Chapter 1 The properties 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 30 ºC 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 pV nRT nRT implying that the pressure would be
nRT nRT
p
V
All quantities on the right are given to us except n, which can be computed from the given mass of Ar.
25 g
n
1
39 .95 gmol
0.62 6 mol 2
so p
-1
-1
( 0 .626 626 mol ) ( 8 .31 10 Lbark mol ) (30 (30
273k)
1 .5L
10 . 5bar
not 2.0 bar. (b) The van der Waals equation is
p
so p
RT V m
b
(8.31
a V m
2
10
2
LbarK
1
)
( 30
(1.5L / 0.626mol )
2
273 )K
10 .4bar
1
1.5(b) A sample 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 pV
nRT nRT so p
and T and
pi
p f
Ti
T f
The final pressure, then, ought to be
p f
pi T f T i
(125 k Pa Pa ) (23 (23
(11
273 273 )K
273)K
120 120 k Pa Pa
2
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 0.500 0.250 000 000 000 -1 V m/Lmol 29.8649 44.8090 89.6384 -1 1.07144 0.714110 0.356975 ρ/(gL ) Solution: pV m / T will give All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pV
the best value of R. R.
m nRT ＝ RT RT M
The molar mass is obtained from pV Which upon rearrangement gives M ＝
m RT RT V p
RT RT ρ p
The best value of M M is obtained from an extrapolation of
RT RT p
p
versus p to p= 0；the intercept is M/RT
Draw up the following table
pV m / T ) /( L atm ( pV atm K -1 mol -1 )
p / atm atm
(
-1
/ p) /( g L atm
0.750 000 0.500 000
0.082 0014 0.082 0227
1.428 59 1.428 22
0.250 000
0.082 0414
1.427 90
From Fig. 1.1(a)
pV m T
-1
0.08 082 2 0615 L atm K mol
p
M
)
-1
p 0 -
From Fig. 1.1(b)
-1
1.42755 g L 1 atm
-1
p 0
RT
p
( 0.082 0615 L atm K -1 mol
-1
)
( 273 .15 K)
-
(1.42755 g L 1 atm
p 0
31 .9987 g mol
1
R deviates from the accepted value by 0.005 per cent. The error results from the fact The value obtained for R
that only three data points are available and that a linear extrapolation was employed. The molar mass, however, however, agrees exactly with the accepted value, probably because of compensating plotting errors. THE PROPERTIES OF GASES
3
-1
)
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
V m＝
is related to the molar volume V m by
M
Where M is the molar mass. Putting this relation into the perfect gas law yields pV m
RT so
pM
RT
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 M
-1
RT
62 .364 LTorr K mol
p
-1
[( 100
120 Torr
273 )K]
(0.6388gL
-1
)
124 gmol
1
The number of atoms per molecule is 124 gmol 31 .0gmol
1 1
4.00
4
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)
V
n J RT
0.225 g
n Ne
p J
20.18gmol 1.11 5
V
(1.115
10
2
mol)
10
-1 2
mol,
pNe
66.5 Torr,
( 62.36LTorr K -1 mol -1 )
T
( 300 K)
66.5Torr
300 K
＝3.137 L
3.14 L
(b) The total pressure is determined from the total amount of gas, n=nCH4+nAr +n Ne.
nCH 4
0.320 g 16.04 g mol
n
(1.995
0.438
p
n J RT [1] V
-2
1
1.995 10 mol 2
1.115) 10 mol 2
( 3.548 10 mol)
n Ar
0.175 g 39.95 g mol
3
1
4.38 10 mol
2
3.548 10 mol -1
-1
(62.36LTorr K mol ) 3.137L
( 300K)
212 Torr
5
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 8 RT
c
1
2
M 1
so
c( He )
M ( Hg )
c ( Hg )
M ( He )
2
200 .59
c
So
1
2
7.079
4.003
(b) The mean kinetic energy of a gas molecule is
3 RT
1
1 2
mc
2
,where c is the root mean square speed
2
M
1 2
mc 2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is 1.
