The Ideal Gas Equation

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / ideal gas equation, content statements covering the ideal gas equation $pV = nRT$pV=nRT in SI units; rearrangement for $n$n, $V$V, $T$T, $p$p and $M$M ($= mRT/pV$=mRT/pV); application to gases produced in chemical reactions; and qualitative treatment of deviations from ideal behaviour (refer to the official OCR H432 specification document for exact wording). This is the gas-phase extension of the mole concept and the bridge between mole-based stoichiometry and the energetics work of Module 3.

The ideal gas equation $pV = nRT$pV=nRT is the gas-phase counterpart of the master mole equation $n = m/M$n=m/M. It connects four bulk-state properties — pressure $p$p, volume $V$V, temperature $T$T and amount $n$n — through the universal gas constant $R$R. The framework is the product of three eighteenth- and nineteenth-century gas-law discoveries: Boyle's law (1662, $pV$pV constant at fixed $T$T), Charles's law (1787, $V/T$V/T constant at fixed $p$p) and Avogadro's law (1811, $V/n$V/n constant at fixed $p, T$p,T). Their combination, first written down by Benoît Clapeyron in 1834 and refined by Rudolf Clausius in the 1850s, gives a single equation describing any ideal gas at any conditions. The OCR A-Level specification expects you to apply it confidently in SI units only — pressure in pascals, volume in cubic metres, temperature in kelvin — and to be aware of why real gases deviate at high pressures and low temperatures. This is OCR's distinctive choice; AQA uses a fixed molar gas volume at RTP, sacrificing flexibility for arithmetic convenience.

Key Equation: $pV = nRT$pV=nRT with $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$R=8.314 J K−1 mol−1, $p$p in Pa, $V$V in m³, $T$T in K.

Derived form: $M = \dfrac{mRT}{pV}$M=pVmRT​.

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Historical Background

The ideal gas equation combines three gas laws discovered between the seventeenth and early nineteenth centuries:

• Boyle's law (1662): at constant $T$T and $n$n, $pV = \text{constant}$pV=constant.
• Charles's law (1787): at constant $p$p and $n$n, $V/T = \text{constant}$V/T=constant.
• Avogadro's law (1811): at constant $p$p and $T$T, $V/n = \text{constant}$V/n=constant.

Together they imply that $pV/(nT)$pV/(nT) is a universal constant — the gas constant $R$R — and the relation $pV = nRT$pV=nRT falls out. Clapeyron's 1834 paper was the first explicit statement of the combined law; Avogadro's hypothesis (equal volumes of gases at the same $T, p$T,p contain equal numbers of particles) is the same Avogadro whose constant you met three lessons ago, and the gas constant $R$R is itself related to the Avogadro constant by $R = Lk_\text{B}$R=LkB​, where $k_\text{B}$kB​ is the Boltzmann constant.

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The Ideal Gas Equation

$pV = nRT$pV=nRT

Symbol	Quantity	SI unit
$p$p	pressure	Pa (pascals)
$V$V	volume	m³
$n$n	amount of substance	mol
$R$R	gas constant	$8.314\text{ J K}^{-1}\text{ mol}^{-1}$8.314 J K−1 mol−1
$T$T	temperature	K

Critical: SI units throughout — Pa, m³, K. Mismatched units are the single biggest source of errors in these problems.

The gas constant $R$R has the value $8.314\text{ J K}^{-1}\text{ mol}^{-1}$8.314 J K−1 mol−1 (equivalent to $8.314\text{ N m K}^{-1}\text{ mol}^{-1}$8.314 N m K−1 mol−1). It is universal — the same for any ideal gas. OCR provides $R$R on the data sheet but you must recognise it on sight.

