The Ideal Gas Equation

Spec mapping: OCR H556 Module 5.2 — Ideal gases (the equation of state of an ideal gas in the mole form $pV = nRT$pV=nRT and the molecule form $pV = NkT$pV=NkT; the molar gas constant $R$R; absolute temperature scale; the limits of the ideal-gas model and deviations of real gases). Refer to the official OCR H556 specification document for exact wording.

Boyle's law, Charles's law and the pressure law each describe one face of the same underlying behaviour: a fixed sample of gas obeys $pV/T = \text{constant}$pV/T=constant under any combination of changes. That constant, however, depends on how much gas is in the container. Double the amount and you double the constant. The ideal gas equation is the single line that promotes this empirical observation into a universal law by making the dependence on amount explicit:

$pV = nRT$pV=nRT

The molar gas constant $R$R that appears here is the same number — $8.31$8.31 J mol$^{-1}$−1 K$^{-1}$−1 — for every gas that behaves ideally, whether it is helium, nitrogen, argon or steam well above its boiling point. This universality is the central message of Lesson 6: at A-Level we treat any sufficiently dilute, sufficiently hot gas as obeying $pV = nRT$pV=nRT exactly, and we extract from a small set of measurements ($p$p, $V$V, $T$T, mass) the amount of gas and ultimately, in Lesson 7, the number of molecules.

This treatment goes well beyond the GCSE "plug-in" of the gas laws: we derive the equation from the combined gas law, give it a clean dimensional check, work eight numerical examples spanning car tyres, scuba tanks, hot-air balloons and divers' lungs, and end with the misconceptions that cost candidates marks in OCR H556 papers.

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Generalising the Combined Gas Law

If you double the amount of gas in a container without changing p or T, you need twice the volume. This is intuitive: gas molecules occupy space, and more molecules need more space. Therefore the constant in the combined gas law pV/T = const must be proportional to the amount of gas.

We define the amount of gas as the number of moles n, and write

$$pV / T = n R$$pV/T=nR

where R is a universal constant — the molar gas constant — that does not depend on which gas we are considering. Rearranging:

$$pV = nRT$$pV=nRT

This is the ideal gas equation.

Note the structure:

• p is in pascals (Pa).
• V is in cubic metres (m³).
• n is in moles.
• R is the molar gas constant.
• T is in kelvin (K).

The units of the left-hand side are Pa × m³ = N m⁻² × m³ = N m = J. So pV has units of energy (joules), and nRT must too. This gives us the units of R: J mol⁻¹ K⁻¹. The numerical value is

$$R = 8.31\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1} \quad \text{(OCR data sheet value)}$$R=8.31Jmol−1K−1(OCR data sheet value)

(More precisely 8.314462618..., but 8.31 is what OCR uses.)

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The Mole

The mole is the SI unit for the amount of substance. It is defined (since 2019) so that one mole contains exactly N_A = 6.022 × 10²³ elementary entities, where N_A is the Avogadro constant. For our purposes:

• 1 mole of any gas contains 6.022 × 10²³ molecules.
• 1 mole of hydrogen (H_2) has a mass of 2.0 g.
• 1 mole of nitrogen (N_2) has a mass of 28.0 g.
• 1 mole of oxygen (O_2) has a mass of 32.0 g.
• 1 mole of air (a mixture, averaged) has a mass of about 29 g.

To convert between the number of moles n, the mass m and the molar mass M, use

$$n = m / M$$n=m/M

where M is in kg mol⁻¹ (or g mol⁻¹, as long as the units match). For instance, 14 g of nitrogen is n = 14/28 = 0.50 mol.

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The Ideal Gas Model

An ideal gas is a theoretical construct in which:

1. The gas obeys pV = nRT exactly, at all p and T.
2. Intermolecular forces are zero (except during collisions).
3. The molecules themselves have zero volume.
4. Collisions are perfectly elastic.

No real gas is ideal in this sense, but all real gases approach ideal behaviour when they are dilute (low density, high molar volume) and hot (well above their condensation temperature). Air at room temperature and atmospheric pressure is very close to ideal — the error in pV = nRT is well under 1%.

Real gases deviate most noticeably:

• At high pressure, when the molecules are packed closely enough for their own volume to matter.
• At low temperature, when intermolecular attractions pull the molecules together and the gas begins to liquefy.

The van der Waals equation and more sophisticated models correct for these effects, but they lie beyond A-Level. For this course, assume pV = nRT holds unless told otherwise.