6
1.14(b) The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25ºC and assuming that air consists of N 2 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 1
8 RT
c
M
2
8 （8.314 JK mol -1
= c
( 28 .02
-1
3
10
1
)
kg mol
kT (b) The mean free path is calculated from
2 With
2
d
( 3.95
10
10
m)
2
1
2
4.90
( 298 K)
2
4.75
-1
10
2
ms
1
[33]
p 10
19
m
2
Then,
（1.381 10 2
1
2
(4.90 10
19
m2 )
23
-1
JK )
(1 10 9 Torr)
( 298 K)
4
1 atm
1.013
760 Torr
5
10 Pa
4 10 m
1 atm
(c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32, sin ce and c have already been calculated z
c
4.75c 4
1
10
2
s
1
4.46 10 m 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.
7
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
kT 2
1
2
（1.381 10
＝
p
2
1
2
23
JK 1 ) (217K )
0.43 (10 9 m)2
4.1 10 7 m
(12.1 10 3 Pa atm 1 )
1.17(b) How many collisions per second does an N 2 molecule make at an altitude of 15 km?(See Exercise 1.16b for data.) Solution: Obtain data from Exercise 1.17(a) is z Substituting
2
0.43 nm , p
12 .1
1
16 mkT 3
10 Pa , m
2
p
( 28 .02 u), andT
217 K
we
obtain
4 (0.43 10
z
( 28.02) (1.6605 10
27
18
2
3
m ) (12.1 10 Pa)
kg)
1.381 10
23
J K
1
( 217 K )
1
2
=9.9×108s-1
8
1.21(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
V c pc
3b
3( 0 .0436 Lmol
1
)
0 .131 Lmol
a
1 .32 atm L2 mol -2
27 b 2
27 ( 0 .08206L atm K 1 )2
25 .7 atm
8(1.32atmL 2 mol
8a
and T c
1
1
27 Rb
27(0.08206 L atm K )
2
)
(0.0436 L mol
1
)
109 K
1.22(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 Z
pV m
V m
RT
V m . perfect
(a) Beacuse Z
V m
1 .12 Repulsive
(b) The molar V V
V m . perfect
volume
(1.12) V m . perfect
(1.12)
(0.08206
（0.12 ）V m . perfect ， we have
0 .12 V m . perfect
forces
dominate.
is (1.12)
RT p
L atm K -1 mol 12 atm
-1
)
(350 K)
2 .7 L mol
1
9
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 V m 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:
V m Z
(a)
M ρ
1
0 .5678 g mol
1
pVm
(0.083145 L bar K mol
V m
RT V m
-
b
a V m
2
1
(31.72 8 L mol 1
RT
V m
31 .728 L mol
(1.00 bar)
(b) Using p Z
18 .015 g mol
1
)
1
)
(383 K)
0.9963
and subsititut ing into the expression for Z above we get
a
b V m RT
31.72 8 L mol -1 -1 31 .728 L mol
-1 0 .03049 L mol
5 .536L 2 atm mol 31.72 8 L mol -1
2
0 .03049L mol -1
0 .995 4
Comment. Both values of Z are very close to the perfect gas value of 1.000, indicating that water vapour is
essentially perfect at 1.00 bar pressure.
10
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: The molar volume is obtained by solving Z V m
0.08206 L atm K 1 )
ZRT p
(300 K)
20 atm
(a) Then, V
nV m
(8.2
10 - 3 mol)
pV m RT
34 , for V m，which yields
1 .059 L mol
(1.0 59 L mol -1 )
8.7
1
10 -3 L
8.7 mL
(b) An approximat e value of B can be obtained from eqn 36 by truncation of the series expansion B
V m
pV m RT
1 -1
(1.0 59 L mol )
V m
Z-1
0 .86
1
0 .15 L mol
-1
11
1.27(b) The critical constants of ethane are pc=45.6 atm, V c=148 cm3 mol-1, and T c=305.4 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules. Solution: The critical volume of a van der Waals gas is 3b
V c
1
so b
1
V c
148 cm 3 mol
1
49 .3cm 3 mol
1
0 .0493 L mol
1
3 3 By interpreti ng b as the cxcluded 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 is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b
r
4 π 2r
N A
3
1
3
so r
3(49.3 cm 3 mol
1
2 4ππ(6.02
23
1
3b
3
2 4 πN A 1
1
)
10 mol
1
3
1 .347
)
10
8
cm
1.94
10 -10 m
The critical pressure is pc
so a
a
27 b 2 27 pc b 2
27 48 .20 atm
0 .0493 L mol
1 2
3 .16 L2 atm mol
2
But this problem is overdeterm ined. We have another piece of informatio n T c
8a
27 Rb According to the constants we have already determined , T c should be
T c
8 3.16 L2atm mol 27 0.08206 L atm K 1mol
1
2
(0.0493 L mol
1
)
231 K
However, the reported T c is 305.4 K, suggesting our computed a/b is about 25 per cent lower than it should be.
