Pressure from collisions · Gas pressure and temperature · Absolute zero · Kelvin scale
AQA GCSE Physics 4.3 | Year 10 | Foundation & Higher
🔵 Explain how gas pressure arises from particle collisions with container walls
🔴 Describe how increasing temperature affects particle motion and pressure
🟢 State what is meant by absolute zero and its value in °C
🟡 Convert between the Celsius scale and the Kelvin scale
🔵 Explain the relationship between gas pressure and temperature (constant volume)
🔴 Use the pressure–temperature relationship to solve quantitative problems
🔵 How Gases Exert Pressure
A gas is made up of a very large number of tiny particles (atoms or molecules) that are in continuous, random motion. Unlike solids or liquids, gas particles are widely spaced and move freely in all directions at high speeds — typically hundreds of metres per second at room temperature.
Gas pressure is the force exerted per unit area on the walls of a container, caused by gas particles repeatedly colliding with those walls.
Each time a particle hits a wall, it exerts a tiny force on the wall during the collision (Newton's Third Law — the wall pushes back on the particle, reversing its direction). Because there are an enormous number of particles, and they collide with the walls billions of times per second, the combined effect is a steady, measurable pressure.
Pressure is defined as:
P = F ÷ A
where P is pressure in Pascals (Pa), F is the force in Newtons (N), and A is the area in metres squared (m²).
For a gas in a sealed container at a fixed volume, the pressure depends on two things:
How frequently particles collide with the walls
How much force each collision exerts (which depends on particle speed)
More collisions per second + harder collisions = higher gas pressure.
Quantity
Symbol
SI Unit
Pressure
P
Pascal (Pa)
Force
F
Newton (N)
Area
A
metre² (m²)
🔴 Temperature and Particle Speed
Temperature is a measure of the average kinetic energy of the particles in a substance. When you heat a gas, you supply energy to the particles — they move faster on average.
For a gas in a fixed container (constant volume):
Higher temperature → particles move faster
Faster particles hit the walls more often
Each collision exerts a greater force on the wall
Therefore, pressure increases
At constant volume, increasing the temperature of a gas increases its pressure.
Conversely, if you cool the gas, particles slow down, collisions become less frequent and less forceful, and the pressure drops. This relationship between pressure and temperature (at constant volume) is one of the fundamental gas laws studied in physics and chemistry.
The relationship is directly proportional — but only when temperature is measured on the Kelvin scale (see next section). This means if you double the absolute (Kelvin) temperature, the pressure also doubles (provided the volume does not change).
P ÷ T = constant (constant volume)
This can be rearranged to compare two states of the same gas:
P₁ ÷ T₁ = P₂ ÷ T₂
where P₁ and T₁ are the initial pressure and temperature, and P₂ and T₂ are the final values. T must always be in Kelvin.
🟢 Absolute Zero and the Kelvin Scale
Scientists noticed that if you plot pressure against temperature (in °C) for a fixed volume of gas and extend the straight line back, it always crosses the temperature axis at −273 °C (more precisely −273.15 °C). At this point, the pressure would be zero — meaning the particles would have no kinetic energy and would not be moving at all.
Absolute zero is the lowest possible temperature, at which particles have the minimum possible internal energy. It is equal to −273 °C (−273.15 °C precisely).
In practice, absolute zero can never be fully reached, though scientists have come extremely close (within billionths of a degree) in laboratory conditions.
Because −273 °C is the coldest anything can ever be, it makes sense to use this as the zero point of a temperature scale. This gives us the Kelvin scale (also called the absolute temperature scale).
The Kelvin scale starts at absolute zero (0 K = −273 °C). Its divisions are the same size as degrees Celsius, so a change of 1 K equals a change of 1 °C.
Conversion formulae:
T (K) = θ (°C) + 273
θ (°C) = T (K) − 273
Temperature in °C
Temperature in K
Significance
−273 °C
0 K
Absolute zero
0 °C
273 K
Freezing point of water
20 °C
293 K
Room temperature
100 °C
373 K
Boiling point of water
The Kelvin temperature is always 273 more than the Celsius temperature (using the GCSE approximation).
🟡 Using the Pressure–Temperature Law
The pressure–temperature relationship for a fixed mass of gas at constant volume is one of the core quantitative skills for this topic. The key equation is:
P₁ / T₁ = P₂ / T₂
This can be rearranged depending on which quantity you need to find:
P₂ = P₁ × (T₂ / T₁)
T₂ = T₁ × (P₂ / P₁)
Step-by-step method:
Identify P₁, T₁ (initial state) and what is known about the second state
Convert all temperatures to Kelvin by adding 273
Substitute into the equation and rearrange
Calculate and include units in the answer
⚠️ The most common mistake is forgetting to convert Celsius to Kelvin before substituting. Always add 273 to °C values!
This relationship only holds at constant volume. If both pressure and volume change, a more complex combined gas law is needed (studied at A-Level).
Example 1: A sealed gas cylinder has a pressure of 200 000 Pa at a temperature of 27 °C. The cylinder is heated until its temperature reaches 127 °C. Assuming the volume stays constant, calculate the new pressure.
