Test

BMAT Physics Notes: Matter

Advice & Insight From BMAT Specialists

States of Matter

On a microscopic level we view matter to be made of atoms/molecules that have certain properties.
 
SOLID
 
In the solid state atoms are vibrating about fixed positions.
The atoms are close together and can be arranged in a regular pattern for many hundreds of layers (e.g. in a metal).
There are attractive and repulsive forces acting on the atoms which cause them to vibrate.
The atoms have both  kinetic energy ( because they are moving) and potential energy (because they experience forces).
Most solids are difficult to compress because of the strong repulsive forces between them.
They also do not expand a lot when heated.
Heat is transferred through a solid by conduction.
Solids have a fixed shape.

LIQUID
 
In the liquid state the atoms do not vibrate and are not arranged in a regular pattern.
They are able to move around each other and they are slightly further apart, on average, than atoms of a solid.
The atoms move in straight lines, this is called translational motion.
The atoms are continually making and breaking bonds with neighbouring atoms.
Liquids are difficult to compress because of the repulsive forces.
Most liquids are poor conductors. Heat is transferred through a liquid mainly by convection.
Liquids do not have a fixed shape but take the shape of the container in which they are placed.
The atoms have both  kinetic energy and potential energy.

GASES

In a gas the atoms are much further apart than those of a solid or liquid.The atoms/molecules move in straight lines and move with different speeds. They do not experience any forces (except during collisions).
They have kinetic energy only.
Gases are easy to compress or expand because the atoms do not experience forces.
They expand a lot when heated.
Gases are poor conductors of heat.
Heat is transferred through a gas by convection.

Intensive BMAT Course

Comprehensive 5 in 1 Package with a Full Day Intensive BMAT Course, Online BMAT Course Access & Much More…

Online BMAT Course

Online BMAT Tutorials, Expert Techniques & BMAT Mock Examinations With Our Popular BMAT Portal

Private BMAT Tuition Image
Private BMAT Tuition

1-1 BMAT Support To Optimise Your BMAT Score & Convert Your Weakest BMAT Section To Your Strongest.

​Change of State

​Suppose a lump of metal is heated steadily. The temperature rises. Eventually, if it is heated long enough the metal reaches its melting point. If further heating occurs the metal will begin to melt and change state i.e. change from a solid to a liquid.
Heat is required to change the metal from a solid to a liquid but whilst it is melting the temperature remains constant.
 
The amount of heat required to change any mass of a solid into a liquid at its melting point is called latent heat of fusion.

 
The amount of heat required to change 1 kg  of a solid into a liquid at its melting point is called specific latent heat of fusion.

Examples

​To change 2kg of ice into water would require 2 x 334 kJ,
To change 3kg of ice into water would require 3 x 334 kJ, and so on.
 
In general, the thermal energy (E) required to change m kg of a solid into a liquid at its melting point is given by:

Example 1
​

A lump of ice of mass 0.1 kg is at its melting point of 0°C. Heat is supplied to the ice by a small electrical heater of power 50 W.
​
Calculate (i) the heat required to melt all the ice,  (ii) the time taken.
 

Picture

(ii​) ​Power = Energy/time    Hence  time = energy/power  =  33400/50 =  668s

Optimise Your BMAT Performance

Learn time-efficient BMAT strategies and practice with reflective BMAT questions & worked solutions.

Suppose a liquid is heated steadily. The temperature rises. Eventually, if it is heated long enough the liquid reaches its boiling point. If further heating occurs the liquid  will change state i.e. change from a liquid to a gas.
Heat is required to change the liquid into a gas  whilst it is boiling but the temperature remains constant.

The amount of heat required to change any mass of a liquid into a gas at its boiling point is called latent heat of vaporisation.
​
The amount of heat required to change 1 kg  of a liquid into a gas at its boiling point is called specific latent heat of vaporisation.

Example

Picture
 
This means to change 1kg of water at 100° C  to 1kg of vapour (gas) at 100° C requires 2.26 MJ
 
To change 2kg of water (at 100° C) into gas ( 100° C) would require 2 x 2.26 MJ
To change 3kg of water (at 100° C) into gas ( 100° C) would require 3 x 2.26 MJ, and so on.