12
1.29(b) Suggest the pressure and temperature at which 1.0 mol of (a) H 2S, (b)CO2, (c) Ar will be in states that correspond to 1.0 mol N2 at 1.0 atm and 25ºC. Solution:
States that have the same reduced pressure, temperatur e, and volume are said to correspond . The reduced pressure and temperatur e for N2 at 1.0 atm and 25 C are pr
p
1.0 atm
pc
33.54 atm
0 .030
T
and T r
Tc
(25
273) K
126.3 K
2 .36
The correspond ing states are (a) For H 2S p
pr pc
T
T r T c
(0.030) (2.36)
(88.3 atm) (373.2 K)
2.6 atm 881 K
(Critical constants of H 2S obtained from handbook o f chemistr y and phys ics. ) (b)For CO 2 p
pr pc
(0.030)
T
T r T c
(2.36)
p
pr pc
(0.030)
T
T r T c
(2.36)
(72.85 atm) (304.2K)
2.2 atm
718 K
(c) For Ar (48.00atm) (150.72K)
1 .4 atm 356K
13
1.30(b) A certain gas obeys the van der Waals equation with a=0.76 m 6 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:
The van der Waals equation is p
RT V m
b
a V m
2
which can be solved for b b
V m p
RT a V m
4.00
10
4
3
m mol
1
2
The compressio n factor is Z
pV m RT
4.0
6
10 Pa
4.00
8.3145 J K 1 mol
10 1
4
3
m mol 288 K
1
0.67
14
Problems 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:
pV
nRT [12] implies that, with n constant,
p f V f
pi V i
T f
T i
Solving for p f , the pressure at its maximum altitude, yields p f Substituting V i 4 p f
4
3
3
3
r i r f
3
4
3
3
r i
T f T i
4
and V f pi
r i
r f
3 3
r f 1.0m 3.0m
T f T i 3
V i
T f
V f
T i
pi
3
pi 253 K 293 K
1.0 atm
3.2 10
2
atm
15
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: We assume that no H2 remains after the reaction has gone to completion . The balanced equation is N2
3H 2
2NH
3
We can draw up the following
N2 n
Initial amount Final amount
n
p
V
H2 1
0
n'
1.66mol
n' 3 1.33 mol 0.80
10
2
L atm K 1 mol 22.4 L
x (H 2 ) p
0
p(N 2 )
x (N 2 ) p
0.20
x (NH 3 ) p
2
0 0
8.206
p(H 2 ) p(NH 3 )
NH3 0
n'
3 0.33mol 0.20
Specifically Mole fraction
nRT
table
1.66 atm
0.80
1
273.15 K
Total n n' 1 n n' 3 1.66 mol 1.00
1.66 atm
0.33 atm
1.66 atm
0.33 atm
16
1.15 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: (a)V m
RT
(8.206
10 2 L atm K 1 mol
p
(b)From
p
1
)(350 K)
12 .5 L mol
2.30 atm RT V m
b
a V m
2
RT
39b , we obtain V m
1
b rearrange3 9b
a
p
V m
2
Then , with a and b from Table 1.6 V m
(8.206
10 2 L atm K 1 mol
1
)(350 K)
2
(2.30 atm) 28 .7 2L mol 2.34
6.579L atm mol (12.5 L mol
1
2
5 .622
10
2
L mol
1
)2
1
5 .622
10
2
L mol
1
12 .3 L mol
1
Substituti on of 12.3 L mol 1 into the denominato r of the first expression again results in V m
12 .3 L mol 1 , so the cycle of approximat ion may be terminated .
