The Ideal Gas Equation and Work Done Revision Notes | A Level AQA Physics
pV = NkT is the ideal gas equation for N molecules, where k = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant. Free AQA notes and 10 questions answered.
Updated 12 Sept 2026 · Revise Right Team
The Ideal Gas Equation and Work Done is part of Thermal physics (A-level only) in A Level AQA Physics.
6 min
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What this lesson covers
pV = NkT is the ideal gas equation for N molecules, where k = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant. Free AQA notes and 10 questions answered.
State and apply the ideal gas equation in both molar form $pV = nRT$ and molecular form $pV = NkT$
Explain the physical meaning of each variable in the ideal gas equation and identify appropriate SI units
Calculate work done by or on a gas using $W = p\Delta V$ and interpret the sign of the result
Apply both forms of the ideal gas equation to multi-step problems involving changes in pressure, volume, and temperature
Key ideas to keep in view
Use this lesson to stay inside Thermal physics (A-level only) while connecting the detail back to the wider Further mechanics and thermal physics (A-level only) topic.
Ideal gas: A theoretical gas model in which molecules have negligible volume, exert no intermolecular forces, and undergo perfectly elastic collisions. Used when applying $pV = nRT$ or $pV = NkT$.
Molar gas constant $R$: A universal constant equal to 8.31 J mol$^{-1}$ K$^{-1}$, linking the macroscopic properties of pressure, volume, and temperature for one mole of an ideal gas.
Boltzmann constant $k$: A fundamental constant equal to $1.38 \times 10^{-23}$ J K$^{-1}$, linking temperature to the average kinetic energy of a single molecule. Equal to $R/N_A$.
Work done by a gas: The energy transferred when a gas expands or is compressed against a pressure, calculated as $W = p\Delta V$. Measured in joules (J).
Revision notes
Read the core explanations from The Ideal Gas Equation and Work Done before testing yourself from memory.
1. Why Gases Need Their Own Equation
Solids and liquids are relatively predictable — compress them and almost nothing happens. Gases are different. Double the pressure on a gas and its volume halves. Heat it up and it expands dramatically. To handle all three variables — pressure $p$, volume $V$, and temperature $T$ — at once, physicists developed a single, elegant equation.
That equation is the ideal gas equation, and it sits at the heart of thermal physics. Before we write it down, though, we need to be precise about what we mean by an ideal gas — because the equation only works under specific assumptions.
These assumptions matter for exam questions. If a question asks you to evaluate whether a gas behaves ideally, you need to reference these conditions — not just quote the equation.
2. The Ideal Gas Equation — Two Forms
There are two versions of the ideal gas equation, and AQA expects you to know both. The choice between them depends on whether you are working with moles of gas or counting individual molecules.
Notice that the two forms are linked through Avogadro's number $N_A = 6.02 \times 10^{23}$ mol$^{-1}$. Since $N = nN_A$, substituting into $pV = NkT$ gives $pV = nN_A kT$. For this to equal $pV = nRT$, it must be true that $R = N_A k$. This connection is worth knowing — it shows the two equations are not separate laws but two views of the same physical reality.
One variable causes more errors than any other: temperature. The ideal gas equation only works in kelvin. Converting is straightforward — add 273 to any Celsius value — but forgetting to do so is one of the most common ways to drop marks on this topic.
3. Connecting the Two Forms — The Role of $k$ and $R$
The Boltzmann constant $k = 1.38 \times 10^{-23}$ J K$^{-1}$ is the per-molecule equivalent of the molar gas constant $R$. Think of $R$ as the "bulk" version and $k$ as the "single molecule" version. When a question gives you the number of molecules rather than moles, reach for $pV = NkT$. When it gives you moles, use $pV = nRT$.
The relationship $k = R / N_A$ also means you can convert between the two forms mid-calculation if needed. For example, if you know $n$ but need $N$, multiply by $N_A$. This flexibility is tested in multi-step problems where the question deliberately mixes the two scales.
4. Using $pV = nRT$ in Calculations
Exam questions on the ideal gas equation almost always require you to rearrange for one unknown. The structure is always the same: identify the known quantities, convert units, rearrange, substitute, and check. Let's walk through a full example so you can see every step clearly.
