Instructor solution manual to accompany physical chemistry 7th ed

Solutions to exercisesDiscussion questions E1.1b The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mi

Solutions to exercises

Discussion questions

E1.1(b) The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied

alone the same container as the mixture at the same temperature It is a limiting law because it holdsexactly only under conditions where the gases have no effect upon each other This can only be true

in the limit of zero pressure where the molecules of the gas are very far apart Hence, Dalton’s lawholds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation

E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid

and vapour phases disappears We usually describe this situation by saying that above the criticaltemperature the liquid phase cannot be produced by the application of pressure alone The liquid andvapour phases can no longer coexist, though fluids in the so-called supercritical region have bothliquid and vapour characteristics (See Box 6.1 for a more thorough discussion of the supercriticalstate.)

E1.3(b) The van der Waals equation is a cubic equation in the volume,V Any cubic equation has certain

properties, one of which is that there are some values of the coefficients of the variable where thenumber of real roots passes from three to one In fact, any equation of state of odd degree higherthan 1 can in principle account for critical behavior because for equations of odd degree inV there

are necessarily some values of temperature and pressure for which the number of real roots ofV

passes fromn(odd) to 1 That is, the multiple values of V converge from n to 1 as T → Tc Thismathematical result is consistent with passing from a two phase region (more than one volume for agivenT and p) to a one phase region (only one V for a given T and p and this corresponds to the

observed experimental result as the critical point is reached

All quantities on the right are given to us exceptn, which can be computed from the given mass

(b) The van der Waals equation is

p = RT

Vm− b−

a

V2 m

E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature,

and volume Once this is done, the mass of the gas can be computed from the amount and the molarmass using

pV = nRT

son = pV RT =(1.00 atm) × (1.013 × 105Pa atm−1) × (4.00 × 103m3)

(8.3145 J K−1mol−1) × (20 + 273) K = 1.66 × 10

5mol

andm = (1.66 × 105mol) × (16.04 g mol−1) = 2.67 × 106g= 2.67 × 103kg

E1.10(b) All gases are perfect in the limit of zero pressure Therefore the extrapolated value of pVm/T will

give the best value ofR.

The molar mass is obtained frompV = nRT = M m RT

which upon rearrangement givesM = V m RT p = ρ RT p

The best value ofM is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is M/RT

Draw up the following table

p/atm (pVm/T )/(L atm K−1mol−1) (ρ/p)/(g L−1atm−1)

1.0 0.75 0.50 0.25 0

1.42755

Figure 1.1(b)

The value obtained forR deviates from the accepted value by 0.005 per cent The error results from

the fact that only three data points are available and that a linear extrapolation was employed Themolar mass, however, agrees exactly with the accepted value, probably because of compensatingplotting errors

E1.11(b) The mass density ρ is related to the molar volume Vmby

Vm= M ρ

whereM is the molar mass Putting this relation into the perfect gas law yields

pVm= RT so pM ρ = RT

Rearranging this result gives an expression forM; once we know the molar mass, we can divide by

the molar mass of phosphorus atoms to determine the number of atoms per gas molecule

E1.13(b) (a) The volume occupied by each gas is the same, since each completely fills the container Thus

solving forV from eqn 14 we have (assuming a perfect gas)

(b) The total pressure is determined from the total amount of gas,n = nCH 4+ nAr+ nNe.

E1.15(b) This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that

temperature at which the volume of a sample of gas would become zero if the substance remained agas at low temperatures The solution uses the experimental fact that the volume is a linear function

of the Celsius temperature

ThusV = V0+ αV0θ = V0+ bθ, b = αV0

At absolute zero,V = 0, or 0 = 20.00 L + 0.0741 L◦C−1× θ(abs zero)

θ(abs zero) = − 20.00 L

0.0741 L◦C−1 = −270◦Cwhich is close to the accepted value of−273◦C

0.150 L

= 270 atm (2 significant figures)

(b) From Table (1.6) for H2S

a = 4.484 L2atm mol−1 b = 4.34 × 10−2L mol−1

p = V − nb nRT −an2

V2

(i) p = (1.0 mol) × (8.206 × 10−2L atm K−1mol−1) × (273.15 K)

22.414 L − (1.0 mol) × (4.34 × 10−2L mol−1)

−(4.484 L2atm mol−1) × (1.0 mol)2

(22.414 L)2

= 0.99 atm(ii) p = (1.0 mol) × (8.206 × 10−2L atm K−1mol−1) × (500 K)

0.150 L − (1.0 mol) × (4.34 × 10−2L mol−1)

−(4.484 L2atm mol−1) × (1.0 mol)2

(0.150 L)2

= 185.6 atm ≈ 190 atm (2 significant figures).

