# Pressure

#### Prerequisites

## Making sense of pressure from a molecular viewpoint

In a liquid or a gas, the relative positions of the atoms or molecules making up the system are not fixed. At STP, they are continually traveling at a high velocity (about 400 m/s for oxygen molecules in air) and colliding with each other -- and with anything else that might happen to be imbedded in the fluid.

[For the analysis in this webpage and the definition of pressure, we will ignore gravity. Though each molecule in a liquid or gas feels the force of gravity and falls between collisions, they are moving so fast and travel such a short distance between collisions that the effect is hardly noticeable. When we go back to the macro picture, however, the effect of gravity adds up and results in the *Buoyant Force* -- an important phenomenon in many biological situations.]

If we imagine a fluid on one side of a surface, this picture suggests that the molecules of the fluid will bounce off the molecules of the surface. Since when the molecules bounce off, their momentum reverses, it means that they are feeling a force from the surface (that could be estimated using the Impulse-Momentum Theorem). By Newton's 3rd law, they will then exert a force back on any surface they happen to touch.

This is an interesting and difficult picture to think about. In the figure at the right is shown a simulation of a set of gas molecules moving around in a box. The gas molecules that hit the right wall of the box feel a force to the left since their momenta change from right-going to left-going. The molecules that are hitting the left wall of the box feel a force to the right since their momenta change from left-going to right-going. By N3, these molecules therefore exert a force on the right wall of the box to the right and on the left wall of the box to the left. The force of the molecules basically pushes outwards in all directions.

So the direction of the force depends on the surface. The molecules are sort of "exerting forces in all directions at the same time" -- sort of like how the (scalar) tension we define in a chain pulls in both directions at once. In the chain, it's only when you say which end you are looking at that the (scalar) tension turns into a (vector) force. With our fluids, we only get a force and a direction when we select a surface.

We can also see that as long as the molecules have a uniform density and speed, we expect that the force that a surface feels depends on the area of the surface. The bigger the surface the more molecules hit it per second, so the more force it feels. This suggests that we define the magnitude of the *force per unit area* felt by a surface as a property of the fluid -- the ** pressure**.

## Making area a vector

We want to put vectors into a law that gives a force whenever possible so we see from our equation where the direction comes from. In this case, the direction comes from the surface area. Can we make area into a vector so as to include direction in our equation?

We can make area a vector by using a bit of a trick. We live in a 3D space. A small bit of area is a plane -- it specifies two dimensions. This leaves one dimension -- a direction. The perpendicular to an area is a unique direction. Well -- it's almost unique. You could take it in either direction going away from the surface.

Define the * area vector *for a bit of area, $\overrightarrow{A}$, as a vector perpendicular to the surface area being considered, pointing out of the volume that we are considering that is exerting a force on that bit of area. And we will give this vector a length equal to the magnitude of the area. With this, the

**that the fluid exerts is the scalar value (in N/m**

*pressure*^{2}) that would give the measured force on the area $A$.

$$\overrightarrow{F}_{fluid \rightarrow A} = p \overrightarrow{A}$$

Careful! This equation is WAY trickier than it looks. First of all, note where the vectors are: on the force and on the area. This relies on our convention that the vector area points perpendicular to *A* and outward from the fluid. And the force we are calculating is the force that the fluid exerts on the area. The pressure, *p*, has units of N/m^{2}, but is a scalar -- it has no direction. The way the equation is usually written (*p = F/A*) hides all the directional information and the question of what force the equation is talking about.

Joe Redish 10/23/11

#### Follow-ons

Last Modified: February 12, 2019