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2.43
Using the data shown in Table 2.1, calculate the pressure exerted by 2.500 moles of carbon
dioxide confined in a volume of 1.000 L at 450 K. Compare the pressure with that calculated
assuming ideal behavior.
For CO2, a
(p +~)
atmL2 mol-2 and b
= 3.60
Lmol-1.
Rearrange the van der Waals equation,
(V - nb) = nRT to give
an2
nRT
p=-V - nb
=
= 0.0427
V2
(2.500 mol) (0.08206 L atm K-I
mol-I)(450
(3.60 atm L2 mol-2)(2.500
K)
(1.000 L)2
1.000 L - (2.500 mol) (0.0427 L mol-I)
p
nRT
= --
V
(2.500 mol) (0.08206LatmK-1
= -----------------
mol)2
mol-I)
1.000 L
(450K)
= 92.3
atm
The pressure calculated using the van der Waals equation of state is smaller than that calculated
using the ideal gas equation. Thus, under these conditions there are net attractive forces between
molecules.
. ed re I'
Th e d eSlr
atlOns h"Ip IS Vnns ,
in terms of M and T:
He
=
Vnns, He
---=
Vnns 0 '
"Vnns
vnns He
or --'
0- = 1.Write the v nns 's in this relationship
J3~THe
He
J
' ,
=
3RT
THe
2.63
~
O,
MH
(4.003gmol-1)
= Mo~ TO, = 32.00 g mol-I
(273
+ 25) K = 37.3
K
The speeds of 12 particles (in em s-I) are 0.5, 1.5, 1.8, 1.8, 1.8, 1.8,2.0,2.5,2.5,3.0,3.5,
and 4.0. Find (a) the average speed, (b) the root-mean-square speed, and (c) the most probable
speed of these particles. Explain your results.
12
c=
LC;
;=1
N
(0.5
+ 1.5 + 1.8 + 1.8 + 1.8 + 1.8+
•.
2.0 + 2.5
12
+ 2.5 + 3.0 + 3.5 + 4.0)
cm
S-l
12
LCf
c2 = ;=1
N
(0.52
+ 1.52 + 1.82 + 1.82 + 1.82 + 1.82 + 2.ol + 2.52 + 2.52 + 3.ol + 3.52 + 4.02)
cm2 s-z
12
Cnns
(c)
crop
= 1.8 cm
S-l,
=
j7i = 2.4
cm s-l
as this is the speed that appears most frequently.
As expected, Cnns > c. However, because 12 particles do not constitute a macroscopic system,
crop can be greater or smalJer than Cnns or c.
@
2.5
Starting with the ideal-gas equation, show how you can calculate the molar mass of a gas from
a knowledge of its density.
PV =nRT
M=
m RT
V P
2.17
m
= -RT
M
= pRT
P
Nitrogen dioxide (N02) cannot be obtained in a pure form in the gas phase because it exists as
a mixture of N02 and NZ04• At 25°C and 0.98 atm, the density of this gas mixture is 2.7 g L -1.
What is the partial pressure of each gas?
Calculate the average molar mass, Mmix' of the mixture using the relation between M, p, T, and
P derived in Problem 2.5. This average molar mass is related to the mole fraction of N02, xNO"
and the mole fraction ofN204, xN,O.' The mole fractions can then be used to calculate the partial
pressures of the gases.
PmixRT
:Mnux = ---
Pmix
(2.7 g L-I) (0.08206 L atm K-1 mol-I) (273
= ----------------0.98 atm
xNo,MNO,
+ xN,o.MN,o.
xNO, (46.01 g mol-I)
+ (I
+ (I
+ xN,O. = 1, or xN,O. =
-I
1 - xNO,' Using this
- XNO,) MN,o. = 67.4 g mol-I
- XNO,) (92.02 g mol-I)
46.01xNo,
= 67.4 g mol
= 1v1mix = 67.4 g mol-I
The sum of all mole fractions is unity, that is, xNO,
relation, the above equation becomes
xNo,MNO,
+ 25) K
+ 92.02
= 67.4 g mol-I
- 92.02xNO, = 67.4
46.0lxNO, = 24.6
XNO, =0.535
PNO, = XNO,Pmix = (0.535) (0.98 atm) = 0.52 atm
PN,O. = xN,O. Pmix = (0.465) (0.98 atm) = 0.46 atm