(I didn't even have a basic formal education in physics. I'm learning through the internet out of my own interest, so if there are any silly mistakes, kindly bear with and guide me through.)

Everywhere, everyone is saying that pressure will same be for a given height. How is it possible? When volume changes, doesn't pressure change in calculations?

[Pressure will be 2.5 bars approximately (including atmospheric pressure) for a height of 15 metres from ground level. I checked on many sites that 2.5 bars equals 2.5 kg/cm².]

Now let's take base area as 2000 cm² and height of 15 m as constant in three scenarios:


In scenario 1 a straight vertical tube from ground level will have capacity of 3000 litres with a weight of 2.5 kg per cm² [3000 litres / 2000 cm² = 1.5 kg/cm² + 1 kg/cm² atmospheric pressure].

In scenario 2 the tube becomes narrow from the base area, resulting in a volume of 2000 litres.

In scenario 3 the tube becomes wide from the base area, resulting in a volume of 4000 litres


When they say pressure remains the same at a given height, does it mean that weight will be 2.5 kg/cm² in all 3 scenarios, where volume is 3000, 2000, or 4000 litres (base area 2000 cm², height 15 m constant)? How is it possible? Where am I wrong?

  • $\begingroup$ Hi @VividV: Concerning your latest edits: We appreciate your clean-up effort, but try to limit edits of old questions to only a handful at the time not to clog the front page, and try to make every edit count, i.e. as substantial as possible. Minor edits should be avoided. $\endgroup$ – Qmechanic Jul 28 '14 at 23:49

If you replace the fluid scenarios with solids with matching have the same shapes, then indeed the pressure at bottom would be the weight of each solid divided by base area, i.e., they would be different! Then what's different about the fluid case?

I presume in scenario 1 you have a prism, in scenario 2 you have a bottom-heavy frustum, and in scenario 3 you have a top-heavy frustum.

What's special for the fluids is that pressure acts in all direction. It is important to note that the fluid is in a container. At each point of the fluid-container interface, the fluid pushes against the container (normal to surface). But since there is no net acceleration on the fluid, so the container must push back with an equal and opposing force.

In scenario 3, the container side wall pushes diagonally-up, thus supporting the extra weight, and making the pressure at the bottom less than that of the solid case. In scenario 2, the container wall pushes diagonally-down, thus supplying the downward force in addition to the weight of the fluid itself. In the end the force contribution of the container walls make the pressure at the bottom equal for all 3 scenarios.

Now, for an arbitrary container filled with fluid that are connected, a useful intuition is to think of the fluid as being "carved out" from a large pool of fluid. The fluid-container interface acts identically as the fluid-fluid interface in the pool, so that at each internal point, the pressure depends only on depth, and not the container's geometry.

Of course, the effect of gravity means that each small "unit" of fluid presses its own weight downward, thereby increasing pressure as you move downward.

  • $\begingroup$ ya, your frustum part was spot on. So u have clarified that pressure will be same, for a given height, irrespective of shape of container. Thanks a lot. But does it mean that weight exerted on the base area will also be same, i.e 2.5 kg/cm2 since pressure is same i.e 2.5 bar for a height of 15 metres? $\endgroup$ – user52187 Jul 14 '14 at 8:41
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    $\begingroup$ If by base area you mean between the container and the ground, then it would be the solid case again. Neglecting the container's mass, the net pressure on the container (including upward) would conspire to make up the difference. Note that were also ignoring atmospheric pressure. That depends on how we measure (relative) pressure, and may come into play in some phenomena. $\endgroup$ – arccosh Jul 14 '14 at 13:15

Consider this diagram showing the three columns you describe all connected to the same body of water:


Your question asks whether the three pressures $P_1$, $P_2$ and $P_3$ will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if $P_1 > P_2$ then water would flow from the base of column 1 to the base of column 2 until the pressures became equal.

OK, but the next question is whether the heights of all the columns are the same. Again the answer is yes because if they weren't we could connect the tops of the columns, let water flow between them and we would have a perpetual motion machine.

So we conclude that the pressure is only related to the height $h$ and does not depend on the shape of the column. Specifically, the relationship between pressure and height is:

$$ P = \rho g h $$

where $\rho$ is the density of the fluid and $g$ is the gravitational acceleration (9.81 m/sec$^2$).

  • $\begingroup$ Nice mention of perpetual motion there, reminds me of this ridiculous video youtube.com/watch?v=287qd4uI7-E (the first clip is the one pertinent to this question). By the way, another way to explain this would be doing the typical derivation that uses Newton's laws and the definition of pressure to arrive at that equation. Some people don't seem to like it, but it worked for me. $\endgroup$ – Physics Llama Jul 14 '14 at 7:50
  • $\begingroup$ @John Rennie Thanks. But i had asked for separate containers with same base area, same height, but different shape resulting in different volume. $\endgroup$ – user52187 Jul 14 '14 at 8:47
  • $\begingroup$ @User52187: You can redraw John's diagram with equal base areas. Johns remarks then apply just the same. $\endgroup$ – RedGrittyBrick Jul 14 '14 at 8:49
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    $\begingroup$ @user52187: the base area doesn't matter. My argument that the pressure must be the same is independant of the base area. If you really want the base areas to be the same just imagine the right hand colum expanded in width to match the base area of the other two. $\endgroup$ – John Rennie Jul 14 '14 at 8:51
  • $\begingroup$ Oh k. Understood now. ty $\endgroup$ – user52187 Jul 14 '14 at 8:55

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