E1.17(b) The critical constants of a van der Waals gas are

27(0.08206 L atm K−1mol−1) × (0.0436 L mol−1) = 109 K

E1.18(b) The compression factor is

Z = pVm

RT =

Vm

Vm,perfect

(a) BecauseVm = Vm,perfect + 0.12 Vm,perfect = (1.12)Vm,perfect, we haveZ = 1.12

Repulsive forces dominate

(b) The molar volume is

V = (1.12)Vm,perfect = (1.12) ×

RT p



x2+

a p



x − ab

p = 0

withx = Vm/(L mol−1).

The coefficients in the equation are evaluated as

(1.360 L2atm mol−2) × (3.183 × 10−2L mol−1)

(200 bar) × (1.013 atm bar−1) = 2.137 × 10−4(L mol−1)

3

Thus, the equation to be solved isx3− 0.1526x2+ (6.71 × 10−3)x − (2.137 × 10−4) = 0.

Calculators and computer software for the solution of polynomials are readily available In this case

we find

x = 0.112 or Vm = 0.112 L mol−1

The difference is about 15 per cent

E1.20(b) (a) Vm =M ρ =18.015 g mol−1

and substituting into the expression forZ above we get

Comment Both values ofZ are very close to the perfect gas value of 1.000, indicating that water

vapour is essentially perfect at 1.00 bar pressure.

E1.21(b) The molar volume is obtained by solving Z = pV RTm [1.20b], for Vm, which yields

(b) An approximate value ofB can be obtained from eqn 1.22 by truncation of the series expansion

after the second term,B/Vm, in the series Then,

(b) The partial pressures are

pN= xNptot= (0.63) × (4.0 atm) = 2.5 atm

andpH= (0.37) × (4.0 atm) = 1.5 atm

E1.23(b) The critical volume of a van der Waals gas is

Vc= 3b

sob =1

3Vc= 1

3(148 cm3mol−1) = 49.3 cm3mol−1 = 0.0493 L mol−1

By interpretingb as the excluded volume of a mole of spherical molecules, we can obtain an estimate

of molecular size The centres of spherical particles are excluded from a sphere whose radius isthe diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadroconstant is the molar excluded volumeb

pc= a

27b2

soa = 27pcb2= 27(48.20 atm) × (0.0493 L mol−1)2= 3.16 L2atm mol−2

But this problem is overdetermined We have another piece of information

Tc = 8a

27Rb

According to the constants we have already determined,Tcshould be

Tc = 8(3.16 L2atm mol−2)

27(0.08206 L atm K−1mol−1) × (0.0493 L mol−1)= 231 K

However, the reportedTcis 305.4 K, suggesting our computeda/b is about 25 per cent lower than it

should be

E1.24(b) (a) The Boyle temperature is the temperature at which lim

Vm →∞

dZ

d(1/Vm)vanishes According to the

van der Waals equation

(b) By interpretingb as the excluded volume of a mole of spherical molecules, we can obtain an

estimate of molecular size The centres of spherical particles are excluded from a sphere whoseradius is the diameter of those spherical particles (i.e twice their radius); the Avogadro constanttimes the volume is the molar excluded volumeb

E1.25(b) States that have the same reduced pressure, temperature, and volume are said to correspond The

reduced pressure and temperature for N2at 1.0 atm and 25◦C are

The corresponding states are

which can be solved forb

b = Vm−p + RT a

V2 m

Solutions to problems

Solutions to numerical problems

P1.2 Identifyingpexin the equationp = pex+ ρgh [1.4] as the pressure at the top of the straw and p as

the atmospheric pressure on the liquid, the pressure difference is

SubstitutingVi =4

3πr3

i andVf= 4

3πr3 f

pf =



(4/3)πr3 i

(4/3)πr3 f

P1.6 The value of absolute zero can be expressed in terms ofα by using the requirement that the volume

of a perfect gas becomes zero at the absolute zero of temperature Hence

0= V0[1+ αθ(abs zero)]

Thenθ(abs zero) = −1

α

All gases become perfect in the limit of zero pressure, so the best value ofα and, hence, θ(abs zero)

is obtained by extrapolatingα to zero pressure This is done in Fig 1.2 Using the extrapolated value,

α = 3.6637 × 10−3◦C−1, or

θ(abs zero) = − 1

3.6637 × 10−3◦C−1 = −272.95◦Cwhich is close to the accepted value of−273.15◦C

P1.7 The mass of displaced gas isρV , where V is the volume of the bulb and ρ is the density of the gas.

