Liquids - does increased pressure increase "internal energy"?

Question

I do not have much experience with working with liquids so I am basing a lot of my intuition on what I know about gases (which I have a feeling is not good, but gases are the closest thing to liquids that I feel like I understand decently :D).

I have just stumbled upon something that I can not wrap my head around. With gases we have the internal energy $U = \frac{3}{2}nk_BT = \frac{3}{2}pV$ (for monatomic gases) from the ideal gas law. Intuitively I would guess that this finding would generalise into taking a volume integral of the pressure if it is not constant (for example with a very very high cillinder with air inside, the pressure at the bottom would be higher than at the top). And I can imagine this energy transfering into a different kind of energy, for example by pushing a piston.

However, with water or ideal liquids I am not really sure. If I have a tank full of an ideal liquid that I seal on the top and apply a huge pressure on, the seal will not move (as opposed to the gas case), so I do zero work on the system? So the energy increase of the water should be zero? But in the derivation of Bernoulli's principle the pressure plays an important role and is in the end a part of the equation. There we have a term that is clearly based on the kinetic energy of the volume element ($\frac{1}{2}\rho v^2$), another one based on its potential energy ($\rho g h$), and the last one on "pressure energy"(?) - I have seen it called like that.

I also feel like pressure should add some sort of energy even into incompressible liquids - if I once again had a sealed tank, made from for example glass, with the seal pointing towards the floor and the glass bottom pointing towards the sky and I applied some pressure to the seal and gradually increased it (always locking the press in place so it does not jump), eventually the glass top would break - here I would expect some shards to come flying off, with energy being delivered from (I assume) the freed pressure?

What is correct here and/or what am I missing in each of the cases (I feel like some of them contradict others, while all of them make sense to me :D)? Is the situation different in case of an ideal liquid and in the case of water (which is a tiny bit compressible)? Thanks a lot!

Answer 1

With gases, when you apply pressure, the gas compresses, and energy is transferred into the system, often as work done on it. In liquids, since they're almost incompressible, applying pressure doesn’t change their volume, so no work is done in the same way. However, pressure still stores energy in the system, which can be released if, for example, a pressurized container breaks. This energy is reflected in the pressure term of Bernoulli’s principle, even for incompressible liquids. The key point is that pressure can store energy, even if the liquid doesn’t compress much.

Answer 2

However, with water or ideal liquids I am not really sure. If I have a tank full of an ideal liquid that I seal on the top and apply a huge pressure on, the seal will not move (as opposed to the gas case), so I do zero work on the system?

Correct.

So the energy increase of the water should be zero?

Correct, if it was incompressible. In reality water compresses a little, so some small work is stored in the water (much less than the so-called "pressure energy", which is actually not energy of water, but a work term).

But in the derivation of Bernoulli's principle the pressure plays an important role and is in the end a part of the equation. There we have a term that is clearly based on the kinetic energy of the volume element ($\frac{1}{2}\rho v^2$), another one based on its potential energy ($\rho g h$), and the last one on "pressure energy"(?) - I have seen it called like that.

Correct. However, the pressure term in the Bernoulli equation does not represent energy stored in the liquid element, but it represents work that will be done on this element by surrounding liquid, when the element moves from its location to another location with zero pressure. This is clear from the usual derivation of the Bernoulli equation, which considers work of pressure forces on incompressible fluid in a streamtube.

I also feel like pressure should add some sort of energy even into incompressible liquids - if I once again had a sealed tank, made from for example glass, with the seal pointing towards the floor and the glass bottom pointing towards the sky and I applied some pressure to the seal and gradually increased it (always locking the press in place so it does not jump), eventually the glass top would break - here I would expect some shards to come flying off, with energy being delivered from (I assume) the freed pressure?

In case of perfectly incompressible fluid, no - if its temperature does not change, it can't store any internal energy due to increased pressure, irrespective of value of this pressure. The kinetic energy of the shards and fluid comes from work done on them by the surrounding fluid, and this is the same as work of the source of pressure on the fluid moving elsewhere at the same time. The incompressible fluid just transfers this work from the source to the accelerated shards and fluid, the fluid in between stores no energy.

I've tried to explain this in other past answers, see here:

How is Bernoulli's equation a statement of conservation of energy?

What is the term $P$ in bernoulli's equation defined as?

What is Pressure Energy?

Answer 3

It might help to get quantitative.

You’re asking about $\left(\frac{\partial U}{\partial P}\right)_T$, the change in internal energy induced by a pressure change at constant temperature.

From the fundamental relation, this is

$$\begin{align}\left(\frac{\partial (T\,dS-P\,dV)}{\partial P}\right)_T&=T\left(\frac{\partial S}{\partial P}\right)_T-P\left(\frac{\partial V}{\partial P}\right)_T\\&= -T\left(\frac{\partial V}{\partial T}\right)_P-P \left(\frac{\partial V}{\partial P}\right)_T\\&=-TV\alpha+\frac{PV}{K} \end{align},$$

where I’ve used a Maxwell relation and where $\alpha$ is the thermal expansion coefficient and $K$ the bulk modulus.

For the ideal gas, $\alpha=\frac{1}{T}$ and $K=P$, and there is no internal energy increase from isothermal compression; the ideal gas is not an enthalpic spring but an entropic spring, storing not strain energy but low entropy (or free energy).

For other materials, such as condensed matter, strain energy is stored as mediated by both material properties. (Incompressibility implies that $\alpha\to 0$ and $K\to\infty$, but this is an idealization; no material is truly incompressible. The bulk modulus of water, say, is around 2 GPa, much larger than that of familiar gases but much smaller than that of familiar solids.)