Solucionario capítulo 1

1. Before using this manual, please read the preceding note To the Student.
Chapter 1
1.1 (a) F. (b) T. (c) T. (d) F. (e) F. (A mixture of ice and liquid water has
one substance.)
1.2 (a) Closed, nonisolated; (b) open, nonisolated; (c) open; nonisolated.
1.3 (a) Three. (b) Three; solid AgBr; solid AgCl, and the solution.
1.4 So that pressure or composition differences between systems A and B won’t
cause changes in the properties of A and B. Such changes can then result only
from a temperature difference between A and B.
1.5 (a)
3
3
3 3
g 1 kg (100 cm)
19.3 19300 kg/m
cm 1000 g (1 m)
=
(b)
$800 1 troy oz 7000 grains 1 pound
$25.72/g
troy oz 480 grains 1 pound 453.59 g
=
m = ρV = (19.3 g/cm3
)(106
cm3
) = 1.93 × 107
g
(1.93 × 107
)($25.72/g) = $4.96 × 108
1.6 (a) T. (b) F. (c) T. (d) T.
1.7 (a) 32.0. (b) 32.0 amu. (c) 32.0. (d) 32.0 g/mol.
1.8 100 g of solution contains 12.0 g of HCl and 88.0 g of water. HCln =
(12.0 g)(1 mol/36.46 g) = 0.329 mol; OH2
n = (88.0 g)(1 mol/18.015 g) =
4.885 mol. HClx = 0.329/(0.329 + 4.885) = 0.0631; OH2
x = 1 – HClx =
0.9369.
1

2. 1.9 (a)
23
23
12.0 g/mol
1.99 10 g/atom
6.022 10 atoms/mol
−
= ×
×
(b)
23
23
18.0 g/mol
2.99 10 g/molecule
6.022 10 molecules/mol
−
= ×
×
1.10 (a) 3 3
m (209 g/mol) (9.20 g/cm ) 22.7 cm /mol.V M ρ= = =
(b) (22.7 cm3
/mol)/(6.02 × 1023
atoms/mol) = 3.77 × 10–23
cm3
/atom.
(c) The diameter d of the Po atom equals the edge length of the surrounding
cube, so d3
= 3.77 × 10–23
cm3
and d = 3.35 × 10–8
cm.
(d)
3 9 34 4
3 3 (1 10 m)V rπ π −
= = × = 4.2 × 10–27
m3
= 4.2 × 10–21
cm3
. We have
(4.2 × 10–21
cm3
)/(3.77 × 10–23
cm3
/atom) = 111 atoms.
(e)
9 34
3 (50 10 m)V π −
= × = 5.2 × 10–22
m3
= 5.2 × 10–16
cm3
.
(5.2 × 10–16
cm3
)/( 3.77 × 10–23
cm3
/atom) = 1.4 × 107
atoms.
(f) The particle’s volume is (2 × 10–9
nm)3
= 8 × 10–27
m3
= 8 × 10–21
cm3
.
The particle has (8 × 10–21
cm3
)/(3.77 × 10–23
cm3
/atom) = 212 atoms. The
edge length of the particle is 2 nm = 20 × 10–8
cm and use of the atomic
diameter from (c) shows that the number of atoms along an edge of the
particle is (20 × 10–8
cm)/(3.35 × 10–8
cm) = 6 atoms. With 6 atoms along
each edge of the cube, the top face of the cube has 36 atoms and the
cubic particle contains 6 layers of atoms with 36 atoms in each layer.
(This comes to 36 × 6 = 216 atoms, and the slight discrepancy with the
number 212 will be ignored.) The 36 atoms in the top layer and the 36
atoms in the bottom layer lie on the surface of the cube. If you draw a
square array of 36 circles with 6 circles along each of the 4 sides of the
square, you will see that 20 circles lie on the sides of the square and 16
circles lie in the interior of the square. [We have 4(6) – 4 = 20, where 4 is
subtracted to allow for the fact that circles at the corners of the square lie
on two sides and should not be counted twice.] Thus the four layers of
atoms between the top and bottom layers each contain 16 interior atoms
and the total number of atoms not on the surface is 16 × 4 = 64. The
percentage of interior atoms is (64/216) × 100% = 30% and 70% of the
atoms lie on the surface.
(g) Repetition of the reasoning of part (f) shows the particle has a volume of
1 × 10–15
cm3
, contains 2.65 × 107
atoms, and has 299 atoms along each
edge. With 299 atoms on an edge, the top layer of the cube has 2992
=
2

