Exam (elaborations)

Exam (elaborations) TEST BANK FOR Physical Chemistry By Atkins 8th Edition

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Harvard University

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Physical Chemistry

Exam (elaborations) TEST BANK FOR Physical Chemistry By Atkins 8th Edition (Instructor Solution Manual to ISBN 7594) Physical Chemistry, ISBN: 4334 (Instructor Solution Manual to ISBN 7594) Part 1: Equilibrium 1 The properties of gases Solutions to exercises Discussion questions E1.1(b) T...

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Uploaded on November 7, 2021
Number of pages 479
Written in 2021/2022
Type Exam (elaborations)
Contains Questions & answers

physical chemistry
9780716787594
9780716774334
isbn 9780716774334
test bank for physical chemistry by atkins 8th edition

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,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 = pi Vi
pi Vi (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
1.5 L
not 2.0 bar.

, 4 INSTRUCTOR’S MANUAL

(b) The van der Waals equation is
RT a
p= −
Vm − b Vm2

(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/0.626̄ mol)2
E1.6(b) (a) Boyle’s law applies.

pV = constant so pf Vf = pi Vi

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 × 102 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
pi Tf (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 × 105 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.

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