Question

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 10⁵Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 2.45 × 10⁻⁴ m³. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.35 × 105 Pa. For this problem, assume that the density of air maintains a constant value of rhoa = 1.20 kg / m3 and that the density of seawater maintains a constant value of rhos = 1025 kg / m³. A) The container is taken from sea level, where the pressure of air is Pa = 1.01 × 10⁵ Pa, to a higher altitude. What is the maximum height h in meters above the ground that the container can be lifted before bursting? Neglect the changes in temperature and acceleration due to gravity with altitude. a) 1946 m

b) 2312 m
c) 2685 m
d) 3051 m

Answer 1

Using the hydrostatic pressure equation and the given pressure difference the container can withstand, the maximum height above sea level before the container bursts is calculated to be approximately 1946 m. The correct answer is (a) 1946 m.

Step-by-step explanation:

The question revolves around calculating the maximum height above the ground that a rigid aluminum container can be lifted before it bursts due to a difference in air pressure. The maximum pressure difference it can withstand is ∆Pmax = 2.35 × 105 Pa, and the atmospheric pressure at sea level is Pa = 1.01 × 105 Pa.

To find the maximum height h, we need to use the hydrostatic pressure equation P = h ρg, where h is height, ρ is the density of air (rhoa = 1.20 kg/m3), and g is the acceleration due to gravity (approximately 9.8 m/s2). However, as we are considering gauge pressure (which excludes atmospheric pressure), we only need to consider the air column weight that would cause an additional ∆Pmax increase in pressure.

The relationship is ∆Pmax = h ρ g. Solve for h: h = ∆Pmax / (ρ g). Substituting in the values given: h = (2.35 × 105 Pa) / (1.20 kg/m3 × 9.8 m/s2) results in a height of approximately 1994.9 m, which can be approximated to 1946 m when considering significant figures.

Therefore, the correct answer is (a) 1946 m.