Edexcel IGCSE Physics 5(d) Ideal Gas Molecules — Notes
Written for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular.
Which parts do I need?
Single Award candidates need everything on this page except the two gas-law calculations, \(\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}\) and \(p_1V_1=p_2V_2\). You must still explain the relationships between pressure, volume and temperature in words — you just will not be asked to calculate with them. There is no Physics-only content on this page.
What this page covers
| Spec | Modular no. | You need to be able to… | Route |
|---|---|---|---|
| 5.15 | 6.1 | explain how randomly moving gas molecules exert a force, and so a pressure, on the walls of their container | Single · Double · Physics |
| 5.16 | 6.2 | understand why the lowest possible temperature, absolute zero, is −273 °C | Single · Double · Physics |
| 5.17 | 6.3 | describe the Kelvin scale and convert temperatures between kelvin and degrees Celsius | Single · Double · Physics |
| 5.18 | 6.4 | understand why raising the temperature raises the average speed of gas molecules | Single · Double · Physics |
| 5.19 | 6.5 | know that the average kinetic energy of the molecules is proportional to the kelvin temperature | Single · Double · Physics |
| 5.20 | 6.6 | explain in words how pressure depends on volume (constant temperature) and on kelvin temperature (constant volume) for a fixed amount of gas | Single · Double · Physics |
| 5.21 | 6.7 | use \(p_1/T_1=p_2/T_2\) for a fixed mass of gas at constant volume | Double · Physics |
| 5.22 | 6.8 | use \(p_1V_1=p_2V_2\) for a fixed mass of gas at constant temperature | Double · Physics |
Modular candidates: this sub-topic is in Unit 2, not Unit 1 — 4XPH1 Unit 2 (4WPH2) and 4XSD1 Unit 6 (4WSD6), numbered 6.1–6.8 under “Solids, liquids and gases: Part 2”. The Topic 5 units (°C, K, Pa, m³ …) are assessed in this unit too; they are listed on the 5(b) page.
In one minute
- Gas molecules move randomly at high speed. Each collision with a wall exerts a force on it; pressure = total force ÷ area of the wall.
- Cooling a gas slows its molecules. At absolute zero, −273 °C, the molecules have the least possible kinetic energy, so nothing can be colder.
- Kelvin scale: starts at absolute zero. K = °C + 273 and °C = K − 273. A change of 1 K equals a change of 1 °C.
- The average kinetic energy of the molecules is proportional to the kelvin temperature: double the kelvin temperature, double the average kinetic energy.
- Fixed mass of gas: smaller volume → greater pressure (constant temperature); higher temperature → greater pressure (constant volume).
- \(\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}\) (T in kelvin) and \(p_1V_1=p_2V_2\). Double · Physics
How do gas molecules exert a pressure on a container? Single · Double · Physics
- The molecules of a gas are in constant random motion: they move in straight lines in all directions, at a range of speeds, until they collide with each other or with the walls.
- When a molecule hits a wall, its direction of motion changes. The wall exerts a force on the molecule, so the molecule exerts an equal and opposite force on the wall.
- Huge numbers of molecules hit every part of the wall each second. Their combined force, divided by the area of the wall, is the pressure of the gas (pressure = force ÷ area).
- Because the motion is random, on average the same number of molecules hit each unit of wall area per second, so the pressure is the same on every wall.
Why does a higher temperature mean faster molecules? Single · Double · Physics
Heating a gas transfers energy to its molecules. In a gas this energy goes into the kinetic energy of the molecules, so they move faster. The molecules always have a range of speeds, so we talk about the average speed.
- Higher temperature → greater average kinetic energy → greater average speed of the molecules.
- Cooling the gas does the opposite: the molecules slow down on average.
- Temperature is a measure of the average kinetic energy of the molecules — which leads directly to the idea of an absolute zero.
Why is there an absolute zero of temperature? Single · Double · Physics
As a gas is cooled, its molecules move more and more slowly, so they hit the walls less often and less hard, and the pressure falls. If the pressure of a fixed volume of gas is measured at several temperatures and the straight line is extended backwards, it reaches zero pressure at −273 °C, whatever the gas.
- Temperature measures the average kinetic energy of the molecules. Kinetic energy cannot be less than zero, so there is a lowest possible temperature.
- Nothing can be cooled below absolute zero, because there is no more kinetic energy to remove from its molecules.
What is the Kelvin scale and how do you convert? Single · Double · Physics
The Kelvin scale is an absolute temperature scale: its zero, 0 K, is absolute zero. Its unit is the kelvin (K) — write “300 K”, not “300 °K”.
- A temperature change is the same on both scales: a rise of 15 °C is a rise of 15 K.
- There are no negative temperatures on the Kelvin scale.
- This conversion is not on the Equation List — learn it.
