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Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
E1.2(c) A perfect gas undergoes isothermal compression, which reduces its volume by
3.08 dm3. The final pressure and volume of the gas are 6.42 bar and 5.38 dm3,
respectively. Calculate the original pressure of the gas in (a) bar, (b) atm.
E1.2(c)
Boyle’s law [1.6] in the form pf Vf  piVi can be solved for either initial or final pressure, hence
pi 
Vf
 pf
Vi
Vf  5.38dm3 
Vi  5.38dm3  3.08dm 3  8.46 dm 3 
p f  6.42bar
Therefore,
5.38dm3 
(a)
pi  
3   (6.42bar)  4.08bar
8.46dm


1atm
 4.03atm
(b) Since 1 atm = 1.013 bar, pi  (4.08bar)  1.013bar


E1.3(c) A car tyre (i.e. an automobile tire) was inflated to a pressure of 30 lb in –2 (1.00
atm = 14.7 lb in–2) on a winter’s day when the temperature was 0F. What pressure will
be found, assuming no leaks have occurred and that the volume is constant, on a
subsequent hot summer’s day when the temperature is 95F? What complications should
be taken into account in practice?
E1.3(c) The perfect gas law, pV  nRT [1.8], can be rearranged to
constant. Hence,
p nR

 constant , if n and V are
T V
pf pi
T
or, solving for pf , pf  f  pi

Tf Ti
Ti
Internal pressure  pump pressure  atmospheric pressure
2
2
2
pi  30 lb in  147 lb in  447 lb in  Ti  255 K(0F) Tf  308 K(95F)
pf 
308K
2
2
 447 lbin  540 lbin
255K
2
2
Therefore, p(pump)  540 lbin  147 lbin  39 lbin
2
Complications are those factors that destroy the constancy of V or n, such as the change in
volume of the tyre, the change in rigidity of the material from which it is made, and loss of
pressure by leaks and diffusion.
E1.5(c) What pressure difference must be generated along a vertical single stage water
pump tube to extract water from a well 5.0 m deep?
E1.5(c) Identifying pex in the equation p  pex   gh [13] as the pressure at the top of the tube and p as
the atmospheric pressure on the liquid, the pressure difference is
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© Oxford University Press, 2006. All rights reserved.
Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
3
p  pex   gh  (10 103 kg m )  (981m s 2)  (5.0 m)
 5.0 104 Pa ( 0.50 atm)
E1.6(c) A manometer consists of a U-shaped tube containing a liquid. One side is
connected to the apparatus and the other is open to the atmosphere. The pressure inside
the apparatus is then determined from the difference in heights of the liquid. Suppose the
liquid is an oil with density 0.865 g cm-3, the external pressure is 750 Torr, and the open
side is 15.0 cm lower than the side connected to the apparatus. What is the pressure in
the apparatus?
E1.6(c)
The pressure in the apparatus is given by
p  patm   gh[1.3]
 1 atm   1.013105 Pa 
  0.9997  10 5 Pa
p
 750Torr  

atm

1 atm
 760Torr  

 gh  0.865 g cm
-3

 1 kg 
 3 
 10 g 

 106 cm3 


3
 m 
-2
 9.806 m s  0.15 m  1272 Pa
5
3
5
p  0.9997  10 Pa + 1.272  10 Pa = 1.012  10 Pa = 101 kPa
E1.8(c) At 0.000C and 1.000 atm, the mass density of the vapour of a hydrocarbon is
1.97 kg m–3. What is the molecular formula of the hydrocarbon under these conditions?
What is the hydrocarbon?
E1.8(c) Since p  1 atm , the approximation that the vapour is a perfect gas is adequate. Then (as in
Exercise 1.7(b)),
m
pV  nRT 
RT  Upon rearrangement
M
(8.314 Pa m3 K 1 mol1 )  (273K)
 RT 
 (1.97 kg m3 ) 
 0.0441k g mol 1= 44.1 g mol-1

5
 p 
1.013  10 Pa
M  
The formula of the vapour is then C3 H 8 , which is one of the propanes.
E1.9(c) Calculate the mass of water vapour present in a house of volume 1000 m3 that
contains air at 22C on a day when the relative humidity is 48 per cent indoors.
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Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
E1.9(c) The partial pressure of the water vapour in the house is: pH2O  (048)  (`19.827Torr)  9.5Torr
Assuming that the perfect gas equation [1.8] applies, with n  m  pV  m RT or
M
M
3
3
1atm
10
dm