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Unit Conversions

Pressure

Unit	Conversion to Pa
1 kPa	$10^3$103 Pa
1 MPa	$10^6$106 Pa
1 atm	$101\,325$101325 Pa ($\approx 101\text{ kPa}$≈101 kPa)
1 bar	$10^5$105 Pa
1 mmHg (torr)	$\approx 133.3$≈133.3 Pa

At sea level atmospheric pressure is approximately $101\,000\text{ Pa}$101000 Pa. When OCR writes "atmospheric pressure" with no further qualification, use $101\,000$101000 or $101\,325\text{ Pa}$101325 Pa.

Volume

Unit	Conversion to m³
1 dm³	$10^{-3}\text{ m}^3$10−3 m3
1 cm³	$10^{-6}\text{ m}^3$10−6 m3
1 L	$10^{-3}\text{ m}^3$10−3 m3

Temperature

Temperature must be in kelvin:

$T(\text{K}) = \theta(^{\circ}\text{C}) + 273$T(K)=θ(∘C)+273

• $0\,^{\circ}\text{C} = 273\text{ K}$0∘C=273 K
• $20\,^{\circ}\text{C} = 293\text{ K}$20∘C=293 K
• $25\,^{\circ}\text{C} = 298\text{ K}$25∘C=298 K (standard)
• $100\,^{\circ}\text{C} = 373\text{ K}$100∘C=373 K

OCR usually writes the conversion as $+273$+273 at A-Level; the more accurate $+273.15$+273.15 is acceptable.

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Rearrangements

$n = \frac{pV}{RT}, \quad V = \frac{nRT}{p}, \quad T = \frac{pV}{nR}, \quad p = \frac{nRT}{V}$n=RTpV​,V=pnRT​,T=nRpV​,p=VnRT​

Combined with $n = m/M$n=m/M:

$M = \frac{mRT}{pV}$M=pVmRT​

This molar-mass form is the standard way to identify an unknown gas from $m$m, $V$V, $p$p, $T$T data.

Decision tree — choosing the right rearrangement

flowchart TD
    A[Gas-equation question] --> B{What is asked?}
    B -->|n| C[Use n = pV / RT]
    B -->|V| D[Use V = nRT / p]
    B -->|T| E[Use T = pV / nR]
    B -->|p| F[Use p = nRT / V]
    B -->|M molar mass| G[Use M = mRT / pV]
    C --> H[Check SI units: Pa, m^3, K]
    D --> H
    E --> H
    F --> H
    G --> H
    H --> I[Substitute and solve]

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Worked Example 1 — Moles of a gas

What amount of O₂ occupies $2.50\text{ dm}^3$2.50 dm3 at $25.0\,^{\circ}\text{C}$25.0∘C and $100\text{ kPa}$100 kPa?

• $V = 2.50 \times 10^{-3}\text{ m}^3$V=2.50×10−3 m3; $p = 1.00 \times 10^5\text{ Pa}$p=1.00×105 Pa; $T = 298\text{ K}$T=298 K
• $n = pV/RT = (1.00 \times 10^5)(2.50 \times 10^{-3}) / [(8.314)(298)] = 250/2477.6 =$n=pV/RT=(1.00×105)(2.50×10−3)/[(8.314)(298)]=250/2477.6= $0.101\text{ mol}$0.101 mol

Worked Example 2 — Volume

$V$V for $0.250\text{ mol}$0.250 mol N₂ at $101\,000\text{ Pa}$101000 Pa and $20.0\,^{\circ}\text{C}$20.0∘C?

• $V = nRT/p = (0.250)(8.314)(293)/101\,000 = 608.9/101\,000 = 6.03 \times 10^{-3}\text{ m}^3 =$V=nRT/p=(0.250)(8.314)(293)/101000=608.9/101000=6.03×10−3 m3= $6.03\text{ dm}^3$6.03 dm3

Worked Example 3 — Molar mass of unknown gas

$0.500\text{ g}$0.500 g of a gas in $250\text{ cm}^3$250 cm3 at $101\,000\text{ Pa}$101000 Pa and $100\,^{\circ}\text{C}$100∘C.

