Exerc cios resolvidos_-_cap._01-atkins_(a)

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PART 1 Equilibrium

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1 The properties of gases
Answers to discussion questions
D1.1 An equation of state is an equation that relates the variables that define the state of a system to each other.
Boyle, Charles, and Avogadro established these relations for gases at low pressures (perfect gases) by
appropriate experiments. Boyle determined how volume varies with pressure (V ∝ 1/p), Charles how
volume varies with temperature (V ∝ T), and Avogadro how volume varies with amount of gas (V ∝ n).
Combining all of these proportionalities into one we find
AQ: Please
check we have
change ‘P’ to
‘p’.
V ∝
nT
p
.
Inserting the constant of proportionality, R, yields the perfect gas equation
V =
RnT
p
or pV = nRT.
D1.3 Consider three temperature regions:
(1) T < TB. At very low pressures, all gases show a compression factor, Z ≈ 1. At high pressures, all
gases have Z > 1, signifying that they have a molar volume greater than a perfect gas, which implies
that repulsive forces are dominant. At intermediate pressures, most gases show Z < 1, indicating
that attractive forces reducing the molar volume below the perfect value are dominant.
(2) T ≈ TB. Z ≈ 1 at low pressures, slightly greater than 1 at intermediate pressures, and significantly
greater than 1 only at high pressures. There is a balance between the attractive and repulsive forces
at low to intermediate pressures, but the repulsive forces predominate at high pressures where the
molecules are very close to each other.
(3) T > TB. Z > 1 at all pressures because the frequency of collisions between molecules increases
with temperature.
D1.5 The van der Waals equation ‘corrects’ the perfect gas equation for both attractive and repulsive
interactions between the molecules in a real gas. See Justification 1.1 for a fuller explanation.
The Bertholet equation accounts for the volume of the molecules in a manner similar to the van der
Waals equation but the term representing molecular attractions is modified to account for the effect of
temperature. Experimentally one finds that the van der Waals a decreases with increasing temperature.
Theory (see Chapter 18) also suggests that intermolecular attractions can decrease with temperature.

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4 SOLUTIONS MANUAL
This variation of the attractive interaction with temperature can be accounted for in the equation of state
by replacing the van der Waals a with a/T.
Solutions to exercises
E1.1(a) (a) The perfect gas equation [1.8] is: pV = nRT.
Solving for the pressure gives p =
nRT
V
.
The amount of xenon is n =
131 g
131 g mol−1
= 1.00 mol.
p =
(1.00 mol) × (0.0821 dm3
atm K−1 mol−1
) × (298.15 K)
1.0 dm3
= 24 atm .
That is, the sample would exert a pressure of 24 atm if it were a perfect gas, not 20 atm.
(b) The van der Waals equation [1.21a] for the pressure of a gas is p =
nRT
V − nb
−
an2
V2
.
For xenon, Table 1.6 gives a = 4.137 dm6
atm mol−2
and b = 5.16 × 10−2 dm3
mol−1
.
Inserting these constants, the terms in the equation for p become
nRT
V − nb
=
(1.00 mol) × (0.08206 dm3
atm K−1 mol−1
) × (298.15 K)
1.0 dm3
− {(1.00 mol) × (5.16 × 10−2 dm3
mol−1
)}
= 25.8 atm,
an2
V2
=
(4.137 dm6
atm mol−2
) × (1.00 mol)2
(1.0 dm3
)2
= 4.137 atm.
Therefore, p = 25.8 atm − 4.137 atm = 22 atm .
E1.2(a) Boyle’s law [1.5] in the form pfVf = piVi can be solved for either initial or ﬁnal pressure, hence
pi =
Vf
Vi
× pf,
Vf = 4.65 dm3
, Vi = 4.65 dm3
+ 2.20 dm3
= 6.85 dm3
, pf = 5.04 bar.
Therefore,
(a) pi =
4.65 dm3
6.85 dm3
× (5.04 bar) = 3.42 bar .
(b) Since 1 atm = 1.013 bar, pi = (3.42 bar) ×
1 atm
1.013 bar
= 3.38 atm .
E1.3(a) The perfect gas law, pV = nRT [1.8], can be rearranged to
p
T
=
nR
V
= constant, if n and V are constant.
Hence,
pf
Tf
=
pi
Ti
or, solving for pf, pf =
Tf
Ti
× pi.
Internal pressure = pump pressure + atmospheric pressure.
pi = 24 lb in−2
+ 14.7 lb in−2
= 38.7 lb in−2
, Ti = 268 K(−5◦
C), Tf = 308 K(35◦
C).
pf =
308 K
268 K
× 38.7 lb in−2
= 44.5 lb in−2
.
Therefore, p(pump) = 44.5 lb in−2
−14.7 lb in−2
= 30 lb in−2 .

