Chapter 1 The properties of gases
Exercises
1.2(b) (a)Could 25 g of argon gas in a vessel of volume 1.5L exert a pressure of 2.0 bar at 30ºC if it behaved as a perfect gas? If not, what pressure would it exert?(b) what pressure would it exert if it behaved as a van der Waals gas?
Solution: (a) The perfect gas law is
pV
nRT
implying that the pressure would be
V
nRT
p
All quantities on the right are given to us except n, which can be computed from the given mass of Ar.
mol
6
62
0
gmol
95
39
g
25
1.
.
n
so10
5
bar
L
5
1
273k)
(30
)
mol
Lbark
10
31
8
mol
626
0
2 -1 -1.
.
.
(
)
.
(
p
not 2.0 bar.(b) The van der Waals equation is
2 m m
V
a
b
V
RT
p
sobar
mol
barK
p
10
.
4
)
6
62
.
0
/
L
5
.
1
(
K
)
273
30
(
)
L
10
31
.
8
(
2 1 2
2
1.5(b) A sample of hydrogen gas was found to have a pressure of 125 kPa when the temperature was 23ºC, what can its pressure be expected to be when the temperature is 11ºC?
Solution:The relation between pressure and temperature at constant volume can be derived from the perfect gas law
nRT
pV
so f f iT
p
Ti
p
T
p
and
The final pressure, then, ought to be
Pa
Pa
T
T
p
p
i f i f120
k
273)K
(23
K
)
273
11
(
)
k
125
(
1.7(b) The following data have been obtained for oxygen gas at 273.15K. Calculate the best value of the gas constant R from them and the best value of the molar mass of O2.
P/atm 0.750 000 0.500 000 0.250 000 Vm/Lmol -1 29.8649 44.8090 89.6384 ρ/(gL-1 ) 1.07144 0.714110 0.356975 Solution:
All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of
pV
m/
T
will give the best value of R.The molar mass is obtained from
RT
M
m
nRT
pV
＝
Which upon rearrangement gives
p
RT
p
RT
ρ
p
RT
V
m
M
＝
The best value of M is obtained from an extrapolation of
p
versus p to p= 0；the intercept is M/RTDraw up the following table
)
mol
K
atm
L
/(
)
/
(
/
-1 -1T
pV
atm
p
m)
atm
L
g
/(
)
/
(
p
-1 -1 0.750 000 0.082 0014 1.428 59 0.500 000 0.082 0227 1.428 22 0.250 000 0.082 0414 1.427 90From Fig. 1.1(a)
From Fig. 1.1(b) 1 -1 0
atm
L
42755
.
1
g
-p
p
1 1 -1 1 -1 -0mol
g
9987
.
31
)
atm
L
42755
.
1
(
K)
15
.
273
(
)
mol
K
atm
L
0615
082
.
0
(
-
g
p
RT
M
p 1 -1 -0mol
K
atm
L
0615
082
.
0
p mT
pV
4
1.8(b) At 100ºC and 120 Torr, the mass density of phosphorus vapour is 0.6388 Kgm-3 . What is the molecular formula of phosphorus under these conditions?
Solution:The mass density
is related to the molar volume Vm by
M
V
m＝
Where
M
is the molar mass. Putting this relation into the perfect gas law yieldsRT
pM
so
RT
pV
m
Rearranging this result gives an expression for M; 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
1 -1 -1 -1
gmol
124
Torr
120
)
(0.6388gL
K]
)
273
100
[(
mol
K
LTorr
364
.
62
p
RT
M
The number of atoms per molecule is
00
.
4
gmol
0
.
31
gmol
124
1 1
1.10(b) A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300K is 66.5 Torr. Calculate(a) the volume and (b) the total pressure of the mixture. Solution: (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V from eqn 14 we have (assuming a perfect gas)
L
3.14
L
7
13
.
3
66.5Torr
K)
300
(
)
mol
K
LTorr
36
.
62
(
mol)
10
5
11
.
1
(
K
300
Torr,
5
.
66
mol,
10
5
11
.