3. 89401 atoms. For one of the 297 layers between the top and bottom
layers, there are 4(299) – 4 = 1192 atoms on the surface and the total
number of atoms on the surface is 297(1192) = 354024. The percentage
of atoms lying on the surface is [(3.54 × 105
)/(2.65 × 107
)] × 100% =
1.3%. (See also Fig. 7.13.)
1.11 (a) T. (b) T. (c) F. (d) T. (e) F. (f) T.
1.12 (a) (5.5 m3
)(100 cm)3
/(1 m)3
= 5.5 × 106
cm3
.
(b) (1.0 × 109
Pa)(1 bar)/(105
Pa) = 1.0 × 104
bar.
(c) (1 hPa)(100 Pa/1 hPa)(1 bar/105
Pa)(750.062 torr/1 bar) = 0.750062 torr.
(d)
.kg/m105.1
m1
cm10
g10
kg1
cm
g5.1 33
3
36
33
×=
1.13 The system pressure is less than the barometric pressure by 304.3 torr –
202.1 torr = 102.2 torr. So Psystem = 754.6 torr – 102.2 torr = 652.4 torr.
1.14 (a) 2 2 2 2Hg Hg H O H O Hg Hg H O H O, so .P gh gh h hρ ρ ρ ρ= = =
ft33.9in.407
g/cm0.997
in.))(30.0g/cm(13.53
3
3
OH2
===h
where the vapor pressure of water was neglected.
(b) Use of P = ρgh and Eq. (2.8) gives P as
(13.53 g/cm3
)(978 cm/s2
)(30.0 × 2.54 cm) 2
dyn/cm1013250
atm1
=
0.995 atm
1.15 For m constant, n is constant, so (1.18) becomes PV/T = nR = const,
which is (1.17).
1.16 (a) n = (24.0 g)(1 mol/44.0 g) = 0.545 mol. P = nRT/V =
(0.545 mol)(82.06 cm3
-atm/mol-K)(273.1 K)/(5000 cm3
) = 2.44 atm.
3

4. (b) V = nRT/P = (1 mol)(82.06 cm3
-atm/mol-K)(298 K)/(1 atm) =
24500 cm3
. One ft3
= (12 in.)3
= (12 × 2.54 cm)3
= 28300 cm3
.
Percent error = [(28300 – 24500)/24500] × 100% = 16%
1.17 Use of P1V1/T1 = P2V2/T2 gives P2 = (V1/V2)(T2/T1)P1 = (V1/2V1)(3T1/T1)P1 =
1.5P1 = 1.5(0.800 bar) = 1.200 bar.
1.18 P = nRT/V = mRT/MV, so M = mRT/PV and
g/mol30.1
)cm(500atm(24.7/760)
K)K)(298.1-atm/mol-cmg)(82.06(0.0200
3
3
==M
The only hydrocarbon with molecular weight 30 is C2H6.
1.19 At this T and P, N2 is a gas that behaves nearly ideally. From PV = nRT =
(m/M)RT, we get m/V = PM/RT, so
3
3
[(500/760) atm](28.01 g/mol)
0.000766 g/cm
(82.06 cm -atm/mol-K)(293 K)
PM
RT
ρ = = =
since 0.667 bar = 0.667(750 torr) = 500 torr.
1.20 PV/nT 82.025 81.948 81.880 cm3
-atm/mol-K
P 1.0000 3.0000 5.0000 atm
Plotting these data and extrapolating to P = 0, we find limP→0(PV/nT) =
82.06 cm3
-atm/mol-K.
4