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THE PROPERTIES OF GASES 5
Complications are those factors that destroy the constancy of V or n, such as the change in volume of
the tire, the change in rigidity of the material from which it is made, and loss of pressure by leaks and
diffusion.
E1.4(a) The perfect gas law in the form p =
nRT
V
[1.8] is appropriate. T and V are given; n needs to be calculated.
n =
0.255 g
20.18 g mol−1
= 1.26 × 10−2
mol, T = 122 K, V = 3.00 dm3
.
Therefore, upon substitution,
p =
(1.26 × 10−2 mol) × (0.08206 dm3
atm K−1 mol−1
) × (122 K)
3.00 dm3
= 4.20 ×10−2 atm .
E1.5(a) Boyle’s law in the form pfVf = piVi is solved for Vf: Vf =
pi
pf
× Vi.
pi = 1.0 atm,
pf = pex + ρgh[1.3] = pi + ρgh = 1.0 atm + ρgh,
ρgh = (1.025 × 103
kg m−3
) × (9.81 m s−2
) × (50 m) = 5.03 × 105
Pa.
Hence, pf = (1.01 × 105 Pa) + (5.03 × 105 Pa) = 6.04 × 105 Pa.
Vf =
1.01 × 105 Pa
6.04 × 105 Pa
× 3.0 m3
= 0.50 m3
.
E1.6(a) The pressure in the apparatus is given by
p = patm + ρgh [1.3].
patm = 770 Torr ×
1 atm
760 Torr
×
1.013 × 10−5 Pa
760 Torr
= 1.026 × 10−5
Pa
ρgh = 0.99707 g cm−3
×
1 kg
103 g
×
106 cm3
m3
× 9.806 m s−2
= 977 Pa
p = 1.026 × 105
Pa + 977 Pa = 1.036 × 105
Pa = 104 kPa .
E1.7(a) The gas pressure is calculated as the force per unit area that a column of water of height 206.402 cm
exerts on the gas due to its weight. The manometer is assumed to have uniform cross-sectional area, A.
Then force, F = mg, where m is the mass of the column of water and g is the acceleration of free fall.
As in Example 1.1, m = ρ × V = ρ × h × A where h = 206.402 cm and A is the cross-sectional area.
p =
F
A
=
ρhAg
A
= ρhg.

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6 SOLUTIONS MANUAL
p = (0.99707 g cm−3
) ×
1 kg
103 g
×
106 cm3
1 m3
× (206.402 cm) ×
1 m
102 cm
× (9.8067 m s−2
)
= 2.0182 × 104
Pa.
V = (20.000 dm3
) ×
1 m3
103 dm3
= 2.0000 × 10−2
m3
.
n =
m
M
=
0.25132 g
4.00260 g mol−1
= 0.062789 mol.
The perfect gas equation [1.8] can be rearranged to give R =
pV
nT
.
R =
(2.0182 × 104 Pa) × (2.0000 × 10−2 m3)
(0.062789 mol) × (773.15 K)
= 8.3147 JK−1mol−1
.
The accepted value is R = 8.3145 J K−1 mol−1
.
Although gas volume data should be extrapolated to p = 0 for the best value of R, helium is close
to being a perfect gas under the conditions here, and thus a value of R close to the accepted value is
obtained.
E1.8(a) Since p < 1 atm, the approximation that the vapor is a perfect gas is adequate. Then (as in
Exercise 1.7(b)),
pV = nRT =
m
M
RT.
Upon rearrangement,
M = ρ
RT
p
= (3.71 kg m−3
) ×
(8.314 Pa m3 K−1 mol−1
) × (773 K)
9.32 × 104 Pa
= 0.256 kg mol−1
= 256 g mol−1 .
This molar mass must be an integral multiple of the molar mass of atomic sulfur; hence
number of S atoms =
256 g mol−1
32.0 g mol−1
= 8.
The formula of the vapor is then S8 .
E1.9(a) The partial pressure of the water vapor in the room is:
pH2O = (0.60) × (26.74 Torr) = 16 Torr.

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Assuming that the perfect gas equation [1.8] applies, with n =
m
M
, pV =
m
M
RT or
m =
pVM
RT
=
(16 Torr) ×
1 atm
760 Torr
× (400 m3) ×
103 dm3
m3
× (18.02 g mol−1
)
(0.0821 dm3
atm K−1 mol−1
) × (300 K)
= 6.2 × 103
g = 6.2 kg .
E1.10(a) (a) For simplicity assume a container of volume 1 dm3. Then the total mass is
mT = nN2 MN2 + nO2 MO2 = 1.146 .g (1)
Assuming that air is a perfect gas, pTV = nTRT, where nT is the total amount of gas
nT =
PTV
RT
=
(0.987 bar) ×
1 atm
1.013 bar
× (1 dm3
)
(0.08206 dm3
atm K−1 mol−1
) × (300 K)
= 0.03955 mol,
nT = nN2 + nO2 = 0.03955 mol. (2)
Equations (1) and (2) are simultaneous equations for the amounts of gas and may be solved for them.
Inserting nO2 from (2) into (1) we get
(nN2 ) × (28.0136 g mol−1
) + (0.03955 mol − nN2 ) × (31.9988 g mol−1
) = 1.146 g.
(1.2655 − 1.1460) g = (3.9852 g mol−1
) × (nN2 ).
nN2 = 0.02999 mol.
nO2 = nT − nN2 = (0.03955 − 0.02999) mol = 9.56 × 10−3
mol.
The mole fractions are
xN2 =
0.02999 mol
0.03955 mol
= 0.7583 , xO2 =
9.56 × 10−3 mol
0.03955 mol
= 0.2417 .
The partial pressures are pN2 = (0.7583) × (0.987 bar) = 0.748 bar ,
pO2 = (0.2417) × (0.987 bar) = 0.239 bar .
The sum checks, (0.748 + 0.239) bar = 0.987 bar.
(b) The simplest way to solve this part is to realize that nT, pT, and mT remain the same as in part (a)
as these are experimentally determined quantities. However, the simultaneous equations that need
to be solved are modiﬁed as follows:
mT = nN2 MN2 + nO2 MO2 + nArMAr = 1.146 g,
nT = nN2 + nO2 + nAr = 0.03955 mol,
Since xAr = 0.0100, nAr = 0.0003955 mol.