1
20.18gmol
g
225
.
0
n
1 -1 -2 Ne 2 1 -Ne
＝
V
T
p
p
RT
n
V
J J(b) The total pressure is determined from the total amount of gas, n=nCH4+nAr+nNe.
Torr
212
L
7
3.13
K)
300
(
)
mol
K
LTorr
36
.
62
(
mol)
10
8
3.54
(
[1]
p
mol
10
8
3.54
mol
10
)
5
11
.
1
438
.
0
5
1.99
(
mol
10
38
.
4
mol
g
39.95
g
0.175
mol
10
5
1.99
mol
g
16.04
g
0.320
1 -1 -2 2 2 3 1 2 -1 4
V
RT
n
n
n
n
J Ar CH6
1.13(b) Determine the ratios of (a) the mean speeds, (b)the mean kinetic energies of He atoms and Hg atoms at 25ºC.
Solution:(a) The mean speed of a gas molecule is
079
.
7
003
.
4
59
.
200
)
He
(
)
Hg
(
)
Hg
(
)
He
(
8
2 1 2 1 2 1
M
M
c
c
so
M
RT
c
(b) The mean kinetic energy of a gas molecule is 2
2
1
m c
,where c is the root mean square speed2 1
3
M
RT
c
So 22
1
1.14(b) The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25ºC and assuming that air consists of N2 molecules with a collision diameter of 395 pm, calculate (a) the mean speed of the
molecules, (b) the mean free path, (c) the collision frequency in the gas.
Solution: (a) The mean speed can be calculated from the formula derived in Example 1.6 2 1
8
M
RT
c
= 2 1 2 1 1 -3 1 -1-ms
10
75
.
4
mol
kg
10
02
.
28
(
K)
298
(
)
mol
JK
314
.
8
8
（
c
(b) The mean free path is calculated from
[
33
]
2
12p
kT
With
d
2
(
3
.
95
10
10m
)
2
4
.
90
10
19m
2 Then,m
10
4
atm
1
Pa
10
1.013
Torr
760
atm
1
Torr)
10
1
(
)
m
10
90
.
4
(
2
K)
298
(
)
JK
10
381
.
1
5 4 9 2 19 2 1 -1 23
（
(c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32, since
andc
have already been calculated 41
10
2s
1m
10
46
.
4
75
.
4
c
c
z
Thus there are 100s between collisions, which is a very long time compared to the usual timescale apparatus used to generate the very low pressure.
8
1.16(b) At an altitude of 15 km the temperature is 217 K and the pressure 12.1 kPa. What is the mean free path of N2 molecules? (σ=0.43nm2)
Solution: The mean free path is
4
.
1
10
m
)
atm
Pa
10
1
.
12
(
m)
(10
0.43
2
)
(217K
)
JK
10
381
.
1
2
7 1 3 2 9 2 1 1 23 2 1
＝
（
p
kT
1.17(b) How many collisions per second does an N2 molecule make at an altitude of 15 km?(See Exercise
1.16b for data.)
Solution: Obtain data from Exercise 1.17(a) is
p
m kT
z
2 16
1
Substituting
0
.
43
nm
2,
p
12
.
1
10
3Pa
,
m
(
28
.
02
u
),
andT
217
K
we obtain
27 23 1
12 3 2 18)
K
217
(
K
J
10
381
.
1
k g)
10
6605
.
1
(
)
02
.
28
(
)
Pa
10
1
.
12
(
)
m
10
43
.
0
(
4
z
=9.9×108s-11.21(b) Estimate the critical constants of a gas with van der Waals parameters a=1.32 atmL2mol-2 and b=0.0436Lmol-1.
Solution: The critical constants of a van der Waals gas are
K
109
)
mol
L
(0.0436
)
K
atm
L
27(0.08206
)
mol
8(1.32atmL
27
8
and
atm
7
25
K
atm
08206L
0
27
mol
L
atm
32
1
27
Lmol
131
0
Lmol
0436
0
3
3
1 1 2 2 2 1 2 -2 2 1 1 c
Rb
a
T
.