5. 1.21 The P/ρ values are 715.3, 706.2, and 697.1 cm3
atm/g. A plot of P/ρ vs. P is a
straight line with intercept 721.4 cm3
atm/g. We have PV = mRT/M, so
M = RT/(P/ρ), and
g/mol31.0
atm/gcm721.
K)K)(273.15-atm/mol-cm(82.06
73
4
3
==M
The only amine with molecular weight 31 is CH3NH2.
1.22 Use of ntot = PV/RT gives
6 2 3
tot 3
(4.85 10 Pa)[1600(10 m) ]
1.867 mol
(8.314 m -Pa/mol-K)(500 K)
n
−
×
= =
The reaction is 2NH3 = N2 + 3H2. Let x moles of N2 be formed. The numbers
of moles of NH3, N2, and H2 present at equilibrium are 1.60 – 2x, x, and 3x,
respectively. Thus ntot/mol = 1.60 – 2x + x + 3x = 1.867, and x = 0.133. Then
n(N2) = 0.133 mol, n(H2) = 0.400 mol, n(NH3) = 1.33 mol
5
81.85
81.90
81.95
82.00
82.05
82.10
0 1 2 3 4 5 6
P/atm
(PV/nT)/(cm3
-atm/mol-K)
(P/ρ)/(cm3
-atm/g)
P/atm
695
700
705
710
715
720
725
0 0.2 0.4 0.6 0.8 1

6. 1.23 Boyle’s law and Charles’ law apply under different conditions (constant T, m
vs. constant P, m); such equations cannot be combined.
1.24 Consider the processes
(P1, V1, T1, m1)  → a
(P1, Va, T1, m2)  → b
(P2, V2, T2, m2)
For step (a), P and T are constant, so V1/m1 = Va/m2. For step (b), m is constant,
so P1Va/T1 = P2V2/T2. Substitution for Va in this last equation gives P2V2/T2 =
P1V1m2/T1m1 or P2V2/m2T2 = P1V1/m1T1.
1.25 Pi = xiP. 2COn = (30.0 g)(1 mol/44.0 g) = 0.682 mol. 2On = 0.625 mol.
2COx = 0.682/(0.682 + 0.625) = 0.522. 2COP = 0.522(3450 kPa) = 1800 kPa.
1.26 (a) At constant temperature, P2 = P1V1/V2 for each gas. Therefore
2H
(20.0 kPa)(3.00 L)
4.00 L
P = 4CH
(10.0 kPa)(1.00 L)
4.00 L
P =
2H 15.0 kPaP = 4CH 2.5 kPaP =
Ptot = 15.0 kPa + 2.5 kPa = 17.5 kPa
(b) Pi = niRT/V and Ptot = ntotRT/V. Hence Pi/Ptot = ni/ntot = xi.
We get 2Hx = 15.0 kPa/17.5 kPa = 0.857 and 4CHx = 2.5 kPa/17.5 kPa =
0.143.
1.27 P(O2) = 751 torr – 21 torr = 730 torr. The equation P1V1/T1 = P2V2/T2 gives
V2 = V1P1T2/P2T1 and
V2 = 3
3
cm32.3
K)torr)(296(760
K)torr)(273)(730cm(36.5
=
1.28 When a steady state is reached, the pressures in the two bulbs are equal. From
P1 = P2, we get n1RT1/V1= n2RT2/V2. Since V1 = V2, we have n1T1 = n2T2. Thus
n1(200 K) = (1.00 mol – n1)(300 K); solving, we get n1 = 0.60 mole in the
200-K bulb and n2 = 0.40 mole in the 300-K bulb.
1.29 We have PV = nRT = NRT/NA, so N/V = PNA/RT and
atm
cm102.46
K)K)(298-atm/mol-cm(82.06
)mol10(6.02 319
3
123
PP
V
N −
−
×=
×
=
6