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8 SOLUTIONS MANUAL
Solving the equations yields
nN2 = 0.03084, xN2 = 0.7798 ,
nO2 = 0.008314, xO2 = 0.2102 .
The partial pressures are:
pN2 = xN2 pT = 0.7798 × 0.987 bar = 0.770 bar ,
pO2 = xO2 pT = 0.2102 × 0.987 bar = 0.207 bar ,
pAr = xArpT = 0.0100 × 0.987 bar = 0.00987 bar .
E1.11(a) This exercise uses the formula, M = ρ
RT
p
, which was developed and used in Exercises 1.7(b) and
1.8(a). Substituting the data,
M =
(1.23 kg m−3
) × (8.314 dm3
kPa K−1
mol−1
) × (330 K)
20 kPa
×
103 g
kg
×
10−3 m3
dm3
= 169 g mol−1 .
E1.12(a) The easiest way to solve this exercise is to assume a sample of mass 1.000 g, then calculate the volume
at each temperature, plot the volume against the Celsius temperature, and extrapolate to V = 0.
Draw up the following table.
θ/◦C ρ/(g dm−3
) V/(dm3
g−1)
−85 1.877 0.5328
0 1.294 0.7728
100 0.946 1.057
V versus θ is plotted in Fig. 1.1. The extrapolation gives a value for absolute zero close to −273◦C.
Alternatively, one could use an equation for V as a linear function of θ, which is Charles’s law, and solve
for the value of absolute zero. V = V0 × (1 + αθ).
At absolute zero, V = 0, then θ (abs. zero) = −
1
α
. The value of α can be obtained from any one of the
data points (except θ = 0) as follows.
From V = V0 × (1 + αθ),
α =
V
V0
− 1
θ
=
1.057
0.7728
− 1
100◦C
= 0.003678(◦
C)−1
−
1
α
= −
1
0.003678(◦C)−1
= −272◦C .
which is close to the value obtained graphically.

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V/(dm3g–1)
u/ °C Figure 1.1
E1.13(a) (a) p =
nRT
V
[1.8].
n = 1.0 mol, T = 273.15 K (i) or 1000 K (ii).
V = 22.414 dm3
(i) or 100 cm3
(ii).
(i) p =
(1.0 mol) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (273.15 K)
22.414 dm3
= 1.0 atm .
(ii) p =
(1.0 mol) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (1000 K)
0.100 dm3
= 8.2 ×102 atm .
(b) p =
nRT
V − nb
−
an2
V2
[1.21a].
From Table 1.6, a = 5.507 dm6
atm mol−2
and b = 6.51 × 10−2 dm3
mol−1
. Therefore,
(i)
nRT
V − nb
=
(1.0 mol) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (273.15 K)
[22.414 − (1.0) × (6.51 × 10−2)] dm3
= 1.003 atm,
an2
V2
=
(5.507 dm6
atm mol−2
) × (1.0 mol)2
(22.414 dm3
)2
= 1.11 × 10−2
atm,
and p = 1.003 atm − 1.11 × 10−2 atm = 0.992 atm = 1.0 atm .
(ii)
nRT
V − nb
=
(1.0 mol) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (1000 K)
(0.100 − 0.0651) dm3
,
= 2.27 × 103
atm,
an2
V2
=
(5.507 dm6
atm mol−2
) × (1.0 mol)2
(0.100 dm3
)2
= 5.51 × 102
atm,
and p = 2.27 × 103 atm − 5.51 × 102 atm = 1.7 ×103 atm .

