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Pascal's law states that a change in pressure in an enclosed body of a fluid gets equally disturbed throughout the entirety of the fluid without getting diminished.

My question is.. is the "Without getting diminished" part always true?. Like it is true here on earth for a small container where density and g are almost constant.

Imagine a fluid enclosed in a container that is a million kilometer³ in volume in vast empty space, without any nearby stars or anything so it is devoid of Gravitational effects. Let's say the fluid is filled to the brim. If we apply 1Pa pressure on one end, will the atoms in the other end a million km away feel the same pressure? Shouldnt the pressure diminish atleast by a tiny bit cause pushing all the way, some atoms absorb some of the force, as if heat or something. Will the pressure not get diminished, atleast by small amount?

I was just curious if Pascal's law is a universal law or if it just works very good enough in real life purposes here on earth so we call it a law.

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It's an approximation, because formulated for incompressible liquids. All liquids are compressible to some degree. You can calculate compressibility factor with the help of speed of sound $c$ : $$ \beta = \frac {1}{\rho c^2} \tag 1$$

For example water has compressibility factor of $$ \beta_{water} = \frac {1}{1 g/cm^3 \cdot (1481 m/s)^2} = 4.5 ×10^{-10} Pa^{−1} \tag 2,$$

Meaning that by applying $1~\text{Pa}$ to water of some volume $V$ uniformly, it will reduce volume by ratio $\frac {\Delta V}{V} = \frac {1~\text{part}}{2.2~\text{billion}}$. So compressibility of water (or liquids in general) for all practical means is very weak, so in most cases liquids can be considered as incompressible. But, of course there are situations when this can't be ignored anymore, for example if you want to solve Navier-Stokes equations with great precision.

If, we will consider liquid as a compressible thing, then applied pressure goes for compression as well, and so other parts get's a bit less pressure propagated, that's why we need "without getting diminished" assumption part of incompressible liquid in Pascal's law.

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    $\begingroup$ You might want to change "fluid" to "liquid" in your answer as fluids can mean liquids as well as gasses which deficiently are compressible. $\endgroup$
    – M. Enns
    Commented Nov 13, 2023 at 22:50
  • $\begingroup$ Thanks, fixed. My native language is not English, so sometimes I forget ambiguity of some words, or redundancy of some words in English. For example in my native Lithuanian language we don't have separate terms for distance change rate amplitude like "speed" and same thing as vector quantity which then is meant to be "velocity". We just have 1 term "greitis" which depending on context means both things :-) Anyway, wikipedia on Pascal law mentions "incompressible fluid", so probably it's a bug in wiki site too. $\endgroup$ Commented Nov 14, 2023 at 7:04
  • $\begingroup$ On the other hand, according to (1) compressibility factor for air is $\approx 1/100000$, so it can be said that squeezing air- doesn't change it's density much. So in this respect air is also not very compressible gas. $\endgroup$ Commented Nov 14, 2023 at 7:42
  • $\begingroup$ I'm not sure this is correct: How yould elastic compression eat up pressure? It takes some of the strain, but the compressed air or fluid still pushes on the surroundings with the same pressure, doesn't it? $\endgroup$
    – Toffomat
    Commented Nov 15, 2023 at 8:02
  • $\begingroup$ Try to build hydraulic press with air or water inside. Which one is more easy to operate ? When you will apply $1~Pa$ to air-based hydraulic press, will you get 1 Pa in press output immediately ? If not , where thermodynamic work $P \Delta V$ has gone ? Not into compression of gases ? And if you compress gasses, as per (1) you make compressibility factor of them smaller. Until the point that gasses may become liquefied (at extreme cases) or act "more like liquid" overall. Almost anything can be compressed into solid phase, so what ? What's the point of compressibility comparison then ? $\endgroup$ Commented Nov 15, 2023 at 12:02

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