7. (a) For P = 1 atm, we get N/V = 2.5 × 1019
cm–3
;
(b) for P = (1/760)10–6
atm, we get N/V = 3.2 × 1010
cm–3
;
(c) for P = (1/760)10–11
atm, N/V = 3.2 × 105
cm–3
.
1.30 Substitution in PV = ntotRT gives ntot = 0.01456 mol. Also
mtot = m1 + m2 = n1M1 + n2M2
0.1480 g = nHe(4.003 g/mol) + (0.01456 mol – nHe)(20.18 g/mol).
nHe = 0.00902 mol, nNe = 0.00554 mol
xHe = 0.00902/0.01456 = 0.619, mHe = 0.0361 g
1.31 The downward force of the atmosphere on the earth’s surface equals the
weight W of the atmosphere, so P = W/A = mg/A and m = AP/g = 4πr2
P/g,
where r is the earth’s radius and P = 1 atm = 101325 N/m2
. Thus
kg105.3
m/s9.807
)N/m10(1.013m)10(6.374 18
2
2526
×=
××
=
π
m
1.32 (a) Multiplication of both sides of the equation by 10–5
bar gives
P = 9.4 × 10–5
bar.
(b) 460 K. (c) 1.2 × 103
bar. (d) 312 K.
1.33 Take one liter of gas. This volume has m = 1.185 g = 2 2N O .m m+
We have ntot = PV/RT =
(1.000 atm)(1000 cm3
)/(82.06 cm3
-atm/mol-K)(298.1 K) = 0.04087 mol.
ntot = 2 2 2 2 2 2N O N N O O/ /n n m M m M+ = + =
2Nm /(28.01 g/mol) + (1.185 g – 2Nm )/(32.00 g/mol) = 0.0487 mol.
Solving, we get 2Nm = 0.862 g; hence 2Om = 0.323 g. Then 2Nn =
0.0308 mol and 2O
n = 0.0101 mol; 2Ox = 0.0101/0.0409 = 0.247.
1.34 (a) Use of Pi = xiP gives 2NP = 0.78(1.00 atm) = 0.78 atm,
2OP = 0.21 atm, PAr = 0.0093 atm, 2COP = 0.0004 atm.
(b) V = 3000 ft3
. 1 ft = 12 in. = 12 × 2.54 cm = 30.48 cm.
V = (3000 ft3
)(30.48 cm)3
/ft3
= 8.5 × 107
cm3
. ntot = PV/RT =
[(740/760) atm](8.5 × 107
cm3
)/[(82.06 cm3
-atm/mol-K)(293 K)] =
7

8. 3.44 × 103
mol. 2Nn = totN2
nx = 0.78(3.44 × 103
mol) = 2.68 × 103
mol.
2Nm = (2680 mol)(28.0 g/mol) = 75 kg. Similarly, 2Om = 23 kg,
mAr = 1.3 kg, 2COm = 60 g. We have ρ = mtot/V =
(99.4 kg)/(8.5 × 107
cm3
) = 0.00117 g/cm3
.
1.35 )(xf ′ is zero at the two points where f is a local minimum and where f is a
local maximum. )(xf ′ is negative for the portion of the curve between the
maximum and the minimum.
1.36 dy/dx = 2x + 1. At x = 1, the slope is 2(1) + 1 = 3.
1.37 (a) 6x2
e–3x
– 6x3
e–3x
; (b)
2
3
24 x
xe−
− ; (c) 1/x (not 1/2x); (d) 1/(1 – x)2
;
(e) 1/(x + 1) – x/(x + 1)2
= 1/(x +1)2
; (f) 2e–2x
/(1 – e–2x
); (g) 6 sin 3x cos 3x.
1.38 (a) y = 2/(1 – x) and dy/dx = 2/(1 – x)2
.
(b) d(x2
e3x
)/dx = 2xe3x
+ 3x2
e3x
;
d2
(x2
e3x
)/dx2
= 2e3x
+ 6xe3x
+ 6xe3x
+ 9x2
e3x
= 2e3x
+ 12xe3x
+ 9x2
e3x
.
(c) dy = (10x – 3 – 2/x2
) dx.
Reminder: Work the problems before looking up their solutions.
1.39 (a) x 0.1 0.01 0.001 0.0001 0.00001
xx
0.794 0.955 0.9931 0.9991 0.99988
This indicates (but does not prove) that the limit is 1.
(b) x 10–3
–10–3
10–4
–10–4
10–5
–10–5
(1 + x)1/x
2.717 2.720 2.7181 2.7184 2.71827 2.71828
This suggests that the limit is e = 2.7182818. . . .
1.40 (a) Results on a calculator with 8-digit display and 11 internal digits are:
∆y/∆x = 277, 223.4, 218.88, 218.44, 218.398, 218.393, 218.4 for ∆x =
10–1
, 10–2
, 10–3
, 10–4
, 10–5
, 10–6
, 10–7
, respectively. The best estimate is
218.393.
8