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10 SOLUTIONS MANUAL
COMMENT. It is instructive to calculate the percentage deviation from perfect gas behaviour for (i) and (ii).
(i)
0.992 − 1.000
1.000
× 100% = 0.8%.
(ii)
(17 × 102) − (8.2 × 102)
8.2 × 102
× 100% = 107%.
Deviations from perfect gas behavior are not observed at p ≈ 1 atm except with very precise apparatus.
E1.14(a) The conversions needed are as follows:
1 atm = 1.013 × 105 Pa 1 Pa = 1 kg m−1 s−2 1 dm6
= 10−6 m6 1 dm3
= 10−3 m3.
Therefore,
a = 0.751 atm dm6
mol−2
becomes, after substitution of the conversions,
a = 7.61 ×10−2 kg m5 s−2 mol−2 , and
b = 0.0226 dm3
mol−1
becomes
b = 2.26 ×10−5 m3 mol−1 .
E1.15(a) The definition of Z is used Z =
pVm
RT
[1.17] =
Vm
V◦
m
.
Vmis the actual molar volume, V◦
m is the perfect gas molar volume. V◦
m =
RT
p
. Since Vm is 12 per cent
smaller than that of a perfect gas, Vm = 0.88V◦
m, and
(a) Z =
0.88V◦
m
V◦
m
= 0.88 .
(b) Vm =
ZRT
p
=
(0.88) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (250 K)
15 atm
= 1.2 dm3 mol−1 .
Since Vm < V◦
m attractive forces dominate.
E1.16(a) The amount of gas is first determined from its mass; then the van der Waals equation is used to determine
its pressure at the working temperature. The initial conditions of 300 K and 100 atm are in a sense
superfluous information.
n =
92.4 kg
28.02 × 10−3kg mol−1
= 3.30 × 103
mol
V = 1.000 m3
= 1.000 × 103
dm3
p =
nRT
V − nb
−
an2
V2
[1.21a] =
(3.30 × 103 mol) × (0.08206 dm3
atm K−1 mol−1
) × (500 K)
(1.000 × 103 dm3
) − (3.30 × 103mol) × (0.0387 dm3
mol−1
)
−
(1.352 dm6
atm mol−2
) × (3.30 × 103 mol)2
(1.000 × 103 dm3
)2
= (155 − 14.8) atm = 140 atm .
E1.17(a) (a) p =
nRT
V
[1.8] =
(10.0 mol) × (0.08206 dm3
atm K−1 mol−1
) × (300 K)
4.860 dm3
= 50.7 atm .

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(b) p =
nRT
V − nb
− a
n
V
2
[1.21a]
=
(10.0 mol) × (0.08206 dm3
atm K−1 mol−1
) × (300 K)
(4.860 dm3
) − (10.0 mol) × (0.0651 dm3
mol−1
)
− (5.507 dm6
atm mol−2
) ×
10.0 mol
4.860 dm3
2
= 58.49 − 23.32 = 35.2 atm .
The compression factor is calculated from its deﬁnition [1.17] after inserting Vm =
V
n
.
To complete the calculation of Z, a value for the pressure, p, is required. The implication in the deﬁnition
[1.17] is that p is the actual pressure as determined experimentally. This pressure is neither the perfect
gas pressure nor the van der Waals pressure. However, on the assumption that the van der Waals equation
provides a value for the pressure close to the experimental value, we can calculate the compression factor
as follows
Z =
pV
nRT
=
(35.2 atm) × (4.860 dm3
)
(10.0 mol) × (0.08206 dm3
atm K−1 mol−1
) × (300 K)
= 0.695 .
COMMENT. If the perfect gas pressure had been used, Z would have been 1, the perfect gas value.
E1.18(a) n = n(H2) + n(N2) = 2.0 mol + 1.0 mol = 3.0 mol, xJ =
nJ
n
[1.14].
(a) x(H2) =
2.0 mol
3.0 mol
= 0.67 , x(N2) =
1.0 mol
3.0 mol
= 0.33 .
(b) The perfect gas law is assumed to hold for each component individually as well as for the mixture
as a whole. Hence, pJ = nJ
RT
V
.
RT
V
=
(8.206 × 10−2 dm3
atm K−1 mol−1
) × (273.15 K)
22.4 dm3
= 1.00 atm mol−1
.
p(H2) = (2.0 mol) × (1.00 atm mol−1
) = 2.0 atm .
p(N2) = (1.0 mol) × (1.00 atm mol−1
) = 1.0 atm .
(c) p = p(H2) + p(N2)[1.15] = 2.0 atm + 1.0 atm = 3.0 atm .
Question. Does Dalton’s law hold for a mixture of van der Waals gases?
E1.19(a) Equations [1.22] are solved for b and a, respectively, and yield b =
Vc
3
and a = 27b2pc = 3V2
c pc.
Substituting the critical constants,
b =
1
3
× (98.7 cm3
mol−1
) = 32.9 cm3 mol−1 ,
a = 3 × (98.7 × 10−3
dm3
mol−1
)2
× (45.6 atm) = 1.33 dm6 atm mol−2 .
Note that knowledge of the critical temperature, Tc, is not required.