1

  (1000 m3 ) 
(9.5Torr)  

 m3   (1802 g mol )
pVM
 760 Torr 


m
RT

1
(00821 dm3 atm K mol 1)  (295K)
 9.3  10 g  9.3 kg
3
E1.12(c) For dry air at 1.0000 atm pressure, the densities at –50C, 0C, and 69C are
1.5826 g dm–3, 1.2929 g dm–3, and 1.0322 g dm–3, respectively. From these data, and
assuming that air obeys Charles’s law, determine a value for the absolute zero of
temperature in degrees Celsius.
E1.12(c) The easiest way to solve this exercise is to assume a sample of mass 1000g , then calculate the
volume at each temperature, plot the volume against the Celsius temperature, and extrapolate to
V 0.
Draw up the following table
/ C
 /(g dm -3 ) V/(dm3 g -1 )
-50
1.5826
0.6319
0
1.2929
0.7734
69
1.0322
0.9688
Linear regression of the V versus  data gives a value for absolute zero close to 273 C .
Alternatively, one could use an equation for V as a linear function of  , which is Charles’s law,
and solve for the value of absolute zero. V  V0  (1  )
At absolute zero, V  0 , then  (abszero)  
one of the data points (except   0 ) as follows.
From V  V0  (1   ) ,
1

. The value of  can be obtained from any
V
  09688 
 V  1  07734   1

 0 
 0003661(C) 1

100C
1
1
 
 273C

0003661(C) 1
which is the same to within experimental error as the value obtained graphically.
E1.13(c) Calculate the pressure exerted by 1.0 mol C6H6 behaving as (a) a perfect gas, (b)
a van der Waals gas when it is confined under the following conditions: (i) at 373.15 K in
22.414 dm3, (ii) at 1000 K in 300 cm3. Use the data in Table 1.6.
E1.13(c) (a)
p  nRT [18]
V
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Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
n  10 mol
T  37315K (i) or 1000 K (ii)
V  22414 dm (i) or 300 cm 3 (ii)
3
1
p
(i)
(10 mol)  (8206 102 dm3 atm K mol 1)  (37315K)
 1.37 atm
22414dm3
1
(ii) p 
(10 mol)  (8206 102 dm3 atm K mol 1)  (1000 K)
 2.73 102 atm
0300dm3
2
p  nRT  an2 [121a ]
V  nb V
(b)
1
From Table 1.5, a  5507dm6 atmmol2 and b  651  102 dm3 mol . Therefore,
1
2
1
3
nRT  (10 mol)  (8206 10 dm atm K mol )  (37315 K)  137 atm
V  nb
[22414  (10)  (651102 )]dm3
(i)
2
an 2 (5507 dm6 atm mol )  (10 mol) 2

 111102 atm
V2
(22414 dm3 ) 2
and p  137 atm  111102 atm  1.36atm
1
(ii)
2
1
3
nRT  (10 mol)  (8206 10 dm atm K mol )  (1000 K)
V  nb
(0300  00651) dm 3
 349  102 atm
2
an 2 (5507 dm6 atm mol )  (10 mol) 2

 1.83  102 atm
V2
(0300 dm3 ) 2
and p  3.49 102 atm  1.84 102 atm  1.65 102 atm
Comment. It is instructive to calculate the percentage deviation from perfect gas behaviour for (i)
and (ii).
(i)
1.36  1.37 100%  0.7%
1.37
(ii)
(1.65 102 )  (2.73 102 )
100%  39.5%
2.73 102
Deviations from perfect gas behaviour are not observed at p  1atm except with very precise
apparatus.
E1.16(c) Cylinders of compressed gas are typically filled to a pressure of 200 bar. For
nitrogen, what would be the molar volume at this pressure and 25C based on (a) the
perfect gas equation, (b) the van der Waals equation. For nitrogen, a = 1.39 dm6 atm
mol–2, b = 0.0391 dm3 mol–1.
E1.16(c) (a) Vmo 
1
RT (8314 J K mol 1)  (29815 K)

1
p
(200 bar)  (105 Pa bar )
 124 104 m 3 mol 1  0.124dm3 mol
1
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Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
(b) The van der Waals equation is a cubic equation in Vm . The most direct way of obtaining the
molar volume would be to solve the cubic analytically. However, this approach is cumbersome,
so we proceed as in Example 1.6. The van der Waals equation is rearranged to the cubic form