9. (b) dy/dx =
2
2 x
xe , and at x = 2, dy/dx = 218.3926.
A BASIC program for part (a) is
5 CX = 0.1 50 PRINT “DELTAX=”;CX;
10 FOR N = 1 TO 7 “ RATIO=”;R
20 X = 2 60 CX = CX/10
30 CY = EXP((X + CX)^2) – EXP(X^2) 70 NEXT N
40 R = CY/CX 80 END
1.41 (a) 1 + ax cos (axy); (b) –2byz sin (by2
z); (c) –(x2
/y2
)ex/y
;
(d) 0; (e) –ae–a/y
/y2
(e–a/y
+ 1)2
.
1.42 (a) nR/P; (b) –2P/nRT3
.
1.43 Equation (1.30) gives dz = 2axy3
dx + 3ax2
y2
dy.
1.44 (a) P dV + V dP; (b) –T–2
dT; (c) 2cT dT; (d) dU + P dV + V dP.
1.45 Partial differentiation of z = x5
/y3
gives
4 2 3 5 2 5
3 2 3 4 2 5
2 5 4 2 4 4 2
4 4 3 4
5 20 3 12
,
3 15 5 15
z x z x z x z x
, ,
x y x y y y y y
z x x z x x z
,
x y x y y y x y y y x y
∂ ∂ ∂ ∂
= = = − =
∂ ∂ ∂ ∂
 ∂ ∂ ∂ ∂ ∂
= − = − = = − = ÷
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 
1.46 (a) P is a function of n, T, and V, so dP = (∂P/∂n)T,V dn + (∂P/∂T)V,n dT +
(∂P/∂V)T,n dV. Partial differentiation of P = nRT/V gives (∂P/∂n)T,V =
RT/V = P/n (where PV = nRT was used), (∂P/∂T)V,n = nR/V = P/T, and
(∂P/∂V)T,n = –nRT/V2
= –P/V. Substitution into the above equation for dP
gives the desired result. (Note that from P = nRT/V, we have
ln P = ln n + ln R + ln T – ln V, from which d ln P = d ln n + d ln T –
d ln V follows at once.)
(b) Approximating small changes by infinitesimal changes, we have
dn ≈ ∆n = 0, dt ≈ ∆T = 1.00 K, dV ≈ ∆V = 50 cm3
.
9

