4.2 Particle Model & Gases
What is Density?
Density (ρ) is a measure of how much mass is packed into a certain volume. It tells us how "compact" a substance is.
Density (ρ) is a measure of how much mass is packed into a certain volume. It tells us how "compact" a substance is.
⚡ Density Formula:
$$\rho = \frac{m}{V}$$
Where:
• $\rho$ (rho) = density in kg/m³ or g/cm³
• $m$ = mass in kg or g
• $V$ = volume in m³ or cm³
Where:
• $\rho$ (rho) = density in kg/m³ or g/cm³
• $m$ = mass in kg or g
• $V$ = volume in m³ or cm³
Solids
Particles close together. Generally highest density.
Liquids
Particles close but can move. Medium density.
Gases
Particles far apart. Lowest density.
💡 Key Exception: Water and Ice
Water is unusual - ice (solid) is less dense than liquid water
• Ice density: 917 kg/m³
• Water density: 1,000 kg/m³
This is why ice floats, and why lakes freeze from the top down.
• Ice density: 917 kg/m³
• Water density: 1,000 kg/m³
This is why ice floats, and why lakes freeze from the top down.
Example 1: Calculating Density
A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density.
Step 1: Write the formula
$\rho = \frac{m}{V}$
Step 2: Substitute values
$\rho = \frac{540}{200}$
Step 3: Calculate
$\rho = 2.7$ g/cm³
Answer: This is aluminium (density = 2.7 g/cm³)
Step 1: Write the formula
$\rho = \frac{m}{V}$
Step 2: Substitute values
$\rho = \frac{540}{200}$
Step 3: Calculate
$\rho = 2.7$ g/cm³
Answer: This is aluminium (density = 2.7 g/cm³)
Example 2: Finding Mass
A gold bar has a volume of 50 cm³. What is its mass?
(Density of gold = 19.3 g/cm³)
Step 1: Rearrange the formula
$m = \rho \times V$
Step 2: Substitute values
$m = 19.3 \times 50$
Step 3: Calculate
$m = 965$ g = 0.965 kg
(Density of gold = 19.3 g/cm³)
Step 1: Rearrange the formula
$m = \rho \times V$
Step 2: Substitute values
$m = 19.3 \times 50$
Step 3: Calculate
$m = 965$ g = 0.965 kg
| Substance | Density (kg/m³) | Density (g/cm³) |
|---|---|---|
| Air | 1.2 | 0.0012 |
| Water | 1,000 | 1.0 |
| Aluminium | 2,700 | 2.7 |
| Iron | 7,800 | 7.8 |
| Gold | 19,300 | 19.3 |
🎯 Density Practice
Gas Particles and Pressure
Gas particles are far apart and move randomly at high speeds. Gas pressure is caused by particles colliding with the walls of their container.
• Particles gain kinetic energy
• Particles move faster
• Hit walls harder and more often
• Pressure increases ↑
• Volume decreases
• Particles hit walls more often
• Same speed, less space
• Pressure increases ↑
Gas particles are far apart and move randomly at high speeds. Gas pressure is caused by particles colliding with the walls of their container.
⚡ Pressure Formula:
$$P = \frac{F}{A}$$
Where:
• $P$ = pressure in pascals (Pa)
• $F$ = force in newtons (N)
• $A$ = area in square metres (m²)
1 Pa = 1 N/m²
Where:
• $P$ = pressure in pascals (Pa)
• $F$ = force in newtons (N)
• $A$ = area in square metres (m²)
1 Pa = 1 N/m²
💡 How Gas Pressure Works:
• Billions of particles collide with container walls
• Each collision exerts a tiny force
• All these forces add up to create pressure
• More collisions = higher pressure
• Each collision exerts a tiny force
• All these forces add up to create pressure
• More collisions = higher pressure
Heating a Gas
(at constant volume)• Particles gain kinetic energy
• Particles move faster
• Hit walls harder and more often
• Pressure increases ↑
Compressing a Gas
(at constant temperature)• Volume decreases
• Particles hit walls more often
• Same speed, less space
• Pressure increases ↑
Example 1: Calculating Pressure
A force of 500 N acts on an area of 0.02 m². Calculate the pressure.
Step 1: Write the formula
$P = \frac{F}{A}$
Step 2: Substitute values
$P = \frac{500}{0.02}$
Step 3: Calculate
$P = 25,000$ Pa = 25 kPa
Step 1: Write the formula
$P = \frac{F}{A}$
Step 2: Substitute values
$P = \frac{500}{0.02}$
Step 3: Calculate
$P = 25,000$ Pa = 25 kPa
Example 2: Finding Force
Atmospheric pressure is 101,000 Pa. What force does the atmosphere exert on a window of area 2 m²?