Worked example: Applying the Ideal Gas Equation — Finding Volume
A cylinder contains 2.00 mol of an ideal gas at a pressure of $2.50 \times 10^5$ Pa and a temperature of 127 °C. Calculate the volume of the gas. [3 marks]
$T = 127 + 273 = 400$ K — The ideal gas equation requires absolute temperature. This is the most common unit error — always convert before substituting. [B1]
$pV = nRT \Rightarrow V = \frac{nRT}{p}$ — Identify the unknown ($V$) and rearrange algebraically before substituting numbers. [M1]
$\frac{\text{mol} \times \text{J mol}^{-1} \text{K}^{-1} \times \text{K}}{\text{Pa}} = \frac{\text{J}}{\text{N m}^{-2}} = \frac{\text{N m}}{\text{N m}^{-2}} = \text{m}^3$ ✓ — Unit analysis confirms the answer is in m$^3$, as expected for volume.
Notice how the unit check in the final step is not optional — it is a genuine verification tool. If your units do not cancel to m$^3$, you have made an error somewhere in the rearrangement or substitution.
5. Work Done by a Gas — $W = p\Delta V$
Now that we have the ideal gas equation, we can ask a deeper question: when a gas expands, where does the energy go? The answer is that the gas does work on its surroundings. Picture a gas inside a cylinder with a movable piston. As the gas expands, it pushes the piston outward — force acting over a distance. That is work.
The sign of $W$ carries physical meaning. When a gas expands, $\Delta V > 0$ and $W > 0$ — the gas does positive work on its surroundings, transferring energy outward. When a gas is compressed, $\Delta V < 0$ and $W < 0$ — the surroundings do work on the gas, transferring energy into it. This sign convention connects directly to the first law of thermodynamics, which you will meet in the broader thermal physics topic.
A quick derivation makes the formula intuitive. Pressure is defined as force per unit area: $p = F/A$. Work is force times distance: $W = F \times d$. For a piston of cross-sectional area $A$ moving a distance $d$, the volume change is $\Delta V = A \times d$. Substituting: $W = F \times d = (pA) \times d = p \times (A \times d) = p\Delta V$. The formula is not arbitrary — it follows directly from the definitions of pressure and work.
6. Worked Example — Calculating Work Done
The formula $W = p\Delta V$ looks straightforward, but exam questions often hide the volume change inside a description of the physical situation. The skill is extracting $\Delta V$ correctly before applying the formula.
Worked example: Calculating Work Done During a Gas Expansion
A gas expands at a constant pressure of $1.20 \times 10^5$ Pa. Its volume increases from $3.00 \times 10^{-3}$ m$^3$ to $8.00 \times 10^{-3}$ m$^3$. Calculate the work done by the gas. [2 marks]
$\Delta V = V_{\text{final}} - V_{\text{initial}} = 8.00 \times 10^{-3} - 3.00 \times 10^{-3} = 5.00 \times 10^{-3}$ m$^3$ — Always compute $\Delta V$ as final minus initial. A positive result confirms the gas is expanding. [M1]
$W = p\Delta V = 1.20 \times 10^5 \times 5.00 \times 10^{-3}$ — Both quantities are already in SI units (Pa and m$^3$), so no conversion is needed.
$W = +600$ J — Positive work done means the gas transferred 600 J of energy to its surroundings during the expansion.
Answer: $W = 600$ J (work done by the gas on its surroundings)
7. Linking the Two Equations — A Combined Problem
The ideal gas equation and the work done formula are not isolated tools — they appear together in multi-step problems. A typical exam question might describe a gas being heated at constant pressure, ask you to find the new volume using $pV = nRT$, and then ask for the work done using $W = p\Delta V$. Recognising this two-step structure is the key to full marks.
Worked example: Combined Problem — Heating at Constant Pressure
A sample of 0.500 mol of an ideal gas is held at a constant pressure of $1.00 \times 10^5$ Pa. The gas is heated from 300 K to 500 K. (a) Calculate the initial and final volumes. (b) Calculate the work done by the gas during this process. [5 marks]
$V_1 = \frac{nRT_1}{p} = \frac{0.500 \times 8.31 \times 300}{1.00 \times 10^5}$ — Rearrange for $V$ and substitute $T_1 = 300$ K. [M1]
$0.500 \times 8.31 = 4.155$; $4.155 \times 300 = 1246.5$; $V_1 = \frac{1246.5}{1.00 \times 10^5} = 1.247 \times 10^{-2}$ m$^3$ — Carrying full precision through intermediate steps avoids rounding errors in the final answer.