10. The original pressure is P = nRT/V = 0.8206 atm. Then ∆P ≈ dP ≈
(0.8206 atm)[0 + (1.00 K)/(300 K) – (50 cm3
)/(30000 cm3
)] =
0.00137 atm.
(c) The accurate final pressure is (1.0000 mol)(82.06 cm3
-atm/mol-K) ×
(301.00 K)/(30050 cm3
) = 0.82197 atm. The accurate ∆P is
0.82197 atm – 0.8206 atm = 0.00137 atm.
1.47 1.000 bar = 750 torr = (750 torr)(1 atm/760 torr) = 0.987 atm.
Vm = V/n = (nRT/P)/n = RT/P =
(82.06 cm3
-atm/mol-K)(293.1 K)/(0.987 atm) = 2.44 × 104
cm3
/mol.
1.48 (a) Division by n gives (P + a/ 2
mV )(Vm – b) = RT.
(b) The units of b are the same as those of Vm, namely, cm3
/mol.
P and 2
m/Va have the same units, so the units of a are bar ⋅ cm6
/mol2
.
1.49 α = (1/Vm)(∂Vm/∂T)P = (1/Vm)(c2 + 2c3T – c5P), where Vm is given by (1.40).
κ = – (1/Vm)(∂Vm/∂P)T = –(1/Vm)(–c4 – c5T) =
(c4 + c5T)/(c1 + c2T + c3T2
– c4P – c5PT).
1.50 (a) ρ = m/V = (m/n)/(V/n) = M/Vm, so Vm = M/ρ =
(18.0153 g/mol)/(0.98804 g/cm3
) = 18.233 cm3
/mol.
(b) κ = –(1/Vm)(∂Vm/∂P)T and dVm/Vm = –κ dP at constant T. Integration gives
ln (Vm2/Vm1) = –κ(P2 – P1) at constant T.
κ = (4.4 × 10–10
Pa–1
)(101325 Pa/1 atm) = 4.46 × 10–5
atm–1
and
ln[Vm2/(18.233 cm3
/mol)] = –(4.46 × 10–5
atm–1
)(100 atm – 1 atm) =
–0.0044, so Vm2/(18.233 cm3
/mol) = e–0.0044
= 0.9956 and Vm2 =
18.15 cm3
/mol.
1.51 (a) At constant P, the equation PVm = RT gives Vm = aT, where a = R/P is a
positive constant. The isobars on a Vm vs. T diagram are straight lines that
start at the origin and have positive slopes. (As P increases, the slope
decreases.)
10

11. (b) For Vm constant, PVm = RT gives P = bT, where b = R/Vm is constant. The
isochores on a P vs. T diagram are straight lines that start at the origin
and have positive slopes.
1.52 (a) Partial differentiation of V = nRT(1 + aP)/P gives (∂V/∂T)P,n =
nR(1 + aP)/P. The equation of state gives nR(1 + aP) = PV/T, so
(∂V/∂T)P,n = V/T. Then α = (1/V)(∂V/∂T)P,n = 1/T.
Partial differentiation of V = nRT(1/P + a) gives (∂V/∂P)T,n =
–nRT/P2
= –[PV/(1 + aP)]/P2
= –V/P(1 + aP), where the equation of state
was used. Then κ = –(1/V)(∂V/∂P)T = 1/P(1 + aP).
(b) Solving the equation of state for P, we get P = nRT/(V – anRT); partial
differentiation gives (∂P/∂T)V = nR/(V – anRT) + an2
R2
T/(V – anRT)2
=
P/T + aP2
/T, where P = nRT/(V – anRT) was used. From (a), we have α/κ
= P(1 + aP)/T = P/T + aP2
/T, which agrees with Eq. (1.45).
1.53 For small ∆T, we have
1 1
P P
V V
V T V T
α
∂ ∆   
= ≈ ÷  ÷
∂ ∆   
Since α is an intensive property, we can take any quantity of water. For 1 g,
the equation V = m/ρ gives V = 1.002965 cm3
at 25°C, 1 atm and
V = 1.003227 cm3
at 26°C, 1 atm.
Hence
3 3
1
3
1 1.003227 cm 1.002965 cm
0.00026 K
1.003 cm 26 C 25 C
α −−
≈ =
° − °
Similarly, κ = –(1/V)(∂V/∂P)T ≈ –(1/V)(∆V/∆P)T. At 25°C and 2 atm,
we calculate V = 1.002916 cm3
for 1 g of water.
Thus
3 3
5 1
3
1 1.002916 cm 1.002965 cm
4.9 10 atm
1.003 cm 2 atm 1 atm
κ − −−
≈ − = ×
−
Eq. (1.45) gives
m
4 1
5 1
2.6 10 K
5.3 atm/K
4.9 10 atmV
P
T
α
κ
− −
− −
∂ × 
= ≈ = ÷
∂ × 
1.54 (a) Drawing the tangent line to the 500-bar Vm-vs.-T curve at 100°C, one
finds its slope to be (20.8 cm3
/mol – 17 cm3
/mol)/(300°C – 0°C) = 0.013
cm3
/mol-K = (∂Vm/∂T)P at this T and P. The figure gives Vm = 18.2
cm3
/mol at 500 bar and 100°C, so α = (1/Vm) (∂Vm/∂T)P =
(0.013 cm3
/mol-K)/(18.2 cm3
/mol) = 0.0007 K–1
.
11

