Ideal gas molecules

Q: Do I need to remember the gas law formulas for the exam?

Yes, neither nor are provided on an exam formula sheet, so you must recall both. Always convert temperature to Kelvin for calculations.

Q: What is absolute zero?

Absolute zero is the lowest possible temperature, equal to or K. At this temperature, the average kinetic energy of gas molecules is zero, so they stop moving.

Edexcel International GCSE Physics· Section 5, 5.15–5.22· 25 min read

1. Kinetic Theory of Gases and Gas Pressure★★☆☆☆⏱ 5 min

Gas molecules are in constant, random motion at high speeds. As they move, they collide with the walls of their container, exerting a force on the walls. The total force exerted per unit area of the container walls is the gas pressure.

📘 Definition

Gas pressure

Force exerted per unit area by gas molecules colliding with the walls of their container, arising from the rate and magnitude of molecular collisions.

📐 Worked Example

Explain why inflating a balloon increases the pressure inside it until the balloon stretches.

1
When you add air to the balloon, you increase the number of gas molecules inside the fixed initial volume.
2
More molecules mean more frequent collisions with the inner walls of the balloon.
3
The increased number of collisions raises the total force exerted on the balloon walls, increasing the internal pressure.

2. Absolute Zero and the Kelvin Temperature Scale★★☆☆☆⏱ 5 min

As temperature decreases, the average speed of gas molecules decreases, so their average kinetic energy also decreases. Absolute zero is the lowest theoretical temperature, where molecular motion stops completely, so average kinetic energy is zero.

📘 Definition

Kelvin scale

An absolute temperature scale where 0 K is absolute zero, and each degree increment is the same size as a 1°C increment. Convert between Celsius and Kelvin using the formula below:

T(\text{K}) = \theta(\degree\text{C}) + 273

📐 Worked Example

Convert 27°C to Kelvin, and convert 120 K to Celsius.

1
For Celsius to Kelvin: add 273 to the Celsius value.
2
$27 + 273 = 300 \text{ K}$
3
For Kelvin to Celsius: subtract 273 from the Kelvin value.
4
$120 - 273 = -153 \degree \text{C}$

The Kelvin temperature of a gas is directly proportional to the average kinetic energy of its molecules. This means if you double the Kelvin temperature of a gas, you double the average kinetic energy of its molecules (this relationship does not hold for Celsius temperature).

Exam tip:

Never use Celsius temperature in the $p_{1} / T_{1} = p_{2} / T_{2}$ gas law calculation — marks are deducted for this mistake. Always convert to Kelvin first.

3. Qualitative Gas Relationships for Fixed Mass of Gas★★★☆☆⏱ 7 min

For a fixed amount (mass) of gas, three variables are linked: pressure, volume, and Kelvin temperature. You need to explain the relationship between pairs when the third is kept constant.

Pressure and volume (constant temperature): If you decrease the volume of a gas, molecules have less space to move, so they collide with the walls more frequently, increasing pressure. This is an inverse relationship: as volume decreases, pressure increases, and vice versa.
Pressure and Kelvin temperature (constant volume): If you increase the temperature of a fixed volume of gas, molecules gain kinetic energy and move faster. They collide with the walls more often and with greater force, increasing pressure. This is a direct proportional relationship.

📐 Worked Example

A sealed syringe containing air is left in sunlight. Explain why the pressure inside the syringe increases as the air warms up, assuming the plunger is fixed so volume does not change.

1
The fixed plunger means volume of the air is constant, and no gas escapes so mass is fixed.
2
As the air warms, its Kelvin temperature increases, so average kinetic energy of the molecules increases.
3
Faster molecules collide with the syringe walls more frequently and with greater force.
4
The increased rate and force of collisions leads to higher pressure inside the syringe.

4. Gas Law Calculations★★★★☆⏱ 8 min

✓ Calculator OK

You need to recall and apply two separate gas laws for fixed mass of gas. No formulas are provided in the exam, so memorise both.

📘 Definition

Boyle's Law (pressure-volume law)

For a fixed mass of gas at constant temperature:

p_1 V_1 = p_2 V_2

$p_{1}$ and $p_{2}$ are initial and final pressure, $V_{1}$ and $V_{2}$ are initial and final volume. Pressure and volume can use any consistent units, no conversion needed as long as units match for initial and final values.

📘 Definition

Pressure-Temperature Law

For a fixed mass of gas at constant volume:

\frac{p_1}{T_1} = \frac{p_2}{T_2}

$p_{1}$ and $p_{2}$ are initial and final pressure, $T_{1}$ and $T_{2}$ are initial and final temperature in Kelvin.

📐 Worked Example

A fixed mass of gas at 150 kPa has a volume of 30 cm³. The gas is compressed at constant temperature to a volume of 10 cm³. Calculate the new pressure of the gas.