1Write down what we know: P₁ = 200 000 Pa, θ₁ = 27 °C, θ₂ = 127 °C
2Convert temperatures to Kelvin: T₁ = 27 + 273 = 300 K T₂ = 127 + 273 = 400 K
3Write the pressure–temperature equation: P₁ / T₁ = P₂ / T₂ So: P₂ = P₁ × (T₂ / T₁)
The gas must be heated to 75 °C (348 K) to reach a pressure of 180 000 Pa.
Example 3: Convert the following temperatures: (a) 0 °C to Kelvin, (b) 350 K to Celsius, (c) −100 °C to Kelvin.
1(a) T = 0 + 273 = 273 K
2(b) θ = 350 − 273 = 77 °C
3(c) T = −100 + 273 = 173 K
(a) 273 K | (b) 77 °C | (c) 173 K
Example 4: Explain, in terms of particle motion, why the pressure of a gas in a sealed rigid container increases when the gas is heated.
1When the gas is heated, energy is transferred to the gas particles.
2The particles gain kinetic energy and move faster on average.
3Faster particles collide with the container walls more frequently.
4Each collision also exerts a greater force on the wall because the particles are moving faster.
5Since pressure = force ÷ area, and the area of the container walls has not changed, the pressure increases.
Higher temperature → faster particles → more frequent and harder collisions → greater force on walls → increased pressure.
Question 1: What is the value of absolute zero in degrees Celsius?
Question 2: A gas is heated at constant volume. Which of the following best explains why the pressure increases?
Question 3: Convert 47 °C to Kelvin.
Question 4: A gas has pressure 100 000 Pa at 300 K. What is its pressure at 600 K (constant volume)?
Question 5: Convert 500 K to degrees Celsius.
Challenge 1 (Higher): A car tyre contains air at a pressure of 250 000 Pa at a temperature of 7 °C. After a long motorway journey, the temperature of the air inside rises to 57 °C. Assuming the volume of the tyre does not change, calculate the new pressure inside the tyre.
Step 1: Convert temperatures to Kelvin:
T₁ = 7 + 273 = 280 K T₂ = 57 + 273 = 330 K
Step 3: P₂ = 250 000 × 1.1786… = 294 643 Pa ≈ 295 000 Pa
The pressure increases from 250 000 Pa to approximately 295 000 Pa. This explains why tyre manufacturers recommend checking pressures when tyres are cold.
Challenge 2 (Higher): A sealed gas container has a pressure of 120 000 Pa at 27 °C. A student claims that if the container is cooled to −123 °C, the pressure will drop to zero. Is the student correct? Calculate the expected pressure at −123 °C and explain your answer.
Step 1: Convert temperatures to Kelvin:
T₁ = 27 + 273 = 300 K T₂ = −123 + 273 = 150 K
Conclusion: The student is incorrect. The pressure drops to 60 000 Pa, which is half the original pressure (because the Kelvin temperature halved). Pressure would only be zero at absolute zero (0 K = −273 °C), which cannot be reached in practice.
Challenge 3 (Extended Answer): A scientist measures the pressure of a fixed volume of gas at various temperatures and plots a graph of pressure (y-axis) against temperature in Kelvin (x-axis). Describe the shape of the graph and explain what happens to the line when extended back to the x-axis. What is the significance of this x-intercept?
Graph shape: The graph is a straight line through the origin (0 K, 0 Pa). Pressure is directly proportional to temperature in Kelvin.
Extending the line: When plotted against Celsius, the line, when extended backwards, crosses the temperature axis at −273 °C. On the Kelvin graph, it passes through the origin (0 K).
Significance of the x-intercept: The x-intercept at 0 K (−273 °C) represents absolute zero — the theoretical lowest possible temperature, at which the gas would have zero pressure, meaning particles would have zero kinetic energy and no motion. This is the basis for the Kelvin temperature scale.
Mark scheme points: straight line / through origin (on Kelvin graph) / passes through −273 °C on Celsius graph / at 0 K particles have no kinetic energy / pressure = 0 / this temperature is absolute zero and defines the start of the Kelvin scale.
Challenge 4 (6-mark exam style): A fixed mass of gas is stored in a rigid steel container at a pressure of 300 000 Pa and a temperature of 17 °C. The gas is heated. At a temperature of 77 °C, the safety valve opens to release gas when the pressure exceeds 350 000 Pa. (a) Calculate the pressure at 77 °C. (b) Does the safety valve open? Justify your answer.
(a) Calculating pressure at 77 °C:
T₁ = 17 + 273 = 290 K
T₂ = 77 + 273 = 350 K
P₂ = P₁ × (T₂/T₁) = 300 000 × (350/290)
P₂ = 300 000 × 1.2069… = 362 069 Pa ≈ 362 000 Pa
(b) Does the valve open?
Yes — the calculated pressure (362 000 Pa) exceeds the safety valve threshold of 350 000 Pa, so the valve will open.
This makes sense because the Kelvin temperature increased by a factor of 350/290 ≈ 1.21, so the pressure also increased by about 21%, from 300 000 Pa to approximately 362 000 Pa, which is above the 350 000 Pa limit.