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Worked Example 1: Basic Use of pV = nRT

A sealed cylinder of volume V = 0.020 m³ contains n = 0.80 mol of nitrogen gas at T = 293 K. Find the pressure.

$$\begin{aligned} p = nRT / V \\ = (0.80)(8.31)(293) / (0.020) \\ = (0.80)(8.31)(293) / 0.020 \\ = 1947.9 / 0.020 \\ \approx 9.7 \times 10^{4}\,\text{Pa} \end{aligned}$$p=nRT/V=(0.80)(8.31)(293)/(0.020)=(0.80)(8.31)(293)/0.020=1947.9/0.020≈9.7×104Pa​

So the pressure is about 97 kPa, comparable with atmospheric.

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Worked Example 2: Finding the Amount

A steel container of volume V = 0.50 m³ contains helium at p = 2.5 × 10⁵ Pa and T = 300 K. How many moles of helium are present?

$$\begin{aligned} n = pV / (RT) \\ = (2.5 \times 10^{5})(0.50) / ((8.31)(300)) \\ = 125 000 / 2493 \\ \approx 50.1\,\text{mol} \end{aligned}$$n=pV/(RT)=(2.5×105)(0.50)/((8.31)(300))=125000/2493≈50.1mol​

The mass of helium is n × M = 50.1 × 0.004 = 0.20 kg (since helium has molar mass 4.0 g mol⁻¹).

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Worked Example 3: Volume of an Ideal Gas at STP

What volume does 1 mol of an ideal gas occupy at standard temperature and pressure (STP), taken as T = 273 K and p = 1.01 × 10⁵ Pa?

$$\begin{aligned} V &= nRT / p \\ &= (1)(8.31)(273) / (1.01 \times 10^{5}) \\ &= 2268 / 101000 \\ &\approx 0.0225\,\text{m}^{3} \\ &= 22.5\,\text{L} \end{aligned}$$V​=nRT/p=(1)(8.31)(273)/(1.01×105)=2268/101000≈0.0225m3=22.5L​

This value — 22.4 or 22.5 litres per mole at STP — is worth remembering. It gives you a sense of how much space a mole of gas occupies. A 1-litre bottle of air at STP contains about 1/22.4 ≈ 0.045 moles, or about 2.7 × 10²² molecules.

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Worked Example 4: A Car Tyre

A car tyre with internal volume V = 0.020 m³ is pumped up to an absolute pressure of p = 3.0 × 10⁵ Pa at a temperature of T = 290 K. How many moles of air are inside? If the molar mass of air is 0.029 kg mol⁻¹, what is the mass of air in the tyre?

$$\begin{aligned} n = pV / (RT) \\ = (3.0 \times 10^{5})(0.020) / ((8.31)(290)) \\ = 6000 / 2410 \\ \approx 2.49\,\text{mol} \\ m = n M = (2.49)(0.029) \approx 0.072\,\text{kg} = 72 g \end{aligned}$$n=pV/(RT)=(3.0×105)(0.020)/((8.31)(290))=6000/2410≈2.49molm=nM=(2.49)(0.029)≈0.072kg=72g​

So a standard car tyre contains about 72 g of air — roughly the mass of a large kiwi fruit.

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Worked Example 5: A Hot-Air Balloon

A hot-air balloon contains 1500 m³ of air at 100 °C (373 K) while the outside air is at 20 °C (293 K). If the pressure is the same inside and outside (1.01 × 10⁵ Pa), what is the mass of air inside the balloon? What is the mass of air that would occupy the same volume if it were at the outside temperature?

Both calculations use n = pV / (RT) followed by m = nM with M = 0.029 kg mol⁻¹.

Inside:

$$\begin{aligned} n_{\text{inside}} &= pV / (RT) \\ &= (1.01 \times 10^{5})(1500) / ((8.31)(373)) \\ &= 1.515 \times 10^{8} / 3099.6 \\ &\approx 4.89 \times 10^{4}\,\text{mol} \\ m_{\text{inside}} &= n_{\text{inside}} \, M = (4.89 \times 10^{4})(0.029) \\ &\approx 1418\,\text{kg} \end{aligned}$$ninside​minside​​=pV/(RT)=(1.01×105)(1500)/((8.31)(373))=1.515×108/3099.6≈4.89×104mol=ninside​M=(4.89×104)(0.029)≈1418kg​