17
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 U
0 so q
-w
For a perfect gas at constant temperature, d H
d( U
H is also zero
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 w
nRT ln
V f V i -1
-(2.00 mol) q
w 1 .62
(8.3145J K mol
-1
)
(22
273) K ln
31.7 L
1 .62
22.8 L
10
3
3
10 J
(b) Expansion against a constant external pressure w
pex V
Where pex in this case can be computed from the perfect gas law
pV
nRT (2.00 mol)
so p
(22
273) K
31.7 L
and w q
(8.3145J K -1 mol -1 )
- (1.55
w 1 .38
5
10 Pa)
(31.7 - 22.8)L
1000 L m
3
(1000 Lm
3
)
1.55
10 5 Pa (c)
1 .38
3
10 J
10 3 J
Free expansion is expansion against no force, so w 0 , and q
-w
0 as well.
18
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 :
a w
pex V
(b) w
-nRT ln -
7.7
3
10 Pa 1000 L m
2.5L
19 J
3
V f V i
6.56 g 39.95 g mol
-1
-1
8 .3145 J K mol
-1
305 K
ln
2.5
18.5 L
18.5 L
-52.8 J
19
2.8(b) A sample of 2.00 mol CH 3OH(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 vap H Θ )
2.00 mol
35 .3K J mol
1
70.6 kJ
Because the condensation also occurs at constant pressure, the work is w
pex dV
p V
The change in volume from a gas to a condensed phase is approximately equal in magnitude to the volume of the gas w ΔU
p（- Vvapour） nRT q
w
(- 70 .6
2 .00mol
5 .60 )kJ
-1
8 .3145 kJ K mol
-1
64
273 K
5 .60
3
10 J
-65 .0kJ
20
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 C p,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
：
(a)At constant pressure q
CpdT 20 .17 ) T
( 20 .17 ) w ΔU
pΔΔ q
w
100
273K
0 273K
1 2
20 .17
(0 .4001
373
273
nRΔR
( 0 .4001 )T / K dT J K 2
T / K 1 2
0 .4001
1 .00 mol
14.9 - 0.831 kJ
J K
373K
1
373
1
273K
2
273
2
J
8 .3145J Ｋ mol 1
14 .9 1
3
10 J
100 K
H 831 J
14.1 kJ
(b) ΔU and ΔH depend only on temperatur e in perfect gases. Thus, ΔH as above. At constant volume, w
0 and ΔU
q , so q
14 .9kJ and ΔU
14 .1kJ
21
14 .1kJ
2.13(b) A sample of nitrogen of mass 3.12 g at 23.0℃ is allowed to expand reversibl y and adiabatically from 400 mL to 2.00 L. What is the work done by the gas? Solution: Reversible adiabatic work is
w C V T
n C p ,m
R
T f
T i
Where the temperatures are related by [solution to Exercise2.12b] 1 c
Tf
T i
V i V f
where c So Tf and w
C v,m
C p,m
R
23.0
R
(29.125
8.3145 J K 1 mol
R
273.15 K
3.12 g 28.0 g mol
-1
8.3145) J K 1 mol
400
29.125
10
2.00 L
3
L
1
2 .503
1
1 2.503
156 K
8.3145 J K 1 mol
1
156
296 K
325 J
22
2.15(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
p f v f
pi V i
so p f
pi
V i V f
We need pi , which we can obtain from the perfect gas law
pV
nRT
so
1.4 g pi p f
18 g mol
1
nRT
p
V
0.08206 L atm K 1 mol 1.0 L
1.9 atm
1.0 L 3.0 L
1
300 K 1.9 atm
1.3
0.46 atm
23
2.19(b) When 2.0 mol CO 2 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 CO 2 at constant pressure is 37.11 JK -1 mol-1, calculate q,△U ,and △ H . Solution: ΔH 2.0
q p
C p ΔT 3
10 J mol
2.0 mol
1
37.11 J K mol
1
277
250 K
1
ΔH
ΔU Δ pV
ΔU
2.0
3
