Gases and the Absolute (Kelvin) Temperature Scale

This subtopic covers two things: converting between °C and kelvin, and the gas law $pV = \text{constant}$. It is Extended (Supplement) only, so Core candidates sitting Papers 1 and 3 can skip it. Examiners test it because it combines a calculation, a conversion and a particle explanation in one compact question.

How do you convert between Celsius and kelvin?

Add 273. Temperature in kelvin equals temperature in degrees Celsius plus 273. So 27 °C is 300 K, and 0 K (absolute zero) is −273 °C. At absolute zero, particles have the least possible kinetic energy, so no lower temperature exists.

Quantity	Symbol	Unit
Temperature (absolute)	$T$	kelvin (K)
Temperature (Celsius)	$\theta$	degrees Celsius (°C)
Pressure	$p$	pascal (Pa)
Volume	$V$	m³ (or cm³)

In words: kelvin temperature = Celsius temperature + 273. In symbols: $T\ (\text{K}) = \theta\ (\text{°C}) + 273$. A kelvin change equals a Celsius change, so a rise of 10 °C is a rise of 10 K.

What does pV = constant mean?

For a fixed mass of gas at constant temperature, $\text{pressure} \times \text{volume}$ stays the same. In symbols: $p_1 V_1 = p_2 V_2$. Halve the volume and the pressure doubles. The particle explanation: squeezing the gas into a smaller volume means particles hit the walls more frequently, so the force per unit area rises. Temperature must stay constant, because the relationship assumes the particles’ average speed does not change.

The Kelvin link matters for explanation marks too: the average kinetic energy of gas particles is proportional to the kelvin temperature, not the Celsius temperature. Doubling 300 K to 600 K doubles average kinetic energy. “Doubling” 20 °C to 40 °C does not.

Worked Exam Question

A diver’s air cylinder holds 12 litres of air at a pressure of 200 kPa. The air is released slowly into a flexible bag at a pressure of 100 kPa. The temperature stays constant. Calculate the volume of the air in the bag. [3]

Worked solution, set out the way we teach:

1. Equation: $p_1 V_1 = p_2 V_2$
2. Substitute: $200 \times 12 = 100 \times V_2$
3. Rearrange: $V_2 = 2400 \div 100$
4. Answer: $V_2 = 24\ \text{litres}$ (2 significant figures)

Units cancel here because both pressures use kPa and both volumes use litres. Consistency is what matters, not conversion to Pa and m³.

Mark scheme:

• M1: $p_1 V_1 = p_2 V_2$ stated or implied by correct substitution
• M1: correct substitution: $200 \times 12 = 100 \times V_2$
• A1: 24 litres (unit required)

Common Mistakes

• Forgetting the question said “constant temperature” and trying to involve kelvin in a Boyle’s law calculation. $pV = \text{constant}$ needs no temperature value at all.
• Converting −5 °C to kelvin as −278 K. Add 273: the answer is 268 K. Negative kelvin temperatures do not exist.
• Mixing units within one calculation, such as $p_1$ in kPa with $p_2$ in Pa. Match the units on both sides first.
• Writing “particles hit harder” in the compression explanation. At constant temperature the particles move at the same average speed; collisions are more frequent, not harder.
• Claiming kinetic energy is proportional to Celsius temperature. The proportionality only works on the kelvin scale.

Exam Technique Tip

Before substituting into $p_1 V_1 = p_2 V_2$, write a two-column table: state 1 and state 2, with $p$ and $V$ in each. This forces you to pair the right pressure with the right volume. The most common arithmetic slip on this question is swapping $p_2$ and $V_1$, and the table makes that impossible.

How This Is Examined

Extended papers only: Paper 2 multiple choice and Paper 4 theory. Paper 2 favours quick ratio versions (“the pressure doubles, so what happens to the volume?”). Paper 4 sets the full calculation, usually for 3 marks, often followed by a 2-mark particle explanation. The conversion $T = \theta + 273$ can appear inside any thermal question, so treat it as a free mark to bank. There is no required practical, so Papers 5 and 6 rarely touch it beyond reading a pressure gauge. This is one of the most formulaic calculations in 0625.