Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap05

The internal pressure results from attractive interactions alone.. For van der Waals gases and liquids with strong attractive forces largea at small volumes, the internal pressure can be

Solutions to exercises

Discussion questions

E5.1(b) See the solution to Exercise 3.14(a) and Example 5.1, where it is demonstrated thatπ T = a/V2

mfor

a van der Waals gas Therefore, there is no dependence onb for a van der Waals gas The internal

pressure results from attractive interactions alone For van der Waals gases and liquids with strong attractive forces (largea) at small volumes, the internal pressure can be very large.

E5.2(b) The relation (∂G/∂T ) p = −S shows that the Gibbs function of a system decreases with T at

constantp in proportion to the magnitude of its entropy This makes good sense when one considers

the definition of G, which is G = U + pV − T S Hence, G is expected to decrease with T in

proportion to S when p is constant Furthermore, an increase in temperature causes entropy to

increase according to

S =

f

i

dqrev/T

The corresponding increase in molecular disorder causes a decline in th Gibbs energy (Entropy is always positive.)

E5.3(b) The fugacity coefficient,φ, can be expressed in terms of an integral involving the compression factor,

specifically an integral ofZ − 1 (see eqn 5.20) Therefore, we expect that the variation with pressure

of the fugacity coefficient should be similar, in a very qualitative sense, to the variation with pressure

of the compression factor itself Comparison of figures 1.27 and 5.8 of the text shows this to be roughly the case, though the detailed shapes of the curves are necessarily different becauseφ is an integral

function ofZ −1 over a range of pressures So we expect no simple proportionality between φ and Z.

But we findφ < 1 in pressure regions where attractive forces are expected to predominate and φ > 1

when repulsive forces predominate, which in behavior is similar to that ofZ See Section 5.5(b) for

a more complete discussion

Numerical exercises

E5.4(b) α = 1

V ×

∂V

∂T



p κT= −

 1

V



×

∂V

∂p



T



∂S

∂p



T = −



∂V

∂T



p = −αV

E5.5(b) G = nRT ln



pf

pi



at constant temperature, pf

pi

= Vi

Vf

= nRT ln

V

i

Vf



= (2.5 × 10−3mol) × (8.314 J K−1mol−1) × (298 K) × ln

 72 100



= −2.035 = −2.0 J



∂G

∂T



p = −S



∂Gf

∂T



p = −Sf and



∂Gi

∂T



p = −Si

S = Sf− Si= −

∂G

f

∂T



p+

∂G

i

∂T



p

= −

∂(G

f− Gi)

∂T



p = −

∂G

∂T



p

= − ∂

∂T {−73.1 + 42.8 T /K} J

= −42.8 J K−1

E5.7(b) See the solution to Exercise 5.7(a) Without knowledge of the compressibility of methanol we can

only assume thatV = V1(1 − κ T p) ≈ V1 Then

G = V p

ρ = m V so V = m ρ = 25 g

0.791 g cm−3 = 31.61 cm3

G = (31.61 cm3) ×



1 m3

106cm3



× (99.9 × 106Pa)

= +3.2 kJ

E5.8(b) (a) S = nR ln



Vf

Vi



= nR ln



pi

pf

 [Boyle’s Law]

Taking inverse logarithms

pf= pie−S/nR = (150 kPa) exp −



−(−15.0 J K−1) (3.00 mol) × (8.314 J K−1mol−1)



= 274 kPa

(b) G = nRT ln



pf

pi



= −T S [H = 0, constant temperature, perfect gas]

= −(230 K) × (−15.0 J K−1)

= +3450 J = 3.45 kJ

E5.9(b) µ = µf− µi = RT ln



pf

pi



= (8.314 J K−1mol−1) × (323 K) × ln



252.0

92.0



= 2.71 kJ mol−1

E5.10(b) µ0= µ−−+ RT ln

 p

p−  −



µ = µ−−+ RT ln



f

p−  −



µ − µ0= RT ln



f p

 

f

p ≡ φ

µ − µ0= RT ln φ

= (8.314 J K−1mol−1) × (290 K) × ln(0.68)

= −929.8 J mol−1

= −930 J mol−1 or −0.93 kJ mol−1

E5.11(b) B = B

RT = −

(160.0 cm3mol−1) × 1 m 3

10 6 cm 3

(8.314 J K−1mol−1) × (100 K)

