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
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
= 0.626 mol
39.95 g mol
(0.626 mol) × (8.31 × 10−2 dm 3 bar K −1 mol−1 ) × (30 + 273) K
so p= = 10.5bar
1.5 dm 3
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:
pV (1.97 bar) × (2.14 dm 3 )
(i) pi = f f = = 1.07 bar
Vi (2.14 + 1.80) dm 3
(ii) The original pressure in Torr is
1 atm 760 Torr
pi = (1.07 bar) × × = 803 Torr
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]
pi pf
so p ∝ T and =
Ti Tf
The final pressure, then, ought to be
pT (125 kPa) × (11 + 273)K
pf = i f = = 120 kPa
Ti (23 + 273)K
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
pV (1.00 atm) × (1.013 × 105 Pa atm −1 ) × (4.00 × 103 m 3 )
so n= = = 1.66 × 105 mol
RT (8.3145 J K −1mol−1 ) × (20 + 273)K
Once this is done, the mass of the gas can be computed from the amount and the molar
mass:
−1
m = (1.66 × 105 mol) × (16.04 g mol ) = 2.67 × 106 g = 2.67 × 103 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the
total pressure
1
, 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
−3 1 kg 1 cm
p − pex = ρ gh = (1.0 g cm ) × 3 × −2 × (9.81 m s −2 ) × (0.15m)
10 g 10 m
= 1.5 × 103 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,
3
1 kg 1 cm
and ρ gh = 13.55 g cm −3 × × × 0.100 m × 9.806 m s −2 = 1.33 × 104 Pa
103 g 10−2 m
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
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)/(dm3 atm K–1 mol–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
pV
From Figure 1A.1(a), R = lim m = 0.082 062 dm 3 atm K −1 mol−1
p→0
T
Figure 1A.1
(a)
2
, (b)
ρ
From Figure 1A.1(b), lim = 1.427 55 g dm -3 atm −1
p→0 p
ρ
M = lim RT = (0.082062 dm 3 atm K −1 mol−1 ) × (273.15 K) × (1.42755 g dm -3 atm −1 )
p→0 p
= 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
V V m M
Vm = = × =
n m n ρ
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.
−1
RT ρ (8.3145 Pa m 3 mol ) × [(100 + 273) K] × (0.6388 kg m −3 )
M= =
p 1.60 × 104 Pa
= 0.124 kg mol−1 = 124 g mol−1
The number of atoms per molecule is
−1
124 g mol
−1
= 4.00
31.0 g mol
suggesting a formula of P4 .
1A.9(b) Use the perfect gas equation [1A.5] to compute the amount; then convert to mass.
pV
pV = nRT so n=
RT
We need the partial pressure of water, which is 53 per cent of the equilibrium vapour
pressure at the given temperature and standard pressure. (We must look it up in a handbook
like the CRC or other resource such as the NIST Chemistry WebBook.)
p = (0.53) × (2.81 × 103 Pa) = 1.49 × 103 Pa
3
, (1.49 × 103 Pa) × (250 m 3 )
so n= = 151 mol
(8.3145 J K −1 mol−1 ) × (23 + 273) K
−1
and m = (151 mol) × (18.0 g mol ) = 2.72 × 103 g = 2.72 kg
1A.10(b) (i) The volume occupied by each gas is the same, since each completely fills the container.
Thus solving for V we have (assuming a perfect gas, eqn. 1A.5)
n RT
V= J
pJ
We have the pressure of neon, so we focus on it
0.225 g
nNe = = 1.115 × 10−2 mol
20.18 g mol−1
Thus
1.115 × 10−2 mol × 8.3145 Pa m 3 K −1 mol−1 × 300 K
V= = 3.14 × 10−3 m 3 = 3.14 dm 3
8.87 × 103 Pa
(ii) The total pressure is determined from the total amount of gas, n = nCH + nAr + nNe .
4
0.320 g 0.175 g
nCH = −1
= 1.995 × 10−2 mol nAr = = 4.38 × 10−3 mol
4
16.04 g mol 39.95 g mol−1
( )
n = 1.995 + 0.438 + 1.115 × 10−2 mol = 3.55 × 10−2 mol
nRT 3.55 × 10−2 mol × 8.3145 Pa m 3 K −1 mol−1 × 300 K
and p= =
V 3.14 × 10−3 m 3
= 2.82 × 104 Pa = 28.2 kPa
ρ RT
1A.11(b) This exercise uses the formula, M = , which was developed and used in Exercise
p
1A.8(b). First the density must first be calculated.
33.5 × 10−3 g 103 cm 3
ρ= × = 0.134 g dm −3
250 cm 3
dm 3
−1
(0.134 g dm −3 ) × (62.36 dm 3 torr K mol −1) × (298 K)
M= = 16.4 g mol−1
152 torr
1A.12(b) This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as
that temperature at which the volume of a sample of gas would become zero if the substance
remained a gas at low temperatures. The solution uses the experimental fact that the volume
is a linear function of the Celsius temperature:
V = V0 + αθ where V0 = 20.00 dm3 and α = 0.0741 dm3 °C–1 .
At absolute zero, V = 0 = V0 + αθ
V 20.00 dm 3
so θ (abs.zero) = − 0 = − = Ğ270°C
α 0.0741 dm 3 ¡C−1
which is close to the accepted value of –273C.
