Introduction to the Ideal Gas Equation Revision Notes | A Level AQA Chemistry

1. Why One Equation Describes Every Gas

Picture a sealed syringe of air. Push the plunger in, and the pressure rises. Heat it up, and the pressure rises again. Add more gas, and the pressure rises once more. Three completely different actions, yet they all change the same measurable quantity.

Chemists noticed that pressure, volume, temperature, and amount of gas are not independent — they are locked together by a single relationship. That relationship is the ideal gas equation: $pV = nRT$. Master this one equation and you can calculate any of those four quantities from the other three.

2. The Ideal Gas Equation and Its Variables

Each variable has one correct SI unit, and using the wrong unit is the single most common source of lost marks in this topic. The table below fixes those units in your memory before you touch a single calculation.

The word 'ideal' carries a specific physical meaning here. An ideal gas has particles with no volume and no attraction or repulsion between them. Real gases deviate from this model when pressure is very high (particles are forced close together, so their own volume matters) or when temperature is very low (particles move slowly enough for intermolecular forces to become significant).

3. Unit Conversions: The Non-Negotiable First Step

AQA exam questions almost always give data in non-SI units — pressure in kPa, volume in cm$^3$ or dm$^3$, temperature in $^\circ$C. Your first job before writing $pV = nRT$ is to convert every value into SI units. Skipping this step and substituting directly is the most reliable way to get the wrong answer.

• Temperature: $T(\text{K}) = T(^\circ\text{C}) + 273$ — for example, $25^\circ\text{C} = 298\text{ K}$
• Pressure: $1\text{ kPa} = 1000\text{ Pa}$, so multiply kPa by 1000 to get Pa
• Volume: $1\text{ dm}^3 = 1 \times 10^{-3}\text{ m}^3$, so divide dm$^3$ by 1000 to get m$^3$
• Volume: $1\text{ cm}^3 = 1 \times 10^{-6}\text{ m}^3$, so divide cm$^3$ by $10^6$ to get m$^3$

Notice that the gas constant $R = 8.314$ J mol$^{-1}$ K$^{-1}$ is defined using SI units throughout. If you feed it SI values for $p$, $V$, and $T$, the units cancel correctly and you get $n$ in mol. Feed it kPa or cm$^3$ and the units do not cancel — the number you get is meaningless.

4. Rearranging and Applying the Equation

Because $pV = nRT$ is a simple product on both sides, rearranging for any variable is straightforward. Divide both sides by whatever surrounds the unknown, and you have your working formula.

The worked examples below show the full method for two common question types. Work through each step yourself before revealing the next — this is where the real learning happens.

Worked example: Calculating the Amount of Substance from $pV = nRT$

A sealed container holds a gas at a pressure of 150 kPa and a temperature of 27 $^\circ$C. The volume of the container is 2.00 dm$^3$. Calculate the amount, in mol, of gas present. ($R = 8.314$ J mol$^{-1}$ K$^{-1}$)

1. $p = 150 \times 10^3 = 1.50 \times 10^5 \text{ Pa}$ $V = 2.00 \div 1000 = 2.00 \times 10^{-3} \text{ m}^3$ $T = 27 + 273 = 300 \text{ K}$ — Every value must be in SI units before substitution. Pressure: multiply kPa by 1000. Volume: divide dm$^3$ by 1000. Temperature: add 273.
2. $n = \dfrac{pV}{RT}$ — Divide both sides of $pV = nRT$ by $RT$ to isolate $n$.
3. $n = \dfrac{(1.50 \times 10^5) \times (2.00 \times 10^{-3})}{8.314 \times 300}$ — Each value carries its SI unit. The units will cancel to give mol.
4. $1.50 \times 10^5 \times 2.00 \times 10^{-3} = 1.50 \times 2.00 \times 10^{5-3} = 3.00 \times 10^2 = 300$ — Multiply the coefficients and add the exponents: $10^5 \times 10^{-3} = 10^{5+(-3)} = 10^2$.
5. $8.314 \times 300 = 2494.2$ — $8.314 \times 300$: multiply $8.314 \times 3 = 24.942$, then $\times 100 = 2494.2$.
6. $n = \dfrac{300}{2494.2} = 0.1202...\approx 0.120 \text{ mol}$ — Round to 3 significant figures to match the data given. Always state the unit.

Answer: $n = 0.120$ mol (3 s.f.)

Worked example: Calculating the Pressure of a Gas from $pV = nRT$

0.250 mol of nitrogen gas occupies a volume of 5.00 dm$^3$ at a temperature of 100 $^\circ$C. Calculate the pressure of the gas in Pa. ($R = 8.314$ J mol$^{-1}$ K$^{-1}$)

1. $n = 0.250 \text{ mol}$ (already in mol) $V = 5.00 \div 1000 = 5.00 \times 10^{-3} \text{ m}^3$ $T = 100 + 273 = 373 \text{ K}$ — Volume in dm$^3$ must be divided by 1000 to give m$^3$. Temperature in $^\circ$C must have 273 added to give K.
2. $p = \dfrac{nRT}{V}$ — Divide both sides of $pV = nRT$ by $V$.
3. $p = \dfrac{0.250 \times 8.314 \times 373}{5.00 \times 10^{-3}}$ — All values are now in SI units, so the answer will be in Pa.
4. $0.250 \times 8.314 = 2.0785$ $2.0785 \times 373 = 775.28...$ — Work left to right: $0.250 \times 8.314$ first, then multiply by 373. Keeping extra figures avoids rounding errors mid-calculation.
5. $p = \dfrac{775.28}{5.00 \times 10^{-3}} = 775.28 \times \dfrac{1}{5.00 \times 10^{-3}} = 775.28 \times 200 = 155056 \approx 1.55 \times 10^5 \text{ Pa}$ — Dividing by $5.00 \times 10^{-3}$ is the same as multiplying by 200. The answer is in Pa because all inputs were SI.