The balance condition for the two gases ism(bulb) = ρV (bulb), m(bulb) = ρV (bulb)

which implies thatρ = ρ Because [Problem 1.5]ρ = pM RT

the balance condition ispM = pM

which implies thatM= p p × M

This relation is valid in the limit of zero pressure (for a gas behaving perfectly)

In experiment 1,p = 423.22 Torr, p= 327.10 Torr; hence

M =423.22 Torr

327.10 Torr × 70.014 g mol−1= 90.59 g mol−1

In experiment 2,p = 427.22 Torr, p= 293.22 Torr; hence

M =427.22 Torr

293.22 Torr × 70.014 g mol−1= 102.0 g mol−1

In a proper series of experiments one should reduce the pressure (e.g by adjusting the balancedweight) Experiment 2 is closer to zero pressure than experiment 1; it may be safe to conclude that

M ≈ 102 g mol−1 The molecules CH2FCF3 or CHF2CHF2 haveM ≈ 102 g mol−1

P1.9 We assume that no H2remains after the reaction has gone to completion The balanced equation is

[1.25b], we obtain Vm=  RT

p + V a2 m

Substitution of 12.3 L mol−1into the denominator of the first expression again results inVm=

12.3 L mol−1, so the cycle of approximation may be terminated

P1.13 (a) SinceB(TB) = 0 at the Boyle temperature (section 1.3b): B(TB) = a + b e −c/TB2 = 0

P1.15 From Table 1.6Tc=

23

12bpc

R

 Thus

Tc =

23



×

12pcb R



=

83

4π × (8.86 × 10−29m3)

1/3

= 0.28 nm

Solutions to theoretical problems

P1.18 This expansion has already been given in the solutions to Exercise 1.24(a) and Problem 1.17; the

and hence find B = b − a

RT and C = b2

SinceC = 1200 cm6mol−2, b = C1/2 = 34.6 cm3mol−1

a = RT (b − B) = (8.206 × 10−2) × (273 L atm mol−1) × (34.6 + 21.7) cm3mol−1

= (22.40 L atm mol−1) × (56.3 × 10−3L mol−1) = 1.26 L2atm mol−2

P1.22 For a real gas we may use the virial expansion in terms ofp [1.21]

Therefore, the limiting slope of a plot of p



From the equation of state (a + bT )

V2 m

P1.27 The two masses represent the same volume of gas under identical conditions, and therefore, the same

number of molecules (Avogadro’s principle) and moles,n Thus, the masses can be expressed as

[1.29]

Z − 27p2r

512T3 r

AtTr= 1.2 and pr= 3 eqn 3 predicts that Z is the root of

Z3−

3

The real root is Z = 0.611 and this prediction is independent of the specific gas.

Figure 1.27 indicates that the experimental result for the listed gases is closer to 0.55.

Figure 1.3

The buoyant force on the cylinder is

Fbuoy = Fbottom− Ftop

= A(pbottom− ptop)

according to the barometric formula

ptop = pbottome−Mgh/RT

whereM is the molar mass of the environment (air) Since h is small, the exponential can be expanded

in a Taylor series aroundh = 0

e−x = 1 − x + 1

2!x2+ · · ·

 Keeping the first-order term onlyyields

ptop = pbottom

1−Mgh RT



The buoyant force becomes



g = nMg



n = pbottomV RT



n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass

of the displaced environment ThusFbuoy= mg The net force is the difference between the buoyant

force and the weight of the balloon Thus

Fnet= mg − mballoong = (m − mballoon)g

This is Archimedes’ principle

Solutions to exercises

Discussion questions

E2.1(b) Work is a transfer of energy that results in orderly motion of the atoms and molecules in a system;

heat is a transfer of energy that results in disorderly motion See Molecular Interpretation 2.1 for amore detailed discussion

E2.2(b) Rewrite the two expressions as follows:

(1) adiabaticp ∝ 1/V γ (2) isothermalp ∝ 1/V

The physical reason for the difference is that, in the isothermal expansion, energy flows into thesystem as heat and maintains the temperature despite the fact that energy is lost as work, whereas inthe adiabatic case, where no heat flows into the system, the temperature must fall as the system doeswork Therefore, the pressure must fall faster in the adiabatic process than in the isothermal case.Mathematically this corresponds toγ > 1.