)
.
(
.
b
a
p
.
)
.
(
b
V
c c1.22(b) A gas at 350K and 12 atm has a molar volume 12 per cent larger than calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces?
Solution: The compression factor is
mol
L
7
2
atm
12
K)
(350
)
mol
K
atm
L
(0.08206
(1.12)
(1.12)
(1.12)
is
volume
molar
The
(b)
dominate.
forces
Repulsive
12
1
have
we
0.12
12
0
Beacuse
(a)
1 1 -1 -m
.
V
p
RT
V
V
.
Z
V
V
.
V
V
V
V
RT
pV
Z
perfect . m perfect . m perfect . m perfect . m perfect . m m m，
）
（
10
1.24(b) The density of water vapour at 1.00 bar and 383 K is 0.5678 kg m-3.(a) Determine the molar volume Vm of water and the compression factor Z, from these data. (b) Calculate Z from the van der Waals
equation with a=5.536L2atm mol-2 and b=0.03049L mol-1.
Solution: (a)
4
995
0
mol
03049L
0
mol
L
8
31.72
mol
atm
536L
5
mol
L
03049
0
mol
L
728
31
mol
L
8
31.72
get
we
above
for
expression
the
into
ing
subsititut
and
p
Using
(b)
0.9963
K)
(383
)
mol
K
bar
L
(0.083145
)
mol
L
8
(31.72
bar)
(1.00
mol
L
8
72
31
mol
g
5678
0
mol
g
015
18
1 -1 -2 2 1 -1 -1 -2 1 1 1 1 1 1.
.
.
.
.
RT
V
a
b
V
V
Z
Z
V
a
b
V
RT
RT
pVm
Z
.
.
.
ρ
M
V
m m m m m m
Comment. Both values of Z are very close to the perfect gas value of 1.000, indicating that water vapour is
1.25(b) At 300 K and 20 atm, the compression factor of a gas is 0.86. Calculate (a) the volume occupied by 3.2 mmol of the gas under these conditions and (b) an approximate value of the second virrial coefficient B at 300 K. Solution:
-1 1 -3 -1 -3 -1 1mol
L
15
0
1
86
0
)
mol
L
59
(1.0
1
1
expansion
series
the
of
truncation
by
36
eqn
from
obtained
be
can
of
value
e
approximat
An
(b)
mL
8.7
L
10
8.7
)
mol
L
59
(1.0
mol)
10
(8.2
Then,
(a)
mol
L
59
0
1
atm
20
K)
(300
)
K
atm
L
0.08206
yields
which
for
34
solving
by
obtained
is
volume
molar
The
.
.
Z-V
RT
pV
V
B
B
nV
V
.
p
ZRT
V
V
,
RT
pV
Z
m m m m m m m
，
12
1.27(b) The critical constants of ethane are pc=45.6 atm, Vc=148 cm 3
mol-1, and Tc=305.4 K. Calculate the
van der Waals parameters of the gas and estimate the radius of the molecules. Solution:
be.
should
it
than
lower
cent
per
25
about
is
computed
our
suggesting
K,
305.4
is
reported
the
However,
K
231
)
mol
L
(0.0493
mol
K
atm
L
0.08206
27
mol
atm
L
3.16
8
be
should
,
determined
already
have
we
constants
the
to
According
27
8
n
informatio
of
piece
another
have
We
ined.
overdeterm
is
problem
this
But
mol
atm
L
16
3
mol
L
0493
0
atm
20
48
27
27
so
27
is
pressure
critical
The
m
10
1.94
cm
10
347
1
)
mol
10
4ππ(6.02
)
mol
cm
3(49.3
2
1
4
3
2
1
so
3
2
4
volume
excluded
molar
the
is
constant
Avogadro
the
times
volume
that
radius);
their
twice
(i.e.,
particles
spherical
those
of
diameter
the
is
radius
whose
sphere
from
excluded
are
particles
spherical
of
centres
The
size.
molecular
of
estimate
an
obtain
can
we
molecules,
spherical
of
mole
a
of
volume
cxcluded
the
as
ng
interpreti
By
mol
L
0493
0
mol
3cm
49
mol
cm
148
3
1
3
1
b
so
3
is
gas
Waals
der
van
a
of
volume
critical
The
1 1 1 2 2 2 2 2 1 2 2 10 -8 3 1 1 23 1 3 3 1 3 1 1 3 1 3a/b
T
T
T
Rb
a
T
.
