Gas Pressure and Temperature

This lesson explains how the particle model accounts for gas pressure, how temperature affects gas pressure at constant volume, and the relationship $P \propto T$P∝T (in kelvin). This topic is part of the Particle Model section of the Edexcel GCSE Combined Science specification (1SC0).

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What Causes Gas Pressure?

Gas particles are in constant random motion. They collide with each other and with the walls of their container. Each collision exerts a force on the container wall. The total effect of billions of collisions per second produces a measurable pressure.

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

Symbol	Quantity	Unit
$P$P	Pressure	pascals (Pa) or newtons per square metre (N/m²)
$F$F	Force	newtons (N)
$A$A	Area	square metres (m²)

graph TD
    A["Gas particles move randomly"] --> B["Particles collide with container walls"]
    B --> C["Each collision exerts a tiny force"]
    C --> D["Total force over area = Pressure"]

Exam Tip: When explaining gas pressure, always mention: (1) particles are in constant random motion, (2) they collide with the walls, and (3) each collision exerts a force on the wall. These three points are needed for full marks.

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Factors That Affect Gas Pressure (Constant Volume)

If a gas is kept in a sealed container (constant volume), the pressure can be changed by altering the temperature or the number of particles.

Increasing Temperature at Constant Volume

When the temperature increases:

1. The kinetic energy of the particles increases — they move faster.
2. The particles collide with the walls more frequently.
3. Each collision is harder (greater force per collision).
4. The pressure increases.

Temperature	Particle speed	Collision frequency	Force per collision	Pressure
Low	Slow	Less frequent	Smaller	Lower
High	Fast	More frequent	Greater	Higher

graph LR
    A["Temperature increases"] --> B["Particles gain kinetic energy"]
    B --> C["Move faster"]
    C --> D["More frequent collisions"]
    C --> E["Harder collisions"]
    D --> F["Pressure increases"]
    E --> F

Exam Tip: Many students only mention that the particles move faster. For full marks you must also state that collisions are more frequent and each collision exerts a greater force.

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The Kelvin Scale

For the pressure–temperature relationship to work mathematically, temperature must be measured in kelvin (K), not degrees Celsius.

Converting Between Celsius and Kelvin

$T(\text{K}) = T(°\text{C}) + 273$T(K)=T(°C)+273

$T(°\text{C}) = T(\text{K}) - 273$T(°C)=T(K)−273

Celsius (°C)	Kelvin (K)	Significance
−273	0	Absolute zero — particles have minimum possible energy; no further cooling is possible
0	273	Melting point of water
100	373	Boiling point of water
20	293	Typical room temperature

Absolute Zero

Absolute zero (0 K or −273 °C) is the lowest possible temperature. At absolute zero, particles have the minimum possible kinetic energy (they essentially stop moving). The pressure of a gas would theoretically be zero.

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Pressure and Temperature: The Relationship

For a fixed mass of gas at constant volume:

$P \propto T \quad \text{(where T is in kelvin)}$P∝T(where T is in kelvin)

This means that if you double the absolute temperature (in kelvin), the pressure doubles.

Equivalently:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$T1​P1​​=T2​P2​​

Worked Example 1

A sealed container of gas has a pressure of 100 kPa at 27 °C. The gas is heated to 327 °C. Calculate the new pressure.

Step 1 — Convert to kelvin:

$T_1 = 27 + 273 = 300 \text{ K}$T1​=27+273=300 K $T_2 = 327 + 273 = 600 \text{ K}$T2​=327+273=600 K

Step 2 — Apply the relationship:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$T1​P1​​=T2​P2​​

$P_2 = P_1 \times \frac{T_2}{T_1} = 100 \times \frac{600}{300} = 200 \text{ kPa}$P2​=P1​×T1​T2​​=100×300600​=200 kPa

The temperature doubled (in kelvin), so the pressure doubled.

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Worked Example 2

A tyre contains air at a pressure of 250 kPa and a temperature of 20 °C. After a long journey the temperature rises to 50 °C. Calculate the new pressure (assume constant volume).

$T_1 = 20 + 273 = 293 \text{ K}$T1​=20+273=293 K $T_2 = 50 + 273 = 323 \text{ K}$T2​=50+273=323 K

$P_2 = P_1 \times \frac{T_2}{T_1} = 250 \times \frac{323}{293} = 275.6 \text{ kPa (to 4 s.f.)}$P2​=P1​×T1​T2​​=250×293323​=275.6 kPa (to 4 s.f.)

Exam Tip: Always convert temperatures to kelvin before using the pressure–temperature relationship. If you use degrees Celsius, you will get the wrong answer.

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Pressure–Temperature Graph

A graph of pressure (y-axis) against temperature in kelvin (x-axis) for a fixed mass of gas at constant volume is a straight line through the origin.

Feature	Detail
Shape	Straight line
Passes through	The origin (0 K, 0 Pa)
Gradient	Depends on the amount of gas and volume
Relationship shown	Direct proportion ($P \propto T$P∝T)

If the x-axis is in °C, the line is still straight but the intercept on the temperature axis is at −273 °C (absolute zero), not at the origin.

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Increasing the Number of Particles

If more gas is pumped into a sealed container at constant temperature and volume:

• There are more particles colliding with the walls.
• The collision frequency increases.
• The pressure increases.

This is why a tyre pump increases the pressure — it forces more air particles into the tyre.

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Decreasing Volume (Compression)

If the volume of a gas decreases at constant temperature:

• The same number of particles occupy a smaller space.
• Particles collide with the walls more frequently.
• The pressure increases.

This is why a bicycle pump gets warm — compressing the gas increases both the collision rate and the temperature.

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Common Misconceptions