Density and pressure

In a nutshell

The density of an object is a measure of its 'compactness', in other words, how tightly packed its mass is. Pressure is the force on an object per unit area.

Equations

Constants

Variable Definitions

Density

The density of an object is its mass per unit volume:

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$$\rho={m\over V}$$.

Example

Gold has a density of $$19\, 300\,kg/m^3$$. Calculate the volume of $$5\,kg$$ of gold.

Write down the relevant information provided in the question (converting to base units where appropriate):

 $$\rho = 19\,300\,kg/m^3\newline m=5\,kg$$   

Write down the relevant formula:

 $$\rho=\frac{m}{V}$$

Rearrange the relevant formula:

 $$V=\frac{m}{\rho}$$

Substitute the relevant information into the correct formula:

 $$V=\frac{5}{19300} \newline \\[0.1in] V=\underline{2.6 \times 10^{-4}\,m^3}$$

So $$5\,kg$$ of gold occupies a volume of $$\underline{2.6 \times 10^{-4}\,m^3}.$$

Density of solids, liquids and gases

​​The particle model of matter states that everything is made of particles. Dense materials have their particles tightly packed together, so the mass of the object occupies a small volume. 

Less dense substances have their particles more far apart, so the mass of the substance is spread out over a large volume. Therefore, two objects of equal volume but different density will weigh different amounts.

For example, particles in a solid tend to be very close together, so solids are dense. Particles in a liquid tend to be more separated, so liquids tend to be less dense than solids. Particles in a gas have much more distance between them, so gases are far less dense than solids and liquids.

A Less dense B More dense

Curiosity: Did you know that solid gold is about 20 times more dense than liquid water? A 500ml bottle of water weighs half a kilogram, but the same volume of solid gold weighs nearly 10kg. Even denser are neutron stars - just a teaspoon of one has a mass of one billion tonnes! 

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Determining density

To determine the density of an object, you must measure its volume and mass. Scales can be used to determine the mass of the object. 

Determining the density of a cuboid:

1. First use vernier callipers to take accurate measurements of side length $$l$$, width $$w$$ and height $$h$$.​
2. Substituting this into the formula for the volume, $$V = lwh$$​. 
3. The mass of the object can be found be placing it on scales and noting down its mass.
4. The mass and volume can then be substituted into the formula for density $$\rho = \dfrac{m}{V}$$.​

Determining the density of an irregular shape:

1. The volume of an irregular shape can be found by submerging it in a eureka can filled with water.
2. The eureka can displaces water from its spout when an object is placed in it. The amount of water displaced is equal to the volume of the object as $$1 \ cm^3 = 1 \ ml$$ of water.
3. The mass of the object can be found by placing it on scales and noting down its mass.
4. The mass and volume can then be substituted into the formula for density $$\rho = \dfrac{m}{V}$$.

Pressure

Pressure is the force exerted on an object per unit cross sectional area:

​$$p = \dfrac{F}{A}$$

The unit of pressure is the pascal $$Pa$$, which is $$Nm^{-2}$$. 

An object immersed in a fluid feels pressure due to the fluid's weight. The pressure at a height, $$h$$​, in a fluid is the same in all directions and is given by:

​$$p = h \rho g$$

To derive this, you can consider a cylindrical column height $$h$$ and cross-sectional area $$A$$. The pressure of the column is equal to its weight divided by its cross-sectional area, where:

$$W = mg$$​​

From the formula for density, the mass is:

​$$m = \rho V\newline \\[0.1in] W =\rho V g$$

As the volume is $$Ah$$:

$$W = \rho Ahg$$​​

The pressure is defined as:

​$$p = \dfrac{F}{A}\newline \\[0.1in] p= \dfrac{\rho Ahg}{A}\newline \\[0.1in] p= \rho hg$$

Note: The pressure difference is due to the change in height of the upper and lower surfaces of the object.

Archimedes' principle

When an object is submerged in fluid, it experiences an opposing force pushing it upward, called upthrust. This occurs as there is a pressure difference between the upper and lower surfaces of the object. 

Archimedes' principle states that 'The upthrust exerted on a body immersed in fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces.'

For an object to float, its weight must equal the force of upthrust opposing it. If the object sinks, the force due to the object's weight is greater than the upthrust, which is:

​$$W = \rho Vg$$​​