# Understanding pressure in a liquid

1. Jun 16, 2014

### Alexander83

I've always found pressure fairly intuitive to understand when dealing with ideal gases, but am struggling a little with a thought experiment concerning pressure changes in a liquid.

Here's a simple thought experiment I'm struggling with. I'm thinking of a still volume of pure water at constant temperature in the Earth's gravitational field. The water has a constant temperature and thus, its density should be constant throughout. Per the hydrostatic equilibrium equation, the pressure in the volume of water should increase as one moves deeper in the water column. Given that water is taken to be incompressible, this change in pressure should not affect its density.

My conceptual difficulty is in understanding what's happening at a molecular level to cause this increase in pressure. I keep wanting to think of the ideal gas case where pressure can be visualized as a net force per unit area exerted over a hypothetical surface in the fluid owing to random molecular collisions. Trying to apply this model to the water would seem to imply that, in order for the pressure to increase with depth one of two things should occur. Either the density would have to increase in order to have more molecules collide with the surface (this violates the incompressibility assumption). Or, the average speed of the molecules at constant density would have to increase (this violates the constant temperature assumption).

I'm left with one of two conclusions.

1. Either my simple model of an incompressible, constant temperature liquid in a gravitational field is impossible, OR

2. My model of what "causes pressure" in a liquid like water is wrong. Can pressure in a liquid still be visualized as due to the effects of random molecular collisions as in an ideal gas? If not, what's the appropriate mental model?

Alex

2. Jun 16, 2014

I don't know why you think that. As the depth increases, the amount of particles above a particular surface will increase. So the force exerted over the Area will increase.
In fact, the pressure equation is derived from this.
We know that $P=\rho g h$
This is derived like this:
$P=\frac{F}{A}$
$P=\frac{\rho \times A_b \times h \times g }{A_b}$
The two $A_b$ cancel out and we get $P=\rho \times g \times h$

3. Jun 16, 2014

### jbriggs444

The point is that the interaction between adjacent parcels in a fluid is purely a local phenomenon. If the state of a parcel 1000 meters deep beneath the ocean's surface is the same as the state of a parcel 1 meter deep beneath the ocean's surface (same temperature, same density, stationary) then one would expect that the pressure they exert on their neighboring parcels should also be identical.

The original poster's question is a sound one.

One answer is that water is not incompressible. It is only close to incompressible. So the state of a parcel 1000 meters deep is different from one at the surface. It is slightly more dense. Liquid water is not an ideal gas (obviously!!). A slight increase in density manifests as a large increase in pressure.

4. Jun 17, 2014

### Alexander83

Thanks, jbriggs444! I think you've got at the answer to the question. Thanks for making the point about local interactions - that got at the core of my question and you managed to express that more eloquently than I did in the original question.

Alex

5. Jun 17, 2014

### 256bits

You can also think of the increase in pressure ( for a gas ) as being as result of not more molecules colliding the surface, but of a molecule colliding more often. Since, if we take a volume with a certain number of molecules and decrease the volume, we still have the same number of molecules. Your description considers keeping the volume constant and adding more molecules.

As an example, a ball representing a molecule bouncing around in a box with collisions elastic, will have a contact with a wall that we can easily determine by knowing the velocity of the ball and the length of sides of the box. If we now move the walls inward and therby decrease the volume of the box, and barring any momentum transfer to the ball from the wall, the frequency of contact with the walls of the ball has to increase.

Question is to what point can the volume of the box be reduced? What is the relationship between volume decrease and frequency of contact? Can we even reduce the volume of the box to a small enough volume that the sides are touching the ball at both radii, or even less?

Molecules in a liquid, or solid, are in the same predicament. They are enclosed within a box of other molecules around them, and the electrons of one molecule repels the electrons of the other and they all really do not have any where to go except to just jostle each other around. So then comes along someone who craftly decides to decrease their volume even further. The molecules aren't too happy and complain by resisting and increase their pressure on one another significantly.