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12 SOLUTIONS MANUAL
As b is approximately the volume occupied per mole of particles
vmol ≈
b
NA
=
32.9 × 10−6 m3 mol−1
6.022 × 1023 mol−1
= 5.46 × 10−29
m3
.
Then, with vmol =
4
3
πr3, r ≈
3
4π
× (5.46 × 10−29 m3)
1/3
= 0.24 nm .
E1.20(a) The Boyle temperature, TB, is the temperature at which B = 0. In order to express TB in terms of a and
b, the van der Waals equation must be recast into the form of the virial equation.
p =
RT
Vm − b
−
a
V2
m
[1.21b].
Factoring out
RT
Vm
yields p =
RT
Vm
1
1 − b/Vm
−
a
RTVm
.
So long as b/Vm < 1, the ﬁrst term inside the brackets can be expanded using (1−x)−1 = 1+x+x2+ · · · ,
which gives
p =
RT
Vm
1 + b −
a
RT
×
1
Vm
+ · · ·
We can now identify the second virial coefﬁcient as B = b −
a
RT
.
Since at the Boyle temperature B = 0, TB =
a
bR
=
27Tc
8
.
(a) From Table 1.6, a = 6.260 dm6
atm mol−2
, b = 5.42 × 10−2 dm3
mol−1
. Therefore,
TB =
6.260 dm6
atm mol−2
(5.42 × 10−2 dm3
mol−1
) × (8.206 × 10−2 dm3
atm K−1 mol−1
)
= 1.41 ×103 K .
(b) As in Exercise 1.19(a), vmol ≈
b
NA
=
5.42 × 10−5 m3 mol−1
6.022 × 1023 mol−1
= 9.00 × 10−29 m3
r ≈
3
4π
× (9.00 × 10−29
m3
)
1/3
= 0.59 nm .
E1.21(a) The reduced temperature and pressure of hydrogen are calculated from the relations
Tr =
T
Tc
and pr =
p
pc
[1.24].
Tr =
298 K
33.23 K
= 8.968 [Tc = 33.23 K, Table 1.5],
pr =
1.0 atm
12.8 atm
= 0.0781 [pc = 12.8 atm, Table 1.5].
Hence, the gases named will be in corresponding states at T = 8.968 × Tc and at p = 0.0781 × pc.

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 13 — #13
(a) For ammonia, Tc = 405.5 K and pc = 111.3 atm (Table 1.5), so
T = (8.968) × (405.5 K) = 3.64 ×103 K ,
p = (0.0781) × (111.3 atm) = 8.7 atm .
(b) For xenon, Tc = 289.75 K and pc = 58.0 atm, so
T = (8.968) × (289.75 K) = 2.60 ×103 K ,
p = 0.0781) × (58.0 atm) = 4.5 atm .
(c) For helium, Tc = 5.21 K and pc = 2.26 atm, so
T = (8.968) × (5.21 K) = 46.7 K ,
p = (0.0781) × (2.26 atm) = 0.18 atm .
E1.22(a) The van der Waals equation [1.21b] is solved for b, which yields
b = Vm −
RT
p +
a
V2
m
.
Substituting the data
b = 5.00 × 10−4
m3
mol−1
−
(8.314 J K−1 mol−1
) × (273 K)
(3.0 × 106 Pa) +
0.50 m6 Pa mol−2
(5.00 × 10−4 m3 mol−1
)2
= 0.46 × 10−4
m3
mol−1
.
Z =
pVm
RT
[1.17] =
(3.0 × 106 Pa) × (5.00 × 10−4 m3)
(8.314 J K−1 mol−1
) × (273 K)
= 0.66.
COMMENT. The deﬁnition of Z involves the actual pressure, volume, and temperature and does not depend
upon the equation of state used to relate these variables.
Solutions to problems
Solutions to numerical problems
P1.1 Since the Neptunians know about perfect gas behavior, we may assume that they will write pV = nRT
at both temperatures. We may also assume that they will establish the size of their absolute unit to be
the same as the ◦N, just as we write 1K = 1◦C. Thus
pV(T1) = 28.0 dm3
atm = nRT1 = nR × (T1 + 0◦
N),
pV(T2) = 40.0 dm3
atm = nRT2 = nR × (T1 + 100◦
N),
or T1 =
28.0 dm3
atm
nR
, T1 + 100◦
N =
40.0 dm3
atm
nR
.

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 14 — #14
14 SOLUTIONS MANUAL
Dividing,
T1 + 100◦N
T1
=
40.0 dm3
atm
28.0 dm3
atm
= 1.429 or T1 + 100◦N = 1.429T1, T1 = 233 absolute units.
As in the relationship between our Kelvin scale and Celsius scale T = θ− absolute zero(◦N) so absolute
zero(◦N) = −233◦N .
COMMENT. To facilitate communication with Earth students we have converted the Neptunians’ units of
the pV product to units familiar to humans, which are dm3 atm. However, we see from the solution that only
the ratio of pV products is required, and that will be the same in any civilization.
Question. If the Neptunians’ unit of volume is the lagoon (L), their unit of pressure is the poseidon (P),
their unit of amount is the nereid (n), and their unit of absolute temperature is the titan (T), what is the
value of the Neptunians’ gas constant (R) in units of L, P, n, and T?
P1.3 The value of absolute zero can be expressed in terms of α 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)].
Then θ (abs. zero) = −
1
α
.
All gases become perfect in the limit of zero pressure, so the best value of α and, hence, θ (abs. zero)
is obtained by extrapolating α 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◦C ,
which is close to the accepted value of −273.15◦C.
3.662
3.664
3.666
3.668
3.670
3.672
0 800200 400 600
p/Torr Figure 1.2

