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In this Physics tutorial, you will learn:

- What is matter?
- How many states of matter are there normally? What are their properties?
- The meaning of fluids
- What is volume? Which are the units (standard and auxiliary of volume?
- What is density? How can we calculate it?
- How to calculate the density of a mixture?

Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

9.1 | Fluids. Density of Fluids |

You may have noticed that water and vinegar mix together while water and oil do not mix. Can you explain this phenomenon?

Do you think a bottle filled with milk has the same weight as the same bottle filled with honey? Why?

Do you remember the tricky question: "Which weighs more, 1 kg of iron or 1 kg of cotton?" Why people often give a wrong answer to this question?

Matter is described as a **"physical substance in general, which occupies space and possesses rest mass, especially as distinct from energy."**

In simpler words, matter is composed by the combination of a very large number of small particles known as "atoms". In a certain sense, atoms are the "building blocks" of matter. The dimensions of atoms vary from 0.1 to 0.5 nm (1 nm = 10-9 m). To give an idea how small atoms are, we can mention the fact that an atom is about one million times smaller than the average thickness of human hair.

Matter can exist in three stable states in nature (in fact, there are five states of matter but three are common to us). They are:

In this state, particles are "packed" in stable positions, which they cannot leave. In other words, particles in solid state cannot change their neighbours. They always have the same particles around them.

If we use an energetic approach to explain the states of matter, we must stress that in solid state the potential energy that keeps the particles together, which is also known as the **"binding energy"** is much greater than kinetic energy of particles. As a result, atoms in solids can only vibrate around their equilibrium position but they cannot leave their actual place and find new neighbouring atoms.

The most suitable pattern used to describe the solid state is a cube in whose vertexes some spheres are placed and whose sides are made of elastic springs as shown in the figure below.

If you kick one of spheres with your finger, energy is transferred from the hand to the first sphere and then to all the other spheres through the elastic springs. As a result, all spheres will start vibrating around their equilibrium position. This is more or less what happens with particles of solids in the micro-world.

In this state, particles are still very close but they are not "attached" together. Their potential energy is still a bit greater than kinetic energy but the difference between them is very narrow. As a result, particles can slide easily over each other, like plastic balls when poured on a playground. Look at the figure.

The balls distribute throughout the lower base of container due to gravity. They cannot accumulate on top of each other until the first layer is completely filled with balls. Then, the second layer start to form and so on. The same thing occurs in liquids as well. This is the reason why the surface of a liquid is flat.

In this state, particles move freely in space as their kinetic energy is greater than the potential (binding) energy. They look like balloons flying in the sky or like bubbles flowing in water. Look at the figure.

The dictionary describes the meaning of word **"fluid"** as **"something able to flow easily"**. In this regard, only **liquids** and **gases** qualify to belong in this category as they can flow easily. For example, if you pour some water on the floor, it immediately spreads along the entire surface due to the flowing process. Likewise, if you press the nozzle of a perfume bottle, all people present in that room will smell almost immediately its scent because the perfume's particles flow easily in the air.

You may ask yourself why solids (especially when crushed) do not belong to the "family" of fluids. As an argument to question the above definition you may use the example of flour when poured on a cooking utensil. Apparently, the flour seems as flowing. However, unlike in liquids, it is quite impossible for such substances distribute evenly throughout the utensil. The flour will stay more or less in the position shown below.

Therefore, despite powder-like solids can flow at a certain extent they cannot be considered as fluids, because their flow is not so easy. Moreover, their upper surface is not flat like in liquids.

From definition of matter provided in the first paragraph, we can outline two physical features. They are:

- The
**rest mass**, or simply**mass**of the substance. - The
**space**it occupies. This feature is related to the**volume**of substance, as by definition**"Volume, V, is the amount of space an object occupies."**

Obviously, not all objects which have the same mass, occupy the same space. As an example, we can mention the situation described in the last question in the "Introduction" paragraph, i.e. when comparing 1 kg of iron and 1 kg of cotton. They have both the same mass but cotton occupies a much larger space than iron. For this reason, people tend to give a wrong answer when asked which object weighs more. In most cases, they chose iron as an option because they are confused by the fact that for the same volume, iron weighs more than cotton.

