Microscopic source of pressure in an incompressible fluid

Question

The pressure of a fluid can be explained microscopically in terms of molecules bouncing off the walls of a container. The molecules have a certain mass and speed, so when they bounce they transfer a certain momentum to the wall; and since a certain number of bounces occur per unit area and time, the averaged-out result of all the bounces is pressure. So the pressure depends on the speed (temperature) and density of the molecules.

If we consider an incompressible fluid, however, we can apparently alter the pressure without changing the temperature or density. Suppose we have a container of fluid and apply force to it using a piston. Since it's incompressible, the piston doesn't move, so the density doesn't change; no work is done on the fluid, so we can't heat it up. Nevertheless the force is transmitted through the fluid "somehow" and results in an increase in pressure. A large force will cause a large increase in pressure.

Evidently the approximation of a perfectly incompressible fluid breaks down here. Pushing on it with a piston must slightly change the density, temperature, or both. How does this work? How do slight changes in these properties result in a large change in pressure, microscopically?

Answer 1

How do slight changes in these properties result in a large change in pressure, microscopically?

Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the force is great, also the pressure is great.

One way to explain this is the following. In solids, the atoms are so close to each other that the inter-atomic repulsion forces become close to comparable to intra-atomic forces between charged particles that constitute them.

The electric force between particles gets stronger at small distances, since the Coulomb law says it changes with distance of particles $r$ as $1/r^2$. To change the volume of a solid appreciably, the force per area associated with one atom would have to be comparable to the Coulomb electric force of the nucleus on the electron. Its value for atoms is so immense that it is unattainable by common standards. It would be also hard to find a container that would keep its shape and resist expansion under such great forces. Hence changes in volume by forces of common magnitude are commonly negligible.

Answer 2

Yes, the pressure increases - but that does not require any work - you moved the piston with the force f by 0mm - and you have spend fN * 0mm = 0J on it.
You could as well use a static wall instead of a piston.

Assume we have the higher pressure now.
Can we use the pressure to drive something with it - to extract work?
We open the container and make use of all the fluit that streams out of it until the original pressure is reached. The volume of that is 0, as the pressure drops immediately, without any expansion.

An interesting aspect is that it did not matter what the actual higher pressure was - or whether it existed in the first place.

We can say, as you wrote, it's true

• "we can apparently alter the pressure"

but it turned out it's not true

• "we could alter the appearent pressure".

also, as intuition may sugest.

Answer 3

First, the picture you have of the pressure can be applied both on physical boundaries (container walls, where the wall resists the pressure force) but also on any fictitious boundary you may draw inside the liquid, there it will be balanced by the same pressure on the other side.

Now, as you mention, incompressibility is a theoretical limit of extremely high compression modulus. Which means that a tiny change of volume leads to an enormous change of pressure, that is, the force that any small volume of liquid exerts on the rest of the liquid across its boundaries.

Now why does the work go to zero in this limit? Take the analogy of a spring $k$ which is clamped at one end. Apply a constant force $F$ at the other, you get a displacement $\delta$ which goes to zero as $k$ goes to infinity, so that $F$ remains the same. But the work is $F\delta = k\delta^2$, and will go to zero -- even though you were able, through this spring that you barely deform, to apply force $F$ also on the wall it is clamped to.