12. (b) Drawing the tangent line to the 300°C Vm-vs.-P curve at 2000 bar, one
finds its slope to be –0.0011 cm3
/mol-bar. The figure gives Vm =
20.5 cm3
/mol at this T and P, so κ = –(1/Vm)(∂Vm/∂P)T =
(0.0011 cm3
/mol-bar)/(20.5 cm3
/mol) = 5 × 10–5
bar–1
.
1.55 Equation (1.45) gives α/κ = (∂P/∂T)V ≈ (∆P/∆T)V, so
4 1
5 1
1.7 10 K
(6 K) 22 atm; 23 atm
4.7 10 atm
P T P
α
κ
− −
− −
×
∆ ≈ ∆ = = ≈
×
1.56 (a) As P increases, the molecules are forced closer together; the decrease in
empty space between the molecules makes it harder to compress the
substance, and κ is smaller.
(b) Most substances expand as T increases. The increased space between
molecules makes it easier to compress the substance, and κ increases.
1.57 (a) κ = –(1/V)(∂V/∂P)T ≈ –(1/V)(∆V/∆P)T and ∆P ≈ –∆V/Vκ at constant T.
For a 1% volume decrease, ∆V = –0.01V and we have ∆P ≈ 0.01V/Vκ =
0.01/κ = 0.01/(5 × 10–6
atm–1
) = 2000 atm.
(b) ∆P ≈ 0.01/κ = 0.01/(1 × 10–4
atm–1
) = 100 atm.
1.58 (a)
4
0
2 1J
J=
+∑ = (0 + 1) + (2 + 1) + (4 + 1) + (6 + 1) + (8 + 1) = 25.
(b) 1
s
i ii
xV=∑ .
(c)
2
4 5 61
( )i i ii
b b b=
+ +∑ = b14 + b15 + b16 + b24 + b25 + b26.
1.59 ∑=
n
i 1
cai = ca1 + ca2 + ··· + can = c(a1 + a2 + ··· + an) = c ∑=
n
i 1
ai. Q.E.D.
∑=
n
i 1
(ai + bi) = (a1 + b1) + (a2 + b2) + ··· + (an + bn) =
a1 + a2 + ··· + an + b1 + b2 + ··· + bn = ∑=
n
i 1
ai + ∑=
n
i 1
bi. Q.E.D.
The left side of (1.51) is ∑=
n
i 1
∑=
m
j 1 aibj = ∑=
n
i 1
(aib1 + aib2 + ··· + aibm) =
∑=
n
i 1
ai (b1 + b2 + ··· + bm) = ( )∑=
n
i ia1
(b1 + b2 + ··· + bm) =
∑=
n
i ia1 ∑=
m
j 1
bj, which is the right side of (1.51).
12