.
.
b
p
a
b
a
p
.
r
πN
b
r
r
π
N
b
b
a
b
.
.
V
b
V
c c c c c c A A c c
1.29(b) Suggest the pressure and temperature at which 1.0 mol of (a) H2S, (b)CO2, (c) Ar will be in states
that correspond to 1.0 mol N2 at 1.0 atm and 25ºC.
Solution:
K
56
3
(150.72K)
(2.36)
atm
4
1
(48.00atm)
(0.030)
Ar
For
(c)
K
718
(304.2K)
(2.36)
atm
2.2
atm)
(72.85
(0.030)
CO
(b)For
)
from
obtained
S
H
of
constants
(Critical
K
881
K)
(373.2
(2.36)
atm
2.6
atm)
(88.3
(0.030)
S
H
For
(a)
are
states
ing
correspond
The
36
2
K
126.3
K
273)
(25
and
030
0
atm
33.54
atm
1.0
are
C
25
and
atm
1.0
at
N2
for
e
temperatur
and
pressure
reduced
The
.
correspond
to
said
are
volume
and
e,
temperatur
pressure,
reduced
same
the
have
that
States
2 2 2
c r c r c r c r c r c r r c rT
T
T
.
p
p
p
T
T
T
p
p
p
ics.
y and phys
f chem istr
handbook o
T
T
T
p
p
p
.
Tc
T
T
.
p
p
p
14
1.30(b) A certain gas obeys the van der Waals equation with a=0.76 m6 Pa mol-2. Its volume is found to be 4.00×10-4 m3 mol-1 at 288 K and 4.0Mpa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure?
Solution:
0.67
K
288
mol
K
J
8.3145
mol
m
10
4.00
Pa
10
4.0
is
factor
n
compressio
The
mol
m
10
.00
4
for
solved
be
can
which
is
equation
Waals
der
van
The
1 1 3 4 6 1 3 4 2 2 1
RT
pV
Z
V
a
p
RT
V
b
b
V
a
b
V
RT
p
m m m m mProblems
Numerical problems
1.4 A meterological balloon had a radius of 1.0 m when released at sea level at 20°C and expanded to a radius of 3.0 m when it had risen to its maximum altitude where the temperature was -20°C. What is the pressure inside the balloon at that altitude?
Solution:
1.0
atm
3.2
10
atm
K
293
K
253
3.0m
1.0m
3
4
3
4
3
4
and
3
4
ng
Substituti
yields
altitude,
maximum
its
at
pressure
the
,
for
Solving
constant,
n
with
that,
implies
[12]
2 3 3 3 3 3 3
i i f f i i i f f i f f f i i i i f f i f f i i i f f fp
T
T
r
r
p
T
T
r
r
p
r
V
r
V
p
T
T
V
V
p
p
T
V
p
T
V
p
nRT
pV
16
1.10 A vessel of volume 22.4 L contains 2.0 mol H2, and 1.0 mol N2 at 273.15 K initially. All the H2 reacted
with sufficient N2 to form NH3. Calculate the partial pressure and the total pressure of the final mixture.4
Solution:
table
following
the
up
draw
can
We
2NH
3H
N
is
equation
balanced
The
.
completion
to
gone
has
reaction
the
after
remains
H2
no
that
assume
We
3 2 2
N2 H2 NH3 Total Initial amount nn
'
0n
n
'
Final amount'
n
n
3
1
0n
'
3
2
'
n
n
3
1
Specifically 0.33mol 0 1.33 mol 1.66 mol
Mole fraction 0.20 0 0.80 1.00
0.80
1.66atm
0.33atm ) (NH ) (NH atm 0.33 atm 1.66 0.20 ) (N ) (N 0 ) (H ) (H atm 6 6 . 1 L 22.4 K 273.15 mol K atm L 10 8.206 1.66mol 3 3 2 2 2 2 1 1 2 p x p p x p p x p V nRT p1.15 Calculate the molar volume of chlorine gas at 350 K and 2.30 atm using (a) the perfect gas law and (b) the van der Waals equation. Use the answer to (a) to calculate a first approximation to the correction term for attraction and then use successive approximations to obtain a numerical answer for part(b).
