 × 105Pa

Solutions to problems

Solutions to numerical problems

dT =

vapS vapV =

vapH

Tb vapV [6.6, Clapeyron equation]

= 14.4 × 103J mol−1

(180 K) × (14.5 × 10−3− 1.15 × 10−4) m3mol−1 = + 5.56 kPa K−1

(b) dp

dT =

vapH

RT2 × p 11, with d ln p = dp

p

= (14.4 × 103J mol−1) × (1.013 × 105Pa)

(8.314 J K−1mol−1) × (180 K)2 = + 5.42 kPa K−1

The percentage error is 2.5 per cent

∂µ(l)

∂p



T −

∂µ(s)

∂p



T = Vm(l) − Vm(s)[6.13] = M

1

ρ



= (18.02 g mol−1) ×

1

1.000 g cm−3 − 1

0.917 g cm−3



= −1.63 cm3mol−1

(b)

∂µ(g)

∂p



T −

∂µ(l)

∂p



T = Vm(g) − Vm(l)

= (18.02 g mol−1) ×

1

0.598 g L−1 − 1

0.958 × 103g L−1



= + 30.1 L mol−1

At 1.0 atm and 100◦C ,µ(l) = µ(g); therefore, at 1.2 atm and 100◦Cµ(g)−µ(l) ≈ Vvap p = (as in Problem 6.4)

(30.1 × 10−3m3mol−1) × (0.2) × (1.013 × 105Pa) ≈ + 0.6 kJ mol−1

Sinceµ(g) > µ(l), the gas tends to condense into a liquid.

P6.7 The amount (moles) of water evaporated isng=pH 2 OV

RT

The heat leaving the water isq = n vapH

The temperature change of the water is T = −q

nC p,m,n = amount of liquid water

Therefore, T = −pH2 OV vapH

RT nC p,m

= −(23.8 Torr) × (50.0 L) × (44.0 × 103J mol−1)

(62.364 L Torr K−1mol−1) × (298.15 K) × (75.5 J K−1mol−1) × 250 g

18.02 g mol−1

= −2.7 K

The final temperature will be about 22◦C

P6.9 (a) Follow the procedure in Problem 6.8, but note thatTb= 227.5◦C is obvious from the data

(b) Draw up the following table

θ/◦C 57.4 100.4 133.0 157.3 203.5 227.5

T /K 330.6 373.6 406.2 430.5 476.7 500.7

lnp/Torr 0.00 2.30 3.69 4.61 5.99 6.63

The points are plotted in Fig 6.2 The slope is−6.4 × 103K, so − vapH

R = −6.4 × 103K,

implying that vapH = +53 kJ mol−1

2.0 2.2 2.4 2.6 2.8 3.0 0

2

4

6

Figure 6.2 P6.11 (a) The phase diagram is shown in Fig 6.3.

Liquid

Solid

Vapour

2 0 –2

–4

–6

–8

0 100 200 300 400 500 600

Liquid–Vapour Solid–Liquid

Figure 6.3

(b) The standard melting point is the temperature at which solid and liquid are in equilibrium at

1 bar That temperature can be found by solving the equation of the solid–liquid coexistence curve for the temperature

1= p3/bar + 1000(5.60 + 11.727x)x,

So 11 727x2+ 5600x + (4.362 × 10−7− 1) = 0

The quadratic formula yields

x = −5600 ± {(5600)2− 4(11 727) × (−1)}1/2

−1 ±1+4(11 727)

5600 2

1/2

2

11727 5600

The square root is rewritten to make it clear that the square root is of the form{1 + a}1/2, with

a 1; thus the numerator is approximately −1 + (1 + 1

2a) = 1

2a, and the whole expression

reduces to

x ≈ 1/5600 = 1.79 × 10−4

Thus, the melting point is

T = (1 + x)T3= (1.000179) × (178.15 K) = 178.18 K

(c) The standard boiling point is the temperature at which the liquid and vapour are in equilibrium

at 1 bar That temperature can be found by solving the equation of the liquid–vapour coexistence curve for the temperature This equation is too complicated to solve analytically, but not difficult

to solve numerically with a spreadsheet The calculated answer is T = 383.6 K

(d) The slope of the liquid–vapour coexistence curve is given by

dp

dT =

vapH

T vapV− − so vapH−− = T vapV−−dp

dT

The slope can be obtained by differentiating the equation for the coexistence curve

dp

dT = p

d lnp

dT = p

d lnp

dy

dy

dT

dp

dT =

10.418

y2 − 15.996 + 2(14.015)y − 3(5.0120)y2− (1.70) × (4.7224) × (1 − y)0.70

×

p

Tc



At the boiling point,y = 0.6458, so

dp

dT = 2.851 × 10−2bar K−1= 2.851 kPa K−1

and vapH−−= (383.6 K)×



(30.3 − 0.12) L mol−1

1000 L m−3



×(2.851 kPa K−1) = 33.0 kJ mol−1

P6.12 The slope of the solid–vapour coexistence curve is given by

dp

dT =

subH−−

T subV− − so subH−−= T subV−−dp

dT

The slope can be obtained by differentiating the coexistence curve graphically (Fig 6.4)

50

40

30

20

10

144 146 148 150 152 154 156

Figure 6.4

dp

dT = 4.41 Pa K−1

according to the exponential best fit of the data The change in volume is the volume of the vapour

Vm= RT

(8.3145 J K−1mol−1) × (150 K)

