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32. (a) We set up a ratio using Eq. 1925:
()
2
2
2
2
2
N
Ar
Ar
2
NA
r
N
1/
2
/
.
2
/
d
dN
V
d
V
π
§·
λ
==
¨¸
λ
π
©¹
Therefore, we obtain
2
2
6
N
Ar
6
r
27.5 10 cm
1.7.
9.9 10 cm
d
d
−
−
λ
×
=
λ×
(b) Using Eq. 192 and the ideal gas law, we substitute
N
/
V = N
A
n
/
V = N
A
p
/
RT
into Eq.
19–25 and find
2
A
.
2
RT
dpN
λ=
π
Comparing (for the same species of molecule) at two different pressures and
temperatures, this leads to
22
1
11
2
.
Tp
λ
§
·
=
¨
¸
λ
©
¹
With
λ
1
= 9.9
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 Spring '08
 Any
 Physics

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