$V_2 = \frac{nRT_2}{p} = \frac{0.500 \times 8.31 \times 500}{1.00 \times 10^5}$ — Same method, now with $T_2 = 500$ K. [A1 for both volumes]
$0.500 \times 8.31 = 4.155$; $4.155 \times 500 = 2077.5$; $V_2 = \frac{2077.5}{1.00 \times 10^5} = 2.078 \times 10^{-2}$ m$^3$ — Notice $V_2 / V_1 = 500/300$, confirming Charles's Law: volume is proportional to temperature at constant pressure.
There is an elegant shortcut hidden in this result. Since $p$ is constant, $W = p\Delta V = p(V_2 - V_1) = nR(T_2 - T_1)$. Substituting: $W = 0.500 \times 8.31 \times (500 - 300) = 0.500 \times 8.31 \times 200 = 831$ J. This shortcut is worth knowing for constant-pressure problems — it bypasses the need to calculate individual volumes.
8. Exam Technique and Common Mistakes
These two equations generate a predictable set of errors in exam scripts. Knowing the traps in advance is as valuable as knowing the physics itself. The guidance below is drawn directly from the patterns that cost students marks most often on AQA thermal physics questions.
Together, $pV = nRT$, $pV = NkT$, and $W = p\Delta V$ form the quantitative toolkit for AQA thermal physics. The ideal gas equation links the macroscopic state of a gas to the amount of substance and temperature, while the work done formula connects thermodynamic processes to energy transfer — both spec requirements are met when you can apply them individually and in combination.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
pV = NkT is the molecular form of the ideal gas equation, where N is the number of molecules and k = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant. It is used when a question gives the number of molecules rather than moles, and it is equivalent to pV = nRT because R = Nₐk, with Nₐ the Avogadro constant.
What is an ideal gas?
An ideal gas is one whose molecules occupy negligible volume compared to the container, experience no intermolecular forces except during collisions, and undergo perfectly elastic collisions with each other and the container walls. Real gases approximate ideal behaviour at low pressures and high temperatures.
How do you calculate work done by a gas?
Work done by a gas at constant pressure is W = pΔV, where ΔV is the final volume minus the initial volume. A gas expanding at 1.20 × 10⁵ Pa from 3.00 × 10⁻³ m³ to 8.00 × 10⁻³ m³ does W = 1.20 × 10⁵ × 5.00 × 10⁻³ = 600 J of work on its surroundings.
When do real gases behave like ideal gases?
Real gases approximate ideal behaviour at low pressures and high temperatures. Under those conditions the molecules occupy negligible volume compared with the container and experience no intermolecular forces except during collisions, as the ideal gas model assumes.
How do you use pV = nRT to find the volume of a gas?
Convert the temperature to kelvin, rearrange to V = nRT/p, then substitute. For 2.00 mol of gas at 2.50 × 10⁵ Pa and 127 °C (400 K), V = 2.00 × 8.31 × 400 ÷ 2.50 × 10⁵ = 6648 ÷ 250000 = 0.0266 m³.
What does the sign of work done by a gas mean?
When a gas expands, ΔV is positive and W = pΔV is positive: the gas does work on its surroundings, transferring energy outward. When a gas is compressed, ΔV is less than zero, so W is less than zero: the surroundings do work on the gas, transferring energy into it.
How do you derive the work done by a gas?
Work done by a gas, W = pΔV, follows from the definitions of pressure and work. For a piston of cross-sectional area A moving a distance d, the force is F = pA and the volume change is ΔV = A × d, so W = F × d = pA × d = pΔV.
What is the shortcut for work done at constant pressure?
At constant pressure, the work done by an ideal gas can be found as W = nR(T₂ − T₁), since pΔV = nRΔT. For 0.500 mol heated from 300 K to 500 K, W = 0.500 × 8.31 × 200 = 831 J, without calculating the individual volumes.
What are the units of pV?
The product pV has units of joules, because Pa × m³ = N m⁻² × m³ = N m = J. That is why nRT, with n in mol, R in J mol⁻¹ K⁻¹ and T in K, also gives joules, and a unit check like this confirms a rearranged ideal gas equation.
What are the common mistakes with the ideal gas equation?
The most common ideal gas equation mistake is using temperature in Celsius instead of kelvin, so 273 must be added to any Celsius value first. Another is mixing up the two forms: pV = nRT needs the number of moles, while pV = NkT needs the number of molecules.