E2.3(b) Standard reaction enthalpies can be calculated from a knowledge of the standard enthalpies of

forma-tion of all the substances (reactants and products) participating in the reacforma-tion This is an exact methodwhich involves no approximations The only disadvantage is that standard enthalpies of formationare not known for all substances

Approximate values can be obtained from mean bond enthalpies See almost any general chemistry

text, for example, Chemical Principles, by Atkins and Jones, Section 6.21, for an illustration of the

method of calculation This method is often quite inaccurate, though, because the average values ofthe bond enthalpies used may not be close to the actual values in the compounds of interest.Another somewhat more reliable approximate method is based on thermochemical groups whichmimic more closely the bonding situations in the compounds of interest See Example 2.6 for anillustration of this kind of calculation Though better, this method suffers from the same kind ofdefects as the average bond enthalpy approach, since the group values used are also averages.Computer aided molecular modeling is now the method of choice for estimating standard reactionenthalpies, especially for large molecules with complex three-dimensional structures, but accuratenumerical values are still difficult to obtain



= −91 J

E2.7(b) For a perfect gas at constant temperature

so q = −w

For a perfect gas at constant temperature,

dH = d(U + pV )

we have already noted thatU does not change at constant temperature; nor does pV if the gas obeys

Boyle’s law These apply to all three cases below

(a) Isothermal reversible expansion

(c) Free expansion is expansion against no force, sow = 0 , and q = −w = 0 as well.

E2.8(b) The perfect gas law leads to

E2.11(b) vapH−−) = (2.00 mol) × (−35.3 kJ mol−1) = −70.6 kJ

Because the condensation also occurs at constant pressure, the work is

w = −



pexdThe change in volume from a gas to a condensed phase is approximately equal in magnitude to thevolume of the gas

w ≈ −p(−Vvapor) = nRT = (2.00 mol) × (8.3145 kJ K−1mol−1) × (64 + 273) K

E2.24(b) For adiabatic compression, q = 0 and

w = C V −1mol−1) × (255 − 220) K = 2.4 × 103J

3J

= 2.4 × 103J+ (2.5 mol) × (8.3145 J K−1mol−1) × (255 − 220) K = 3.1 × 103JThe initial and final states are related by

E2.27(b) In an adiabatic process, q = 0 Work against a constant external pressure is

w = −pex 3Pa) × (15 cm) × (22 cm2)

(100 cm m−1)3 = −36 J

w n(C p,m − R)

(b) For adiabatic expansion against a constant external pressure

Vf= (20.811 J K−1mol−1) × (0.535 mol) × (200 K) + (110 × 103Pa) × (4.04 × 10−3m3)

110× 103Pa+ (20.811 J K−1mol −1)×(110×103 Pa)

8.3145 J K−1mol−1

Vf= 6.93 × 10−3m3Finally, the temperature is

Tf = pfVf

nR =

(110 × 103Pa) × (6.93 × 10−3m3) (0.535 mol) × (8.3145 J K−1mol−1)= 171 K

E2.29(b) At constant pressure

vapH−−= (0.75 mol) × (32.0 kJ mol−1) = 24.0 kJ

and vapor= −nRT = −(0.75 mol) × (8.3145 J K−1mol−1) × (260 K)

= −1.6 × 103J= −1.6 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

E2.30(b) The reaction is

The enthalpies of formation of all of these compounds are available in Table 2.5 Therefore

hydH−− = [−126.15 − (−0.13)] kJ mol−1= −126.02 kJ mol−1

If we had to, we could find fH−−(C4H8) from information about another of its reactions

fH−−for the reaction

Comment In this case cU−−and cH−−differed by≈ 0.1 per cent Thus, to within 3 significant

figures, it would not have mattered if we had used cH−− instead of cU−−, but for very precisework it would

E2.36(b) The reaction is

AgBr(s) → Ag+(aq) + Br−(aq)

solH−− fH−−(Ag+ fH−−(Br− fH−−(AgBr)

= [105.58 + (−121.55) − (−100.37)] kJ mol−1= +84.40 kJ mol−1

E2.37(b) The difference of the equations is C(gr) → C(d)

transH−− = [−393.51 − (−395.41)] kJ mol−1= +1.90 kJ mol−1

E2.38(b) Combustion of liquid butane can be considered as a two-step process: vaporization of the liquid

followed by combustion of the butane gas Hess’s law states that the enthalpy of the overall process

is the sum of the enthalpies of the steps

(a) cH−−= [21.0 + (−2878)] kJ mol−1= −2857 kJ mol−1

(b) The net ionic reaction is obtained from

H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → Na+(aq) + Cl−(aq) + H2O(l)

and is H+(aq) + OH−(aq) → H2O(l)

rH−− fH−−(H2O fH−−(H+ fH−−(OH−, aq)

