Particle Motion and Pressure in Gases

This lesson covers how the particle model explains gas pressure, the relationship between temperature and pressure, and the relationship between pressure and volume ($pV = \text{constant}$pV=constant for a fixed mass at constant temperature). This is part of the AQA GCSE Combined Science Trilogy specification (8464, section 6.3.2).

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What Causes Gas Pressure?

Gas particles are in constant random motion, travelling at high speeds in all directions. When these particles collide with the walls of their container, they exert a force on the wall. The total force of many billions of collisions per second, spread over the area of the walls, creates pressure.

$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$Pressure=AreaForce​

Exam Tip: When explaining gas pressure, always refer to particles colliding with the walls of the container. Do not say particles collide with each other — while they do, that is not what causes pressure on the container walls.

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Temperature and Gas Pressure (Constant Volume)

When the temperature of a gas in a sealed container (fixed volume) increases:

1. Particles gain more kinetic energy and move faster.
2. They collide with the walls more frequently (more collisions per second).
3. Each collision exerts a greater force (because particles are moving faster).
4. Therefore the pressure increases.

This relationship is:

• Pressure is directly proportional to absolute temperature (in kelvin) at constant volume.

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Absolute Zero and the Kelvin Scale

Absolute zero is the lowest possible temperature — the temperature at which particles have the minimum possible kinetic energy (they cannot lose any more energy). At absolute zero, a gas would theoretically have zero pressure.

Temperature scale	Absolute zero	Boiling point of water
Celsius	−273 °C	100 °C
Kelvin	0 K	373 K

Converting between Celsius and Kelvin

$T(K) = T(°C) + 273$T(K)=T(°C)+273

$T(°C) = T(K) - 273$T(°C)=T(K)−273

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Pressure and Volume (Constant Temperature)

For a fixed mass of gas at a constant temperature:

$p \times V = \text{constant}$p×V=constant

$p_1 V_1 = p_2 V_2$p1​V1​=p2​V2​

This means pressure is inversely proportional to volume. If the volume is halved, the pressure doubles — and vice versa.

Why Does Reducing Volume Increase Pressure?

If the volume of the container decreases (but the temperature stays the same):

• The particles have less space to move in.
• They hit the walls more frequently (more collisions per second).
• The average speed (and therefore the force of each collision) stays the same (constant temperature).
• Therefore the pressure increases.

graph TD
    A["Volume halved"] --> B["Same number of particles<br/>in smaller space"]
    B --> C["More frequent collisions<br/>with the walls"]
    C --> D["Pressure doubles"]

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Worked Example 1

A gas has a volume of 100 cm³ at a pressure of 200 kPa. The gas is compressed to a volume of 50 cm³ at constant temperature. Calculate the new pressure.

$p_1 V_1 = p_2 V_2$p1​V1​=p2​V2​

$200 \times 100 = p_2 \times 50$200×100=p2​×50

$p_2 = \frac{200 \times 100}{50} = 400 \text{ kPa}$p2​=50200×100​=400 kPa

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Worked Example 2

A sealed container of gas is at 300 K and a pressure of 100 kPa. The temperature is increased to 600 K. Calculate the new pressure.

Since pressure is proportional to absolute temperature at constant volume:

$\frac{p_1}{T_1} = \frac{p_2}{T_2}$T1​p1​​=T2​p2​​

$\frac{100}{300} = \frac{p_2}{600}$300100​=600p2​​

$p_2 = \frac{100 \times 600}{300} = 200 \text{ kPa}$p2​=300100×600​=200 kPa

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Worked Example 3

A balloon has a volume of 4 litres at sea level (100 kPa). It rises to a height where the pressure is 50 kPa. What is the new volume (assuming constant temperature)?

$p_1 V_1 = p_2 V_2$p1​V1​=p2​V2​

$100 \times 4 = 50 \times V_2$100×4=50×V2​

$V_2 = \frac{100 \times 4}{50} = 8 \text{ litres}$V2​=50100×4​=8 litres

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Graphs

Pressure vs Volume (at constant temperature)

The graph is a smooth hyperbola — as pressure increases, volume decreases.

Pressure vs 1/Volume

A straight line through the origin — this confirms that pressure is inversely proportional to volume.

Pressure vs Temperature (kelvin, at constant volume)

A straight line through the origin — this confirms that pressure is directly proportional to absolute temperature.

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Common Misconceptions

Misconception	Correction
Gas pressure is caused by particles colliding with each other	Pressure is caused by particles colliding with the walls of the container
Reducing the volume makes the particles move faster	At constant temperature, the particles move at the same speed — they just collide with the walls more often
You can use Celsius in gas law calculations	You must use kelvin for proportionality relationships involving temperature
Gas particles are stationary at low temperatures	Particles are always moving — they only stop at absolute zero (0 K), which cannot actually be reached

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Summary

• Gas pressure is caused by particles colliding with the walls of the container.
• Increasing temperature at constant volume increases pressure (particles move faster, more frequent and harder collisions).
• For a fixed mass of gas at constant temperature: $pV = \text{constant}$pV=constant (pressure is inversely proportional to volume).
• Absolute zero (0 K = −273 °C) is the temperature at which particles have minimum kinetic energy.
• Temperature must be in kelvin for gas law calculations.

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Trilogy scope note: AQA GCSE Combined Science: Trilogy (8464) excludes the full quantitative gas-pressure treatment (6.3.3 — Triple only). Combined Science students still need to understand qualitatively how particle motion creates pressure, how pressure changes with temperature and volume, and how this links back to the particle model. The calculations in this lesson are useful as stretch material and help deepen understanding of the particle model applied to gases.

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Worked Example 4 — Everyday Application: A Car Tyre

On a cold morning (270 K), a car tyre has a pressure of 220 kPa. After driving, the tyre has heated to 320 K. If the volume is roughly constant, what is the new pressure?

$\frac{p_1}{T_1} = \frac{p_2}{T_2}$T1​p1​​=T2​p2​​

$p_2 = p_1 \times \frac{T_2}{T_1} = 220 \times \frac{320}{270} \approx 261 \text{ kPa}$p2​=p1​×T1​T2​​=220×270320​≈261 kPa