= −1.924 × 10−7Pa−1

φ = e B p+···

≈ e−1.924×10−7Pa−1 ×62 ×10 6 Pa

≈ e−11.93

= 7 × 10−6 or of the order of 10−6

E5.12(b) G = nVmp = V p

= (1.0 L) ×



1 m3

103L



× (200 × 103Pa)

= 200 Pa m3= 200 J

E5.13(b) Gm = RT ln



pf

pi



= (8.314 J K−1mol−1) × (500 K) × ln



100.0 kPa

50.0 kPa



= +2.88 kJ mol−1

E5.14(b) V =



∂G

∂p



T [5.10] = RT

p + B + C p + D p2

which is the virial equation of state

E5.15(b)



∂S

∂V



T =



∂p

∂T



V

For a Dieterici gas

p = RT e V −a/RT Vm

∂p

∂T



Vm

= R 1+

a

RVmT

e−a/RVmT

Vm− b

dS =



∂S

∂Vm



T dVm =



∂p

∂T



V dVm

S =

Vm,f

Vm,i

dS = R



1+ a

RVmT



e−a/RVmT 

Vm,f − b

Vm,i − b



For a perfect gasS = R ln



Vm,f

Vm,i



S for a Dieterici gas may be greater or lesser than S for a perfect gas depending on T and the

magnitudes ofa and b At very high T , S is greater At very low T , S is less.

Solutions to problems

Solutions to numerical problem

P5.2 For the reaction

N2(g) + 3H2(g) → 2NH3(g) rG−−= 2fG−−(NH3, g)

(a) rG−−(500 K) = τrG−−(Tc) + (1 − τ)rH−−(Tc)

 Problem 5.1,τ = T T

c

=



500 K

298.15 K



× (2) × (−16.45 kJ mol−1)

+



1− 500 K

298.15 K



× (2) × (−46.11 kJ mol−1)

= −55.17 + 62.43 kJ mol−1= +7 kJ mol−1

(b) rG−−(1000 K) =



1000 K

298.15 K



× (2) × (−16.45 kJ mol−1)

+



1− 1000 K

298.15 K



× (2) × (−46.11 kJ mol−1)

= (−110.35 + 217.09) kJ mol−1= +107 kJ mol−1

Solutions to theoretical problems

P5.5 We start from the fundamental relation

dU = T dS − p dV [2]

But, sinceU = U(S, V ), we may also write

dU =

∂U

∂S



VdS +

∂U

∂V



SdV

Comparing the two expressions, we see that

∂U

∂S



V = T and

∂U

∂V



S = −p

These relations are true in general and hence hold for the perfect gas We can demonstrate this more explicitly for the perfect gas as follows For the perfect gas at constant volume

dU = C VdT

dS =dqrev

T =

C V dT T

Then



dU

dS



V =

∂U

∂S



V = C V dT

C VdT T

= T

For a reversible adiabatic (constant-entropy) change in a perfect gas

dU = dw = −p dV

Therefore,



∂U

∂V



S = −p

P5.8

∂p

∂S



V = −

∂T

∂V



S [Maxwell relation]

∂S

∂T

V

∂V

∂S

T

[chain relation]=

∂S

∂V

T

∂S

∂T

V

[inversion]

=

∂p

∂T V

∂S

∂U

V

∂U

∂T

V

[Maxwell relation]=−

∂p

∂V T ∂V

∂T

p

∂S

∂U

V

∂U

∂T

V

[chain relation]

= −

∂V

∂T

p

∂U

∂S

V

∂V

∂p T ∂U

∂T

V

[inversion twice]= αT

κ T C V



∂U

∂S



V = T

P5.10

∂H

∂p



T =

∂H

∂S



p

∂S

∂p



T +

∂H

∂p



S [Relation 1, Further information 1.7]

dH = T dS + V dp [Problem 5.6]

dH =



∂H

∂S



p dS +



∂H

∂p



S dp (H = H (p, S))





compare Thus,

∂H

∂S



p = T ,

∂H

∂p



S = V [dH exact]

Substitution yields,



∂H

∂p



T = T



∂S

∂p



T + V = −T



∂V

∂T



p + V [Maxwell relation]

(a) ForpV = nRT

∂V

∂T



p= nR p , hence

∂H

∂p



T = −nRT p + V = 0

(b) Forp = nRT

V − nb−

an2

V2 [Table 1.6]