RT  2  a 
ab
RT  2  a 
ab
Vm3   b 
 0 or x3   b 
0
 Vm    Vm 
x  x
p 
p
p 
p

 p

 p
1
with x  Vm  (dm3 mol ) .
The coefficients in the equation are evaluated as
b+
(8.206´ 10- 2 dm3 mol- 1 )´ (298.15K )
RT
= (3.91´ 10- 2 dm3 mol- 1 ) +
p
(200 bar )´ (1.013 atm bar - 1 )
= (3.91´ 10- 2 + 0.1208)dm3 mol- 1 = 0.1599 dm 3mol- 1
a
1.39 dm6 atm mol- 2
=
= 6.86´ 10- 3 (dm3 mol- 1 ) 2
- 1
p (200bar) ´ (1.013 atm bar )
ab (1.39 dm 6 atm mol- 2 ) ´ (3.91´ 10- 2 dm3 mol- 1 )
=
= 2.68´ 10- 4 (dm3 mol- 1 )3
p
(200 bar)´ (1.013 atm bar- 1 )
Thus, the equation to be solved is x3 - 0.1599x2 + (6.86´ 10- 3 ) x - (2.68´ 10- 4 ) = 0.
Calculators and computer software for the solution of polynomials are readily available. In this
case we find
x  0122 or Vm  0.122 dm3 mol 1
Thedifferenceis about 2 percent.
E1.18(c) A vessel of volume 24.4 dm3 contains 1.0 mol H2 and 2.5 mol N2 at 298.15 K.
Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c)
their total pressure.
n  n(H 2 )  n(N 2 )  10 mol  2.5 mol  35 mol
E1.18(c)
(a)
(b)
xJ 
nJ
[114]
n
10 mol
2.5mol
 029
x(N 2 ) 
 071
35mol
35mol
The perfect gas law is assumed to hold for each component individually as well as for the
RT
mixture as a whole. Hence, pJ  nJ
V
x(H 2 ) 
RT
V
1
(8206 102 dm3 atm K mol 1)  (29815K)
1

 100 atm mol
3
24.4dm
1
p(H2 )  (10 mol)  (100 atm mol )  10atm
1
p(N2 )  (2.5mol)  (100 atm mol )  2.5atm
(c)
p  p(H 2 )  p(N 2 )[115]  10 atm  2.5atm  35atm
E1.19 (c)The critical constants of carbon dioxide are pc = 72.85 atm, Vc = 94.0 cm3 mol–1,
and Tc = 304.2 K. Calculate the van der Waals parameters of the gas and estimate the
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Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
radius of the molecules. Compare to the values in Table 1.6. What might your
comparison suggest?
E1.19(c) Equations [1.22] are solved for b and a, respectively, and yield b 
Vc
and a  27b 2 pc  3Vc2 pc
3
Substituting the critical constants
1
b   (94.0 cm3 mol 1)  31.3 cm3 mol 1
3
1
a  3  (94.0 103 dm3 mol ) 2  (72.85atm)  1.93dm6 atm mol
2
Note that knowledge of the critical temperature, Tc , is not required.
As b is approximately the volume occupied per mole of particles
31.3  106 m3 mol 1
b
vmol 

 520  1029 m3
NA
6022 1023 mol 1
1 3
4
 3

 (520  1029 m3 )   023nm
Then, with vmol   r 3  r  
3
 4

The values of a and b given in Table 1.6 differ substantially from the values calculated here.
Consequently, the value of the radius of the molecule given above cannot be taken too seriously.
E1.20(c) Use the van der Waals parameters for carbon dioxide to calculate approximate
values of (a) the Boyle temperature of the gas and (b) the radius of a CO2 molecule
regarded as a sphere (a = 3.610 dm6 atm mol-2, b = 0.0429 dm3 mol-1).
E1.20(c) (a) The Boyle temperature is the temperature at which lim
Vm 
van der Waals equation
Z
so
pVm

RT

RT
Vm  b

 Va2 Vm
m
RT

dZ
vanishes. According to the
d(1Vm )
Vm
a

Vm  b Vm RT
 dZ
dZ

d(1Vm )  dVm
  dVm 


  d(1Vm ) 
 dZ 
Vm
1
a 
2
 Vm2 

 2

  Vm 
2
 dVm 
 (Vm  b) Vm  b Vm RT 
Vm2 b
a


(Vm  b) 2 RT
In the limit of large molar volume, we have
lim
Vm 
dZ
a
b
 0 so
d(1Vm )
RT
a
b
RT
2
and T 
(3.610dm6 atm mol )
a

 1025K
Rb (008206 dm3 atm K 1 mol 1)  (00429 dm3 mol 1)
(b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an
estimate of molecular size. The centres of spherical particles are excluded from a sphere
whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro
constant times the volume is the molar excluded volume b.
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Higher Education
Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases
Additional end of chapter problems
1 3
 4 (2r )3 
1  3b 
b  NA 

 so r  
3
2  4 N A 


1 3
1  3(00429 dm 3 mol 1) 
r 

2  4 (6022 1023 mol 1) 
 1286  109 dm  129  1010 m  0129 nm
E1.22(c) A certain gas obeys the van der Waals equation with a = 0.580 m6 Pa mol–2. Its
volume is found to be 3.50  10–4 m3 mol–1 at 273 K and 5.0 MPa. From this information
calculate the van der Waals constant b. What is the compression factor for this gas at the
prevailing temperature and pressure?
E1.22(c) The van der Waals equation [121b] is solved for b, which yields
b  Vm 
RT

a 
p 2 
Vm 

Substituting the data
1
b  3.50 104 m3 mol 1 
(8314 J K mol 1)  (273K)

 0580 m 6 Pa mol 2 
6
(5

0

10
Pa)



4
3
1 2 
 (3.50 10 m mol ) 

 1.17 104 m3 mol 1
Z
pVm
(50  106 Pa)  (3.50  104 m3 mol-1 )
[1.17] 
 077
1
RT
(8314 J K mol 1)  (273K)
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.
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