nC p,m ΔT
10 J mol
ΔU 1
nR ΔR 2.0 mol
so
ΔU
ΔH 1
8.3145 J K mol
1
nR ΔR 277
250 K
1.6
3
10 J mol
24
1
2.21 (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 w
C V T
2.5 mol
w
q
2.4
ΔH
ΔU Δ pV 3
10 J
mol
1
255
220 K
2.4
3
10 J
3
ΔU
2.4
1
27.6 K J
10 J ΔU
2.5 mol
nRΔR 1
8.3145 J K mol
-1
255
220 K
3.1
3
10 J
The initial and final states are related by c
V f T f
c
c
V i T i
where c V i
V f p f
so
V f
T f -1
1
C V,m
27.6 J K mol
R
8.314 J K mol
nRT i
1
8.3145 J K mol 200
0.022 9 m nRT f
3
3.32
1 1
2.5 mol
pi
V f
V i
T i
220 K 255 K
2.5 mol
1
220 K
3
10 Pa
0.022 9 m
3
3.32
0.014 m 1
8.3145 J K mol 0.014 m
3
3
1
14 L 255 K
3 .8
5
10 Pa
25
2.24(b) Consider a system consisting of 3.0 mol O 2 (assumed to be a perfect gas) at 25℃ confined to a cylinder of cross-section 22 cm 2 at 820 kPa. The gas is allowed to expand adiabatically and irreversibly against a constant pressure of 1.0 atm. Calculate q,w,△U ,and △ H and △T when the piston has moved 15 cm. Solution: In an adiabatic process, q=0. Work against a constant external pressure is w
pex V
ΔU
q
w ΔT
110
w
3
10 Pa
15 cm
22 cm
(100 cm m
)
3
36 J
36 J
C V ΔT
n C p,m -R ΔT
so
w n C p,m -R 36 J
3.0 mol H
1
2
29 .355
ΔU Δ pV
36 J
3.0 mol
-1
8 .3145 J K mol ΔU
-1
0.57 K
)
0.57 K
nRΔR -1
(8.3145 J K mol
-1
50 J
26
2.27(b) The standard enthalpy of formation of phenol -165.0 kJ mol -1. Calculate its standard enthalpy of combustion. Solution: The reaction is
C 6 H 5 OH c
H 6
6
7O 2
6CO 2
H CO2
f
393 .51
285 .83
3
3053.6 kJ mol
3
3H 2 O H H 2 O
f
165 .0
H C 6 H 5 OH
f
7 0 K J mol
7
H O 2
f
1
1
27
Θ 2.29(b) From the following data, determine Δ f H for diborane, B2H6(g), at 298 K:
1 B2H6 g
3O 2 g 3
2 2B s
2 1
3 H2 g
2
O2 g
B 2O3 s
O2 g
H 2O g H
Solution: We need 4 2B s
f
3H 2 g
reaction(4 ) Thus,
H 3
H θ
3H 2 O g
r
H θ
r
H θ
r
1941 kJ mol
1
2368 kJ mol
1
241 .8 kJ mol
1
for the reaction
B 2 H6 g
reaction(2 )
f
2368
B 2O3 s
3
reaction(3 ) - reaction(1 )
H reaction(2 )
r
241 .8
3
H reaction(3 )
r
1941 K J mol
1
1152 K J mol
H reaction(1 )
r 1
28
2.32(b) When 2.25 mg of anthracene, C 14H10(s), was burned in a bomb calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant. By how much will the temperature rise when 135 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? Solution: For anthracene the reaction is 33
C14 H10 s Θ
2
O2 g
Θ
ΔcU
Δc H
Θ
Δc H
n g RT 26 1
7163 kJ mol
Θ
ΔcU
n g
-
2.25
nΔcU
-
5 2
mol
5 2
3
-1
1
8 .3 10 kJ K mol
298 K (assume T
298 K)
1
Θ
qV
5H 2O l
( Hand book of chemist ry and phy sics )
1
7163 kJ mol -7157 kJ mol
q
14CO 2 g
3
10 g
172.23 g mol
1
7157 kJ mol
1
0.0935kJ q
C
0.0935 kJ
T
0.0693 kJ K
1.35 K
1
69.3 J K
1
When phenol is used the reaction is C6 H 5 OH s Θ
Θ
ΔcU
Θ
Δc H
n g RT ,
3054 kJ mol 3050 kJ mol q
ΔT
1
3054 kJ mol
Δc H
135
3
1
2
2
O2 g
6CO 2 g
3H 2O l
Table2.5 -
n g
3 2 3
8 .314
mol -1
10 kJ K mol
1
298 K
1
3
10 g
94.12 g mol
1
3050 kJ mol
q
4.37 5 kJ
C
0.0693 kJ K
Comment.