Solution:
.
terminated
be
may
ion
approximat
of
cycle
the
so
mol
L
3
12
in
results
again
expression
first
the
of
r
denominato
the
into
mol
L
12.3
of
on
Substituti
mol
L
3
12
mol
L
10
622
5
2.34
mol
L
2
7
28
mol
L
10
622
5
)
mol
L
(12.5
mol
atm
6.579L
atm)
(2.30
K)
)(350
mol
K
atm
L
10
(8.206
1.6
Table
from
b
and
a
with
Then
9b
rearrange3
obtain
we
,
39b
(b)From
mol
L
5
12
atm
2.30
K)
)(350
mol
K
atm
L
10
(8.206
(a)
1 1 1 1 2 1 1 2 2 1 2 2 1 1 2 2 2 1 1 1 2,
.
V
.
.
.
.
V
,
b
V
a
p
RT
V
V
a
b
V
RT
p
.
p
RT
V
m m m m m m m
18
Chapter 2 The First Law：the concepts
Exercises
2.4 (b) A sample consisting of 2.00 mol He is expanded isothermally at 22℃ from 22.8 L to 31.7 L (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely(against zero external pressure). For the three processes calculate q,w,△U,and △H.
Solution：For a perfect gas at constant temperature
-w
q
U
0
so
For a perfect gas at constant temperature,
H
is also zero)
d(
d
H
U
pV
we have already noted that U 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
J
10
62
1
10
62
1
L
22.8
L
31.7
ln
K
273)
(22
)
mol
K
(8.3145J
mol)
-(2.00
ln
3 3 1 -1-
.
w
q
.
V
V
nRT
w
i f(b) Expansion against a constant external pressure
V
p
w
ex
Where
p
ex in this case can be computed from the perfect gas lawJ
10
38
1
J
10
38
1
m
L
1000
22.8)L
-(31.7
Pa)
10
(1.55
-and
Pa
10
1.55
)
Lm
1000
(
L
31.7
K
273)
(22
)
mol
K
(8.3145J
mol)
(2.00
p
3 3 3 5 5 3 1 -1
-
.
w
q
.
w
so
nRT
pV
(c)2.6(b) A sample of argon of mass 6.56 g occupies 18.5 L at 305 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 2.5 L. (b) Calculate the work that would be done if the same expansion occurred reversibly.
Solution :
J
-52.8
L
18.5
L
18.5
2.5
ln
K
305
mol
K
J
3145
8
mol
g
39.95
g
6.56
ln
(b)
J
19
m
L
1000
2.5L
Pa
10
7.7
1 -1 -1 -3 3
.
V
V
-nRT
w
V
p
w
a
i f ex20
2.8(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64℃. The
standard enthalpy of vaporization of methanol at 64℃ is 35.3 kJ mol-1. Find q,w,△U,and △H for this process.
Solution:
q
ΔH
n
(
Δ
vapH
Θ)
2.00
mol
35
.
3K
J
mol
1
70
.
6
kJ
Because the condensation also occurs at constant pressure, the work isV
p
V
p
w
exd
The change in volume from a gas to a condensed phase is approximately equal in magnitude to the volume of the gas
k J
0
65
k J
60
5
6
70
J
10
60
5
K
273
64
mol
K
kJ
3145
8
00mol
2
V
-
vapour -1 -1 3.
-)
.
.
(-w
q
ΔU
.
.
