At 10 m under water, pressure is 2 atmospheres. As temperature is not different, and density is minimally different, how does pressure increase at the molecular level (collisions per unit area)?

  • 1
    $\begingroup$ What are your thoughts on it? $\endgroup$ – Kyle Kanos Jan 26 '15 at 4:05
  • $\begingroup$ There is more force behind each individual collision despite the fact that each individual water molecule is not moving faster. $\endgroup$ – Brinn Belyea Jan 26 '15 at 15:29

Your intuition about pressure is coming from $PV=nRT$, the ideal gas law. In that, more molecules (n) or smaller volume (V) means a proportional higher pressure because there are more molecules bouncing per second off a given bit of wall: More collisions per second per cm^2.

But that's the ideal gas law. Liquid water is far from ideal, and it's certainly not a gas.

With only unusual exceptions, liquids and solids have their molecules continually in contact. For small compressions, not enough to change the phase, the compression is compressing the molecules themselves. When you compress liquid water, you're trying to compress the electron "clouds" that form the orbitals and bonds. Electrons don't like being forced closer together...


The increase of pressure with depth is due to the pull of the Earth's gravity on the liquid above. This is apparent from the presence of $g$ in the formula for excess liquid pressure at depth $h$, namely $p=h\rho g.$

The pull of the Earth's gravity on a water column is balanced by the contact force of the layer of water below the column. This layer exerts the force because it is slightly compressed.

Now think about a jar of ball-bearings. Under the pull gravity (a downward pull, note) the balls will similarly exert contact forces on each other and, the sides of the jar. If the ball bearings were perfectly smooth, it's possible to show that the pressure the mean balls exert on the sides of the jar, as well as the bottom, is given by $p=h\rho g.$

But water molecules are bonded to each other rather than being in smooth contact with each other. True, but their vibrational motion, and the presence of vacancies (missing molecules) allow them to move relative to each other over a short period of time, a sort of quasi-sliding. So we have ball-bearing-like behaviour, and $p=h\rho g.$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.