A more useful interpretation of pressure, especially in regards to fluids (gases and liquids) is to think of it as an energy per unit volume. Notice how the units of pascals equals N/m2, and when both numerator and denominator by are multiplied by meters (m), these units become N/m2 = (N·m)/(m3) = J/m3.
Pressure and gravitational potential energy per unit volume are then terms in an energy density "conservation" equation, and they are allowed to "exchange" Pa provided the fluid is static and there is no external work being put in or taken out of the fluid. In this form, then by picking two locations in the same static fluid, an increase or decrease in the ρ·g·∆y must have a corresponding decrease or increase in pressure (∆P).
To explain what's going on here, we are going to analyze the static fluid that exists at both ground level and at a higher elevation: the air surrounding the balloon (i.e., the entire atmosphere), and not the contents of the balloon, which do not simultaneously exist at both locations (as it is "transported," and technically not a "static" fluid.)
Let's compare the air surrounding the balloon at ground level, and compare it to the air at the final higher elevation. Since the gravitational potential energy density depends on elevation, as the elevation of the balloon increases, the gravitational potential energy density of the surrounding air increases.
Looking at the energy density "conservation" equation for static fluids:
0 = ΔP + ρ·g·∆y,
Since the gravitational potential energy density of the air surrounding the balloon increases as it moves to higher elevations, then ρ·g·∆y is positive. In order to balance out this equation to equal zero on the left-hand side, the pressure of the air surrounding the balloon must have a corresponding decrease, making ΔP negative, such that:
0 = (–) + (+),
meaning that there is more air pressure at ground level than at a higher elevation. Essentially the pressure within the balloon remains constant, and because it is surrounded with lower pressure air at a higher elevation, the balloon will expand in size, and eventually "pop."
FB = ρ·g·V,
where ρ is the density of the surrounding fluid (water), and volume V is the entire volume of the diver, as he is fully submerged.
Here, since the object (the submerged diver) is floating underwater, Newton's first law applies, and the downwards weight force and the upwards buoyant force balance out.
For the red ship, which Newton's law applies to its motion? How do the magnitudes of the downwards weight force and the upwards buoyant force compare? What fluid density should be put into the ρ in the buoyant force calculation?