Fluids and Density

Any substance that flow is called as fluid. They may either occupy the shape of their container or change the container’s shape. This term can be used to both gases and liquids, since the two both have no orderly long-range arrangement compared to that of a solid. However, gas and liquid differs in the ability of their molecules to attract each other. The molecules of a liquid are nearer to each other compared to the gas molecules, thus, liquid molecules could exert attractive force to each other and have tendency to be staying together whereas gas molecules are too far from each other, and may attract each other but of very weak attractive force. This is the difference between gas and liquid, liquid has cohesion while gas does not have.

Two of the most important quantities in studying fluids are density and pressure. These two are very useful to compare different fluids as density and pressure vary in any type or kind of fluid.


Density is an important property of any material, not just fluids but also in solids. It is defined as the mass per unit volume of the substance. To find the density of a fluid at any point, a small volume element is isolated around that point and the mass of the fluid present within the isolated element is then measured. Density can be expressed in an equation as

\(\rho = {\Delta m \over \Delta v}\).

For uniform density, the above equation can be simply written as

\(\rho = {m \over v}\).

Relative density of a substance is the ratio of the substance’s density to the density of pure water, at 4C. Relative density can be written mathematically as,

\(\text{relative density} = {\text{density of substance} \over \text{density of water}} = {\rho_{substance} \over \rho_{water}}\)

Since the density of pure water at 4C is 1.0 x 103 kg/m3, thus, relative density can be written mathematically as,

\({\text{density of substance} \over 1.0 \times 10^3\; kg/m^3}\).

The table below shows a list of the densities of some common substances.


Example 1.

An object has a mass of 5.6 kg and a volume of 4.2 x 10-3 m3. What is the object’s:

  1. density and
  2. relative density?


a. In finding the density of the object we use the equation \(\rho = {m \over V}\).

\(\rho = {5.6\;kg \over 4.2 \times 10^-3 m^3} = 1.33 \times 10^3\;kg/m^3\)

The object’s density is 1.33 x 103 kg/m3.


b. We use the equation for relative density,

\(\text{relative density} = {\text{density of substance} \over 1.0 \times 10^3\; kg/m^3}\)

Substitute the value of the density of the object solved in a to the above equation,

\(\text{relative density} = {1.33 \times 10^3\;kg/m^3 \over 1.0 \times 10^3\; kg/m^3} = 1.33\).

The relative density of the object is 1.33.

Example 2.

What mass of lead has a volume of 3.7 x 10-3 m3? The density of lead is 11.3 x 103 kg/m3.

 \(\rho = {m \over V} \implies m = \rho V\)

Substitute the given values,

\(m = (11.3 \times 10^3\; kg/m^3)( 3.7 \times 10^{-3}\; m^3) = 41.81\;kg\).

The mass of lead is 41.81 kg.

Example 3.

The relative density of the seawater is 1.03.

  1. What is its density?
  2. If the mass of the seawater is 10.9 kg, what is its volume?


  1. \(\text{relative density} = {\rho_{seawater} \over \rho_{water}} \implies \rho_{seawater} = (\text{relative density}) (\rho_{water})\)

\(\rho_{seawater} = (1.03)( 1.0 \times 10^3\; kg/m^3) = 1.03 \times 10^3\;kg/m^3\).


  1. \(\rho = {m \over V} \implies V = {m \over \rho}\)

\(V = {10.9\;kg \over 1.03 \times 10^3\;kg/m^3} = 1.06 \times 10^{-2}\;m^3\)