Solutions Manual For Atkins’ Physical
Chemistry, 12th Edition, By Peter Atkins
| All Chapters 1-19 Covered.
1 The properties of gases 1A The perfect gas
Answers to discussion questions
1A.2 The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied
alone the same container as the mixture at the same temperature. Dalton’s law is a limiting
law because it holds exactly only under conditions where the gases have no effect upon each
other. This can only be true in the limit of zero pressure where the molecules of the gas are
very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real
gases, the law is only an approximation.
Solutions to exercises
1A.1(b) The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be
p = nRT
V
All quantities on the right are given to us except n, which can be computed from the given
mass of Ar.
n= 25 g −1 = 0.626 mol
39.95 g mol
so p = (0.626 mol) × (8.31×10−2 dm1.53 dm bar3 K−1 mol−1) × (30 + 273) 10.5bar K =
So no , the sample would not exert a pressure of 2.0 bar.
1A.2(b) Boyle’s law [1A.4a] applies.
pV = constant so pfVf = piVi
Solve for the initial pressure: pVfVi f = (1.97(2. bar14
1.07 bar
+)1×.80)(2.14dmdm3 3) =
(i) pi =
(ii) The original pressure in Torr is
803 Torr
1
, 1.013 bar 1 atm
1A.3(b) The relation between pressure and temperature at constant volume can be derived from the
perfect gas law, pV = nRT [1A.5]
p p
so p ∝T and i= f
Ti Tf
The final pressure, then, ought to be
pT
pf =Ti i f = (125 kPa(23)+×273)(11+K273)K = 120 kPa
1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure,
temperature, and volume.
pV = nRT
so n = pV = (1.00 atm) × (1.013×−110mo5 Pa atml−1) × (20−1)+× 273)(4.00K×103m3) =
1.66 ×105 mol
RT (8.3145 J K
Once this is done, the mass of the gas can be computed from the amount and the molar mass:
m = (1.66 ×105 mol) × (16.04 g mol−1) = 2.67 ×106 g = 2.67 ×10 3 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the
total pressure
pi = (1.07 bar) × 1 atm × 760 Torr =
p = pex + ρgh .
Let pex be the pressure at the top of the straw and p the pressure on the surface of the liquid
(atmospheric pressure). Thus the pressure difference is
3
1 kg 1 cm
p − pex = ρgh = (1.0gcm−3) × 10 3 g × 10 − 2 m × (9.81 m s−2 ) × (0.15m)
= 1.5×10 3 Pa = 1.5×10−2 atm
1A.6(b) The pressure in the apparatus is given by
p = pex + ρgh [1A.1]
where pex = 760 Torr = 1 atm = 1.013×105 Pa, and ρgh = 13.55 g cm−3 × 101 kg3 g
× 101 cm−2 m 3 × 0.100 m × 9.806 m s−2 = 1.33×104 Pa
p = 1.013×105 Pa +1.33×104 Pa = 1.146 ×105 Pa = 115 kPa
pV pVm
1A.7(b) Rearrange the perfect gas equation [1A.5] to give R = =
nT T
All gases are perfect in the limit of zero pressure. Therefore the value of pVm/T extrapolated
to zero pressure will give the best value of R. The molar mass can be introduced through
2
, m
pV = nRT = RT
M
m RT RT
which upon rearrangement gives M = =ρ
V p p
The best value of M is obtained from an extrapolation of ρ/p versus p to zero pressure; the
intercept is M/RT.
Draw up the following table:
p/atm (pVm/T)/(dm atm K mol ) 3 –1
–1 (ρ/p)/(g dm–3 atm–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 90
From Figure 1A.1(a), R = limp→ pVT 0.082 062 dm3 atm K−1 mol−1
m =
0
Figure 1A.1
(a)
(b)
3
, From Figure 1A.1(b), lim ρ = 1.427 55 g dm -3 atm −1
p→ 0 p
ρ
M = limp→0 RT p = (0.082062 dm3 atm K−1 mol−1) × (273.15 K) × (1.42755 g dm-3
atm−1)
= 31.9988 g mol−1
The value obtained for R deviates from the accepted value by 0.005 per cent, better than can
be expected from a linear extrapolation from three data points.
1A.8(b) The mass density ρ is related to the molar volume Vm by
Vm = Vn = Vm × mn = Mρ
where M is the molar mass. Putting this relation into the perfect gas law [1A.5] yields
pM
pVm = RT so = RT ρ
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 = RTρ= (8.3145 Pa m3 mol−1) ×[(100 + 4273)K] Pa × (0.6388kg m−3)
p 1.60 ×10
= 0.124 kg mol−1 = 124 g mol−1
The number of atoms per molecule is
124gmol−−11 = 4.00
31.0gmol suggesting
a formula of P4 .
