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