물리화학 7판 / Atkins 정답 입니다.
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Tf
(250 cm3)
than 1 can in principle account for critical behavior because for equations of odd degree in V there
E1.5(b) (a) The perfect gas law is
RT
Ti
of Ar.
p=0
Ti
1.013 bar
E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation has certain
Vi
= (150 cm3) × (35 + 273)K
vapour phases can no longer coexist, though fluids in the so-called supercritical region have both
E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid
(8.3145 JK−1 mol−1) × (20 + 273)K
The best value of M is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is
= pf
pV = nRT so p ∝ T and
(1.5L/0.626 mol) − 3.20 × 10−2 L mol−1
p
From Fig. 1.1(b),
0.250 000 0.082 0414 1.427 90
pVm
= (104 kPa) × (2000 cm3)
= 92.4K
T
(23 + 273)K
= 1.42755 g L−1 atm−1
Vi
p/atm (pVm/T )/(L atmK−1 mol−1) (ρ/p)/(gL−1 atm−1)
(b) The van der Waals equation is
760 Torr
) = 2.67 × 106 g = 2.67 × 103 kg
= 832 kPa
implying that the pressure would be
ρ
0.750 000 0.082 0014 1.428 59
(2.14 + 1.80) dm3
Draw up the following table
= ρ
pV = constant so pfVf = piVi
number of real roots passes from three to one. In fact, any equation of state of odd degree higher
= 0.082 061 5 L atmK−1 mol−1
E1.4(b) Boyle’s law applies.
and Tf = VfTi
alone the same container as the mixture at the same temperature. It is a limiting law because it holds
not 2.0 bar.
M
are necessarily some values of temperature and pressure for which the number of real roots of V
솔루션
= (125 kPa) × (11 + 273)K
×
= 1.66 × 105 mol
RT
exactly only under conditions where the gases have no effect upon each other. This can only be true
Solutions to exercises
pf = piVi
− a
p
properties, one of which is that there are some values of the coefficients of the variable where the
pi = (8.04 × 102 Torr) ×
p
Ti
500 cm3
물리화학 7판 / Atkins 정답 입니다. 1 atm
E1.6(b) (a) Boyle’s law applies.
and vapour phases disappears. We usually describe this situation by saying that above the critical
= 0.626 mol
V
= Vf
39.95 g mol−1
Vm − b
holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation.
and pi = pfVf
pV = nRT
= (1.00 atm) × (1.013 × 105 Pa atm−1) × (4.00 × 103 m3)
Vf
(1.5L/0.62¯6 mol)2
Tf
Vi
= 10.5 bar
THE PROPERTIES OF GASES 5
liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical
= 120 kPa
so n = pV
물리화학 7판 / Atkins 솔루션 입니다. 다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다.
give the best value of R.
RT
1.5L
given T and p) to a one phase region (only one V for a given T and p and this corresponds to the
pV = constant so pfVf = piVi
p=0
and volume. Once this is done, the mass of the gas can be computed from the amount and the molar
V2m
= 8.04 × 102 Torr
Discussion questions
The molar mass is obtained from pV = nRT = m
E1.7(b) Charles’s law applies.
in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law
− (1.337 L2 atm mol−2) × (1.013 bar atm−1)
so p = (0.626 mol) × (8.31 × 10−2 L barK−1 mol−1) × (30 + 273K)
레포트 > 공학,기술계열
다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다.
mass using
mathematical result is consistent with passing from a two phase region (more than one volume for a
which upon rearrangement gives M = m
E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature,
Numerical exercises
From Fig. 1.1(a),
순서
p = nRT
pi
E1.1(b) The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied
M/RT .
= (1.48 × 103 Torr) × (2.14 dm3)
물리화학 7판 / Atkins 정답 입니다. 정답 보고 공부 대박나길 바랍니다. 솔루션 보고 공부 대박나길 바랍니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다.
The final pressure, then, ought to be
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pf = piTf
RT
All quantities on the right are given to us except n, which can be computed from the given mass
0.500 000 0.082 0227 1.428 22
설명
p = RT
pV = nRT
E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will
and m = (1.66 × 105 mol) × (16.04 g mol−1
= 10.4 bar
n = 25 g
gas law
passes from n(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc. This
E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect
1 The properties of gases
V
4 INSTRUCTOR’S MANUAL
1 atm
(b) The original pressure in bar is
temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and
observed experimental result as the critical point is reached.
V ∝ T so
state.)
= 1.07 bar
so p = (8.31 × 10−2 L barK−1 mol−1) × (30 + 273)K
다.