To avoid such confusion, it is better to introduce a new quantity that takes into consideration both **mass** and **volume** of objects. This quantity is known as **"density"** and it is calculated by the formula,

Density = *Mass**/**Volume*

The above formula helps us give the definition of density. Thus, by definition,

In formulae, we express the density by the letter ρ the mass by m and the volume by V. Therefore, the equation of density becomes

ρ = *m**/**V*

Obviously, the SI unit of density is [** kg/m^{3}**] as mass is measured in [kg] and volume in [m

1000 *kg**/**m*^{3}

=*1000 kg**/**1 m*^{3}

=*1 000 000 g**/**1 000 000 cm*^{3}

= 1*g**/**cm*^{3}

=

=

= 1

In the above conversion, we have considered the fact that units of volume change 1000 by 1000. Thus, 1 m^{3} = 1000 dm^{3} = 1 000 000 cm^{3}. Also, 1 kg = 1000 g.

A liquid sample has a mass of 360 g and it occupies 400 cm^{3} of space in its vessel. Calculate the density of liquid in ** g/cm^{3}** and then convert it into

We have m = 320 g and V = 400 cm^{3}. Thus,

ρ = *m**/**V*

=*360 g**/**400 cm*^{3}

= 0.8*g**/**cm*^{3}

=

= 0.8

When converted into ** kg/m^{3}**, the above result becomes

0.9 *g**/**cm*^{3}

=*0.9 g**/**1 cm*^{3}

=*0.009 kg**/**0.000001 m*^{3}

= 900*kg**/**m*^{3}

=

=

= 900

As mentioned above, the units of volume change 1000 by 1000. Thus, jumping from one unit of volume into another using the SI system of units is a pretty hard task. It is like trying to climb the stairs which have very high steps as shown in the figure below.

To avoid such issues, an auxiliary unit known as "Litre, L" is introduced. Litre is used only in fluids, i.e. to express the volume of liquids and gases. Litre is numerically equal to cubic decimetre (1 L = 1 dm^{3}) but it offers an important advantage compared to standard SI units of volume: its multiples and sub-multiples change 10 by 10 instead of 1000 by 1000, facilitating a lot in this way the process of calculations. It is like building some auxiliary steps between the already built ones in the stairs shown in the figure above. Look at the figure.

**Remark!** The auxiliary units of volume exist only for amounts we use in daily life, i.e. they extend from cm^{3} (mL) to m^{3} (kL). No auxiliary units exist out of this range.

A graduated cylinder weighs 83.0 g when empty and 158.0 kg when 90 mL of an unknown liquid is poured in it. Calculate the density of liquid in ** kg/m^{3}**.

Mass of the liquid is obtained by subtracting the two values shown in the balance. Thus,

m_{liq} = m_{final} - m_{initial}

= 158.0 g - 83.0 g

= 75.0 g

= 158.0 g - 83.0 g

= 75.0 g

The volume of liquid is written as 90 cm^{3} instead of 90 mL. Therefore, we obtain for density of liquid,

ρ_{liq} = *m*_{liq}*/**V*_{liq}

=*75.0 g**/**90 cm*^{3}

= 0.8*g**/**cm*^{3}

= 800*kg**/**m*^{3}

=

= 0.8

= 800

To calculate the density of a gas, we simply divide its mass with the volume of the whole container as gases fill up all the space of the closed container in which they are.

An empty balloon weighs 4.6 g. Calculate the mass of balloon in grams after filling it with 4 L of air. Take the density of air equal to 1.3 ** kg/m^{3}**.

Given that

m_{empty ballon} = 4.6 g

ρair = 1.3 ** kg/m^{3}** = 0.0013

V_{balloon} = V_{air} = 4 L = 4000 mL = 4000 cm^{3}

we obtain for mass of air

m_{air} = ρ_{air} × V_{air}

= 0.0013*g**/**cm*^{3} × 4000 cm^{3}

= 5.2 g

= 0.0013

= 5.2 g

Therefore, the total mass of balloon after filling it with air is

m_{tot} = m_{empty balloon} + m_{air}

= 4.6 g + 5.2 g

= 9.8 g

= 4.6 g + 5.2 g

= 9.8 g

In most cases, materials we use in daily life are not made of pure substances but by mixing two or more substances. As an example in this regard, we can mention beverages we drink every day. Thus, alcoholic beverages are not made of pure alcohol as people could not consume them in such a state, but they are mixed with water instead. The percentage of alcohol contained in a certain beverage is usually written in the etiquette stamped on the bottle. Thus, if a 333 ml bottle of beer contains 10% of alcohol, this means there are 333 / 10 = 33.3 mL of alcohol and 333 - 33.3 = 299.7 mL of water in that bottle.

