Learning Objectives

Learning Objectives

By the end of this section, you will be able to:

  • Define pressure
  • Explain the relationship between pressure and force
  • Calculate force given pressure and area

The information presented in this section supports the following AP® learning objectives and science practices:

  • 7.A.1.1 The student is able to make claims about how the pressure of an ideal gas is connected to the force exerted by molecules on the walls of the container, and how changes in pressure affect the thermal equilibrium of the system. (S.P. 6.4, 7.2)

You have no doubt heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. Pressure PP size 12{P} {} is defined as

11.6 P=FAP=FA size 12{P= { {F} over {A} } } {}

where FF size 12{P} {} is a force applied to an area A,A, size 12{P} {} that is perpendicular to the force.


Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or

11.7 P=FA.P=FA. size 12{P= { {F} over {A} } } {}

A given force can have a significantly different effect depending on the area over which the force is exerted, as shown in Figure 11.6. The SI unit for pressure is the pascal, where

11.8 1 Pa=1N/m2.1 Pa=1N/m2. size 12{1`"Pa"=1`"Nm" rSup { size 8{2} } } {}

In addition to the pascal, there are many other units for pressure that are in common use. In meteorology, atmospheric pressure is often described in units of millibar (mb), where

11.9 100 mb=1×105Pa .100 mb=1×105Pa . size 12{"100"`"mb"=1 times "10" rSup { size 8{5} } `"Pa"} {}

Pounds per square inch lb/in2orpsilb/in2orpsi size 12{ left ("lb/in" rSup { size 8{2} } `"or"``"psi" right )} {} is still sometimes used as a measure of tire pressure, and millimeters of mercury (mm Hg) is still often used in the measurement of blood pressure. Pressure is defined for all states of matter but is particularly important when discussing fluids.

In figure a, the person is poked with a finger exerting a small pressure due to the large area of contact and, in b, he is poked with a syringe exerting a large pressure due to the small area of contact.
Figure 11.6 (a) While the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the same force applied to an area the size of the sharp end of a needle is great enough to break the skin.

Making Connections: Gas Particles in a Container

A rectangular figure represents a closed container a closed container with dark-blue spheres representing particles. Each particle has an arrow pointing in a different direction. Two particles show arrows pointing toward the edge of the container. Two particles point toward each other.
Figure 11.7 Gas particles vibrate within a closed container.

Imagine a closed container full of quickly vibrating gas particles. As the particles rapidly move around the container, they will repeatedly strike each other and the walls of the container.

When the particles strike the walls, a few interesting changes will occur.

Gas2a is identical yet smaller to the previous Gas1 figure above with five particles each with arrows pointing in different directions. Gas2b has the same size rectangle but many more particles.
Figure 11.8 The figure on the left shows gas particles traveling within a closed container. As the particles travel, they collide with each other and with the container walls. The figure on the right shows an increased number of gas particles within the container, which will result in an increased number of collisions.

Each time the particles strike the walls of this container, they will apply a force to the container walls. An increase in gas particles will result in more collisions, and a greater force will be applied. The increased force will result in an increased pressure on the container walls, as the areas of the container walls remain constant.

Gas3a is a rectangle with four particles pointing in different directions. For three of the arrows, they touch the edge of the rectangle and a short red arrow is reflected from the wall with an F label. Gas3b also has 4 particles with blue arrows and three of the arrows point to the edge of the rectangle. Each of the blue arrows pointing away from the blue spheres is longer than the arrows in figure Gas3a. Each of the three red arrows in figure Gas3b reflecting off the wall are also longer than the blue a
Figure 11.9 A force F is placed upon the particles each time they strike the container walls. These forces result in new trajectories, as shown by the red arrows. Because the particles in the figure on the right are moving more quickly, they experience a larger force than those shown in the figure on the left.

If the speed of the particles is increased, then each particle will experience a greater change in momentum when it strikes a container wall. Just like a fast-moving tennis ball recoiling off a hard surface, the greater the particle's momentum, the more force it will experience when it collides. For verification, see the impulse-momentum theorem described in Chapter 8.

However, the more interesting change will be at the wall itself. Due to Newton's third law, it is not only the force on the particle that will increase, but the force on the container will increase as well! While not all particles will move with the same velocity, or strike the wall in the same way, they will experience an average change in momentum upon each collision. The force that these particles impart to the container walls is a good measure of this average change in momentum. Both of these relationships will be useful in Chapter 12, as you consider the ideal gas law. For now, it is good to recognize that laws commonly used to understand macroscopic phenomena can be applied to phenomena at the particle level as well.

Example 11.2 Calculating Force Exerted by the Air: What Force Does a Pressure Exert?

An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads 6.90×106Pa6.90×106Pa size 12{6 "." "90" times "10" rSup { size 8{6} } `"Pa"} {}. What force does the air inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter?


We can find the force exerted from the definition of pressure given in P=FAP=FA size 12{P= { {F} over {A} } } {}, provided we can find the area AA size 12{A} {} acted upon.


By rearranging the definition of pressure to solve for force, we see that

11.10 F=PA.F=PA. size 12{F= ital "PA"} {}

Here, the pressure PP size 12{P} {} is given, as is the area of the end of the cylinder AA size 12{A} {}, given by A=πr2A=πr2 size 12{A=πr rSup { size 8{2} } } {}. Thus,

11.11 F=6.90×106N/m23.140.0750 m2=1.22×105N.F=6.90×106N/m23.140.0750 m2=1.22×105N.alignl { stack { size 12{F= left (6 "." "90" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } right ) left (3 "." "14" right ) left (0 "." "0750"`m right ) rSup { size 8{2} } } {} # =1 "." "22" times "10" rSup { size 8{5} } `N "." {} } } {}


Wow! No wonder the tank must be strong. Since we found F=PAF=PA size 12{F= ital "PA"} {}, we see that the force exerted by a pressure is directly proportional to the area acted upon as well as the pressure itself.

The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: They are always exerted perpendicular to any surface. (See the tire in Figure 11.10, for example.) Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides. (See Figure 11.11.)

The forces inside a tire are shown by arrow lines. An inset shows an enlarged view of the valve in the tire. Air pressure in the tire keeps the valve closed.
Figure 11.10 Pressure inside this tire exerts forces perpendicular to all surfaces it contacts. The arrows give representative directions and magnitudes of the forces exerted at various points. Note that static fluids do not exert shearing forces.
A man swimming underwater has many arrows pointing toward him to represent the directions and magnitudes of the forces exerted on him at various points.
Figure 11.11 Pressure is exerted on all sides of this swimmer, since the water would flow into the space he occupies if he were not there. The arrows represent the directions and magnitudes of the forces exerted at various points on the swimmer. Note that the forces are larger underneath, due to greater depth, giving a net upward or buoyant force that is balanced by the weight of the swimmer.

Making Connections: Pressure

Figure 11.10 and Figure 11.11 both show pressure at the barrier between an object and a fluid. Note that this pressure also exists within the fluid itself. Just as particles will create a force when colliding with the swimmer in Figure 11.11, they will do the same each time they strike each other. These forces can be represented by arrows, whose vectors show the resulting direction of particle movement. The same factors that determine the magnitude of pressure upon the fluid barrier will determine the magnitude of pressure within the fluid itself. These factors will be discussed in Chapter 13.

PhET Explorations: Gas Properties

Pump gas molecules to a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other.

This icon links to a P H E T Interactive activity when clicked.
Figure 11.12 Gas Properties