P10: Particle Motion and Pressure
Particle motion in gases and gas pressure
Particle motion in gases and gas pressure
In 1827, Robert Brown observed pollen grains in water moving in random, zigzag paths. This random motion is called Brownian motion.
The explanation: tiny, fast-moving water or air particles collide with larger, visible particles (like smoke or pollen). These collisions are uneven and random, causing the larger particles to move in unpredictable zigzag paths.
Explain why smoke particles in air show Brownian motion.
Air particles are much smaller and move very rapidly. They collide with the larger smoke particles from all sides, but unevenly. At any instant, more air particles may hit one side of a smoke particle than the other, causing it to change direction randomly. This gives the smoke particle a random, zigzag path.
Gas particles are in constant random motion. When they collide with the walls of their container, they exert a force on the wall. The pressure is this force per unit area.
p = F / A
p = pressure (Pa), F = force (N), A = area (m²)
If a gas is heated at constant volume, the temperature increases. This means:
A sealed gas cylinder is at 20°C and 200 kPa. It is heated to 40°C. Explain what happens to the pressure.
The temperature increases, so gas particles gain kinetic energy and move faster. They collide with the cylinder walls more frequently and with greater force. Since the volume is constant (sealed cylinder), the pressure increases. (Note: the pressure doesn't double because the temperature hasn't doubled in Kelvin — see the Kelvin scale section.)
If the temperature of a gas is kept constant and the volume is decreased:
p₁V₁ = p₂V₂ (at constant temperature, for a fixed mass of gas)
p₁, p₂ = initial and final pressure; V₁, V₂ = initial and final volume
A gas has a pressure of 100 kPa and a volume of 5.0 m³. If the volume is compressed to 2.0 m³ at constant temperature, what is the new pressure?
p₁V₁ = p₂V₂
100 × 5.0 = p₂ × 2.0
500 = 2.0 p₂
p₂ = 500 / 2.0 = 250 kPa
A fixed mass of gas has a pressure of 200 kPa and a volume of 3.0 m³. The pressure is changed to 150 kPa at constant temperature. Calculate the new volume.
p₁V₁ = p₂V₂
200 × 3.0 = 150 × V₂
600 = 150 V₂
V₂ = 600 / 150 = 4.0 m³
If you cool a gas, the particles slow down. At some point they would stop moving completely — this is absolute zero.
K = °C + 273
To convert from Celsius to Kelvin: add 273
To convert from Kelvin to Celsius: subtract 273
| Celsius (°C) | Kelvin (K) | Description |
|---|---|---|
| −273 | 0 | Absolute zero — particles have minimum energy |
| 0 | 273 | Freezing point of water |
| 20 | 293 | Room temperature |
| 100 | 373 | Boiling point of water |
Convert: (a) 25°C to K, (b) 373 K to °C, (c) −196°C to K
(a) 25 + 273 = 298 K
(b) 373 − 273 = 100°C
(c) −196 + 273 = 77 K
Q1: Foundation Describe what causes gas pressure in a sealed container.
Q2: Foundation Explain how Brownian motion provides evidence for the particle model.
Q3: Higher A gas at 150 kPa has a volume of 4.0 m³. The volume is reduced to 1.5 m³ at constant temperature. Calculate the new pressure.
Q4: Higher Convert: (a) 37°C to K, (b) 300 K to °C.
Q5: Higher Explain, in terms of particles, why decreasing the volume of a gas increases its pressure at constant temperature.
A gas at 300 K and 200 kPa is heated to 450 K at constant volume. Find the new pressure: p₂ = p₁ × T₂/T₁ = 200 × 450/300 = 300 kPa. A separate sample at 120 kPa and 0.5 m³ is compressed to 0.2 m³ at constant temperature: p₂ = p₁V₁/V₂ = 120 × 0.5 / 0.2 = 300 kPa.
1. Wrong: Doubling the Celsius temperature doubles the pressure. Correct: Pressure is proportional to Kelvin temperature, not Celsius. 20°C (293 K) to 40°C (313 K) is not a doubling of temperature.
2. Wrong: Gas pressure is caused by particles pushing each other. Correct: Gas pressure is caused by particles colliding with the walls of the container and exerting a force on them.
3. Wrong: Absolute zero is when particles move very slowly. Correct: At absolute zero (0 K = −273°C), particles have zero kinetic energy and stop moving completely — it is the lowest possible temperature.
6 marks: A sealed balloon contains a fixed mass of gas at constant temperature. Describe and explain, in terms of particles, what happens to the pressure when the balloon is compressed to half its original volume.
When the balloon is compressed to half its volume, the same number of gas particles are now in a space half the size. The particles travel shorter distances between collisions with the walls, so they collide with the walls more frequently. Each collision still exerts the same force on average because the temperature is constant (so the average speed of particles is unchanged). The increased frequency of collisions means a greater force per unit area on the walls. Since pressure = force / area, and the area of the walls has also decreased, the pressure increases. Boyle's Law tells us that p₁V₁ = p₂V₂, so halving the volume doubles the pressure.
Mark scheme: 1 mark for same number of particles in smaller space; 1 mark for shorter distance between collisions; 1 mark for more frequent collisions with walls; 1 mark for same force per collision (constant temperature); 1 mark for greater force per unit area; 1 mark for pressure doubles (p₁V₁ = p₂V₂). (6 marks total)
A student investigates Boyle's Law using a sealed syringe connected to a pressure gauge. They compress the syringe and record the volume and pressure. The theoretical values (calculated using p₁V₁ = p₂V₂) are shown alongside their measurements.
| Volume (cm³) | Measured pressure (kPa) | Theoretical pressure (kPa) |
|---|---|---|
| 20 | 101 | 101 |
| 15 | 130 | 135 |
| 10 | 186 | 202 |
| 5 | 340 | 404 |
(a) At 20 cm³ the measured and theoretical values agree. Explain why.
(b) As volume decreases, the measured values become increasingly lower than theoretical values. Suggest why.
(c) Explain why Boyle's Law assumes constant temperature and how failing to maintain this would affect the results.
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