1A.9(b) Use the perfect gas equation [1A.5] to compute the amount; then convert to mass.
pV = nRT
so
4
Chemistry, 12th Edition, By Peter Atkins
| All Chapters 1-19 Covered.
1 The properties of gases 1A The perfect gas
Answers to discussion questions
1A.2 The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied
alone the same container as the mixture at the same temperature. Dalton’s law is a limiting
law because it holds exactly only under conditions where the gases have no effect upon each
other. This can only be true in the limit of zero pressure where the molecules of the gas are
very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real
gases, the law is only an approximation.
Solutions to exercises
1A.1(b) The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be
p = nRT
V
All quantities on the right are given to us except n, which can be computed from the given
mass of Ar.
n= 25 g −1 = 0.626 mol
39.95 g mol
so p = (0.626 mol) × (8.31×10−2 dm1.53 dm bar3 K−1 mol−1) × (30 + 273) 10.5bar K =
So no , the sample would not exert a pressure of 2.0 bar.
1A.2(b) Boyle’s law [1A.4a] applies.
pV = constant so pfVf = piVi
Solve for the initial pressure: pVfVi f = (1.97(2. bar14
1.07 bar
+)1×.80)(2.14dmdm3 3) =
(i) pi =
(ii) The original pressure in Torr is
803 Torr
1
, 1.013 bar 1 atm
1A.3(b) The relation between pressure and temperature at constant volume can be derived from the
perfect gas law, pV = nRT [1A.5]
p p
so p ∝T and i= f
Ti Tf
The final pressure, then, ought to be
pT
pf =Ti i f = (125 kPa(23)+×273)(11+K273)K = 120 kPa
1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure,
temperature, and volume.
pV = nRT
so n = pV = (1.00 atm) × (1.013×−110mo5 Pa atml−1) × (20−1)+× 273)(4.00K×103m3) =
1.66 ×105 mol
RT (8.3145 J K
Once this is done, the mass of the gas can be computed from the amount and the molar mass:
m = (1.66 ×105 mol) × (16.04 g mol−1) = 2.67 ×106 g = 2.67 ×10 3 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the
total pressure
pi = (1.07 bar) × 1 atm × 760 Torr =
p = pex + ρgh .
Let pex be the pressure at the top of the straw and p the pressure on the surface of the liquid
(atmospheric pressure). Thus the pressure difference is
3
1 kg 1 cm
p − pex = ρgh = (1.0gcm−3) × 10 3 g × 10 − 2 m × (9.81 m s−2 ) × (0.15m)
= 1.5×10 3 Pa = 1.5×10−2 atm
1A.6(b) The pressure in the apparatus is given by
p = pex + ρgh [1A.1]
where pex = 760 Torr = 1 atm = 1.013×105 Pa, and ρgh = 13.55 g cm−3 × 101 kg3 g
× 101 cm−2 m 3 × 0.100 m × 9.806 m s−2 = 1.33×104 Pa
p = 1.013×105 Pa +1.33×104 Pa = 1.146 ×105 Pa = 115 kPa
pV pVm
1A.7(b) Rearrange the perfect gas equation [1A.5] to give R = =
nT T
All gases are perfect in the limit of zero pressure. Therefore the value of pVm/T extrapolated
to zero pressure will give the best value of R. The molar mass can be introduced through
2
, m
pV = nRT = RT
M
m RT RT
which upon rearrangement gives M = =ρ
V p p
The best value of M is obtained from an extrapolation of ρ/p versus p to zero pressure; the
intercept is M/RT.
Draw up the following table:
p/atm (pVm/T)/(dm atm K mol ) 3 –1
–1 (ρ/p)/(g dm–3 atm–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 90
From Figure 1A.1(a), R = limp→ pVT 0.082 062 dm3 atm K−1 mol−1
m =
0
Figure 1A.1
(a)
(b)
3
, From Figure 1A.1(b), lim ρ = 1.427 55 g dm -3 atm −1
p→ 0 p
ρ
M = limp→0 RT p = (0.082062 dm3 atm K−1 mol−1) × (273.15 K) × (1.42755 g dm-3
atm−1)
= 31.9988 g mol−1
The value obtained for R deviates from the accepted value by 0.005 per cent, better than can
be expected from a linear extrapolation from three data points.
1A.8(b) The mass density ρ is related to the molar volume Vm by
Vm = Vn = Vm × mn = Mρ
where M is the molar mass. Putting this relation into the perfect gas law [1A.5] yields
pM
pVm = RT so = RT ρ
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 = RTρ= (8.3145 Pa m3 mol−1) ×[(100 + 4273)K] Pa × (0.6388kg m−3)
p 1.60 ×10
= 0.124 kg mol−1 = 124 g mol−1
The number of atoms per molecule is
124gmol−−11 = 4.00
31.0gmol suggesting
a formula of P4 .
1A.9(b) Use the perfect gas equation [1A.5] to compute the amount; then convert to mass.
pV = nRT
so
4