Answer: $p = 1.55 \times 10^5$ Pa (3 s.f.)

5. Linking the Ideal Gas Equation to Molar Mass

One powerful extension of $pV = nRT$ is finding the molar mass of an unknown gas. Because $n = m \div M$ (where $m$ is mass in grams and $M$ is molar mass in g mol$^{-1}$), you can substitute this into the ideal gas equation to get $pV = \dfrac{m}{M}RT$, then rearrange for $M$.

This approach appears in AQA questions where you are given the mass of a gas sample along with its pressure, volume, and temperature. The method is a direct application of the same equation — the only extra step is expressing $n$ in terms of mass and molar mass before rearranging.

Worked example: Finding Molar Mass Using $pV = nRT$

A 0.800 g sample of a volatile liquid is vaporised and found to occupy 312 cm$^3$ at a pressure of 101 kPa and a temperature of 100 $^\circ$C. Calculate the molar mass of the compound. ($R = 8.314$ J mol$^{-1}$ K$^{-1}$)

1. $p = 101 \times 10^3 = 1.01 \times 10^5 \text{ Pa}$ $V = 312 \div 10^6 = 3.12 \times 10^{-4} \text{ m}^3$ $T = 100 + 273 = 373 \text{ K}$ — Volume is in cm$^3$: divide by $10^6$ to convert to m$^3$. Temperature: add 273.
2. $n = \dfrac{pV}{RT} = \dfrac{(1.01 \times 10^5) \times (3.12 \times 10^{-4})}{8.314 \times 373}$ — Calculate the number of moles first, then use it to find molar mass.
3. $1.01 \times 10^5 \times 3.12 \times 10^{-4} = 1.01 \times 3.12 \times 10^{5-4} = 3.1512 \times 10^1 = 31.512$ — $10^5 \times 10^{-4} = 10^{5+(-4)} = 10^1 = 10$. Then $1.01 \times 3.12 = 3.1512$, so the product is $31.512$.
4. $8.314 \times 373 = 3101.122$ — $8.314 \times 373$: $8.314 \times 300 = 2494.2$; $8.314 \times 73 = 606.922$; sum $= 3101.122$.
5. $n = \dfrac{31.512}{3101.122} = 0.010161...\text{ mol}$ — Keep extra significant figures here to avoid rounding errors in the next step.
6. $M = \dfrac{m}{n} = \dfrac{0.800}{0.010161} = 78.7...\approx 78.7 \text{ g mol}^{-1}$ — Molar mass = mass $\div$ moles. The answer is close to 78 g mol$^{-1}$, consistent with benzene ($\text{C}_6\text{H}_6$, $M_r = 78$).

Answer: $M \approx 78.7$ g mol$^{-1}$ (3 s.f.)

6. Ideal vs Real Gas Behaviour

The ideal gas equation works remarkably well for most gases under everyday conditions, but it rests on two assumptions that break down at extremes: that gas particles have no volume of their own, and that there are no intermolecular forces between them.

At high pressures, particles are compressed close together. Their own physical volume becomes a significant fraction of the total container volume, so the gas occupies more space than the ideal equation predicts. At low temperatures, particles move slowly and intermolecular attractions pull them together, reducing the pressure below the ideal prediction.

7. Exam Technique and Common Mistakes

AQA mark schemes for ideal gas equation questions award marks at three distinct stages: the correct rearrangement, the correct substitution with SI units, and the correct final answer with units. Losing a mark at the substitution stage — because a temperature was left in $^\circ$C or a volume was left in cm$^3$ — is avoidable with a disciplined conversion step before every calculation.

When a question asks you to 'calculate', show every step of your working. Even if your final answer is wrong due to an arithmetic slip, you can still earn method marks for a correct rearrangement and a correct substitution. A bare answer with no working scores nothing if it is incorrect.

That question captures the most conceptually demanding trap in this topic: the difference between a Celsius ratio and a Kelvin ratio. Celsius is an arbitrary scale with its zero at the freezing point of water; Kelvin starts at absolute zero, where particle motion theoretically ceases. Only Kelvin ratios are physically meaningful in the ideal gas equation.

Every calculation in this topic reduces to the same disciplined sequence: convert to SI units, identify the unknown, rearrange $pV = nRT$, substitute, and state the answer with its unit. Practise that sequence until it is automatic, and the ideal gas equation becomes one of the most reliable sources of marks in the Physical Chemistry paper.