15
1
1
4.37 5 kJ
63 .1 K
In this case ΔcU Θ and Δc H Θ differed by
0.1 per cent. Thus, to within 3 significan t
figures, it would not have mattered if we had used Δc H Θ instead of ΔcU Θ , but for very precise work it would.
29
2.35(b) Given that the standard enthalpy combustion of graphite is -393.51 kJ mol -1 and that of diamond is -395.41kJ mol-1, calculate the enthalpy of the graphite diamond transition. Solution: The difference of the equations is C gr
Δtrans H Θ
393.51
395.41 kJ mol
1
Cd
1 .90 kJ mol
1
30
2.39(b) Use standard enthalpies of formation to calculate the standard enthalpies of the following reactions: a Cyclopropa ne g b ) HCl ( aq
propene g
NaOH aq
NaCl aq
H 2O l
Solution: Θ
( a ) Δr H
Θ
Θ
Δ f H propene , g
Δ f H cycloropan e , g
20 .42
53 .30 kJ mol
1
1
32.88 kJ mol
b The met ionic reaction is obtained from H
aq
Cl
and is H Θ
Δr H
aq
aq
Na
OH Θ
aq
Δ f H H2 O, l
285 .83
0
55 .84 kJ mol
aq
OH
aq
Na
aq
Cl
aq
H2 O l
H2O l Θ
Θ
Δ f H H , aq 229 .99
kJ mol
Δ f H OH , aq 1
1
31
2.40(b) Given the reactions(1) and (2) below, determine (a) Δr H Θ and ΔrU Θ for reaction (3), (b) Δ f H Θ for both HI(g) and H2O(g) all at 298 K. Assume all gases are perfect.
1) H 2 ( g
I2 g
θ
2HI g
52 .96 kJ mol
H
r
2 2H 2 g
O2 g
2H 2 O g
3 4HI g
O2 g
I2 s
H θ
1
483 .64 kJ mol
r
1
2H 2 O g
Solution: reaction 3
reaction 2
Θ
( a ) Δr H 3
Θ
Δr H 2
2 reaction 1 Θ
2 Δr H 1
483 .64 kJ mol
Θ
Θ
Δr H
1
-589.56kJ mol
1
Θ
Θ
Δ f H H 2O
1 2 1 2
1
n g RT
-589.56kJ mol
b Δ f H HI
2 52 .96 kJ mol
1
-589.56kJ mol Δr U
1
3
-1
8 .314 J K mol
7 .43 kJ mol
52 .96 kJ mol
1
483 .64 kJ mol
1
298 K
582 .13 kJ mol
26 .48 kJ mol 1
1
1
1
241 .82 kJ mol
1
32
2.44(b) Calculate Δr H Θ and Δr U Θ at 298 K and Δr H Θ at 348 K for the hydrogenation of ethyne (acetylene)to ethane (ethylene) from the enthalpy of combustion and heat capacity data in Tables 2.5 and 2.6. Assume the heat capacities to be constant over the temperature range involved. Solution: The hydrogenation reaction is
1 C2H 2 g
H2 g
Δ r H Θ T c
C2H 4 g
?
