[해결책]유체역학 frankwhite 5판 solution
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작성일 20-11-29 14:51
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g kg
π
4.81E 11 4.81E5
Chapter 1 • Introduction
Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent?
m dVol (Air Vol) 4 R (Air thickness)
Then the density of air containing 1012 molecules per mm3 is, in SI units,
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3
12
순서
(0.6 kg/m )4 (6.377E6 m) (20E3 m) Ans.
kg m
Avogadro’s number 6.023E23 molecules/g mol
= ≈
[해결책]유체역학 frankwhite 5판 solution
ρ = −
and total number of molecules of air in the entire atmosphere of the earth.
ρ ρ ρ π
= = = −
atmosphere is
= = − = ⋅
Solution: The mass of one molecule of air may be computed as
mm m
m s K
10 4.81E 23
molecules g
t avg avg e
Finally, from the perfect gas law, Eq. (1.13), at 20°C = 293 K, we obtain the pressure:
−
Solution: Let Re be the earth’s radius ≈ 6377 km. Then the total mass of air in the
−
3 3
6.1E18 kg
p RT 4.81E 5 287 (293 K) .
Molecular weight 28.97 mol 1
2
⋅
Download : 유체역학 frankwhite 5판 s.zip( 53 )
m(one molecule) 4.8E 23 gm/molecule
1.1 A gas at 20°C may be rarefied if it contains less than 1012 molecules per mm3. If
다.
1.3E44
3 2
molecules
= = ≈ Ans.
m(atmosphere) 6.1E21 grams
= = ≈
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[해결책]유체역학 frankwhite 5판 solution
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Dividing by the mass of one molecule ≈ 4.8E−23 g (see Prob. 1.1 above), we obtain the
and average density 0.6 kg/m3 (see Table A-6). Use these values to estimate the total mass
1.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km
mm molecule
N
= − = −
total number of molecules in the earth’s atmosphere:
설명
m 4.81E 23 g
4.0 Pa ns
2
3 2
ρ Α
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