Here is a bit on Van der Waals forces between molecules and atoms.
http://www.thermopedia.com/content/1233/
which is self descriptive.
Note though that as the distance beteen particles decreases, the force increases dramatically.

And another on degeneracy pressure
http://www.eng.fsu.edu/~dommelen/quantum/style_a/cboxdp.html
where quantum physics comes to the rescue.

Hopefully that gives you a bit of thought and means for doing some research on aspects of thermal pressure, that we normally associate with, and degeneracy pressure.

6. Jun 18, 2014

### mic*

Adjacent gives the better answer. Water having a higher "head of pressure" as depth measurement increases should not be thought of as a result of water becoming slightly more compressed.

The pressure is due to gravity. If you have five people on your shoulders, you feel more potential energy above you than if you have one. Your height does not change in either circumstance.

7. Jun 18, 2014

### jbriggs444

All other things being equal, yes your height does change when the weight on your shoulders increases.

8. Jun 18, 2014

### 256bits

A better answer in which way. To calculate the pressure at a depth of water is the easy part.

9. Jun 18, 2014

Must agree with 256 and Alexander. You can create high water pressures without a huge head of water above. Take a sideways piston or something, just so it is clear the "push" does not have to be from above. Also with increasing pressure gas will at some point liquefy and be "incompressible", but you can still increase its pressure. This is just because most of us do not have the equipment to measure its compression accurately on that scale, where the slope of the potential energy gets very steep.

Long answer short: no such thing as incompressible fluids, they are a macroscopic approximation.

10. Jun 18, 2014

### Staff: Mentor

Not if it's above its critical temperature.

Chet

11. Jun 18, 2014

Caught me there Chet! Still I believe at some point the gas will become "incompressible" on the macroscape, but will retain the property of expanding to fill its container if at supercritical temperature. Am I wrong? The supercritical condition as I recall just means that liquid and gas are indistinguishable since their densities coincide at that temperature and above, I.e. Single phase.

12. Jun 18, 2014

### Staff: Mentor

Yes. Right on.

Chet

13. Jun 19, 2014

### mic*

The above quote is the part that I highlight in the OP that seems a bit back to front. Anything happening at a molecular level is an effect from the height of the water column - ie gravity. It is not the cause of any pressure.

The piston scenario provided by crador highlights this well. For simplicity lets call it hydraulics. If you push a (essentially incompressible) fluid through a tube/line, the pressure (force x area) applied by the piston is equal to the pressure (force x area) that can be recaptured at the other end of the line - discounting any loss due to friction and bends in the hydraulic line.

While it may be argued that hydraulic fluids are compressible, any degree to which they are is actually detracts from the apparent pressure that can be recaptured at the end of the line. Certainly any such compression does not cause the pressure at the end of the line. The cause is the piston force.

In the case of the water column, more pressure at a greater depth is caused by more Newtons of water.

14. Jun 19, 2014

I believe the piston force causes the compression, which effects a higher pressure. That is: the piston force effects a compression until the pressure due to said compression balances said piston force. Or else compression would continue.

To my knowledge the main idea is that a sample of water in a certain state will have a certain temp, pressure, volume etc. how you get this sample to this state is immaterial, and the properties of the sample as named are direct results of molecular motion (statistical mechanics I believe). So the pressure of the sample is indeed an effect of the molecular motions and other such minutiae. But you need some forces to bring the water sample into this state and hold it there, unless you are god (which many of us are in our own thought experiments) and can set that state as an initial condition without any explanation of how it got there (and it will decompress as time starts).

I think the difficulty here is some of us argue as physicists (little gods), while some of us are engineers (practical mortals). The former say the piston/water column isn't necessary, it is just an experimental element external to the sample of interest which has properties regardless of its state's origin. The latter say the piston/water column or some such thing is always necessary practically, so that is what is causing said pressure.

15. Jun 20, 2014

### 256bits

Good explanation. I see your point of view.

16. Jun 20, 2014

Nicely put.