Page 714 - 9780077418427.pdf

Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile
tiL12214_appe_643-698.indd Page 691 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 691 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles

)
(
3
Substitute this expression for mass in w = mg. lb _ 1.013 × 10 millibar
__

P = 14.7

in 2 lb _

w = ρ (L × W × H)g 14.7
2
in
−3
)
_
1.013 × 10
Since the force is the weight of the air, __ lb _ millibar

= 14.7 ( 2 ( )

w _ 14.7 in lb _

pressure = 2
A in
3
ρ (L × W × H)g
__ = 1.01 × 10 millibar

∴ pressure =

3
A = 1.01 × 10 hPa
Also A = L × W
22.6. Convert millibar to inches of mercury by using a conversion ratio.
Hence
ρ (L × W × H)g
__
pressure = P = 507 millibar

(L × W) 1,013 millibar = 29.92 in mercury
ρ Hg (L × W)
__
= P = ? in of mercury

(L × W)
(
)
29.92 in of mercury
= ρ Hg P = 507 millibar __

1,013 millibar
2
Note: To determine pressure in units of N/m , mass units must 29.92 in of mercury
)
(

be kilograms and length units must be meters. = 507 _ millibar ( __ )
1,013
millibar
H = 32 km pressure = ρ Hg = 14.97 in of mercury
kg
_

ρ = 1.2 Convert H to meters: 22.7. Rearrange the equation to solve for volume at the pressure at
m 3
3
pressure = ? _ ) 5.6 km altitude.
1 × 10 m
32 km (

km N _ _

P 1 V 1
4
3.2 × 10 m P 1 = 2.5 P 1 V 1 = P 2 V 2 ∴ V 2 = P 2
2
m
kg
_ 4 m _ V 1 = 1.0 m 3 N _
2)
(
2)
3

pressure = 1.2 (3.2 × 10 m) 9.8 ( 2.5 (1.0 m )

m 3 s N _ __
cm

P 2 = 10.0 V 2 =
2
N _
kg ⋅ m
2)
_ m ( 5.0

2
s
4

= 1.2(3.2 × 10 )(9.8)(m) _ V 2 = ? cm

m 3 2 ( N _ 2) 3
5 N _ (cm )
_
_ cm

= 3.8 × 10 = (2.5)(1.0)

m 2 (10.0) ( N _ 2)

22.4. Pressure is defined as a force per unit area. The force is due to
cm
the weight of a column of air overlying the point being = 0.25 m 3
considered. If density is constant, the mass of the air can be
22.8. Rearrange the equation to solve for volume at the pressure at
determined from its density. Use the relationship between
5.6 km altitude.
pressure, density, and height from the equations for density and
weight derived in exercise 22.3. Then, rearrange this P 1 = 1,013 hPa _

P 1 V 1
relationship to solve height. V 1 = 250.0 cm 3 P 1 V 1 = P 2 V 2 ∴ P 2 = V 2
2
Note: To determine pressure in units of N/m , mass units V 2 = 450.0 cm 3 __
3
(1,013 hPa)(250.0 cm )

must be kilograms and length units must be meters. P 2 = ? P 2 = (450.0 cm )
3
3
(1,013)(250.0) (hPa)(cm )
N _ pressure __ _

pressure = 10.0 pressure = ρ Hg ∴ H = _ = 3

m 2 ρ g (450.0) (cm )
N _
kg
_ 10.0 = 563 hPa

ρ = 1.2 __
2
m
m 3 H = 22.9. Use Wein’s displacement law equation to solve for wavelength.
kg
m _
_
H = ? 1.2 9.8
3 (
2)

m s T = 6,000 K λ Peak = __
−3
2.897 × 10 K ⋅ m

N _ λ peak = ? T
−3
2.897 × 10 K ⋅ m
2
_ _ = __
10.0

m

= 6,000 K
(1.2)(9.8) kg
_ m _
3 ( 2 )

2.897 × 10 _

−3
_ K ⋅ m

m s =
6,000 K
−1
= 8.5 × 10 m = 4.8 × 10 m
−7
2
22.5. Convert lb/in to millibars, using the conversion factor inside 22.10. Determine the reflected solar radiation by multiplying the

the front cover of the textbook. percentage reflected by clouds from Figure 22.6 by the

lb _ lb _ incoming solar radiation. Divide the total incoming solar
P = 14.7 1.013 bar = 14.7
in 2 in 2
lb _
1.013 bar = 14.7 Convert bar to millibar:
in 2
P = ? millibar _
(
1 millibar

1.013 bar −3 )
1 × 10 bar
3
1.013 × 10 millibar
lb _
3
∴ 1.013 × 10 millibar = 14.7
in 2
E-49 APPENDIX E Solutions for Group A Parallel Exercises 691