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 15 — #15
P1.5
p
T
=
nR
V
= constant, if n and V are constant. Hence,
p
T
=
p3
T3
, where p is the measured pressure at
temperature, T, and p3 and T3 are the triple point pressure and temperature, respectively. Rearranging,
p =
p3
T3
T.
The ratio
p3
T3
is a constant =
6.69 kPa
273.16 K
= 0.0245 kPa K−1. Thus the change in p, p, is proportional to
the change in temperature, T : p = (0.0245 kPa K−1) × ( T).
(a) p = (0.0245 kPa K−1) × (1.00 K) = 0.0245 kPa .
(b) Rearranging, p =
T
T3
p3 =
373.16 K
273.16 K
× (6.69 kPa) = 9.14 kPa .
(c) Since
p
T
is a constant at constant n and V, it always has the value 0.0245 kPa K−1; hence
p = p374.15 K − p373.15 K = (0.0245 kPa K−1
) × (1.00 K) = 0.0245 kPa .
P1.7 (a) Vm =
RT
p
=
(8.206 × 10−2 dm3
atm K−1 mol−1
) × (350 K)
2.30 atm
= 12.5 dm3 mol−1 .
(b) From p =
RT
Vm − b
−
a
V2
m
[1.21b], we obtain Vm =
RT
p +
a
V2
m
+ b[rearrange1.21b]
Then, with a and b from Table 1.16,
Vm ≈
8.206 × 10−2 dm3
atm K−1 mol−1
× (350 K)
(2.30 atm) + (6.260 dm6
atm mol−2
)/ 12.5 dm3
mol−1 2
+ (5.42 × 100−2
dm3
mol−1
)
≈
28.7¯2 dm3
mol−1
2.34
+ 5.42 × 10−2
dm3
mol−1
≈ 12.3 dm3 mol−1 .
Substitution of 12.3 dm3
mol−1
into the denominator of the ﬁrst expression again results in
Vm = 12.3 dm3
mol−1
, so the cycle of approximation may be terminated.
P1.9 As indicated by eqns 1.18 and 1.19 the compression factor of a gas may be expressed as either a virial
expansion in p or in
1
Vm
. The virial form of the van der Waals equation is derived in Exercise 1.20(a)
and is p =
RT
Vm
1 + b −
a
RT
×
1
Vm
+ · · ·
Rearranging, Z =
pVm
RT
= 1 + b −
a
RT
×
1
Vm
+ · · ·
On the assumption that the perfect gas expression for Vm is adequate for the second term in this expansion,
we can readily obtain Z as a function of p.
Z = 1 +
1
RT
× b −
a
RT
p + · · ·

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 16 — #16
16 SOLUTIONS MANUAL
(a) Tc = 126.3 K.
Vm =
RT
p
× Z =
RT
p
+ b −
a
RT
+ · · ·
=
(0.08206 dm3
atm K−1 mol−1
) × (126.3 K)
10.0 atm
+ (0.0387 dm3
mol−1
) −
1.352 dm6
atm mol−2
(0.08206 dm3
atm K−1mol−1
) × (126.3 K)
= (1.036 − 0.092) dm3
mol−1
= 0.944 dm3 mol−1 .
Z =
p
RT
× (Vm) =
(10.0 atm) × (0.944 dm3
mol−1
)
(0.08206 dm3
atm K−1 mol−1
) × (126.3 K)
= 0.911.
(b) The Boyle temperature corresponds to the temperature at which the second virial coefficient is zero,
hence correct to the first power in p, Z = 1, and the gas is close to perfect. However, if we assume
that N2 is a van der Waals gas, when the second virial coefficient is zero,
b −
a
RTB
= 0, or TB =
a
bR
.
TB =
1.352 dm6
atm mol−2
(0.0387 dm3
mol−1
) × (0.08206 dm3
atm K−1 mol−1
)
= 426 K.
The experimental value (Table 1.5) is 327.2 K. The discrepancy may be explained by two
considerations.
1. Terms beyond the first power in p should not be dropped in the expansion for Z.
2. Nitrogen is only approximately a van der Waals gas.
When Z = 1, Vm =
RT
p
, and using TB = 327.2 K
=
(0.08206 dm3
atm K−1mol−1
) × 327.2 K
10.0 atm
= 2.69 dm3 mol−1
and this is the ideal value of Vm. Using the experimental value of TB and inserting this value into
the expansion for Vm above, we have
Vm =
0.08206 dm3
atm K−1mol−1
× 327.2 K
10.0 atm
+ 0.0387 dm3
mol−1
−
1.352 dm6
atm mol−2
0.08206 dm3
atm K−1mol−1
× 327.2 K
= (2.685 − 0.012) dm3
mol−1
= 2.67 dm3 mol−1
and Z =
Vm
V◦
m
=
2.67 dm3
mol−1
2.69 dm3
mol−1
= 0.992 ≈ 1.