13. 1.60 (a)
2
3
−
∫ (2V + 5V2
) dV = (V2
+ 5V3
/3) 2
3|−
= (4 – 40/3) – (9 + 45) = –
190/3.
(b)
4
2∫ V–1
dV = ln V
4
2 = ln 4 – 1n 2 = ln 2 = 0.693.
(c) 1
∞
∫ V–3
dV = –½V–2 ∞
1| = 0 – (–½) = ½.
(d) Let z = 3
x . Then dz = 3x2
dx and
/2
0
π
∫ x2
cos x3
dx = (1/3) /8
0
3
π
∫ cos
z dz = (1/3) sin z
/8
0
3
π
= (1/3)[sin(π3
/8) – 0] = –0.2233.
1.61 (a) –a–1
cos ax + C.
(b) –a–1
cos ax π
0| = (1 – cos aπ)/a.
(c) Differentiation of the (b) answer gives –a–2
+ a–2
cos aπ + a–1
π sin aπ.
(d) –a/T + C.
1.62 (a) α = (1/Vm)(∂Vm/∂T)P; dVm/Vm = α dT; 2 1 2
1 m m 1 ;V dV dTα−
∫ = ∫
ln(Vm2/Vm1) = α(T2 – T1) at constant P, where the T dependence of α was
neglected over the short range of T. ln(Vm2/18.2334 cm3
mol–1
) =
(4.576 × 10–4
/K)(2.00 K). ln(Vm2/cm3
mol–1
) = ln 18.2334 + 0.0009152
and Vm2 = 18.2501 cm3
/mol.
(b) κ = –(1/Vm)(∂Vm/∂P)T; dVm/Vm = –κ dP; ln(Vm2/Vm1) = –κ(P2 – P1) at
constant T, where the P dependence of κ was neglected.
ln(Vm2/18.2334 cm3
mol–1
) = –(44.17 × 10–6
bar–1
)(199 bar) and Vm2 =
18.074 cm3
/mol.
1.63 (a) Function; (b) number; (c) number.
1.64 In (b) and (c).
1.65 (a) 4 5
/ 2 3 /5 ;x
x e C+ + (b) 7
24x .
13

14. 1.66 (a)
3 2 2
2 1
Δ .
n
ii
x dx x x=
∫ ≈ ∑ For ∆x = 0.1 and xi at the left end of each
subinterval,
2
1
n
ii
x x=
∆∑ = 0.1[22
+ (2.1)2
+ (2.2)2
+ ··· + (2.9)2
] = 6.085.
For ∆x = 0.01, we get 6.30835. For ∆x = 0.001, we get 6.33083. The
exact value is (x3
/3)
3
2 = 27/3 – 8/3 = 6.33333. . . .
(b)
2 2 2 2 2
1 0 (0.01) (0.02) (0.99)
0 0.01[ ]x
e dx e e e e− − − − −
≈ + + + +∫ L = 0.74998.
A BASIC program for part (a) is
10 N = 10 45 X = X + DX
15 FOR J = 1 TO 3 50 NEXT I
20 X = 2 55 PRINT “DELTAX=”; DX; “SUM=”;S
25 DX = 1/N 60 N = 10*N
30 S = 0 65 NEXT J
35 FOR I = 1 TO N 70 END
40 S = S + X*X*DX
1.67 (a) log (4.2 × 101750
) = log 4.2 + log 101750
= 0.62 + 1750 = 1750.62.
(b) ln (6.0 × 10–200
) = 2.3026 log (6.0 × 10–200
) =
2.3026 log 6.0 + 2.3026 log 10–200
= 1.79 – 460.52 = –458.73.
(c) log y = –138.265; y = 10–138.265
= 10–0.265
10–138
= 0.543 × 10–138
.
(d) ln z = 260.433 = 2.3026 log z; log z = 113.10;
z = 100.10
10113
= 1.26 × 10113
.
1.68 (a) 5, since 25
= 32.
(b) 0.
1.69 (a) intensive; (b) extensive; (c) intensive; (d) intensive; (e) intensive;
(f) intensive;
1.70 One finds that a plot of PVm vs. P is approximately linear with an intercept of
PVm = 58.90 L atm/mol at P = 0. The ideal-gas law PVm = RT applies to O2 in
14

15. the limit of zero pressure, so
3
m
30
58900 cm -atm/mol
lim 717.8 K
82.06 cm -atm/mol-KP
PV
T
R→
= = =
1.71 (a) T. (b) F. (c) F. (d) T. (e) F. (f) T. (g) T. (h) T. (i) T. (j) F. (k) F.
15
58.88
58.90
58.92
58.94
58.96
58.98
59.00
59.02
59.04
0 200 400 600 800 1000 1200
PVm/(L-atm/mol)
P/torr