fluid_mechanics

1.5 Ideal Gas Law 13
In the ideal gas law,
absolute pressures
and temperatures
must be used.
an ideal gas. It is known to closely approximate the behavior of real gases under normal conditions
when the gases are not approaching liquefaction.
Pressure in a fluid at rest is defined as the normal force per unit area exerted on a plane surface
1real or imaginary2 immersed in a fluid and is created by the bombardment of the surface with the fluid
molecules. From the definition, pressure has the dimension of FL 2 , and in BG units is expressed as
lbft 2 1psf2 or lbin. 2 1psi2 and in SI units as Nm 2 . In SI, 1 Nm 2 defined as a pascal, abbreviated as
Pa, and pressures are commonly specified in pascals. The pressure in the ideal gas law must be expressed
as an absolute pressure, denoted (abs), which means that it is measured relative to absolute
zero pressure 1a pressure that would only occur in a perfect vacuum2. Standard sea-level atmospheric
pressure 1by international agreement2 is 14.696 psi 1abs2 or 101.33 kPa 1abs2. For most calculations these
pressures can be rounded to 14.7 psi and 101 kPa, respectively. In engineering it is common practice
to measure pressure relative to the local atmospheric pressure, and when measured in this fashion it is
called gage pressure. Thus, the absolute pressure can be obtained from the gage pressure by adding the
value of the atmospheric pressure. For example, as shown by the figure in the margin on the next page,
a pressure of 30 psi 1gage2 in a tire is equal to 44.7 psi 1abs2 at standard atmospheric pressure. Pressure
is a particularly important fluid characteristic and it will be discussed more fully in the next chapter.
E XAMPLE 1.3
Ideal Gas Law
GIVEN The compressed air tank shown in Fig. E1.3a has a
volume of 0.84 ft 3 . The temperature is 70 °F and the atmospheric
pressure is 14.7 psi 1abs2.
FIND When the tank is filled with air at a gage pressure of 50 psi,
determine the density of the air and the weight of air in the tank.
SOLUTION
The air density can be obtained from the ideal gas law 1Eq. 1.82
so that
r p
RT
r 150 lb in. 2 14.7 lbin. 2 21144 in. 2 ft 2 2
11716 ft # lbslug # °R23170 4602°R4
0.0102 slugsft 3
(Ans)
Note that both the pressure and temperature were changed to absolute
values.
F I G U R E E1.3a
Jenny Products, Inc.)
(Photograph courtesy of
W, lb
0.5
0.4
0.3
0.2
0.1
0
–20 0 20 40
p, psi
F I G U R E E1.3b
(50 psi, 0.276 lb)
60 80 100
The weight, w, of the air is equal to
w rg 1volume2
10.0102 slugft 3 2132.2 ft s 2 210.84 ft 3 2
0.276 slug # fts 2
so that since 1 lb 1 slug # ft s 2
w 0.276 lb
(Ans)
COMMENT By repeating the calculations for various values of
the pressure, p, the results shown in Fig. E1.3b are obtained. Note
that doubling the gage pressure does not double the amount of air
in the tank, but doubling the absolute pressure does. Thus, a scuba
diving tank at a gage pressure of 100 psi does not contain twice the
amount of air as when the gage reads 50 psi.