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 17 — #17
(c) TI = 621 K [Table 2.9].
Vm =
0.08206 dm3
atm K−1mol−1
× 621 K
10.0 atm
+ 0.0387 dm3
mol−1
−
1.352 dm6
atm mol−2
0.08206 dm3
atm K−1mol−1
× 621 K
= (5.096 + 0.012) dm3
mol−1
= 5.11 dm3 mol−1
and Z =
5.11 dm3
mol−1
5.10 dm3
mol−1
= 1.002 ≈ 1.
Based on the values of TB and TI given in Tables 1.4 and 2.9 and assuming that N2 is a van der Waals
gas, the calculated value of Z is closest to 1 at TI, but the difference from the value at TB is less than
the accuracy of the method.
P1.11 (a) Vm =
molar mass
density
=
M
ρ
=
18.02 g mol−1
1.332 × 102 g dm−3
= 0.1353 dm3 mol−1 .
(b) Z =
pVm
RT
[1.17b] =
(327.6 atm) × (0.1353 dm3
mol−1
)
(0.08206 dm3
atm K−1 mol−1
) × (776.4 K)
= 0.6957 .
(c) Two expansions for Z based on the van der Waals equation are given in Problem 1.9. They are
Z = 1 + b −
a
RT
×
1
Vm
+ · · ·
= 1 + (0.0305 dm3
mol−1
) −
5.464 dm6
atm mol−2
(0.08206 dm3
atm K−1 mol−1
) × (776.4 K)
×
1
0.1353 dm3
mol−1
= 1 − 0.4084 = 0.5916 ≈ 0.59.
Z = 1 +
1
RT
× b −
a
RT
× (p) + · · ·
= 1 +
1
(0.08206 dm3
atm K−1 mol−1
) × (776.4 K)
× (0.0305 dm3
mol−1
) −
5.464 dm6
atm mol−2
(0.08206 dm3
atm K−1 mol−1
) × (776.4 K)
× 327.6 atm
= 1 − 0.2842 ≈ 0.72 .
In this case the expansion in p gives a value close to the experimental value; the expansion in
1
Vm
is not as good. However, when terms beyond the second are included the results from the two
expansions for Z converge.
P1.13 Vc = 2b, Tc =
a
4bR
[Table 1.7]

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 18 — #18
18 SOLUTIONS MANUAL
Hence, with Vc and Tc from Table 1.5, b =
1
2
Vc =
1
2
× (118.8 cm3 mol−1
) = 59.4 cm3 mol−1 .
a = 4bRTc = 2RTcVc
= (2) × (8.206 × 10−2
dm3
atm K−1
mol−1
) × (289.75 K) × (118.8 × 10−3
dm3
mol−1
)
= 5.649 dm6 atm mol−2 .
Hence
p =
RT
Vm − b
e−a/RTVm =
nRT
V − nb
e−na/RTV
=
(1.0 mol) × (8.206 × 10−2 dm3
atm K−1 mol−1
) × (298 K)
(1.0 dm3
) − (1.0 mol) × (59.4 × 10−3 dm3
mol−1
)
× exp
−(1.0 mol) × (5.649 dm6
atm mol−2
)
(8.206 × 10−2 dm3
atm K−1 mol−1
) × (298 K) × (1.0 dm6
atm mol−1
)
= 26.0 atm × e−0.231
= 21 atm .
Solutions to theoretical problems
P1.15 This expansion has already been given in the solutions to Exercise 1.20(a) and Problem 1.14; the
result is
p =
RT
Vm
1 + b −
a
RT
1
Vm
+
b2
V2
m
+ · · · .
Compare this expansion with p =
RT
Vm
1 +
B
Vm
+
C
Vm2
+ · · · [1.19]
and hence find B = b −
a
RT
and C = b2 .
Since C = 1200 cm6 mol−2
, b = C1/2 = 34.6 cm3 mol−1
a = RT(b − B) = (8.206 × 10−2
) × (273 dm3
atm mol−1
) × (34.6 + 21.7) cm3
mol−1
= (22.40 dm3
atm mol−1
) × (56.3 × 10−3
dm3
mol−1
) = 1.26 dm6 atm mol−2 .
P1.17 The critical point corresponds to a point of zero slope that is simultaneously a point of inflection in a
plot of pressure versus molar volume. A critical point exists if there are values of p, V, and T that result
in a point that satisfies these conditions.
p =
RT
Vm
−
B
V2
m
+
C
V3
m
.

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 19 — #19
∂p
∂Vm T
= −
RT
V2
m
+
2B
V3
m
−
3C
V4
m
= 0
∂2p
∂V2
m T
=
2RT
V3
m
−
6B
V4
m
+
12C
V5
m
= 0



at the critical point.
That is,
−RTcV2
c + 2BVc − 3C = 0
RTcV2
c − 3BVc + 6C = 0
which solve to Vc =
3C
B
, Tc =
B2
3RC
.
Now use the equation of state to ﬁnd pc
pc =
RTc
Vc
−
B
V2
c
+
C
V3
c
=
RB2
3RC
×
B
3C
− B
B
3C
2
+ C
B
3C
3
=
B3
27C2
.
It follows that Zc =
pcVc
RTc
=
B3
27C2
×
3C
B
×
1
R
×
3RC
B2
=
1
3
.
P1.19 For a real gas we may use the virial expansion in terms of p [1.18]
p =
nRT
V
(1 + B p + · · · ) = ρ
RT
M
(1 + B p + · · · )
which rearranges to
p
ρ
=
RT
M
+
RT B
M
p + · · · .
Therefore, the limiting slope of a plot of
p
ρ
against p is
B RT
M
. From Fig. 1.3 the limiting slope is
B RT
M
=
(5.84 − 5.44) × 104 m2 s−2
(10.132 − 1.223) × 104 Pa
= 4.4 × 10−2
kg−1
m3
.
From Fig. 1.2,
RT
M
= 5.40 × 104 m2 s−2; hence
B =
4.4 × 10−2 kg−1
m3
5.40 × 104 m2 s−2
= 0.81 × 10−6
Pa−1
,
B = (0.81 × 10−6
Pa−1
) × (1.0133 × 105
Pa atm−1
) = 0.082 atm−1 .
B = RTB [Problem 1.18]
= (8.206 × 10−2
dm3
atm K−1
mol−1
) × (298 K) × (0.082 atm−1
)
= 2.0 dm3 mol−1 .
P1.21 The critical temperature is that temperature above which the gas cannot be liqueﬁed by the application
of pressure alone. Below the critical temperature two phases, liquid and gas, may coexist at equilibrium,
and in the two-phase region there is more than one molar volume corresponding to the same conditions