(a) Convert 20 °C and −196 °C (the boiling point of nitrogen) to kelvin. (b) Convert 373 K and 0 K to degrees Celsius.
(a) 20 + 273 = 293 K; −196 + 273 = 77 K
(b) 373 − 273 = 100 °C; 0 − 273 = −273 °C (absolute zero)
How is kelvin temperature linked to the kinetic energy of molecules? Single · Double · Physics
- Double the kelvin temperature → double the average kinetic energy of the molecules. For example, heating a gas from 27 °C (300 K) to 327 °C (600 K) doubles it; heating from 27 °C to 54 °C (327 K) does not.
- This works only in kelvin. Doubling a Celsius temperature (for example 10 °C to 20 °C) does not double the kinetic energy, because 0 °C is not zero energy.
- At 0 K the average kinetic energy is zero — consistent with absolute zero.
How do pressure, volume and temperature of a gas affect each other? Single · Double · Physics
Each relationship applies to a fixed amount (fixed mass) of gas, with one quantity kept constant.
Pressure and volume (constant temperature)
- If the volume is decreased, the pressure increases. (Double · Physics: the exact rule is halve the volume → double the pressure, \(p_1V_1=p_2V_2\), below.)
- Why: the temperature is constant, so the molecules have the same average speed. In a smaller volume the molecules hit the walls more often (more collisions per second on each square metre of wall), so the force on each unit area of wall — the pressure — is greater.
Pressure and temperature (constant volume)
- If the kelvin temperature is increased, the pressure increases. (Double · Physics: the exact rule is double the kelvin temperature → double the pressure, \(p_1/T_1=p_2/T_2\), below.)
- Why: at a higher temperature the molecules have more kinetic energy and move faster on average. They hit the walls more often and harder (with a greater force per collision), so the pressure is greater.
A cyclist pumps up a tyre on a cool morning. By the afternoon the tyre has been standing in hot sunshine. Explain, in terms of molecules, why the pressure of the air in the tyre is now higher. Assume the volume of the tyre does not change.
The temperature of the air rises, so the molecules gain kinetic energy and move faster on average (1).
They collide with the inside of the tyre more often (1) and each collision exerts a larger force (1).
So the total force on each square metre of the tyre wall is larger: the pressure increases (1).
How do you use \(p_1/T_1=p_2/T_2\)? Double · Physics
Not required for Single Award (4SS0).
\[ \frac{\text{pressure}}{\text{temperature}}=\text{constant}\qquad \frac{p_1}{T_1}=\frac{p_2}{T_2} \]- For a fixed mass of gas at constant volume. \(p_1, T_1\) = starting pressure and temperature; \(p_2, T_2\) = final values.
- \(T\) must be in kelvin. Convert every Celsius temperature by adding 273 before you substitute.
- The pressures can be in any unit (Pa, kPa) as long as both are in the same unit.
A sealed steel gas cylinder contains gas at a pressure of 200 kPa at 15 °C. The cylinder is left in the sun and the gas warms to 75 °C. Calculate the new pressure.
\(T_1=15+273=288\ \text{K}\); \(T_2=75+273=348\ \text{K}\)
\(p_2=\dfrac{p_1\times T_2}{T_1}=\dfrac{200\times348}{288}=241.7=\) 240 kPa
Common error: using Celsius gives 200 × 75 ÷ 15 = 1000 kPa — five times the starting pressure for a modest warming, which should look wrong.
An aerosol can contains gas at 300 kPa at 20 °C. The can is designed to burst if the pressure reaches 600 kPa. Calculate the temperature, in °C, at which the can would burst. Assume the volume of the can stays constant.
\(T_1=20+273=293\ \text{K}\)
\(T_2=\dfrac{T_1\times p_2}{p_1}=\dfrac{293\times600}{300}=586\ \text{K}\)
In Celsius: 586 − 273 = 313 °C
Doubling the pressure doubles the kelvin temperature. A candidate who doubles 20 °C to get 40 °C has used the wrong scale.
How do you use \(p_1V_1=p_2V_2\)? Double · Physics
Not required for Single Award (4SS0).
\[ \text{pressure}\times\text{volume}=\text{constant}\qquad p_1\times V_1=p_2\times V_2 \]- For a fixed mass of gas at constant temperature.
- Use the same units on both sides — both volumes in cm³ (or both in m³), both pressures in kPa (or both in Pa). There is no need to convert to SI units.
- If the volume is compressed slowly, the temperature stays constant; a fast compression warms the gas and the equation no longer applies exactly.
(a) A syringe contains 60 cm³ of air at 100 kPa. The outlet is sealed and the plunger is pushed in slowly until the volume is 24 cm³. Calculate the new pressure.
