14
$\begingroup$

Microscopically, the pressure exerted by a fluid on a surface in contact with it is caused by collisions of molecules of fluid with the surface. As a result of a collision, the component of a molecule's momentum perpendicular to the surface is reversed. The surface must exert an impulsive force on the molecule, and by Newton's Third Law the molecule exerts an equal force perpendicular to the surface. The net result of reaction force exerted by many molecules on the surface gives rise to the pressure on the surface.

The above is an extract from Physics by Resnick, Halliday & Krane.

I've a few questions, conceptual in nature, which stemmed from the above paragraph -

  1. All that is mentioned is that the component of molecule's momentum perpendicular to surface is reversed; nothing is mentioned about its magnitude. If, they wish to tell us that the collision is elastic (as in the case of kinetic theory of gases), why is this a valid assumption? Maybe, the right question to ask is, to what extent is it a valid assumption?

  2. According to the aforementioned extract, pressure arises due to collisions between molecules and the surface. It is also a well known fact that pressure in a static fluid increases with depth. How can we explain that using this collision model? I'm confused because, the nature of collisions should be a property of the fluid, and should not vary with depth.

  3. Is this model - the one that talks about pressure arising due to collisions, sufficient to explain pressure related phenomena in all possible situations; or is it a mere approximation?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ No, the collisions of molecules cannot be considered to be elastic. The collisions can cause the molecules to vibrate. Thus, initial kinetic energy is not necessarily equal to the final kinetic energy. Some energy can be stored as potential energy. $\endgroup$
    – Apoorv Potnis
    Commented Mar 11, 2018 at 15:21
  • 1
    $\begingroup$ For a system in thermal and mechanical equilibrium, the non-elastic nature of the collisions (both with the walls and between molecules) is of no significance because the energy gains and losses come out in the wash (get it? get it?). But in general you need to be looking at the time evolution of the system to understand the importance of any inelasticity that may be present (it is quite absent in the most spherical and bovine models). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 11, 2018 at 17:55

3 Answers 3

Reset to default
17
$\begingroup$

The quoted paragraph from the textbook talks about fluids which usually includes gases, liquids, and plasmas. However, it would not be right to say that for liquids (e.g., consider water for concreteness) the pressure is the kinetic pressure $P_k=nkT$. First of all, we know that we can put water under a piston and increase the pressure isothermically at nearly constant density. If the pressure is due to particle collisions then why does it increase without any increase of temperature and density? Furthermore, using the numbers for water at normal conditions, $n=33e27 m^{-3}$, T=300 K, we'd get the kinetic pressure $P_k$ at about 10 million atmospheres, but we don’t see it!

We don't see this huge pressure because it is largely compensated by intermolecular attraction forces. So the total pressure in a liquid is $P = P_k + P_f$, where $P_f$ (negative at normal conditions) is the component of the pressure due to intermolecular forces, strongly dependent on the density. If water is compressed (at a constant temperature) the resulting pressure increase is due to the change of $P_f$.

So, for water compressed under a piston at a constant temperature, the total observed pressure increases; the thermal pressure caused by water molecules bouncing off the surface does not change in this process but the intermolecular forces respond to the compression changing the total pressure.

Given that the thermal pressure in a liquid is almost entirely compensated by the intermolecular forces, one can model a liquid as a large number of slippery almost incompressible balls lumped together, essentially excluding thermal motion from the picture. This model would have the properties of a real fluid (weakly compressible, isotropic pressure, Pascal law, Archimedes law). If we put such a "liquid" in a vertical column then we'd observe that those balls deeper down from the surface are compressed more (because there is a larger weight above them), and a body embedded in this “liquid” deeper would experience a larger external pressure.