26.1 Pa = 47.8 m3

So

subH−−= (150 K) × (47.8 m3) × (4.41 Pa K−1) = 3.16 × 104J mol−1= 31.6 kJ mol−1

Solutions to theoretical problems

P6.14

∂ G

∂p



T =

∂G

β

∂p



T −

∂G

α

∂p



T = V β − V α

Therefore, ifV β = V α , G is independent of pressure In general, V β α , so that G is nonzero,

though small, sinceV β − V αis small

P6.16 Amount of gas bubbled through liquid= pV RT

(p = initial pressure of gas and emerging gaseous mixture)

Amount of vapour carried away= M m

Mole fraction of vapour in gaseous mixture= m M m

M +RT pV Partial pressure of vapour= p = m M m

M +RT pV × p =

pP V M mRT

mRT

P V M + 1 =

mP A

mA + 1 , A =

RT

P V M

For geraniol,M = 154.2 g mol−1, T = 383 K, V = 5.00 L, p = 1.00 atm, and m = 0.32 g, so

A = (8.206 × 10−2L atm K−1mol−1) × (383 K)

(1.00 atm) × (5.00 L) × (154.2 × 10−3kg mol−1) = 40.76 kg−1

Therefore

p = (0.32 × 10−3kg) × (760 Torr) × (40.76 kg−1)

(0.32 × 10−3kg) × (40.76 kg−1) + 1 = 9.8 Torr

P6.17 p = p0e−Mgh/RT [Box 1.1]

p = p∗e−χ χ = vapR H ×

1

T −

1

T∗

 [6.12]

Let T∗ = Tb the normal boiling point; thenp∗ = 1 atm Let T = Th, the boiling point at the altitudeh Take p0 = 1 atm Boiling occurs when the vapour (p) is equal to the ambient pressure,

that is, whenp(T ) = p(h), and when this is so, T = Th Therefore, sincep0 = p∗, p(T ) = p(h)

implies that

e−Mgh/RT = exp



− vapR H ×

1

Th −T1

b



It follows that 1

Th = 1

Tb + Mgh

T vapH

whereT is the ambient temperature and M the molar mass of the air For water at 3000 m, using

M = 29 g mol−1

1

Th = 1

373 K +(29 × 10−3kg mol−1) × (9.81 m s−2) × (3.000 × 103m)

(293 K) × (40.7 × 103J mol−1)

1.397 × 104K Hence,Th= 363 K (90◦C)

P6.20 Sm = Sm(T , p)

dSm=

∂Sm

∂T



p dT +

∂Sm

∂p



T dp

∂S

m

∂T



p = C p,m

T [Problem 5.7]

∂S

m

∂p



T = −

∂V

m

∂T



p [Maxwell relation]

dqrev= T dSm= C p,mdT − T

∂Vm

∂T



p dp

C S =

∂q

∂T



s = C p,m − T Vmα

∂p

∂T



s = C p,m − αVm× Hm

Vm

[6.7]

P6.22 C(graphite)( ) C(diamond) rG− − = 2.8678 kJ mol−1at TC

We want the pressure at which rG = 0; above that pressure the reaction will be spontaneous.

Equation 5.10 determines the rate of change of rG with p at constant T

(1)

∂ rG

∂p



T = rV = (VD− VG)M

whereM is the molar mas of carbon; VD andVG are the specific volumes of diamond and graphite, respectively

rG(TC, p) may be expanded in a Taylor series around the pressure p−−= 100 kPa at TC.

(2) rG(TC, p) = rG−−(TC, p−−) +

∂

rG− −(TC, p− −)

∂p



T (p − p−−)

+1 2



∂2 rG−−(TC, p−−)

∂p2



T (p − p−−)2+ θ(p − p−−)3

We will neglect the third and higher-order terms; the derivative of the first-order term can be calculated with eqn 1 An expression for the derivative of the second-order term can be derived with eqn 1

(3)



∂2 rG

∂p2



T

=



∂VD

∂p



T −

∂VG

∂p



T

M = {VGκ T (G) − VDκ T (D)}M [3.13]

Calculating the derivatives of eqns 1 and 2 at TCandp−−

(4)

∂

rG(TC, p− −)

∂p



T = (0.284 − 0.444) ×



cm3 g



×

12.01 g

mol



= −1.92 cm3mol−1

(5)



∂2 rG(TC, p−−)

∂p2



T

= {0.444(3.04 × 10−8) − 0.284(0.187 × 10−8)}

×



cm3kPa−1

g



×

12.01 g

mol



= 1.56 × 10−7cm3(kPa)−1mol−1

It is convenient to convert the value of rG−−to the units cm3kPa mol−1

rG−−= 2.8678 kJ mol−1



8.315 × 10−2L bar K−1mol−1

8.315 J K−1mol−1



×



103cm3 L



×



105Pa bar



(6) rG−− = 2.8678 × 106cm3kPa mol−1

Settingχ = p − p−−, eqns 2 and 3–6 give

2.8678 × 106

cm3kPa mol−1− (1.92 cm3

mol−1)χ + (7.80 × 10−8cm3kPa−1mol−1)χ2= 0 when rG(TC, p) = 0 One real root of this equation is

χ = 1.60 × 106kPa= p − p−−or

p = 1.60 × 106kPa− 102kPa

= 1.60 × 106kPa= 1.60 × 104bar Above this pressure the reaction is spontaneous The other real root is much higher: 2.3×107kPa

Question What interpretation might you give to the other real root?

to solve numerically... root is rewritten to make it clear that the square root is of the form{1 + a}1/2, with

a 1; thus the numerator is approximately...

2a) = 1

2a, and the whole expression

reduces to

x ≈ 1/5600 = 1.79 × 10−4

Thus, the melting point is

T