Question

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 4.4 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.26 × 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.What is the maximum height h in meters above the ground that the container can be lifted before bursting?

Answer 1

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

• initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

• atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

• initial volume, $V_0=4.4\times 10^{-4}\ m^3$

• maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

• density of the air, $\rho_a=1.2\ kg.m^{-3}$

• density of sea water, $\rho_s=1.2\ kg.m^{-3}$

The relation between the change in pressure with height is given as:

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$= standard value of gravity

Now putting the respective values:

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (Since the density of air and the density of sea water are assumed to be constant.)