Step 1: Rearrange the formula
$F = P \times A$
Step 2: Substitute values
$F = 101,000 \times 2$
Step 3: Calculate
$F = 202,000$ N = 202 kN
Step 1: Rearrange the formula
$F = P \times A$
Step 2: Substitute values
$F = 101,000 \times 2$
Step 3: Calculate
$F = 202,000$ N = 202 kN
🎯 Pressure Practice
Pressure in Fluids
A fluid is any substance that can flow - this includes liquids and gases. Pressure in a fluid increases with depth because of the weight of fluid above.
Weight > Upthrust
Object is denser than the fluid
Example: Stone in water
Weight < Upthrust
Object is less dense than the fluid
Example: Wood in water
A fluid is any substance that can flow - this includes liquids and gases. Pressure in a fluid increases with depth because of the weight of fluid above.
⚡ Pressure in a Fluid Column:
$$P = h \times \rho \times g$$
Where:
• $P$ = pressure (Pa)
• $h$ = height/depth of fluid column (m)
• $\rho$ = density of fluid (kg/m³)
• $g$ = gravitational field strength (N/kg)
Where:
• $P$ = pressure (Pa)
• $h$ = height/depth of fluid column (m)
• $\rho$ = density of fluid (kg/m³)
• $g$ = gravitational field strength (N/kg)
Surface
Low pressure
↓ Depth increases ↓
Deep water
High pressure
💡 Upthrust and Archimedes' Principle:
Because pressure increases with depth:
• Pressure at bottom of object > pressure at top
• This creates a net upward force called upthrust
Archimedes' Principle:
The upthrust on an object equals the weight of fluid displaced.
• Pressure at bottom of object > pressure at top
• This creates a net upward force called upthrust
Archimedes' Principle:
The upthrust on an object equals the weight of fluid displaced.
Object Sinks
When:Weight > Upthrust
Object is denser than the fluid
Example: Stone in water
Object Floats
When:Weight < Upthrust
Object is less dense than the fluid
Example: Wood in water
Wood
Floats (less dense)
Stone
Sinks (more dense)
Example 1: Pressure at Depth
Calculate the pressure at 15 m depth in water.
(Density of water = 1,000 kg/m³, g = 10 N/kg)
Step 1: Write the formula
$P = h \times \rho \times g$
Step 2: Substitute values
$P = 15 \times 1000 \times 10$
Step 3: Calculate
$P = 150,000$ Pa = 150 kPa
(Density of water = 1,000 kg/m³, g = 10 N/kg)
Step 1: Write the formula
$P = h \times \rho \times g$
Step 2: Substitute values
$P = 15 \times 1000 \times 10$
Step 3: Calculate
$P = 150,000$ Pa = 150 kPa
Example 2: Will It Float?
An object has density 800 kg/m³. Will it float in water (density 1,000 kg/m³)?
Answer: Yes. The object is less dense than water, so it will float.
Rule:
• Object density < fluid density → floats
• Object density > fluid density → sinks
Answer: Yes. The object is less dense than water, so it will float.
Rule:
• Object density < fluid density → floats
• Object density > fluid density → sinks
💡 Atmospheric Pressure:
The air in our atmosphere also exerts pressure:
• At sea level: ~101,000 Pa (101 kPa)
• Caused by the weight of air above us
• Decreases with altitude (less air above)
• This is why your ears pop on a plane.
• At sea level: ~101,000 Pa (101 kPa)
• Caused by the weight of air above us
• Decreases with altitude (less air above)
• This is why your ears pop on a plane.
🧮 Fluid Pressure Calculator:
Depth:
m
Density:
kg/m³
🎯 Float or Sink?
Real Life Applications:
• Ships: Float because their average density (including air inside) is less than water
• Submarines: Control depth by adjusting water in ballast tanks
• Hot air balloons: Hot air is less dense than cold air, creating upthrust
• Diving: Divers must equalise pressure as they descend
• Weather: High/low pressure systems drive wind and weather patterns
• Ships: Float because their average density (including air inside) is less than water
• Submarines: Control depth by adjusting water in ballast tanks
• Hot air balloons: Hot air is less dense than cold air, creating upthrust
• Diving: Divers must equalise pressure as they descend
• Weather: High/low pressure systems drive wind and weather patterns