T = p(V − nb)

na(V − nb)

RV2

∂T

∂V



p = p

nR +

na

RV2 −2na(V − nb)

RV3

Therefore,



∂H

∂p



T = −T

∂T

∂V

p

+ V [inversion] = p −T

nR+RV na2 −2na(V −nb)

RV3

+ V

which yields after algebraic manipulation



∂H

∂p



T = nb −

2na RT

λ2

1− 2na

RT V

λ2, λ = 1 − nb

V

When b

Vm

2na

RT V =

2na

RT ×

1

V ≈

2na

RT ×

p nRT =

2pa

R2T2

Therefore,



∂H

∂p



T ≈ nb −

2na

RT

1− 2pa

R2T2 For argon,a = 1.337 L2atm mol−2, b = 3.20 × 10−2L mol−1,

2na

RT =

(2) × (1.0 mol) × (1.337 L2atm mol−2) (8.206 × 10−2L atm K−1mol−1) × (298 K) = 0.11 L

2pa

R2T2 = (2) × (10.0 atm) × (1.337 L2atm mol−2)

(8.206 × 10−2L atm K−1mol−1) × (298 K)2 = 0.045

Hence,



∂H

∂p



T ≈ {(3.20 × 10−2) − (0.11)} L

1− 0.045 = −0.0817 L = −8.3 J atm−1

H ≈



∂H

∂p



T p ≈ (−8.3 J atm−1) × (1 atm) = −8 J

P5.12 π T = T



∂p

∂T



V − p [5.8]

p = RT

Vm +BRT

V2 m

[The virial expansion, Table 1.6, truncated after the term inB]



∂p

∂T



V = R

Vm +BR

V2 m

+RT

V2 m



∂B

∂T



V = p

T +

RT

V2 m



∂B

∂T



V

Hence,π T = RT2

V2 m

∂B

∂T



V ≈ RT2B

V2

mT

Sinceπ T represents a (usually) small deviation from perfect gas behaviour, we may approximateVm

Vm≈ RT

p π T ≈

p2

R ×

B

T

From the dataB = ((−15.6) − (−28.0)) cm3mol−1= +12.4 cm3mol−1

Hence,

(a) π T = (1.0 atm)2× (12.4 × 10−3L mol−1)

(8.206 × 10−2L atm K−1mol−1) × (50 K) = 3.0 × 10−3atm

(b) π T ∝ p2; so at p = 10.0 atm, π T = 0.30 atm

Comment In (a) π T is 0.3 per cent of p; in (b) it is 3 per cent Hence at these pressures the

approximation forVmis justified At 100 atm it would not be

Question How would you obtain a reliable estimate ofπ T for argon at 100 atm?

∂U

∂T



V and C p =

∂H

∂T



p

(a)

∂C

V

∂V



T =∂V ∂T ∂2U = ∂T ∂V ∂2U =

 ∂

∂T

∂U

∂V



T



V = 0 [π T = 0]

∂C

V

∂p



T =∂p∂T ∂2U =∂T ∂p ∂2U =

 ∂

∂T

∂U

∂p



T



V

=

 ∂

∂T

∂U

∂V



T

∂V

∂p



T



V = 0 (π T = 0)

SinceC p = C V + R,



∂C p

∂x



T =



∂C V

∂x



T forx = p or V

C V andC pmay depend on temperature SincedC V

dT =

d2U

dT2, dC V

dT is nonzero ifU depends on

T through a nonlinear relation See Chapter 20 for further discussion of this point However, for

a perfect monatomic gas,U is a linear function of T ; hence C V is independent ofT A similar

argument applies toC p

(b) This equation of state is the same as that of Problem 5.12.

∂C

V

∂V



T =∂T ∂V ∂2U =

∂π

T

∂T



V [Part (a)]

=



∂

∂T

RT2

V2 m

∂B

∂T



V



V

[Problem 5.12]

= 2RT

V2 m



∂B

∂T



V +RT2

V2 m



∂2B

∂T2



V

= RT

V2 m



∂2(BT )

∂T2



V

P5.15 π T = T

∂p

∂T



V − p [5.8]

p = V − nb nRT × e−an/RT V [Table 1.6]

T

∂p

∂T



V =V − nb nRT × e−an/RT V +RT V na ×V − nb nRT × e−an/RT V = p + RT V nap

Hence, π T = RT V nap

π T → 0 as p → 0, V → ∞, a → 0, and T → ∞ The fact that π T > 0 (because a > 0)

is consistent witha representing attractive contributions, since it implies that



∂U

∂V



T > 0 and the

internal energy rises as the gas expands (so decreasing the average attractive interactions)



∂G

∂p



T dp = V dp G(pf) − G(pi) =

pf

pi

V dp

In order to complete the integration,V as a function of p is required.