Outside (same volume, lower temperature):

$$\begin{aligned} n_{\text{outside}} = (1.01 \times 10^{5})(1500) / ((8.31)(293)) \\ = 1.515 \times 10^{8} / 2434.8 \\ \approx 6.22 \times 10^{4}\,\text{mol} \\ m_{\text{outside}} = (6.22 \times 10^{4})(0.029) \\ \approx 1805\,\text{kg} \end{aligned}$$noutside​=(1.01×105)(1500)/((8.31)(293))=1.515×108/2434.8≈6.22×104molmoutside​=(6.22×104)(0.029)≈1805kg​

The outside air displaced by the balloon would weigh about 1805 kg, whereas the hot air inside weighs only 1418 kg. The upthrust (by Archimedes' principle) is the weight of the displaced air, 1805 × 9.81 ≈ 17 700 N, while the weight of the hot air inside is 1418 × 9.81 ≈ 13 900 N. The net lift is therefore about 3800 N, enough to support a basket, pilot and a small amount of equipment.

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Worked Example 6: Gas Laws in Disguise

A rigid container initially holds nitrogen at p_1 = 1.5 × 10⁵ Pa and T_1 = 280 K. It is heated until the pressure reaches p_2 = 2.1 × 10⁵ Pa. What is the new temperature?

Since the container is rigid, V_1 = V_2, and the ideal gas equation gives

$$\begin{aligned} p_1 V &= n R T_1, \quad p_2 V = n R T_2 \\ \therefore \frac{T_2}{T_1} &= \frac{p_2}{p_1} \\ T_2 &= T_1 \times \frac{p_2}{p_1} \\ &= (280) \times \frac{2.1 \times 10^{5}}{1.5 \times 10^{5}} \\ &= (280)(1.4) \\ &= 392\,\text{K} \end{aligned}$$p1​V∴T1​T2​​T2​​=nRT1​,p2​V=nRT2​=p1​p2​​=T1​×p1​p2​​=(280)×1.5×1052.1×105​=(280)(1.4)=392K​

which is about 119 °C.

This example shows how pV = nRT reduces to the pressure law when V and n are held constant. Boyle's law and Charles's law can be recovered the same way.

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Worked Example 7: A Diver's Lungs

At a depth of 30 m in sea water, a diver's lungs are subject to a pressure of about 4.0 × 10⁵ Pa. If the diver takes a full breath (4.0 L) at that depth and then ascends to the surface (where the pressure is 1.0 × 10⁵ Pa) without exhaling, what would the volume of the gas become?

This is Boyle's law in its pure form (assuming temperature constant):

$$\begin{aligned} p_1 V_1 = p_2 V_2 \\ V_2 = p_1 V_1 / p_2 \\ = (4.0 \times 10^{5})(4.0) / (1.0 \times 10^{5}) \\ = 16 L \end{aligned}$$p1​V1​=p2​V2​V2​=p1​V1​/p2​=(4.0×105)(4.0)/(1.0×105)=16L​

The volume would quadruple — an impossible expansion for the human lungs, which would rupture. This is why divers are trained never to hold their breath when ascending: breathe out continuously. Boyle's law can be lethal if ignored.

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The Link to the Combined Gas Law

For a fixed number of moles n of gas undergoing any change,

$$p_1 V_1 / T_1 = nR = p_2 V_2 / T_2$$p1​V1​/T1​=nR=p2​V2​/T2​

so

$$p_1 V_1 / T_1 = p_2 V_2 / T_2$$p1​V1​/T1​=p2​V2​/T2​

which is the combined gas law from Lesson 5. The ideal gas equation extends this by allowing you to compute the constant nR explicitly once you know the number of moles.

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A Quick Check: Dimensions

Let us verify that pV = nRT has the right units.

• p has units N m⁻² = kg m⁻¹ s⁻².
• V has units m³.
• So pV has units N m = J. Good: pV is an energy.
• n has units mol.
• R has units J mol⁻¹ K⁻¹.
• T has units K.
• So nRT has units mol × (J mol⁻¹ K⁻¹) × K = J. Good: nRT is also an energy.

Both sides are in joules. This is a useful sanity check — if you ever get an answer with weird units, you know you have substituted something in the wrong units.

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Worked Example 8: Non-Trivial Pressure Unit

A scuba tank contains gas at 200 bar. Given that 1 bar = 10⁵ Pa, and the tank has a volume of 12 L at 20 °C, how many moles of gas are in the tank?