In general, the thermal energy (E) required to change m kg of a liquid into a gas at its boiling point is given by:                           
Picture
 
Example 2
​
Picture
 
​Calculate the specific latent heat of vaporisation of the liquid.
 
Heat produced = power x time =  15 x 450 =  6750J
 
Picture

​Microscopic explanation of PRESSURE of a gas

The molecules are moving in random directions and collide  with each other and with the walls of the container. After each collision they are moving in different directions. Hence their momentum has changed. From Newton’s 2nd law of motion the force acting on the molecules is equal to the rate of change of momentum. Hence there must be a force acting on the molecules. From Newton’s 3rd law there must be an equal and opposite force acting on the walls. Since many molecules collide with a given area of the container walls there is a force acting on a given area i.e. pressure.

Microscopic explanation of TEMPERATURE of a gas

When a gas is heated the gas molecules gain KE i.e. they move faster. The temperature of the gas rises.
 Theory shows that the average KE of the gas molecules is proportional to the (Kelvin) Temperature.
[You do not need to know this for the exam].
 
This means that when the temperature of a gas rises the molecules move faster, when the temperature decreases the molecules move slower.

​Ideal gas equation

Suppose a gas is enclosed in a piston chamber as shown.
 
If the gas is compressed as shown then the molecules move closer together. This means they collide more frequently and so the pressure (P) increases and the volume (V) decreases.
(The temperature is constant).

Experiment shows that:   
Pressure x volume  = constant  or    P V = constant
[You have to be able to use equation in calculations].
Any gas that behaves in this way is called an ideal gas.

​Example 3
 
Gas leaks slowly out of a cylinder of constant volume. The temperature of the gas in the cylinder does not change. Which of the following is constant for the gas molecules in the cylinder?
 
A  the number striking unit area of surface in unit time.
 
B  the number of the collisions between molecules per unit time.
 
C  the number per unit volume.
 
D  the average speed.

​Density

Density is defined as mass per unit volume.
 
Picture
​Example 4
 
Picture
​Example 5
 
Picture

Measurement of Density

  1. Measure mass by placing object on a scale balance.
  2. If the object is regular (e.g. a cube of a solid substance) its volume can be calculated.
  3. If the object is small and irregular its volume can be found by placing it in a measuring cylinder and measuring the displacement of the liquid.
  4. Using Density = mass/volume the density can  be calculated.

  SOLIDS have the greatest densities and GASES the smallest.
 
Picture

PRESSURE

Pressure is defined as force per unit area where the force is normal to the surface.
 
Picture
​Example 6
 
Picture
 
(ii) the mass and weight of the block,

(iii) the maximum and minimum pressure the block can exert.

Variation of pressure with depth

Suppose a liquid of density r is placed in a tank as shown.
The pressure p of the liquid acting on the base of the tank is given by:
Picture
This equation is valid for a container of any shape:

Example Answers

​Example 3
D (The temperature is constant and so the average speed is constant).

​Example 4
mass =  density x volume = 1000 x 0.2 = 200 kg
Weight  = mg  = 200 x 10 = 2000N  (Taking g = 10 numerically)
 
Example 6
volume  = 20x10x4 = 800 cm3  = 800 x 10-6 m3

Mass = density x vol = 5000 x 800 x 10-6 = 4 kg
​
Weight = mg = 40N

MAXIMUM pressure will occur when the block rest on its SMALLEST face i.e. 4cm x 10cm
 
Pressure = weight / Area = 40 /(40×10-4) = 10,000 N m-2

MINIMUM pressure will occur when the block rests on its LARGEST face i.e. 20cmx10cm
 
Pressure = weight/area = 40/(200×10-4) = 2000 Nm-2

BMAT Physics Notes: Matter

Intensive BMAT (5 in 1) Course

Private BMAT Tuition

Online BMAT Course

Shopping Cart

Intensive BMAT Course

BMAT Timetable

The BMAT Course