# Why is variation of pressure with depth in a static liquid not accompanied by variation of temperature with depth?

Pressure at a depth h exerted by a static liquid of density ρ on the walls of the container is given by ρgh. That is, it increases with depth. This pressure, as is the general opinion, is because of collisions of fluid particles, that are in thermal agitation, with the wall of the container.

However, if this is true, then since pressure increases with depth, so should the magnitude of thermal agitation of of the particles ( is there any other reason to account for the increase in pressure?). Therefore the temperature, a measure of thermal agitation, should also increase with depth. But, this is not observed anywhere.

My opinion is that the pressure exerted by a liquid on the walls of the container is due to the fact that the system of fluid particles is being sheared to the sides, against the wall of container by the forces of gravity (much like grains of sand filled in a tumbler) and not completely due to thermal agitation of particles. The walls of the container are resisting the fluid particles from being displaced and causing the whole system to lose shape. Hence, there is force and thus pressure on the wall. Is this view justified?

If my view is justified then what fraction of total pressure on the wall is contributed by the thermal agitation of particles (since it may not be possible to completely deny that there are some forces on the wall due to thermal agitation) ?

Someone might argue that the horizontal pressure on the walls cannot be due to vertical forces of gravity that have no horizontal component. But I guess a liquid not confined in a vessel (i.e, if the wall were to suddenly disappear) is much like a heap of small particles (as in grains of sand) being squeezed out of shape, due to lack of high order (seen in lattice structures), under the forces of gravity and the normal force provided by the surface on which the heap rests.

Please forgive me for the verbose description. I just wanted to make my point clear.

PS 1 : I came across a question with an answer relevant to mine here:

Pressure in fluids, in particular horizontal pressure

Going back to your first question to sum up. Horizontal pressure in liquid results from two factors - the kinetic energy of molecules (which makes them move chaotically in all directions) and the Earth's gravity (as molecules are pulled down by gravity and look for escape sideways).

But the fraction of contribution of the two factors is not discussed. Also, if there are indeed two factors that contribute to horizontal pressure as mentioned, then why is it that the derivation of expression of pressure at a depth h in a static liquid does not seem to take into account the "kinetic energy factor"?

PS 2: As I pointed out in a comment to @FGSUZ's answer, I am aware that that the layers of liquid are in thermal contact and therefore in thermal equilibrium (which someone might think is the reason there is no variation in temperature). However if the thermal agitation model is applied then since the whole system is at the same temperature, there should be equal pressure everywhere which is not the case. Hence, I just want to know how valid is this thermal agitation model as opposed to the model I proposed?

• It may help your investigation to know that there are two extremes to compressing something: isothermally and isentropically. Isothermal compression involves thermal contact with a large thermal reservoir. Isentropic compression occurs when an thermally isolated system is compressed reversibly; this process does heat up the system. Nov 25, 2017 at 19:48
• It may also help to distinguish heating a system from doing work on (e.g., compressing) it. When you heat a system, you increase the breadth of particle energies; when you do work on it, you elevate all the particle energies. That's why a fast-moving system isn't intrinsically hotter, even though all the particles have more energy. Nov 25, 2017 at 19:48

According to Pascal's law, the pressure affects all directions, so I wouldn't make any distinction between horizontal and vertical pressure. The change (in hydrostatics) is only given by your $\rho gh$ formula.

Okay, more pressure should imply more temperature, but this temeprature change is completely negligible for most practical cases. Just take a book and press it against two other books (all of them at the same temperature). Then measure temperature again. You won't find any significant change, for sure.

That's because you're producing a very macroscopic force that creates a small elastic deformation. Once you stop making the force, the material liberates and gets to the same original shape. This is barely trasnferred to kinetic energy of the single molecules. You would have to somehow excite normal modes of vibration if you want to heat up the material. That's how a microwave works.

• How can a system exist with this kind of temperature difference ( even though negligible) ? Shouldn't it reach thermal equilibrium eventually?
– AVU
Nov 25, 2017 at 12:14
• In your "book-pressing" argument, you have compared the temperatures of the system of books at different times (before and after pressing) while my question addresses the temperature of systems( here, the layers of liquid stacked one above the other and hence in physical(thus thermal) contact) at the same given instant. I hope you get my point. :)
– AVU
Nov 25, 2017 at 12:22
• Okay, consider a single book, do ou expect lower pages be warmer because the weight of the upper pages is pressing them down? OF course not, because, besides the pressure isn't that high (big surface), the deformation is extremely small and it can be seen as potential energy, not kinetic one. Nov 25, 2017 at 12:55
• That exactly was the point I was trying to make in the question! I (who is against the thermal agitation model) said the pressure exerted on the wall is not due to thermal agitation of molecules (which you see as kinetic energy) but rather due to the very weight of the liquid ( which, I guess, you think of as potential energy) .While your answer does say that the variation in temperature is not seen which really is the case ( in my opinion), it doesn't say explicitly or explain why the thermal agitation model is correct or wrong. Let me know if you too are against the thermal agitation model.
– AVU
Nov 25, 2017 at 13:08
• Umm, no, I'm not haha. I admit I've never measured velocity of molecules, they're too small for me... but there are too many experiments, and too many models that have been successfully tested in labs... the thermal agitation model seems to be correct... If you don't believe it, then I cannot do much. By the way, this doesn't mean THIS experiment of yours must be the prove. Nov 25, 2017 at 16:27

Although pressure increases with temperature, not all pressure is related to it. In fact only for an ideal gas pressure is proportional to T. You can exert pressure on a crystal or, in principle, on liquid helium, at $$T=0$$ by compression. The returning force is due to electronic Pauli repulsion. At finite T thermal motion adds to the pressure and causes thermal expansion, which will cause pressure if the volume is kept constant. At higher T vapour pressure takes over.

Hydrostatic pressure results from compression under the effect of gravity. Temperature variation plays no role if the water is free to expand.

Think of fluid in the container as a continuum. Suppose the equation of state of the fluid is $f(p,\rho,T)=0$. Pressure does increase with depth, and equation of state then says that at least one of $\rho,T,$ must change. If the container is exposed to an ambient of constant temperature then at equilibrium the fluid will also have attained the same uniform temperature everywhere. Which means that density of fluid alone changes (it increases with depth). The mistake in your argument is that you have assumed density to be uniform. So at greater depths even though velocity distribution of molecules has remained the same (because temperature is uniform) the number density of molecules has increased which results in larger pressure.

• Is it not true that a liquid is negligibly compressible (as in the case of water). So the variation in density is also negligible, I guess?
– AVU
Nov 25, 2017 at 11:20
• @AVU Negligible but not zero.
– Deep
Nov 26, 2017 at 5:43