Gas Pressure and Volume [H]

This lesson covers the relationship between gas pressure and volume as required by the AQA GCSE Physics specification (4.3.3). This is Higher Tier only content, labelled [H]. You need to understand how changing the volume of a gas affects its pressure, and be able to use the equation linking pressure and volume (Boyle's Law) in calculations.

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Boyle's Law: Pressure and Volume

Boyle's Law states that for a fixed mass of gas at constant temperature:

The pressure of a gas is inversely proportional to its volume.

This means:

• If you decrease the volume (compress the gas), the pressure increases.
• If you increase the volume (expand the gas), the pressure decreases.

The mathematical relationship is:

p x V = constant (at constant temperature for a fixed mass of gas)

or equivalently:

p1 x V1 = p2 x V2

Where:

• p1 = initial pressure (Pa)
• V1 = initial volume (m3)
• p2 = final pressure (Pa)
• V2 = final volume (m3)

Exam Tip: The equation p x V = constant (or p1 x V1 = p2 x V2) is given on the exam equation sheet for Higher Tier. You must understand what "inversely proportional" means: when one quantity doubles, the other halves. If you halve the volume, you double the pressure.

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Why Does Pressure Increase When Volume Decreases?

The particle model explains Boyle's Law:

graph LR
    subgraph Large_Volume["Large Volume"]
        LV1["O     O     O"]
        LV2["  O     O    "]
        LV3["O     O     O"]
    end
    subgraph Small_Volume["Small Volume<br/>(Compressed)"]
        SV1["O  O  O"]
        SV2["O  O  O"]
        SV3["O  O  O"]
    end

    Large_Volume -->|"Decrease<br/>volume"| Small_Volume

    style Large_Volume fill:#2ecc71,color:#fff
    style Small_Volume fill:#e74c3c,color:#fff

When the volume of a container is decreased (at constant temperature):

1. The same number of gas particles are squeezed into a smaller space.
2. The particles have the same speed and same kinetic energy (because temperature is constant).
3. The particles hit the walls more frequently — there is less distance for them to travel between collisions with the walls.
4. More frequent collisions means a greater force on the walls per second.
5. Since the area of the walls is smaller, and the force per unit area increases, the pressure increases.

When the volume is increased:

1. The particles have more space to move around.
2. They hit the walls less frequently — there is more distance between collisions with the walls.
3. The pressure decreases.

Exam Tip: When explaining why pressure changes with volume, always state that the temperature is constant, so the particles have the same kinetic energy and move at the same speed. The change in pressure is due to the change in the FREQUENCY of collisions with the walls, not because the particles move faster or slower.

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The p x V = constant Equation [H]

For a fixed mass of gas at constant temperature:

p1 x V1 = p2 x V2

This can be rearranged to find any unknown:

To Find	Formula
p2	p2 = (p1 x V1) / V2
V2	V2 = (p1 x V1) / p2
p1	p1 = (p2 x V2) / V1
V1	V1 = (p2 x V2) / p1

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Worked Examples [H]

Worked Example 1

A gas has a volume of 0.006 m3 at a pressure of 100,000 Pa. The gas is compressed to a volume of 0.002 m3 at constant temperature. Calculate the new pressure.

Step 1: Write down known values: p1 = 100,000 Pa, V1 = 0.006 m3, V2 = 0.002 m3

Step 2: Use p1 x V1 = p2 x V2

Step 3: Rearrange: p2 = (p1 x V1) / V2

Step 4: Substitute: p2 = (100,000 x 0.006) / 0.002

Step 5: Calculate: p2 = 600 / 0.002 = 300,000 Pa

The volume was reduced to one-third, so the pressure tripled.

Worked Example 2

A sealed syringe contains 60 cm3 of air at a pressure of 100,000 Pa. The plunger is pushed in until the pressure reaches 200,000 Pa. Calculate the new volume.

Step 1: Write down known values: p1 = 100,000 Pa, V1 = 60 cm3, p2 = 200,000 Pa

Step 2: Use p1 x V1 = p2 x V2

Step 3: Rearrange: V2 = (p1 x V1) / p2

Step 4: Substitute: V2 = (100,000 x 60) / 200,000

Step 5: Calculate: V2 = 6,000,000 / 200,000 = 30 cm3

The pressure doubled, so the volume halved.

Worked Example 3

A diver is at a depth where the pressure is 300,000 Pa. She exhales a bubble with a volume of 0.5 cm3. What will be the volume of the bubble at the surface where the pressure is 100,000 Pa? (Assume constant temperature.)

Step 1: Write down known values: p1 = 300,000 Pa, V1 = 0.5 cm3, p2 = 100,000 Pa

Step 2: Use p1 x V1 = p2 x V2

Step 3: Rearrange: V2 = (p1 x V1) / p2

Step 4: Substitute: V2 = (300,000 x 0.5) / 100,000

Step 5: Calculate: V2 = 150,000 / 100,000 = 1.5 cm3

The pressure decreased to one-third, so the volume tripled.

Exam Tip: In Boyle's Law calculations, the volume units must be the SAME on both sides of the equation, but they do not have to be m3. If both volumes are in cm3, the calculation still works. However, pressure should ideally be in Pa for consistency. If you are given mixed units, convert before calculating.

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Graphical Representation [H]

Pressure vs Volume Graph

When you plot pressure (y-axis) against volume (x-axis) for a fixed mass of gas at constant temperature:

• The graph is a smooth curve (a rectangular hyperbola).
• As volume decreases, pressure increases.
• The curve never touches either axis (as volume approaches zero, pressure approaches infinity, and vice versa).

Pressure vs 1/Volume Graph

When you plot pressure (y-axis) against 1/volume (x-axis):

• The graph is a straight line through the origin.
• This confirms that pressure is directly proportional to 1/volume (i.e. inversely proportional to volume).

Graph	Shape	What It Shows
p vs V	Curved (hyperbola)	Inverse relationship
p vs 1/V	Straight line through origin	Direct proportionality between p and 1/V

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Real-Life Applications of Boyle's Law [H]