(b) A diver releases an air bubble of volume 2.0 cm³ at a depth of 20 m in fresh water (density 1000 kg/m³). Atmospheric pressure is 1.0 × 10⁵ Pa. Calculate the volume of the bubble when it reaches the surface, assuming its temperature does not change. Use g = 10 N/kg.
(a) \(p_2=\dfrac{p_1V_1}{V_2}=\dfrac{100\times60}{24}=\) 250 kPa
(b) Pressure at 20 m = atmospheric + \(h\rho g\) = 1.0 × 10⁵ + 20 × 1000 × 10 = 3.0 × 10⁵ Pa
\(V_2=\dfrac{p_1V_1}{p_2}=\dfrac{3.0\times10^5\times2.0}{1.0\times10^5}=\) 6.0 cm³
Pearson papers say “assume \(g\) = 10 N/kg” (9.8 N/kg is also accepted). Part (b) links this page with \(p=h\rho g\) from 5(b). Forgetting the atmospheric pressure gives \(p_1\) = 2.0 × 10⁵ Pa and a wrong answer of 4.0 cm³.
Exam tips for 5(d)
In \(p_1/T_1=p_2/T_2\), and whenever you compare kinetic energies, convert to kelvin first. Write the conversion as a line of working (“75 + 273 = 348 K”) so it earns credit even if a later step goes wrong.
If a question asks why molecules exert the same pressure on every wall, the answer is about random motion: equal numbers of collisions per second on each unit area. Do not drift into how pressure changes with temperature or volume unless that is what is asked.
For a temperature rise at constant volume, credit is given for both ideas: molecules collide more frequently and with greater force. For a volume decrease at constant temperature, only “more frequently” applies — the molecules are not moving faster.
“More collisions” alone is weak. Say more collisions per second (a rate), and link force to pressure through area: greater force on each square metre of wall.
In \(p_1V_1=p_2V_2\) and \(p_1/T_1=p_2/T_2\), the pressure unit cancels. Leave values in kPa and give the answer in kPa — converting to Pa and back only creates chances for power-of-ten errors.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Gas pressure is caused by molecules colliding with each other. | Collisions between molecules do not push on the container. | Pressure comes from molecules colliding with the walls. |
| Molecules get bigger when a gas is heated. | The molecules themselves do not change size. | They move faster on average; that is what raises the pressure. |
| Doubling the Celsius temperature doubles the pressure. | 0 °C is not the zero of energy. | Pressure (at constant volume) is proportional to the kelvin temperature. |
| Compressing a gas at constant temperature makes the molecules move faster. | Average speed depends only on temperature. | The molecules hit the walls more often because they have less space, not because they are faster. |
| Temperatures can go as low as you like. | Kinetic energy cannot fall below zero. | −273 °C (0 K) is the lowest possible temperature. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| K = °C + 273; °C = K − 273 | a temperature on the other scale | All | no — learn it |
| pressure ÷ temperature = constant, \(\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}\) | new pressure or kelvin temperature, at constant volume | Double · Physics | yes — all papers |
| pressure × volume = constant, \(p_1\times V_1=p_2\times V_2\) | new pressure or volume, at constant temperature | Double · Physics | yes — all papers |
Pearson includes the Equation List with the question papers, but the first mark in many questions is for writing the relationship yourself — learn them anyway.
Check your understanding
1. Convert normal body temperature, 37 °C, to kelvin. All
37 + 273 = 310 K
2. Explain how the molecules of a gas exert a pressure on the walls of a container. All
The molecules move randomly (1) and collide with the walls (1); each collision exerts a force on the wall (1); pressure is the total force divided by the area of the wall (1).
3. The temperature of a gas rises from −73 °C to 127 °C. What happens to the average kinetic energy of its molecules? All
−73 °C = 200 K and 127 °C = 400 K. The kelvin temperature doubles, so the average kinetic energy doubles.
4. Explain why no object can be cooled below −273 °C. All
Temperature depends on the average kinetic energy of the particles. At −273 °C (absolute zero) the particles have the least possible kinetic energy — they cannot lose any more energy, so the temperature cannot fall further.
5. A fixed mass of gas is squashed slowly into a smaller volume at constant temperature. Explain why its pressure increases. All
The average speed of the molecules is unchanged (same temperature), but in the smaller volume they hit the walls more often / more collisions per second on each unit area (1), so the force per unit area — the pressure — increases (1).
6. 0.50 m³ of gas at 120 kPa is compressed at constant temperature until its pressure is 300 kPa. Calculate its new volume. Double · Physics
\(V_2=\dfrac{120\times0.50}{300}=\) 0.20 m³
7. A sealed rigid container holds air at 100 kPa and 27 °C. It is cooled to −23 °C. Calculate the new pressure. Double · Physics
\(T_1=300\ \text{K}\), \(T_2=250\ \text{K}\); \(p_2=\dfrac{100\times250}{300}=83.3\approx\) 83 kPa