SOLUTION MANUAL
All Chapters Included
PHYSICAL CHEMISTRY
, Part 1: Equilibrium
1 The properties of gases
Solutions to exercises
Discussion questions
E1.1(b) 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.
It 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.
E1.2(b) The critical constants represent the state of a system at which the distinction
between the liquid and vapour phases disappears. We usually describe this
situation by saying that above the critical temperature the liquid phase cannot be
produced by the application of pressure alone. The liquid and vapour phases can no
longer coexist, though fluids in the so-called supercritical region have both liquid
and vapour characteristics. (See Box 6.1 for a more thorough discussion of the
supercritical state.)
E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation
has certain properties, one of which is that there are some values of the coefficients
of the variable where the number of real roots passes from three to one. In fact, any
equation of state of odd degree higher than 1 can in principle account for critical
behavior because for equations of odd degree in V there are necessarily some values
of temperature and pressure for which the number of real roots of V passes from
n(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc. This
mathematical result is consistent with passing from a two phase region (more than
one volume for a given T and p) to a one phase region (only one V for a given T and
p and this corresponds to the observed experimental result as the critical point is
reached.
Numerical exercises
E1.4(b) Boyle’s law applies.
pV = constant so pf Vf = piVi
p iV i (104 kPa) × (2000 cm3)
pf = = = 832 kPa
Vf (250 cm3)
E1.5(b) (a) The perfect gas law 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= = 0.626 mol
39.95 g mol —1
(0.626 mol) × (8.31 × 10—2 L bar K—1 mol—1) × (30 + 273 K)
so p = = 10.5 bar
1.5L
not 2.0 bar.
, 4 INSTRUCTOR’S MANUAL
(b) The van der Waals equation is
RT a
p= — 2
Vm — b Vm
(8.31 × 10—2 L bar K—1 mol—1) × (30 + 273) K
so p =
(1.5 L/0.626 mol) — 3.20 × 10—2 L mol—1
(1.337 L2 atm mol—2) × (1.013 bar atm—1)
— = 10.4 bar
(1.5 L/0.626¯ mol)2
E1.6(b) (a) Boyle’s law applies.
pV = constant so pf Vf = piVi
pf Vf (1.48 × 103 Torr) × (2.14 dm3)
and pi = = = 8.04 × 102 Torr
Vi (2.14 + 1.80) dm3
(b) The original pressure in bar is
1 atm 1.013 bar
pi = (8.04 × 10 2 Torr) × × = 1.07 bar
760 Torr 1 atm
E1.7(b) Charles’s law applies.
Vi Vf
V ∝T so =
Ti Tf
Vf Ti (150 cm3) × (35 + 273) K
and Tf = = = 92.4 K
Vi 500 cm3
E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect
gas law
pi pf
pV = nRT so p ∝ T and =
Ti Tf
The final pressure, then, ought to be
piTf (125 kPa) × (11 + 273) K
pf = = = 120 kPa
Ti (23 + 273) K
E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature,
and volume. Once this is done, the mass of the gas can be computed from the amount and the molar
mass using
pV = nRT
pV (1.00 atm) × (1.013 × 10 5 Pa atm —1) × (4.00 × 103 m3)
so n = = = 1.66 × 105 mol
RT (8.3145 J K—1 mol—1) × (20 + 273) K
and m = (1.66 × 105 mol) × (16.04 g mol—1) = 2.67 × 106 g = 2.67 × 103 kg
E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm /T will
give the best value of R.
All Chapters Included
PHYSICAL CHEMISTRY
, Part 1: Equilibrium
1 The properties of gases
Solutions to exercises
Discussion questions
E1.1(b) 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.
It 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.
E1.2(b) The critical constants represent the state of a system at which the distinction
between the liquid and vapour phases disappears. We usually describe this
situation by saying that above the critical temperature the liquid phase cannot be
produced by the application of pressure alone. The liquid and vapour phases can no
longer coexist, though fluids in the so-called supercritical region have both liquid
and vapour characteristics. (See Box 6.1 for a more thorough discussion of the
supercritical state.)
E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation
has certain properties, one of which is that there are some values of the coefficients
of the variable where the number of real roots passes from three to one. In fact, any
equation of state of odd degree higher than 1 can in principle account for critical
behavior because for equations of odd degree in V there are necessarily some values
of temperature and pressure for which the number of real roots of V passes from
n(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc. This
mathematical result is consistent with passing from a two phase region (more than
one volume for a given T and p) to a one phase region (only one V for a given T and
p and this corresponds to the observed experimental result as the critical point is
reached.
Numerical exercises
E1.4(b) Boyle’s law applies.
pV = constant so pf Vf = piVi
p iV i (104 kPa) × (2000 cm3)
pf = = = 832 kPa
Vf (250 cm3)
E1.5(b) (a) The perfect gas law 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= = 0.626 mol
39.95 g mol —1
(0.626 mol) × (8.31 × 10—2 L bar K—1 mol—1) × (30 + 273 K)
so p = = 10.5 bar
1.5L
not 2.0 bar.
, 4 INSTRUCTOR’S MANUAL
(b) The van der Waals equation is
RT a
p= — 2
Vm — b Vm
(8.31 × 10—2 L bar K—1 mol—1) × (30 + 273) K
so p =
(1.5 L/0.626 mol) — 3.20 × 10—2 L mol—1
(1.337 L2 atm mol—2) × (1.013 bar atm—1)
— = 10.4 bar
(1.5 L/0.626¯ mol)2
E1.6(b) (a) Boyle’s law applies.
pV = constant so pf Vf = piVi
pf Vf (1.48 × 103 Torr) × (2.14 dm3)
and pi = = = 8.04 × 102 Torr
Vi (2.14 + 1.80) dm3
(b) The original pressure in bar is
1 atm 1.013 bar
pi = (8.04 × 10 2 Torr) × × = 1.07 bar
760 Torr 1 atm
E1.7(b) Charles’s law applies.
Vi Vf
V ∝T so =
Ti Tf
Vf Ti (150 cm3) × (35 + 273) K
and Tf = = = 92.4 K
Vi 500 cm3
E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect
gas law
pi pf
pV = nRT so p ∝ T and =
Ti Tf
The final pressure, then, ought to be
piTf (125 kPa) × (11 + 273) K
pf = = = 120 kPa
Ti (23 + 273) K
E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature,
and volume. Once this is done, the mass of the gas can be computed from the amount and the molar
mass using
pV = nRT
pV (1.00 atm) × (1.013 × 10 5 Pa atm —1) × (4.00 × 103 m3)
so n = = = 1.66 × 105 mol
RT (8.3145 J K—1 mol—1) × (20 + 273) K
and m = (1.66 × 105 mol) × (16.04 g mol—1) = 2.67 × 106 g = 2.67 × 103 kg
E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm /T will
give the best value of R.
