An Introduction to Matter and Elasticity

This page deals with the basic ideas about Matter, i.e different type of forms, intermolecular potential energy, including simple curves and the terms Matter and Elasticity. This to topics shall be divided into many series, this is part 1 of the series, i shall attached a link to the next part of the series below.

PS: It’s important to stress that this series are for students doing either gce or gcse A level physics.

 

MATTER

As we all know Matter is anything that has mass and occupies space. It exist in 3 forms that is Solid, Liquid and Gas.

SOLID

By definition it is the physical state of matter in which the constitution molecule, atoms or ions have no translation. Although they may vibrate about a fixed position which they occupy in a crystal lattice. Solids exist in crystalline form and are solid to be amorphous solids have a fixed shape. 3 kinds of solids exist

1. Crystallized Solids: Most metals are crystallized solids (granite and diamond)
2. Amorphous solids : for example glass and wax
3. Polymeric solids: for example rubber

 

CRYSTALLINE SOLIDS

They are characterized by the following

1. The particles are in a regular in a regular repeating pattern
2. They conduct electricity
3. They have a definite melting point
4. It maybe transparent or opaque depending on the type of crystal(e.g graphite is opaque, diamond is transparent)
5. Their crystals are called grains an can be as small as 10^-2

 

AMORPHOUS SOLIDS

They are characterized by the following

1. Their particles are in no regular pattern.
2. Their particles have no regular melting point
3. They have no regular outline, no directional properties no cleavage plane
4. Their atoms have no long regular arrangement.

 

POLYMERIC SOLIDS

1. They have large molecules in the forms of long chains
2. They are usually organic and maybe either Natural ( cellulose and proteins), Synthetic(Polythene and various and various forms of carbon.
3. They may be linear, branded or cross-linked.
4. Many polymers may be classified as thermoplastics or thermosetting.

 

With this introduction in mind, we shall go into some calculations i.e  estimating the molecular spacing, this is  are something often ask in the gce or gcse A level, But before jumping into it, one must first have some basics such as Molar mass  and atomic mass. I have attached a link to Wikipedia for does who don’t yet graphs the concepts. Click here

 

Using density to estimate molecular spacing (Atomic spacing)

For this section we shall be using an example to explain how to go about, so please follow alone carefully.

 

 

Using water as an example, 1 mole of water has a mass of 0.018kg and the density of water is 1000kgm^-3

• 1m³ of water has mass of 1000kg
• H2o contains (1000 / 0.018) moles
• 1m3 of water contains (1000na/0.018) molecules

therefore Number of molecules =moles x Na

=  (1000/0.018) X Na = 1000/0.018 molecules.

therefore average volume of a molecule = mass/density = o.o18/1000Na

=  3.0 X 10^-29m³

Now the volume of each molecule is proportional to d³. Now assuming that the molecule is a cubic of side d that is

d³ = 3.0×10^-29m³

d = ∛3.0 X 10^-29m³

d= 3.1×10^-29m³ (which is  of the order quoted at )

N.B: this answer would be different if we assume that this molecules were spherical but it is still accepted as a good estimate.

 

Assuming molecules a spherical shape

Here we are assuming in our calculation that the molecule is spherical in shape.

 

Since volume of sphere is  4/3Πr³

but volume from  our above calculation is equal to 3.1×10^-29m³

therefore   3.1×10^-29m³ =4/3Πr³

making r the subject we have  r = ∛3V /4Π

Summary spacing can be determined if values are given for density, relative atomic mass M and avogardrous constant Na. since the mass of a single atom of solid element is given by M/Na , then the volume occupied by an atom is M/ρNa.

 

Hence the appropriate spacing between atoms of a solid elements is given by d3 = M/ρNa

Where distance between the centers of adjacent atoms

• Number of atoms =Na/M

Therefore Average volume occupied by an atom is  V= M/ρNa.

Average spacing  d = ∛ volume

It’s important  to understand the following before moving forward, but to better explain we shall  have an example.

Example.

Copper has  molar mass of 0.65kg and a density of 8820kgm3

Calculate

1. The number of atoms in 1kg of copper.
2. The average volume occupied by a copper atom.
3. The average spacing of the copper atom.

Solution

1)

n moles = mass/molar mass 

=  1kg /0.65kgmo-1

= 1.538 moles.

therefore number of atoms = moles x Na

= 1.538 x 6.02×10²³

2)

average volume =  M/ρNa.

= (0.65)/(8920 x 6.02×10²³)

= 1.18 x 6.02×10^-29 m3

3)

d³ = 1.98 x 10^-29 m3

d= ∛1.98 x 10^-29 m3

d = 2.28 x 10^-10 m

 

Ouff that was all for today, we shall be posting the next chapter of the  series  in few days  , but before we go we shall be giving a assignment, please try to  do it  and post your answer below so that  i can check out your answers :).

 

Assignment

The density of copper is  8.92 x 10^-3m3 and 1 mole of copper has a mass of  6.35 x 10^-2kg.Calculate

1. The number of atoms in a cubic meter of copper,
2. The separation nearest neighbor atoms given that  the effective volume occupied by an atom is (1/√ 2 )a³. The Specific latent heat of sublimation  of copper  at 35 degree is 5.34x 10⁵kjkg-1

Good luck

 

In the Next Article we are going to explore Variation of Intermolecular force with separation between particles