∂V

∂p



T = −κ T V (given), so d ln V = −κ dp

Hence, the volume varies with pressure as

V

V0

d lnV = −κ T

p

pi

dp

orV = V0e−κ T (p−pi) (V = V0whenp = pi)

Hence,

pf

pi

dG =

V dp = V0

pf

pi

e−κ T (p−p i )dp G(pf) = G(pi) + (V0) ×



1− e−κ T (pf−pi)

κ T



= G(pi) + (V0) ×



1− e−κ T p

κ T



Ifκ T −κ T p≈ 1 −1− κ T p +1

2κ2

T p2 = κ T p −1

2κ2

T p2

Hence, G = G + V0p



1−1

2κ T p

 For the compression of copper, the change in molar Gibbs function is

Gm = Vmp



1−1

2κ T p



=



Mp ρ



×



1−1

2κ T p



=



63.54 g mol−1

8.93 × 106g m−3



× (500) × (1.013 × 105Pa) ×



1−1

2κ T p



= (360.4 J) × 1−1

2κ T p

If we takeκ T = 0 (incompressible), Gm = +360 J For its actual value

1

2κ T p =1

2 × (0.8 × 10−6atm−1) × (500 atm) = 2 × 10−4

1−1

2κ T p = 0.9998

Hence,Gmdiffers from the simpler version by only 2 parts in 104 (0.02 per cent)

P5.19 κ S = −

 1

V



×



∂V

∂p



S = − 1

V ∂V ∂p

S

The only constant-entropy changes of state for a perfect gas are reversible adiabatic changes, for which

pV γ = const

Then,



∂p

∂V



S =



∂

∂V

const

V γ



S = −γ ×

 const

V γ +1



= −γp

V

Therefore,κ S = −1

V −γp V =+1γp Hence, pγ κ S = +1

P5.21 S = S(T , p)

dS =



∂S

∂T



p dT +



∂S

∂p



T dp

T dS = T

∂S

∂T



p dT + T

∂S

∂p



T dp

Use

∂S

∂T



p =

∂S

∂H



p

∂H

∂T



p =T1 × C p

∂H

∂S



p = T , Problem 5.6





∂S

∂p



T = −



∂V

∂T



p [Maxwell relation]

Hence,T dS = C pdT − T

∂V

∂T



p dp = C pdT − αT V dp

For reversible, isothermal compression,T dS = dqrev, dT = 0; hence

dqrev= −αT V dp

qrev=

pf

pi

−αT V dp = −αT V p [α and V assumed constant]

For mercury

qrev= (−1.82 × 10−4K−1) × (273 K) × (1.00 × 10−4m−3) × (1.0 × 108Pa) = −0.50 kJ

P5.25 When we neglectb in the van der Waals equation we have

p = RT

Vm − a

V2 m

and hence

Z = 1 − RT V a

m

Then substituting into eqn 5.20 we get

lnφ =

p

o



Z − 1 p



dp =

p

o

−a

p RT Vmdp

In order to perform this integration we must eliminate the variableVmby solving for it in terms ofp.

Rewriting the expression forp in the form of a quadratic we have

V2

m−RT p Vm+a p = 0 The solution is

Vm= 1

2



RT /p ± 1

p



(RT )2− 4ap



applying the approximation(RT )2 4ap we obtain

Vm= 1

2

RT

p ±

RT p



Choosing the+ sign we get

Vm= RT

p which is the perfect volume

Then

lnφ =

p

0

− a

RT2dp = − ap

(RT )2

For ammoniaa = 4.169 atm L2mol−2

lnφ = − 4.169 atm L2mol−2× 10.00 atm

(0.08206 L atm K−1mol−1× 298.15 K)2

= −0.06965

φ = 0.9237 = f

p

f = φp = 0.9237 × 10.00 atm = 9.237 atm

P5.27 The equation of statepVm

RT = 1 +

qT

Vm

is solved forVm =

RT

2p

 