To calculate the density of a mixture, we use the formula

ρ_{mix} = *m*_{tot}*/**V*_{tot}

Thus, for two substances (1) and (2), we can write

ρ_{mix} = *m*_{1} + m_{2}*/**V*_{1} + V_{2}

For three substances (1), (2) and (3), we can write

ρ_{mix} = **m**_{1} + m_{2} + m_{3}*/**V*_{1} + V_{2} + V_{3}

and so on.

- How much space does a mixture made of 24 g water and 40 g alcohol occupy if the density of water is 1
and that of alcohol is 0.8*g**/**cm*^{3}.*g**/**cm*^{3} What is the density of this mixture in

**a)** We have

V_{water} = *m*_{water}*/**ρ*_{water}

=*24 g**/**1 **g**/**cm*^{3}

= 24 cm^{3}

V_{alcohol} = *m*_{alcohol}*/**ρ*_{alcohol}

=*40 g**/**0.8 **g**/**cm*^{3}

= 50 cm^{3}

=

= 24 cm

V

=

= 50 cm

Therefore, the total volume of the mixture is

V_{tot} = V_{water} + V_{alcohol}

= 24 cm^{3} + 50 cm^{3}

= 74 cm^{3}

= 24 cm

= 74 cm

This result means the mixture occupies 74 cm^{3} of space in the container.

**b)** The density of the given mixture is

ρ_{mix} = *m*_{water} + m_{alcohol}*/**V*_{water} + V_{alcohol}

=*24 g + 40 g**/**24 cm*^{3} + 50 cm^{3}

=*64 g**/**74 cm*^{3}

= 0.865*g**/**cm*^{3}

=

=

= 0.865

Now, let's consider an example with gases.

The atmosphere contains 78% nitrogen, 21% oxygen and 1% carbon dioxide. If densities of the above gases in certain conditions are 1.25 ** kg/m^{3}** for nitrogen, 1.40

- The total mass of air inside a 60 m
^{3}room - The density of air in the given conditions

**a)** First, let's calculate the volumes of each gas in the room. We have:

V_{nitrogen} = 78% of V_{tot}

= 0.78 × 60 m^{3}

= 46.8 m^{3}

V_{oxygen} = 21% of V_{tot}

= 0.21 × 60 m^{3}

= 12.6 m^{3}

V_{carbon dioxide} = 1% of V_{tot}

= 0.01 × 60 m^{3}

= 0.6 m^{3}

= 0.78 × 60 m

= 46.8 m

V

= 0.21 × 60 m

= 12.6 m

V

= 0.01 × 60 m

= 0.6 m

Now, let's calculate the mass of each gas inside the room. We have

m_{nitrogen} = ρ_{nitrogen} × V_{nitrogen}

= 1.25*kg**/**m*^{3} × 46.8 m^{3}

= 58.5 kg

m_{oxygen} = ρ_{oxygen} × V_{oxygen}

= 1.40*kg**/**m*^{3} × 12.6 m^{3}

= 17.64 kg

m_{carbon dioxide} = ρ_{carbon dioxide} × V_{carbon dioxide}

= 1.98*kg**/**m*^{3} × 0.6 m^{3}

= 1.12 kg

= 1.25

= 58.5 kg

m

= 1.40

= 17.64 kg

m

= 1.98

= 1.12 kg

Therefore, the total mass of air inside the room is

m_{air} = m_{nitrogen} + m_{oxygen} + m_{carbon dioxide}

= 58.5 kg + 17.64 kg + 1.12 kg

= 77.26 kg

= 58.5 kg + 17.64 kg + 1.12 kg

= 77.26 kg

**b)** Since the volume of air is equal to the volume of room (air fills up the entire space of the room), we obtain for the density of air

ρ_{air} = *m*_{air}*/**V*_{air}

=*77.26 kg**/**60 m*^{3}

= 1.29*kg**/**m*^{3}

=

= 1.29

Enjoy the "Fluids. Density of Fluids" physics tutorial? People who liked the "Fluids. Density of Fluids" tutorial found the following resources useful:

- Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
- Density and Pressure Revision Notes: Fluids. Density of Fluids. Print the notes so you can revise the key points covered in the physics tutorial for Fluids. Density of Fluids
- Density and Pressure Practice Questions: Fluids. Density of Fluids. Test and improve your knowledge of Fluids. Density of Fluids with example questins and answers
- Check your calculations for Density and Pressure questions with our excellent Density and Pressure calculators which contain full equations and calculations clearly displayed line by line. See the Density and Pressure Calculators by iCalculator™ below.
- Continuing learning density and pressure - read our next physics tutorial: Pressure. Solid Pressure

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