The reactions and accompanying data which are to be combined in order to yield reaction (1) and Are 1
2 H2 g
2
O2 g
3 C2 H4 g
H 2O l
3O 2 g 5
4 C2H 2 g
ΔCH
2H 2O l
O2 g
H 2O l
Θ
2
2CO
2
Δ CH Θ 3
g
2CO 2 g
2 reaction(2 ) - reaction(3 )
reaction(1 )
1
285.83 kJ mol
ΔCH
Θ
4
1411 kJ mol
1
1300 kJ mol
1
reaction(4 )
Hence, a Δ r HΘ T
Δr H Θ 2
285 .83 175 kJ mol Θ
1411
Δr H T
- 175kJ mol b Δr H
Θ
1
348 K
Δr C p
1300
Δc HΘ 4 kJ mol
1
1 Θ
Δr U T
Δc HΘ 3
Δn g RT 26 1
2 .48 kJ mol Δr H
v J C p ,m J 47
Θ
298 K
Δ
g
-1
173 kJ mol Δ r C p 348 K
C p ,m C 2H 4 , g
1
298 K Example2.7
C p ,m C 2 H 2 , g
C p ,m H 2 , g
J
43 .56 Δr H
Θ
43 .93
348 K
176 kJ mol
28 .82
10
175 kJ mol
3 1
kJ K -1 mol 29 .19
1
10
29 .19 3
-1
10
kJ K mol
3 1
kJ K -1 mol
1
50 K
1
33
Problems 2.2 An average human produces about 10MJ of heat each day through metabolic activity. If a human body were an isolated system of mass 65 kg with the heat capacity of water, what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water. What mass of water should be evaporated each day to maintain constant temperature? Solution: Good approximate answers can be obtained from the data for the heat capacity and molar heat of vaporization of water at 25℃.[Table 2.6 and 2.3] C p ,m H 2O
75 .3J K
0.018 kg mol
From ΔH
vap
H
H 2O
44 .0kJ mol
1
3
10 mol
nC p,m ΔT , we obtain 1.0
nC p,m
3.6 n
M ΔH vap H
1
3 .6
1
ΔH
From ΔH m
mol
65 kg
n H 2O
ΔT
1
vap
H
3
-1
10 mol
0.0753 kJ K mol
m M
0 .018 kg mol
4
10 kJ
vap 1
1
37 K
H 1 .0
44.0kJ mol
1
4
10 kJ
4 .09 kg
Comment: This estimate would correspond to about 30 glasses of water per day, which is much higher than the average consumption. The discrepancy may be a result of our assumption that evaporation of water is the main mechanism of heat loss.
34
Chapter 3 The First Law: the machinery Exersises 3.10(b) A vapour at 22 atm and 5℃ was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the Joule-Thomson coefficient, µ, at 5℃, assuming it remains constant over this temperature range. Solution: The Joule-Thomson coefficient
is the ratio of temperature change to pressure change under conditions
of isenthalpic expansion. So μ
T p
ΔT H
Δp
10 K 1.00
22 atm
0.48 K atm
1
35
a
3.11(b) For a van der Waals gas, π T
V m
U m for the isothermal reversible expansion of
. Calculate
2
argon from an initial volume of 1.00 L to 22.1 L at 298 K. What are the values of q and w ? Solution: U m
U m T,V m
dT
dU m
dU m
T
dV m
V m
V m
0 in an isothermal process, so U m
dU m U m
V m V m2
V m
T
V m1 m
-1
2
a
2
5
-1
RT b
dV m V m
22 .1L mol
a
2
V m
22 .1L mol 2
-1
1.00 L mol
0 .954 75 L mol
-1
2
1m
1
-1
-1
5
dV m
1
10 Pa atm
3
3
1
RT V m
2
10 L
130 J mol
pex d V m and p
V m
-1
1 .01325
10 Pa atm 3
-1
1.345 atm L mol
2
0 .28 41 atm L mol
130P am mol
1.00 L mol
-1
21.1 a
1
0 .95475 mol L
1 .01325
22.1 L mol
2
1.345 atm L mol
so w
V m
dV m
1.00 L mol
2
w
dV m
a
22 .1L mol
U m
2
a
V m2
dU m
V m1 m
a
dV m
a a
U m
dT
a
b
V m a
V m
2
2
dV m
for a van der Waals gas
q
ΔU m
Thus q
22.1 L mol 1.00 L mol
-1
-1
RT V m
b
dV m
RT ln V m
b
22 .1 L mol
-1
1 .00 L mol
-1
-1
8.314 J K mol 7.74 69 kJ mol w
774 7 J mol
-1
130 J mol
-1
-1
298 K
ln
22 .1
3 .22
10
2
1 .00
3 .22
10
2
-1
761 7 J mol
-1
7 .62 kJ mol
-1
36