1
This uses Boyle's Law, as temperature is constant: $p_{1} V_{1} = p_{2} V_{2}$
2
Substitute known values:
3
$150 \text{ kPa} \times 30 \text{ cm}^3 = p_2 \times 10 \text{ cm}^3$
4
Rearrange to solve for $p_{2}$ :
5
$p_2 = \frac{150 \times 30}{10} = 450 \text{ kPa}$

📐 Worked Example

A fixed volume of gas has a pressure of 200 kPa at a temperature of 27°C. The gas is heated to 127°C at constant volume. Calculate the new pressure of the gas.

1
First convert both temperatures to Kelvin:
2
$T_1 = 27 + 273 = 300 \text{ K}, T_2 = 127 + 273 = 400 \text{ K}$
3
Use pressure-temperature law: $\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}$
4
Substitute values:
5
$\frac{200 \text{ kPa}}{300 \text{ K}} = \frac{p_2}{400 \text{ K}}$
6
Rearrange:
7
$p_2 = \frac{200 \times 400}{300} = 267 \text{ kPa (3 sig figs)}$

Exam tip:

Always show full working for calculation questions. Even if you get the final answer wrong, you can get marks for correct conversion to Kelvin and correct substitution into the formula.

5. Common Pitfalls

Wrong move:

Using Celsius temperature in $p_{1} / T_{1} = p_{2} / T_{2}$ calculations

Why:

The proportionality only applies to absolute (Kelvin) temperature, so using Celsius gives an incorrect result and loses marks

Correct move:

Always convert temperature from Celsius to Kelvin by adding 273 before using the pressure-temperature gas law

Wrong move:

Mixing units for pressure or volume in Boyle's Law calculations (e.g. using kPa for $p_{1}$ and Pa for $p_{2}$ )

Why:

The ratio only holds if units are the same for initial and final values

Correct move:

Check that both pressure values use the same unit, and both volume values use the same unit before substituting into $p_{1} V_{1} = p_{2} V_{2}$

Wrong move:

Explaining gas pressure without referencing collisions between molecules and container walls

Why:

Examiners require explicit mention of collisions to award full marks for explanation questions

Correct move:

Always link changes in pressure/volume/temperature to changes in the frequency or force of molecular collisions with container walls

Wrong move:

Stating that doubling the Celsius temperature of a gas doubles its average kinetic energy

Why:

Only Kelvin temperature is proportional to average kinetic energy, as 0°C is not zero kinetic energy

Correct move:

Confirm temperature is in Kelvin before stating proportionality between temperature and average kinetic energy of gas molecules

Wrong move:

Using the combined gas law ( $p_{1} V_{1} / T_{1} = p_{2} V_{2} / T_{2}$ ) for all gas calculations

Why:

The Edexcel specification only requires the two separate gas laws, so using the combined law is unnecessary and may lead to mistakes if misremembered

Correct move:

Use only $p_{1} V_{1} = p_{2} V_{2}$ (constant T) or $p_{1} / T_{1} = p_{2} / T_{2}$ (constant V) as required by the question

6. Quick Reference Cheatsheet

Concept	Key Formula / Rule	Exam Note
Kelvin conversion	$T (K) = θ (° C) + 273$	Absolute zero = -273°C = 0 K
Boyle's Law (constant T, fixed mass)	$p_{1} V_{1} = p_{2} V_{2}$	Units for p and V only need to be consistent
Pressure-Temperature Law (constant V, fixed mass)	$p_{1} / T_{1} = p_{2} / T_{2}$	T must be in Kelvin, no Celsius allowed
Kelvin temperature & kinetic energy	$T (K) \propto$ average KE of molecules	Doubling T(K) doubles average molecular KE
Gas pressure cause	Collisions of molecules with container walls	Always reference collisions in explanation answers

7. Frequently Asked

Do I need to remember the gas law formulas for the exam?

Yes, neither $p_{1} V_{1} = p_{2} V_{2}$ nor $p_{1} / T_{1} = p_{2} / T_{2}$ are provided on an exam formula sheet, so you must recall both. Always convert temperature to Kelvin for $p_{1} / T_{1} = p_{2} / T_{2}$ calculations.

What is absolute zero?

Absolute zero is the lowest possible temperature, equal to $- 273° C$ or $0$ K. At this temperature, the average kinetic energy of gas molecules is zero, so they stop moving.

Going deeper

official_documentEdexcel IGCSE Physics 4PH1 SpecificationRefer to Section 5 for full topic guidance

What's Next

Now that you have mastered ideal gas molecules and the core gas laws, you can move on to other topics in the Solids, Liquids and Gases unit. If you are taking the separate Physics award, next revise change of state and specific heat capacity content for higher tier papers. You should also practice past paper questions on gas law calculations and kinetic theory explanations to familiarise yourself with exam phrasing and mark scheme requirements. Remember to always convert temperature to Kelvin for pressure-temperature calculations, and show all working to maximise your marks. This topic is frequently tested in both Paper 1 and Paper 2, so regular practice will help you secure easy marks on calculation and explanation questions.

→Change of State and Specific Heat Capacity