Improve this answer
$\endgroup$
12
  • 1
    $\begingroup$ How do you explain using this model that the pressure in a liquid is isotropic? $\endgroup$
    – Apoorv Potnis
    Commented Mar 11, 2018 at 13:04
  • $\begingroup$ This (p. 81) paper by Michael Berry says that the pressure is a sum of (1)the pressure due to collisions(due to thermal agitations) (2) the intermolecular attractive forces. Is he wrong? I'm not able to understand. $\endgroup$
    – Apoorv Potnis
    Commented Mar 11, 2018 at 15:16
  • $\begingroup$ @Apoorv Potnis Well, that paper by Michael Berry discusses surface tension. That's why he talks there about attractive forces. I was talking about repulsive forces that are responsible for liquids' small compressibility. Thermal motion is a component of pressure in a liquid but I don't think it is anything significant for water at room temperature, its pressure can be increased by orders of magnitude by compression, without any significant increase of temperature or density. $\endgroup$
    – Maxim Umansky
    Commented Mar 11, 2018 at 16:24
  • $\begingroup$ But isn't the equation that he states for total pressure always valid? $\endgroup$
    – Apoorv Potnis
    Commented Mar 11, 2018 at 16:33
  • $\begingroup$ @Apoorv Potnis Well, this is a strongly non-ideal gas situation. The actual equation of state is something complex. But the total pressure can be always formally split into a sum of ideal-like thermal components and the rest, like that paper by Michael Berry paper does. $\endgroup$
    – Maxim Umansky
    Commented Mar 11, 2018 at 17:16
6
$\begingroup$

  1. The model is a very good approximation as long as the mass of the molecules colliding with the surface have negligible individual mass and cross section with respect to the surface they are hitting. The number density of molecules also matters in some cases because more molecules means more internL forces, which might have an impact on their collision rates and momentum change.

  2. Fluid pressure increases with depth because there are more particles above than below. Consider any surface in the fluid parallel to the base. There is a certain distribution of particles above and below it. As you move the surface down, the number of particles above it increase whereas there is a decrease in the number below it. This causes more net pressure ro be exerted downwards and hence the variation with depth.

  3. As you have already mentioned in the first part, and as I have answered, it is an approximation. The Real Gas equation is a better model (for gases) and there are several others. But for most situations involving low density, low pressure and moderate temperatures, such approximations are quite valid.

Cheers!

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ I disagree with the fact that number of particles increase; don't we assume the density to be fairly constant in our usual derivations? $\endgroup$
    – stoic-santiago
    Commented Mar 11, 2018 at 5:07
  • $\begingroup$ @schrodinger_16 The number of particles increases in the sense that since the volume above the surface increases, the number of particles – which is the number density times the volume – increases ( since the number density is fairly constant as you point out). $\endgroup$
    – Yuzuriha Inori
    Commented Mar 11, 2018 at 5:11
  • $\begingroup$ Also, I understand the fact that fluid pressure increases with depth due to more particles above than below; however, I was looking for a purely collision based answer and my doubt still remains - if the collisions are characteristic to the fluid, there should be no variation. $\endgroup$
    – stoic-santiago
    Commented Mar 11, 2018 at 5:14
  • $\begingroup$ The increase in the number of particles would directly imply more collisions and hence the increased pressure. And collision is not characteristic to a fluid (At least in our approximations it is not). $\endgroup$
    – Yuzuriha Inori
    Commented Mar 11, 2018 at 5:41
  • 1
    $\begingroup$ For a liquid like water under a piston, the pressure can be increased isothermally without any significant change of density. This suggests that collisions are not responsible for it. The pressure is due to Van der Waals forces that prevent molecules from coming close to each other. $\endgroup$
    – Maxim Umansky
    Commented Mar 11, 2018 at 6:00
0
$\begingroup$

Some models look at the energy required to change a solid to a liquid and the energy required to change a liquid to a gas. And conclude that liquids are similar to solids. Molecules are connected by molecular bonds. The bond strength in solids is greater than the bond strength for liquids. EG... water at 32 F solid to gas hig = 1218.5 btu/lbm and liquid to gas hfg = 1075.15 btu/lbm. And thus solid to liquid hif 143.35 btu/lbm. i.e.strong bond in liquids and stronger bond in solids

Therefore hydrostatic pressure is due to weight of mass not due to individual molecular collisions of free gas molecules.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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