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Why is pressure in liquids not volume dependent?

  1. Oct 30, 2012 #1
    For example, if you have really large pond and a small pond, and you swim down 3 meters in each of them, the pressure will be the same. Why?

    (Is it possible to describe this using kinetic theory or molecular motion?). It doesn't really make sense to me; if there's more water, and water pressure spreads equally in all directions, then technically shouldn't the volume/mass of water be factored in as well? i.e. in the larger pond, there would be more water crushing you/exerting pressure on you?
     
  2. jcsd
  3. Oct 30, 2012 #2
    consider a vertical cylindrical water vessel, with height h and cross-sectional area A

    the volume of the water is h*A
    the mass of the water in this vessel is h*rho*A, where rho is the density of water
    the weight(downward force) of this mass is g*h*rho*A
    the pressure at the bottom is defined as the force divided by the area

    now, the area of the bottom is A

    so the pressure at the bottom is p = (g*h*rho*A)/(A) = g*h*rho

    ie. it is independent of the radius of the container due to a simple cancellation.
     
  4. Oct 30, 2012 #3

    russ_watters

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    Forget water for a sec: try starting with ice cubes. A 1 cubic meter ice cube is 10 kN. Or 10kN/sq m. Put two next to each other on the ground and they cover 2 sq m and weigh 20kN. Then build a tank around them and let them melt: what has changed?

    If you lie down and rest one of the cubes on your chest, but the second one doesn't fit so you have it set next to you, does it still hurt more just having it near you?
     
    Last edited: Oct 30, 2012
  5. Oct 30, 2012 #4
    Great example, but doesn't this only works with solids, since water particles can flow and move? In a solid the particles of the ice cube are hold firm/tight with strong intermolecular bonds but in water they can jostle against you. I get sort of what you mean, but in water pressure spreads equally in all directions, so the water next to you would still exert pressure on you?

    Great math solution, but I was hoping for a logical explanation using the idea of particles.
     
  6. Oct 30, 2012 #5

    russ_watters

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    Water next to you does exert a force on you: on your side. Melting the ice can't change the magnitude of the force/pressure, it just allows it to come at you from all directions.
     
  7. Oct 30, 2012 #6

    mfb

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    It works with liquids in the same way - horizontal forces cancel if the water is in equilibrium.
     
  8. Oct 30, 2012 #7
    I still don't quite understand this... how does this relate to the fact that pressure in liquids is not volume dependent? If water next to you does exert a force/pressure on you, then why is pressure in liquids not volume dependent?

    Pressure is a scalar, not a vector, so they can't just cancel out, I think. You still experience pressure from both sides.
     
  9. Oct 30, 2012 #8
    I'm not sure one exists, since fluid dynamics is based on the continuum hypothesis which completely ignores the fact that fluids are composed of discrete particles.
     
  10. Oct 30, 2012 #9
    If I understand it correctly, liquids such as water, in general have a natural state of distributing itself evenly over the largest possible surface area. While certain effects such as hydrogen bonding make "clumps" or pools of water, it is understood when you poor water out of a cup, it distributes straight down to the floor and not up to the ceiling. The rate, the force, the impact, are all depending on the acceleration of gravity. In the glass container each of these molecules are often macroscopically treated as a single body as true liquid dynamics aren't necessary to derive the pressure. Again pressure simply being the Force over an Area. In this case, Force is dependent on some molecular forces, but generally its just the Force of Gravity necessary to take into account.

    Gas on the other hand has a different natural state to occupy as much space as possible in all directions. And when you observe Gas in a container, Pressure, which is still Force / Area, has a lot of different variables working on it, and that pressure is now omni-directional. Gravity is still there, but Temperature, Volume, number of Moles etc., change the state of the Gas. So basically pressure for a Gas would stipulate that you are observing an average of pressure on all the walls of the container, taking into the account of Atomic / Molecular Collision, Gravity, Avg. Kinetic Energy etc etc.

    One more thing is that most liquids do not compress, there is displacement. There are a finite number of molecules at every layer in each of these ponds. No more, no less. They are not pushing on each other, they simply fill the gaps. The water you occupy in your 3 meter depth, is diverted upward at the surface since it can't compress downwards or sideways. You sit in the bathtub and water rises. What this correlates to, there is no more pressure replacing that water with your body. However, having 3 meters of x kg water on top of you, will weigh down on you. You also have y kg water below you countering that weight. At the very bottom of the pond you have all of x + y kg of water applying a downward force on top of you. As you can see it isn't pressure via compression on you, but the weight of the water creating pressure. Which relates back to just the Force of Gravity on you. Now most of these basic problems never force you to calculate the area of water or the weight of the water, they simply give you usually an initial Pressure, P0, and then the density of water, gravity, and height, and then you have to determine the new pressure.

    P = P0 + Rho * g * h

    I find that starting off with the equation above is a little bit silly as it takes work on the user to understand the formula / units. Where as if you explain things first by a simple concept as mass or weight, on top of something else, its much easier to relate to. The equation above skips find Mass, Force, and then area, it just goes straight to Pressure, which is Force over an Area. I guess its just a different approach in teaching the subject matter.
     
    Last edited: Oct 30, 2012
  11. Oct 30, 2012 #10

    A.T.

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    Consider just a single particle, moving only horizontally, bouncing back and forth between your body and the pond wall.

    YOU |<--O-->| WALL

    Now move the pond wall further away and place a second particle in:

    YOU |<--O--><--O-->| WALL

    The impulse per time transfered to you by the first particle is still the same. It doesn't matter how much water you align horizontally there.
     
    Last edited: Oct 30, 2012
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