Ideal Gas Laws and Absolute Zero Revision Notes | A Level AQA Physics

1. Why Gas Behaviour Matters

Squeeze a bicycle pump with your thumb over the nozzle and you feel the pressure build. Heat a sealed container and the pressure rises dangerously. These everyday observations are not random — they follow precise, quantitative laws that connect pressure ($p$), volume ($V$), temperature ($T$), and the mass of gas present.

At A Level, your task is to move beyond qualitative description and use these relationships to make accurate predictions and calculations. The gas laws are also the gateway to one of the most profound ideas in physics: the existence of an absolute minimum temperature, below which nothing can go.

2. The Three Experimental Gas Laws

Each gas law isolates one pair of variables while holding everything else constant. Think of it as a controlled experiment: change one thing, measure one thing, fix the rest. The three laws were established experimentally long before anyone understood molecules — which makes them a beautiful example of empirical science.

Picture a sealed syringe. Push the plunger in — the volume halves, the pressure doubles. The molecules are now confined to a smaller space, so they collide with the walls more frequently per unit area. That increased collision rate is what we measure as higher pressure.

A graph of $p$ against $\frac{1}{V}$ gives a straight line through the origin for a fixed mass at constant temperature. A graph of $p$ against $V$ gives a hyperbola. Both shapes are characteristic of an inverse proportionality and are worth recognising in exam data questions.

Warm a balloon and it expands; cool it and it shrinks. Higher temperature means molecules move faster on average, so they push the walls outward more forcefully. At constant pressure, the walls move until the pressure is restored — giving a larger volume.

The critical detail here is that $T$ must be in kelvin, not degrees Celsius. A graph of $V$ against $T$ (in kelvin) is a straight line that, when extrapolated back, passes through the origin — meaning zero volume at zero kelvin. That extrapolation is one experimental route to absolute zero.

This is the law behind the warning labels on aerosol cans: 'Do not expose to temperatures above 50 °C.' At constant volume, raising the temperature increases the average kinetic energy of the molecules, so they strike the walls harder and more frequently — pressure rises. If the container cannot expand, the pressure can become dangerously high.

A graph of $p$ against $T$ (in kelvin) for a fixed mass at constant volume is a straight line through the origin. Extrapolating this line to $p = 0$ gives the same intercept every time: $T = 0$ K. This is the experimental evidence for absolute zero.

3. The Role of Mass and the Combined Gas Law

Each law above holds for a fixed mass of gas. If you add more gas to a container, you increase the number of molecules available to collide with the walls, so pressure rises even if $V$ and $T$ are unchanged. Mass (or equivalently, the number of moles) is therefore a fourth variable that must be controlled or accounted for.

When temperature, pressure, and volume all change simultaneously for a fixed mass, the three laws combine into a single relationship known as the combined gas law:

Notice that each individual law is a special case of this combined form. Set $T_1 = T_2$ and you recover Boyle's Law. Set $p_1 = p_2$ and you recover Charles's Law. Set $V_1 = V_2$ and you recover the Pressure Law. One equation, three laws — that is elegant physics.

Worked example: Applying the Combined Gas Law

A gas occupies $2.0 \times 10^{-3}$ m³ at a pressure of $1.0 \times 10^5$ Pa and a temperature of 300 K. The gas is compressed to a volume of $5.0 \times 10^{-4}$ m³ and heated to 450 K. Calculate the new pressure. [3 marks]

1. $$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$$ — [M1] Selecting the correct relationship for a fixed mass of gas with p, V, and T all changing.
2. $$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}$$ — Multiply both sides by $T_2$ and divide both sides by $V_2$. Check: units of Pa × m³ × K / (K × m³) = Pa. Consistent.
3. $$p_2 = \frac{(1.0 \times 10^5) \times (2.0 \times 10^{-3}) \times 450}{300 \times (5.0 \times 10^{-4})}$$ — [M1] All values substituted with consistent SI units. Temperature is already in kelvin — no conversion needed here.
4. Numerator: $1.0 \times 10^5 \times 2.0 \times 10^{-3} \times 450 = 200 \times 450 = 90000 = 9.0 \times 10^4$ Denominator: $300 \times 5.0 \times 10^{-4} = 0.15$ — Breaking into numerator and denominator reduces arithmetic errors. Note: $1.0 \times 10^5 \times 2.0 \times 10^{-3} = 200$ exactly.
5. $$p_2 = \frac{9.0 \times 10^4}{0.15} = 6.0 \times 10^5 \text{ Pa}$$ — [A1] $9.0 \times 10^4 \div 0.15 = 6.0 \times 10^5$. Sanity check: volume decreased by a factor of 4 and temperature increased by a factor of 1.5, so pressure should increase by $4 \times 1.5 = 6$. Indeed $1.0 \times 10^5 \times 6 = 6.0 \times 10^5$ Pa.