Converting to SI:

$$\begin{aligned} p &= 200 \times 10^{5} = 2.0 \times 10^{7}\,\text{Pa} \\ V &= 12 \times 10^{-3} = 0.012\,\text{m}^{3} \\ T &= 293\,\text{K} \end{aligned}$$pVT​=200×105=2.0×107Pa=12×10−3=0.012m3=293K​

Then

$$\begin{aligned} n = pV / (RT) \\ = (2.0 \times 10^{7})(0.012) / ((8.31)(293)) \\ = 2.4 \times 10^{5} / 2434.8 \\ \approx 98.6\,\text{mol} \end{aligned}$$n=pV/(RT)=(2.0×107)(0.012)/((8.31)(293))=2.4×105/2434.8≈98.6mol​

If this is air, the mass is about 98.6 × 0.029 ≈ 2.86 kg. This is why scuba tanks feel heavy!

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Choosing the Right Form: $pV = nRT$pV=nRT vs $pV = NkT$pV=NkT

Both forms of the ideal gas equation describe the same physics, but each is natural in different contexts. Use the table and decision tree below as a deliberate habit before you reach for a calculator.

Information given	Best form	Why
mass $m$m and molar mass $M$M	$pV = nRT$pV=nRT with $n = m/M$n=m/M	direct line to moles
number of molecules $N$N	$pV = NkT$pV=NkT	$N$N already in the equation
volume per molecule, density at the microscale	$pV = NkT$pV=NkT	molecule-scale natural variables
pressure, density and molar mass of bulk gas	$\rho = pM/(RT)$ρ=pM/(RT) (derived from $pV = nRT$pV=nRT)	density follows immediately

graph TD
    Q[Ideal gas problem]
    Q --> A{Do you know mass<br/>or moles?}
    A -- "Mass / moles given" --> N1[Use pV = nRT]
    A -- "Number of molecules<br/>given or asked" --> N2[Use pV = NkT]
    N1 --> O1[answer in mol]
    N2 --> O2[answer in molecules]
    O1 --> X{Need molecules<br/>or vice versa?}
    O2 --> X
    X -- "yes" --> CONV[Convert with<br/>N = n N_A]
    X -- "no" --> END[Done]

Worked examples 1-3 use $pV = nRT$pV=nRT because the questions specify mass or mole count. Examples in Lesson 7 will use $pV = NkT$pV=NkT because the natural quantities there are numbers of molecules.

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Synoptic Links

• Boltzmann constant (Lesson 7) — the per-molecule form $pV = NkT$pV=NkT follows from $pV = nRT$pV=nRT by writing $n = N/N_A$n=N/NA​ and defining $k = R/N_A$k=R/NA​. Mastering the conversion both ways is examined in H556 written papers.
• Kinetic theory pressure equation (Lesson 9) — $pV = (1/3)Nm\langle c^{2}\rangle$pV=(1/3)Nm⟨c2⟩ is equated to $pV = NkT$pV=NkT in Lesson 10 to derive $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT. The ideal gas equation is the macroscopic bridge that makes that derivation possible.
• Internal energy of an ideal monatomic gas (Module 5.1) — $U = \tfrac{3}{2}nRT$U=23​nRT follows because the only energy a monatomic molecule can store is translational kinetic energy. The ideal gas equation supplies the $nRT$nRT piece.
• Specific heat capacity of gases (Modules 5.1, 5.4) — at constant volume $C_V = \tfrac{3}{2}R$CV​=23​R (monatomic) and at constant pressure $C_p = C_V + R$Cp​=CV​+R; the universal $R$R that appears here is the same $R$R as in the ideal gas equation.
• Astrophysical applications (Module 5.5) — stellar interiors are extremely hot, dilute plasmas; $pV = nRT$pV=nRT is the leading-order equation of state used to model main-sequence stars, and deviations matter only in dense degenerate cores (white dwarfs, neutron stars).

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Specimen question modelled on the OCR H556 paper format

Question (9 marks): A laboratory cylinder of internal volume $V = 6.0 \times 10^{-3}$V=6.0×10−3 m$^{3}$3 contains argon gas. A pressure gauge mounted on the cylinder reads $p_{\text{gauge}} = 1.8 \times 10^{6}$pgauge​=1.8×106 Pa when the gas is at room temperature $T_1 = 293$T1​=293 K. Atmospheric pressure is $p_{\text{atm}} = 1.01 \times 10^{5}$patm​=1.01×105 Pa.