1A.13(b) (i) Mole fractions are
n 2.5 mol
xN = N [1A.9] = = 0.63
ntotal (2.5 + 1.5) mol
Similarly, xH = 0.37
According to the perfect gas law
ptotV = ntotRT
4
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
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
= 0.626 mol
39.95 g mol
(0.626 mol) × (8.31 × 10−2 dm 3 bar K −1 mol−1 ) × (30 + 273) K
so p= = 10.5bar
1.5 dm 3
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:
pV (1.97 bar) × (2.14 dm 3 )
(i) pi = f f = = 1.07 bar
Vi (2.14 + 1.80) dm 3
(ii) The original pressure in Torr is
1 atm 760 Torr
pi = (1.07 bar) × × = 803 Torr
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]
pi pf
so p ∝ T and =
Ti Tf
The final pressure, then, ought to be
pT (125 kPa) × (11 + 273)K
pf = i f = = 120 kPa
Ti (23 + 273)K
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
pV (1.00 atm) × (1.013 × 105 Pa atm −1 ) × (4.00 × 103 m 3 )
so n= = = 1.66 × 105 mol
RT (8.3145 J K −1mol−1 ) × (20 + 273)K
Once this is done, the mass of the gas can be computed from the amount and the molar
mass:
−1
m = (1.66 × 105 mol) × (16.04 g mol ) = 2.67 × 106 g = 2.67 × 103 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the
total pressure
1
, 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
−3 1 kg 1 cm
p − pex = ρ gh = (1.0 g cm ) × 3 × −2 × (9.81 m s −2 ) × (0.15m)
10 g 10 m
= 1.5 × 103 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,
3
1 kg 1 cm
and ρ gh = 13.55 g cm −3 × × × 0.100 m × 9.806 m s −2 = 1.33 × 104 Pa
103 g 10−2 m
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
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)/(dm3 atm K–1 mol–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
pV
From Figure 1A.1(a), R = lim m = 0.082 062 dm 3 atm K −1 mol−1
p→0
T
Figure 1A.1
(a)
2
, (b)
ρ
From Figure 1A.1(b), lim = 1.427 55 g dm -3 atm −1
p→0 p
ρ
M = lim RT = (0.082062 dm 3 atm K −1 mol−1 ) × (273.15 K) × (1.42755 g dm -3 atm −1 )
p→0 p
= 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
V V m M
Vm = = × =
n m n ρ
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.
−1
RT ρ (8.3145 Pa m 3 mol ) × [(100 + 273) K] × (0.6388 kg m −3 )
M= =
p 1.60 × 104 Pa
= 0.124 kg mol−1 = 124 g mol−1
The number of atoms per molecule is
−1
124 g mol
−1
= 4.00
31.0 g mol
suggesting a formula of P4 .
1A.9(b) Use the perfect gas equation [1A.5] to compute the amount; then convert to mass.
pV
pV = nRT so n=
RT
We need the partial pressure of water, which is 53 per cent of the equilibrium vapour
pressure at the given temperature and standard pressure. (We must look it up in a handbook
like the CRC or other resource such as the NIST Chemistry WebBook.)
p = (0.53) × (2.81 × 103 Pa) = 1.49 × 103 Pa
3
, (1.49 × 103 Pa) × (250 m 3 )
so n= = 151 mol
(8.3145 J K −1 mol−1 ) × (23 + 273) K
−1
and m = (151 mol) × (18.0 g mol ) = 2.72 × 103 g = 2.72 kg
1A.10(b) (i) The volume occupied by each gas is the same, since each completely fills the container.
Thus solving for V we have (assuming a perfect gas, eqn. 1A.5)
n RT
V= J
pJ
We have the pressure of neon, so we focus on it
0.225 g
nNe = = 1.115 × 10−2 mol
20.18 g mol−1
Thus
1.115 × 10−2 mol × 8.3145 Pa m 3 K −1 mol−1 × 300 K
V= = 3.14 × 10−3 m 3 = 3.14 dm 3
8.87 × 103 Pa
(ii) The total pressure is determined from the total amount of gas, n = nCH + nAr + nNe .
4
0.320 g 0.175 g
nCH = −1
= 1.995 × 10−2 mol nAr = = 4.38 × 10−3 mol
4
16.04 g mol 39.95 g mol−1
( )
n = 1.995 + 0.438 + 1.115 × 10−2 mol = 3.55 × 10−2 mol
nRT 3.55 × 10−2 mol × 8.3145 Pa m 3 K −1 mol−1 × 300 K
and p= =
V 3.14 × 10−3 m 3
= 2.82 × 104 Pa = 28.2 kPa
ρ RT
1A.11(b) This exercise uses the formula, M = , which was developed and used in Exercise
p
1A.8(b). First the density must first be calculated.
33.5 × 10−3 g 103 cm 3
ρ= × = 0.134 g dm −3
250 cm 3
dm 3
−1
(0.134 g dm −3 ) × (62.36 dm 3 torr K mol −1) × (298 K)
M= = 16.4 g mol−1
152 torr
1A.12(b) This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as
that temperature at which the volume of a sample of gas would become zero if the substance
remained a gas at low temperatures. The solution uses the experimental fact that the volume
is a linear function of the Celsius temperature:
V = V0 + αθ where V0 = 20.00 dm3 and α = 0.0741 dm3 °C–1 .
At absolute zero, V = 0 = V0 + αθ
V 20.00 dm 3
so θ (abs.zero) = − 0 = − = Ğ270°C
α 0.0741 dm 3 ¡C−1
which is close to the accepted value of –273C.
1A.13(b) (i) Mole fractions are
n 2.5 mol
xN = N [1A.9] = = 0.63
ntotal (2.5 + 1.5) mol
Similarly, xH = 0.37
According to the perfect gas law
ptotV = ntotRT
4