Answer: $p_2 = 6.0 \times 10^5$ Pa

4. Absolute Zero and the Kelvin Scale

Here is where the gas laws reveal something profound. Take a fixed volume of gas and measure its pressure at several temperatures in °C. Plot $p$ against $T$ (°C) and you get a straight line. Extend that line backwards — extrapolate it — and it crosses the temperature axis at the same point regardless of which gas you use or how much gas you have: approximately −273 °C.

At that temperature, the extrapolated pressure would be zero. No pressure means no molecular collisions, which implies the molecules have ceased all thermal motion. This is the theoretical minimum temperature: absolute zero, defined as 0 K, equivalent to −273.15 °C.

The Kelvin scale is constructed so that 0 K is absolute zero and each kelvin is the same size as one degree Celsius. The conversion is: $T\text{ (K)} = \theta\text{ (°C)} + 273\text{ K}$ (using the AQA approximation of 273 K rather than 273.15 K). This means 0 °C = 273 K and 100 °C = 373 K.

The graph above shows the key experimental idea: different gases, different amounts, same extrapolated intercept. That universality is what makes absolute zero a fundamental physical constant rather than a property of any particular substance.

A common exam question asks you to describe how absolute zero is determined from a pressure–temperature experiment. The key phrase is extrapolation: you extend the straight-line graph beyond the range of measured data to find where it predicts $p = 0$. That intercept on the temperature axis is absolute zero.

5. The Ideal Gas Equation

The combined gas law handles a fixed mass of gas changing state. But what if you want to relate $p$, $V$, and $T$ to the actual amount of gas present? That requires introducing the number of moles, $n$, and the molar gas constant, $R = 8.31$ J mol$^{-1}$ K$^{-1}$.

You can also write the ideal gas equation in terms of the number of molecules $N$ rather than moles: $pV = NkT$, where $k = 1.38 \times 10^{-23}$ J K$^{-1}$ is the Boltzmann constant. The two forms are equivalent because $R = N_A k$, where $N_A = 6.02 \times 10^{23}$ mol$^{-1}$ is the Avogadro constant.

Notice how the ideal gas equation unifies everything covered so far. At constant $T$ and $n$: $pV = \text{constant}$ — Boyle's Law. At constant $p$ and $n$: $V \propto T$ — Charles's Law. At constant $V$ and $n$: $p \propto T$ — the Pressure Law. The ideal gas equation is the master relationship from which all three laws follow.

Worked example: Using the Ideal Gas Equation to Find Moles

A gas is contained in a vessel of volume $4.0 \times 10^{-3}$ m³ at a pressure of $2.5 \times 10^5$ Pa and a temperature of 27 °C. Calculate the number of moles of gas present. ($R = 8.31$ J mol$^{-1}$ K$^{-1}$) [3 marks]

1. $$T = 27 + 273 = 300 \text{ K}$$ — Always convert °C to K before substituting into any gas law. This is the most common source of dropped marks.
2. $$pV = nRT \implies n = \frac{pV}{RT}$$ — [M1] Correct equation selected and rearranged.
3. $$n = \frac{(2.5 \times 10^5) \times (4.0 \times 10^{-3})}{8.31 \times 300}$$ — [M1] All values in SI units: Pa, m³, J mol⁻¹ K⁻¹, K.
4. $$2.5 \times 10^5 \times 4.0 \times 10^{-3} = 1000 \text{ Pa m}^3 = 1000 \text{ J}$$ — Pa × m³ = N m⁻² × m³ = N m = J. Unit check passes.
5. $$8.31 \times 300 = 2493 \text{ J mol}^{-1}$$ $$n = \frac{1000}{2493} = 0.401 \text{ mol}$$ — [A1] $1000 \div 2493 = 0.401$ mol (3 s.f.). Unit check: J ÷ (J mol⁻¹) = mol. Correct.

Answer: $n = 0.401$ mol (to 3 s.f.)

6. Required Practical: Investigating the Pressure Law

The Pressure Law is the most straightforward of the three to investigate in the laboratory because volume is straightforward to keep constant. The experiment also provides a direct experimental route to absolute zero through extrapolation — making it doubly useful for the specification.

When evaluating this experiment, consider two main sources of uncertainty. First, the gas in the connecting tube between the flask and the gauge is not at the same temperature as the gas in the flask — this introduces a systematic error. Second, the gas may not be perfectly ideal, particularly at low temperatures where intermolecular forces become more significant.

7. Exam Technique and Common Mistakes

Gas law questions follow predictable patterns in AQA papers. The most reliable way to lose marks is a unit error — particularly forgetting to convert temperature to kelvin. The second most common error is misidentifying which law applies, or failing to state the conditions under which a law holds.

For extended-response questions asking you to 'explain' a gas law in terms of molecular behaviour, structure your answer as: observation → molecular interpretation → link back to the macroscopic quantity. For example: 'As temperature increases at constant volume, molecules move faster on average, so they collide with the walls more frequently and with greater force per collision, increasing the pressure.' That chain of reasoning — speed, collision frequency, force, pressure — earns the marks.