1+



1+4pq

R

1/2 so

Z − 1

p =

pVm

RT − 1

qT

pVm =

2q R

1+ 1+4pq

R

1/2

lnφ =

p

0



Z − 1 p



dp[24] = 2q

R

p

0

dp

1+ 1+4pq

R

1/2

Defining,a ≡ 1 +



1+4pq

R

1/2

, dp = R(a − 1)

2q da, gives

lnφ =

a

2



a − 1 a



da [a = 2, when p = 0]

= a − 2 − ln1

2a =



1+4pq

R

1/2

− 1 − ln

 1 2



1+4pq

R

1/2 +1 2



Hence, φ = 2e{(1+4pq/R)

1/2−1}

1+ 1+4pq

R

1/2

This function is plotted in Fig 5.1(a) when 4pq

R

1.0 0.8

0.1 0.01 1.0

1.2

–0.01 –0.1 –1.0

Figure 5.1(a)

ex ≈ 1 + x, (1 + x)1/2≈ 1 + 1

2x, and (1 + x)−1

φ ≈ 1 + pq R

Whenφ is plotted against x = 4pq/R on a linear rather than exponential scale, the apparent curvature

seen in Fig 5.1(a) is diminished and the curve seems almost linear See Fig 5.1(b)

2

Figure 5.1(b)

Solution to applications

P5.28 wadd,max = rG [4.38]

rG−−(37◦C) = τrG−−(Tc) + (1 − τ)rH−−(Tc)

 Problem 5.1, τ = T

Tc

=



310 K

298.15 K



× (−6333 kJ mol−1) +



1− 310 K

298.15 K



× (−5797 kJ mol−1)

= −6354 kJ mol−1 The difference isrG−−(37◦C) − rG−−(Tc) = {−6354 − (−6333)} kJ mol−1= −21 kJ mol−1 Therefore, an additional 21 kJ mol−1of non-expansion work may be done at the higher temperature.

Comment As shown by Problem 5.1, increasing the temperature does not necessarily increase the

maximum non-expansion work The relative magnitude ofrG−− andrH−− is the determining factor

P5.31 The Gibbs–Helmholtz equation is

∂

∂T

G

T



= −H

T2

so for a small temperature change





rG−  −

T



=rH−−

T2 T and rG−2−

T2 = rG−1−

T1 −rH−−

T2T

so

drG−  −

T = −



rH−  −dT

T2 andrG−190−

T190 =rG−220−

T220 + rH−−

 1

T190 −T1

220



rG−190− = rG−220− T190

T220 + rH−−



1−T190

T220



For the monohydrate

rG−190− = (46.2 kJ mol−1) ×



190 K

220 K



+ (127 kJ mol−1) ×



1−190 K

220 K



,

rG−190− = 57.2 kJ mol−1

For the dihydrate

rG−190− = (69.4 kJ mol−1) ×



190 K

220 K



+ (188 kJ mol−1) ×



1−190 K

220 K



,

rG−190− = 85.6 kJ mol−1

For the monohydrate

rG−190− = (93.2 kJ mol−1) ×



190 K

220 K



+ (237 kJ mol−1) ×



1−190 K

220 K



,

rG−190− = 112.8 kJ mol−1

P5.32 The change in the Helmholtz energy equals the maximum work associated with stretching the polymer

Then

dwmax= dA = −f dl

For stretching at constantT

f = −

∂A

∂l



T = −

∂U

∂l



T + T

∂S

∂l



T

assuming that(∂U/∂l) T = 0 (valid for rubbers)

f = T



∂S

∂l



T = T



∂

∂l



T



−3kBl2

2Na2 + C



= T



−3kBl

Na2



= −



3kBT

Na2



l

This tensile force has the Hooke’s law formf = −kHl with kH= 3kBT /Na2

Solutions to problems

Solutions to numerical problem

P5.2 For the reaction

N2(g)...

= (−110.35 + 217.09) kJ mol−1= +107 kJ mol−1

Solutions to theoretical problems

P5.5 We start from the fundamental relation... However, for

a perfect monatomic gas,U is a linear function of T ; hence C V is independent ofT A similar

argument applies to< i>C p

(b)