(d) Ideal gas molecules
- Syllabus
- 2024
- Topic
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- Level
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Learning objectives
5.15Gas pressureExplain how molecules in a gas have random motion and that they exert a force, and hence a pressure, on the walls of a container5.16Absolute zeroUnderstand why there is an absolute zero of temperature, which is –273 °C5.17Kelvin scaleDescribe the Kelvin scale of temperature and be able to convert between the Kelvin and Celsius scales5.18Temperature and gas molecule speedUnderstand why an increase in temperature results in an increase in the average speed of gas molecules5.19Kelvin temperature and kinetic energyKnow that the Kelvin temperature of a gas is proportional to the average kinetic energy of its molecules5.20Gas pressure relationshipsExplain, for a fixed amount of gas, the qualitative relationship between:• pressure and volume at constant temperature• pressure and Kelvin temperature at constant volume5.21Pressure-temperature relationshipUse p₁/T₁ = p₂/T₂ for the pressure and kelvin temperature of a fixed mass of gas at constant volume.5.22Pressure-volume relationshipUse p₁V₁ = p₂V₂ for the pressure and volume of a fixed mass of gas at constant temperature.
Explain how gas molecules create pressure
Gas molecules move rapidly and randomly in all directions. When a molecule collides with a container wall, its direction and momentum change, so it exerts a force on the wall. The combined force from many collisions produces gas pressure.
Use the full causal chain: random molecular motion → collisions with the walls → momentum changes and forces on the walls → total force per unit area, which is pressure. Because motion is random, collisions occur on every wall and a gas at rest exerts pressure in all directions.
p=\frac{F}{A}
Do not say that collisions between gas molecules directly produce pressure on the container. Wall pressure comes from molecules colliding with the walls. A single collision gives a tiny force; the measurable pressure is the average effect of very many collisions.
Understand absolute zero
Absolute zero is the lowest possible temperature: 0 K, approximately −273 °C. It is the zero point of the Kelvin scale.
Cooling a gas reduces its molecules' average kinetic energy and average speed. In the ideal particle model, extrapolating this trend reaches its minimum at absolute zero; going below would require a negative average kinetic energy, which is impossible.
On a pressure–Celsius-temperature graph for a fixed mass of gas at constant volume, extending the straight line to zero pressure gives an estimate near −273 °C. Experimental data may give a nearby value rather than exactly −273 °C.
Absolute zero is not 0 °C. Also, treat 'particles stop' as the idealised IGCSE model: the essential claim is that thermal energy and average kinetic energy have reached their minimum, so a lower temperature is not possible.
Convert between Celsius and Kelvin
The Kelvin scale starts at absolute zero. A temperature interval of 1 K is the same size as an interval of 1 °C, but the zero points differ by about 273.
T(\mathrm{K})=\theta(^{\circ}\mathrm{C})+273\qquad\theta(^{\circ}\mathrm{C})=T(\mathrm{K})-273
| Celsius temperature | Kelvin temperature |
|---|---|
| −273 °C | 0 K |
| 0 °C | 273 K |
| 100 °C | 373 K |
Example: 15 °C = 15 + 273 = 288 K. In reverse, 358 K = 358 − 273 = 85 °C.
Write kelvin as K, not °K or degrees K. Add or subtract 273; do not multiply. Use Kelvin temperatures in gas-law ratios even when the question initially gives Celsius values.
Link gas temperature to average molecular speed
Increasing a gas's temperature increases its molecules' average kinetic energy, so their average speed increases. Cooling reverses the change: average kinetic energy and average speed decrease.
Gas molecules do not all travel at one speed. At any instant there is a range of speeds because collisions continually redistribute energy. Heating shifts the distribution toward higher speeds, which is why the word average matters.
At fixed volume, faster molecules reach the walls more often and undergo larger momentum changes in collisions. This connects a temperature increase to a pressure increase, although that pressure relationship is developed separately.
Higher temperature does not mean every molecule has the same greater speed, and it does not mean the molecules become larger. The change is in the distribution and its average speed.
Relate Kelvin temperature to average kinetic energy
For a gas, Kelvin temperature is directly proportional to the average kinetic energy of its molecules.
T\propto\overline{E_k}
If the Kelvin temperature doubles, the average kinetic energy doubles. A graph of average kinetic energy against Kelvin temperature is a straight line through the origin: both reach their ideal-model minimum at 0 K.
For molecules of one gas, Ek=21mv2. A greater average kinetic energy therefore means a greater average molecular speed, but speed is not directly proportional to temperature because kinetic energy depends on speed squared.
The direct proportionality uses Kelvin, not Celsius. Doubling a Celsius temperature does not double average kinetic energy—for example, 20 °C is 293 K, while 40 °C is 313 K, not twice as large on the absolute scale.
Explain pressure changes in a fixed amount of gas
For a fixed amount of gas, pressure changes when volume or Kelvin temperature changes. To isolate one relationship, the other variable must be held constant.
| Change and fixed condition | Molecular explanation | Pressure relationship |
|---|---|---|
| decrease volume at constant temperature | average speed and force per collision stay the same, but molecules travel less far and hit the walls more frequently | pressure increases; p is inversely proportional to V |
| increase Kelvin temperature at constant volume | molecules move faster, hit the walls more frequently and exert a larger force in each collision | pressure increases; p is directly proportional to T |
The reverse changes follow the same mechanisms: increasing volume at constant temperature reduces collision frequency and pressure; decreasing temperature at constant volume produces slower, less frequent and less forceful wall collisions, so pressure falls.
State the controls: the amount of gas is fixed, and either temperature or volume is constant. 'There are fewer collisions' is incomplete—say fewer collisions with the walls per second. At constant temperature, molecular average speed does not decrease merely because volume increases.
Calculate pressure and Kelvin temperature at constant volume
For a fixed mass of gas at constant volume, pressure is directly proportional to Kelvin temperature. Compare two states with:
\frac{p_1}{T_1}=\frac{p_2}{T_2}
Method: 1. Convert every Celsius temperature to kelvin. 2. Substitute matching state-1 and state-2 values. 3. Rearrange for the unknown. 4. Keep pressure units consistent and check that the direction is sensible.
Example: a gas at constant volume has pressure 9.95×104 Pa at 16 °C. At 32 °C, T1=289 K and T2=305 K. Therefore p2=(9.95×104×305)/289=1.05×105 Pa. The pressure rises because the Kelvin temperature rises.
Do not substitute 16 and 32 into the ratio: Celsius does not start at absolute zero. This equation requires a fixed mass and constant volume; it cannot be used unchanged if gas escapes or the container volume changes.
Calculate pressure and volume at constant temperature
For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume. Compare two states with:
p_1V_1=p_2V_2
Method: 1. Pair each pressure with its volume. 2. Substitute into p1V1=p2V2. 3. Rearrange for the unknown. 4. Use one pressure unit and one volume unit consistently on both sides, then check the inverse trend.
Example: gas at 101 kPa expands from 110 cm³ to 140 cm³ at constant temperature. p2=(101×110)/140=79.4 kPa. The larger volume gives a lower pressure; if the volume were halved, the pressure would double.
The temperature and mass of gas must remain constant. Volume units need not be converted when the same unit is used on both sides, but pressure units must also be consistent. Avoid reversing the pairings: p1 must multiply V1, and p2 must multiply V2.