Answer:
0.71 lbf
Step-by-step explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
P = 32 lbf/in² + 14.7 lbf/in²
P = 46.7 lbf/in²
Absolute temperature is in Kelvin or Rankine:
T = 75 + 459.67 R
T = 534.67 R
Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:
PV = nRT
(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)
n = 0.02442 lb-mol
The molar mass of air is 29 lbm/lb-mol, so the mass is:
m = (0.02442 lb-mol) (29 lbm/lb-mol)
m = 0.708 lbm
The weight of 1 lbm is lbf.
W = 0.708 lbf
Rounded to two significant figures, the weight of the air is 0.71 lbf.