.
nRT
p
w
（
）
2.11(b) The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp,m/(JK-1)=20.17+0.4001(T/K). Calculate q,w,△U,and △H for 1.00 mol when
the temperature of 1.00 mol of gas is raised from 0℃ to 100℃ (a) at constant pressure (b) at constant volume. Solution ：
k J
1
14
so
and
0
volume,
constant
At
above.
as
k J
1
14
and
k J
9
14
Thus,
gases.
perfect
in
e
temperatur
on
only
depend
and
(b)
k J
14.1
k J
0.831
-14.9
J
831
100
mol
3145J
8
mol
00
1
J
10
9
14
J
273
373
4001
0
2
1
273
373
17
20
K
J
K
T
4001
0
2
1
T
17
20
K
J
dT
K
T
4001
0
17
20
CpdT
q
pressure
constant
(a)At
1 1 3 2 2 373K 273K 1 2 273K 100 273K 0 1.
q
,
q
ΔU
w
.
ΔU
.
ΔH
ΔH
ΔU
w
q
ΔU
K
.
.
nRΔR
pΔΔ
w
H
.
.
.
/
.
.
/
)
.
(
.
Ｋ
22
2.13(b) A sample of nitrogen of mass 3.12 g at 23.0℃ is allowed to expand reversibly and adiabatically from 400 mL to 2.00 L. What is the work done by the gas?
Solution: Reversible adiabatic work is
p,m
f i
V
T
n
C
R
T
T
C
w
Where the temperatures are related by [solution to Exercise2.12b]
29.125
8.3145
J
K
mol
156
296
K
325
J
mol
g
28.0
g
3.12
and
K
156
L
2.00
L
10
400
K
273.15
23.0
T
So
503
2
mol
K
J
8.3145
mol
K
J
8.3145)
(29.125
where
1 1 1 -2.503 1 3 f 1 1 1 1 1
w
.
R
R
C
R
C
c
V
V
T
Tf
p,m v,m c f i i2.15(b) Calculate the final pressure of a sample of water vapour of mass 1.4 g that expands reversibly and adiabatically from an initial temperature of 300 K and volume 1.0 L to a final volume of 3.0 L. Take γ=1.3. Solution: For reversible adiabatic expansion
f i i f i i f fV
V
p
p
V
p
v
p
so
We need
p
i , which we can obtain from the perfect gas law
0.46
atm
L
3.0
L
1.0
atm
1.9
atm
1.9
L
1.0
K
300
mol
K
atm
L
0.08206
mol
g
18
g
1.4
so
1.3 1 1 1
f ip
p
V
nRT
p
nRT
pV
24
2.19(b) When 2.0 mol CO2 is heated at a constant pressure of 1.25 atm, its temperature increases from 250
K to 277 K. Given that the molar heat capacity of CO2 at constant pressure is 37.11 JK -1 mol-1, calculate q,△U,and △H . Solution:
1 1
3 1 1 3 1 3 1 1mol
J
10
1.6
K
250
277
mol
K
J
8.3145
mol
2.0
mol
J
10
2.0
so
mol
J
10
2.0
K
250
277
mol
K
J
37.11
mol
2.0
ΔU
nRΔR
ΔH
ΔU
nRΔR
ΔU
pV
Δ
ΔU
ΔH
ΔT
nC
ΔT
C
q
ΔH
p p p,m2.21 (b) A sample consisting of 2.5 mol of perfect gas at 220 K and 200 kPa is compressed reversibly and adiabatically until the temperature reaches 255 K. Given that its molar constant-volume heat capacity is 27.6 JK-1 mol-1, calculate q,w,△U,and △H,and the final pressure and volume.
Solution: For adiabatic compression, q=0 and
2.5
mol
8.3145
J
K
mol
255
220
K
3.1
10
J
J
10
2.4
J
10
2.4
J
10
2.4
K
220
255
mol
J
K
27.6
mol
2.5
3 1 -1 3 3 3 1 1
nRΔR
ΔU
pV
Δ
ΔU
ΔH
w
q
ΔU
T
C
w
VThe initial and final states are related by
