A fluid (liquid or gas) exerts pressure on the walls of its container, or on any surface with which it is in contact. Solids can also exert pressure on a surface with which it is in contact.
The pressure in a gas or liquid produces a force perpendicular to any surface.
The force the fluid pressure produces on the walls of a container can be in any direction, because the walls of the container may be horizontal, vertical or at any angle. A big force on a small area produces a high pressure.
Pressure is defined as the normal force acting per unit cross-sectional area.
We can write this as a word equation:

$\begin{array}{l}
pressure = \frac{{normal\,force}}{{cross\, - \,\sec tional\,area}}\\
p = \frac{F}{A}
\end{array}$

The word ‘normal’ in this context means at right angles to the surface.

Force is measured in newtons and area is measured in square metres. The units of pressure are thus newtons per square metre ($N\,{m^{ - 2}}$), which are given the special name of pascals (Pa).

$1Pa = 1N\,{m^{ - 2}}$

Pressure in a fluid

The pressure in a fluid (a liquid or gas) increases with depth. Divers know this – the further they dive down, the greater the water pressure acting on them. The pressure acts at right angles to every part of their body and acts to crush them. Pilots know this – the higher they fly, the lower is the pressure of the atmosphere. The atmospheric pressure we experience on the surface of the Earth is due to the weight of the atmosphere above us, pressing downwards on the surface of the Earth or at right angles to every surface of our bodies.
The pressure in a fluid depends on three factors:
- the depth h below the surface
- the density ρ of the fluid
- the acceleration due to gravity, g.
In fact, change in pressure p is proportional to each of these and we have:

$\begin{array}{l}
change\,in\,pressure = density \times acceleration\,due\,to\,gravity \times depth\\
\Delta p = \rho gh
\end{array}$

We can derive this relationship using Figure 7.2.

The force acting on the shaded area A on the bottom of the tank is caused by the weight of water above it, pressing downwards. We can calculate this force and hence the pressure as follows:

$\begin{array}{l}
volume\,of\,water = A \times h\\
mass\,off\,water = density \times volume = \rho  \times A \times h\\
weight\,of\,water = mass \times g = \rho  \times A \times h \times g\\
change\,in\,pressure = \frac{{force}}{{area}}\\
 = \rho  \times A \times h \times \frac{g}{A}\\
 = \rho  \times g \times h
\end{array}$

The equation is written as $\Delta p = \rho gh$ because this formula calculates the difference in pressure between the top and bottom of the water in the tank. There is, of course, atmospheric pressure acting on the water at the top of the tank. The total pressure at the bottom of the tank is atmospheric pressure $ + \Delta p$.

WORKED EXAMPLE

1) Figure 7.3 shows a manometer used to measure the pressure of a gas supply. Calculate the pressure difference between the gas inside the pipe and atmospheric pressure.
Step 1: Determine the difference in height h of the water on the two sides of the manometer.
$h = 60 - 30 = 30cm$
Step 2: Because the level of water on the side of the tube next to the gas pipe is lower than on the side open to the atmosphere, the pressure in the gas pipe is above atmospheric pressure.
$\begin{array}{l}
pressure\,difference = \rho  \times g \times h\\
 = 1000 \times 9.81 \times 0.30 = 2940Pa
\end{array}$