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 20 — #20
20 SOLUTIONS MANUAL
5.9
5.8
5.7
5.6
5.5
5.4
0 2 4 6
p/(104 Pa)
(p/r)/(104
m2
s–1
)
8 10 12
y = 5.3963 + 0.046074x R = 0.99549
Figure 1.3
of temperature and pressure. Therefore, any equation of state that can even approximately describe this
situation must allow for more than one real root for the molar volume at some values of T and p, but
as the temperature is increased above Tc, allows only one real root. Thus, appropriate equations of state
must be equations of odd degree in Vm.
The equation of state for gas A may be rewritten V2
m − (RT/p)Vm − (RTb/p) = 0, which is a quadratic
and never has just one real root. Thus, this equation can never model critical behavior. It could possibly
model in a very crude manner a two-phase situation, since there are some conditions under which a
quadratic has two real positive roots, but not the process of liquefaction.
The equation of state of gas B is a ﬁrst-degree equation in Vm and therefore can never model critical
behavior, the process of liquefaction, or the existence of a two-phase region.
A cubic equation is the equation of lowest degree that can show a cross-over from more than one real
root to just one real root as the temperature increases. The van der Waals equation is a cubic equation
in Vm.
P1.23 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
nMN = 2.2990 g
for ‘chemical nitrogen’ and
nArMAr + nNMN = n[xArMAr + (1 − xAr)MN] = 2.3102 g
for ‘atmospheric nitrogen’. Dividing the latter expression by the former yields
xArMAr
MN
+ (1 − xAr) =
2.3102
2.2990
so xAr
MAr
MN
− 1 =
2.3102
2.2990
− 1
and xAr =
(2.3102/2.2990) − 1
(MAr/MN) − 1
=
(2.3102/2.2990) − 1
(39.95 g mol−1
)/(28.013 g mol−1
− 1)
= 0.011 .
COMMENT. This value for the mole fraction of argon in air is close to the modern value.

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 21 — #21
Solutions to applications
P1.25 1 t = 103 kg. Assume 300 t per day.
n(SO2) =
300 × 103 kg
64 × 10−3 kg mol−1
= 4.7 × 106
mol.
V =
nRT
p
=
(4.7 × 106 mol) × (0.082 dm3
atm K−1mol−1
) × 1073 K
1.0 atm
= 4.1 × 108 dm3
.
P1.27 The pressure at the base of a column of height H is p = ρgH (Example 1.1). But the pressure at any
altitude h within the atmospheric column of height H depends only on the air above it; therefore
p = ρg(H − h) and dp = −ρg dh.
Since ρ =
pM
RT
[Problem 1.2], dp = −
pMgdh
RT
, implying that
dp
p
= −
Mg dh
RT
This relation integrates to p = p0e−Mgh/RT
For air M ≈ 29 g mol−1
and at 298 K
Mg
RT
≈
(29 × 10−3 kg mol−1
) × (9.81 m s−2)
2.48 × 103 J mol−1
= 1.1¯5 × 10−4
m−1
[1 J = 1 kg m2
s−2
].
(a) h = 15 cm.
p = p0 × e(−0.15 m)×(1.1¯5×10−4 m−1) = 0.99998 p0;
p − p0
p0
= 0.00 .
(b) h = 11 km = 1.1 × 104 m.
p = p0 × e(−1.1×10−4)×(1.15×10−4m−1) = 0.28 p0;
p − p0
p0
= −0.72 .
P1.29 Refer to Fig. 1.3.
h
Ground
Air
(environment)
Figure 1.4

TRAPP: “CHAP01” — 2006/3/8 — 18:03 — PAGE 22 — #22
22 SOLUTIONS MANUAL
The buoyant force on the cylinder is
Fbuoy = Fbottom − Ftop
= A(pbottom − ptop)
according to the barometric formula.
ptop = pbottome−Mgh/RT
where M is the molar mass of the environment (air). Since h is small, the exponential can be expanded
in a Taylor series around h = 0 e−x = 1 − x +
1
2!
x2 + · · · . Keeping the ﬁrst-order term only yields
ptop = pbottom 1 −
Mgh
RT
.
The buoyant force becomes
Fbuoy = Apbottom 1 − 1 +
Mgh
RT
= Ah
pbottomM
RT
g
=
pbottomVM
RT
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. Thus Fbuoy = mg. The net force is the difference between the buoyant force
and the weight of the balloon. Thus
Fnet = mg − mballoon g = (m − mballoon)g
This is Archimedes’ principle.

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