= [(−285.83) − (0) − (−229.99)] kJ mol−1

= −55.84 kJ mol−1

E2.40(b) reaction (3)= reaction (2) − 2(reaction (1))

rH−−(100◦C) = −571.66 + (373 − 298) × (0.06606) +1

2(3732− 2982)

×(−10.76 × 10−6) − (67) ×

1

373− 1298

solnH−−(CaBr2 fH−−(CaBr2 subH−−(Ca)

vapH−−(Br2 dissH−−(Br2 ionH−−(Ca)

ionH−−(Ca+ egH−− hydH−−(Br−)

= [−(−103.1) − (−682.8) + 178.2 + 30.91 + 192.9 +589.7 + 1145 + 2(−331.0) + 2(−337)] kJ mol−1

= 1587 kJ mol−1and hydH−−(Ca2 +) = −1587 kJ mol−1

Electron gain Br

E2.46 (a) 2,2,4-trimethylpentane has five C(H)3(C) groups, one C(H)2(C)2group, one C(H)(C)3group,

and one C(C)4group

(b) 2,2-dimethylpropane has four C(H3)(C) groups and one C(C)4group

Using data from Table 2.7

(a) [5× (−42.17) + 1 × (−20.7) + 1 × (−6.91) + 1 × 8.16] kJ mol−1= −230.3 kJ mol−1

(b) [4× (−42.17) + 1 × 8.16] kJ mol−1= −160.5 kJ mol−1

Solutions to problems

Assume all gases are perfect unless stated otherwise Unless otherwise stated, thermochemical dataare for 298 K

Solutions to numerical problems

P2.4 We assume that the solid carbon dioxide has already evaporated and is contained within a closed

vessel of 100 cm3which is its initial volume It then expands to a final volume which is determined

by the perfect gas equation

= −272 Pa m3= −0.27 kJ

Even if there is no physical piston, the gas drives back the atmosphere, so the work is also −8.9 kJ

P2.7 The virial expression for pressure up to the second coefficient is

Vm



= nRTB

1

vapH vapH = 8.18 kJ mol−1(Table 2.3)

The volume occupied by the methane gas at a pressurep is V = nRT

p ; therefore

V = qRT

vapH =

(32.5 kJ) × (8.314 J K−1mol−1) × (112 K) (1.013 × 105Pa) × (8.18 kJ mol−1)

= 3.65 × 10−2m3= 36.5 L

P2.14 Cr(C6H6)2(s) → Cr(s) + 2C6H6 g= +2 mol

rH−− rU−−+ 2RT , from [26]

= (8.0 kJ mol−1) + (2) × (8.314 J K−1mol−1) × (583 K) = +17.7 kJ mol−1

In terms of enthalpies of formation

whereTbis the boiling temperature of benzene(353 K) We shall assume that the heat capacities of

graphite and hydrogen are approximately constant in the range of interest, and use their values fromTable 2.6

rH−−(benzene, 583 K) = (49.0 kJ mol−1) + (353 − 298) K × (136.1 J K−1mol−1)

P2.17 We must relate the formation of DyCl3to the three reactions for which we have information

P2.21 When necessary we assume perfect gas behaviour, also, the symbolsw, V, q, U, etc will represent

molar quantities in all cases

Vfn−1−

1

Vin−1

[becausepV n = C]

= piViVin−1

n − 1 ×

1

− 1

forn = 0 and n = 1

In the case for whichn = 0, eqn 1 gives

w = C

0− 1 ×

1

To derive the equation for heat, note that, for a perfect gas, V (Tf− Ti) So

− 1

[using eqn 3(n = 0)]

− 1

[using eqn 4(n = 0, n = 1)]

− 1

[2.31]

− 1



=

1

− 1

[2.37]

− 1



Using the symbols ‘1’ and ‘2’ this becomes

− 1

forn = 0, n = 1

In the case for whichn = 1, (the isothermal case) eqns 7 and 6 yield

(9) q = −w = RT ln V2

V1 = RT ln p1

p2

forn = 1, isothermal case

In the case for whichn = 0 (the isobaric case) eqns 7 and 5 yield

− 1



n − γ (n − 1)(γ − 1)

Solutions to theoretical problems

w = −F

 x2

x1

sinπx a

(0.0821 L atm K−1mol−1) × (313 K) = 0.389 mol

SinceT is a constant along the isotherm, Boyle’s law applies

(d) Since the initial and final states of all three paths are the same,

Path AB is isothermal; hence

V,m= 3

2R is not needed for the

solution

In each case,path ACB, q = −9.5 × 102J; path ADB, q = −1.9 × 104J;path AB, q = −3.0 × 103J

The heat is different for all three paths; heat is not a state property

Vr,1−1 3