• $V = 2.50 \times 10^{-4}\text{ m}^3$V=2.50×10−4 m3; $T = 373\text{ K}$T=373 K
• $n = pV/RT = (101\,000)(2.50 \times 10^{-4})/[(8.314)(373)] = 25.25/3101.1 = 8.143 \times 10^{-3}\text{ mol}$n=pV/RT=(101000)(2.50×10−4)/[(8.314)(373)]=25.25/3101.1=8.143×10−3 mol
• $M = m/n = 0.500/(8.143 \times 10^{-3}) =$M=m/n=0.500/(8.143×10−3)= $61.4\text{ g mol}^{-1}$61.4 g mol−1

Worked Example 4 — Identifying an unknown gas

A sealed $1.00\text{ dm}^3$1.00 dm3 flask at $298\text{ K}$298 K and $101\,000\text{ Pa}$101000 Pa contains $1.80\text{ g}$1.80 g of gas. Identify it.

• $n = pV/RT = (101\,000)(10^{-3})/[(8.314)(298)] = 101/2477.6 = 0.04077\text{ mol}$n=pV/RT=(101000)(10−3)/[(8.314)(298)]=101/2477.6=0.04077 mol
• $M = 1.80/0.04077 =$M=1.80/0.04077= $44.1\text{ g mol}^{-1}$44.1 g mol−1 → CO₂ ($M_r = 44.0$Mr​=44.0).

Common gas $M_r$Mr​ values worth recognising

Gas	$M_r$Mr​
H₂	2.0
He	4.0
CH₄	16.0
NH₃	17.0
H₂O (vapour)	18.0
Ne	20.2
HF	20.0
N₂	28.0
CO	28.0
O₂	32.0
H₂S	34.1
Ar	39.9
CO₂	44.0
NO₂	46.0
SO₂	64.1

Worked Example 5 — Pressure calculation

$0.100\text{ mol}$0.100 mol He in a rigid $1.00\text{ dm}^3$1.00 dm3 container at $25.0\,^{\circ}\text{C}$25.0∘C.

• $p = nRT/V = (0.100)(8.314)(298)/(10^{-3}) = 247.76/10^{-3} = 2.478 \times 10^5\text{ Pa} =$p=nRT/V=(0.100)(8.314)(298)/(10−3)=247.76/10−3=2.478×105 Pa= $248\text{ kPa}$248 kPa

Worked Example 6 — Gas from a reaction

Volume of CO₂ at $101\,000\text{ Pa}$101000 Pa and $298\text{ K}$298 K from decomposition of $10.0\text{ g}$10.0 g CaCO₃.

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​

• $n(\text{CaCO}_3) = 10.0/100.1 = 0.0999\text{ mol}$n(CaCO3​)=10.0/100.1=0.0999 mol → $n(\text{CO}_2) = 0.0999\text{ mol}$n(CO2​)=0.0999 mol
• $V = nRT/p = (0.0999)(8.314)(298)/101\,000 = 247.51/101\,000 = 2.45 \times 10^{-3}\text{ m}^3 =$V=nRT/p=(0.0999)(8.314)(298)/101000=247.51/101000=2.45×10−3 m3= $2.45\text{ dm}^3$2.45 dm3

Worked Example 7 — Temperature

At what $T$T does $0.500\text{ mol}$0.500 mol of an ideal gas occupy $10.0\text{ dm}^3$10.0 dm3 at $150\text{ kPa}$150 kPa?

• $V = 0.0100\text{ m}^3$V=0.0100 m3; $p = 1.50 \times 10^5\text{ Pa}$p=1.50×105 Pa
• $T = pV/(nR) = (1.50 \times 10^5)(0.0100)/[(0.500)(8.314)] = 1500/4.157 =$T=pV/(nR)=(1.50×105)(0.0100)/[(0.500)(8.314)]=1500/4.157= $361\text{ K}$361 K ($88\,^{\circ}\text{C}$88∘C)

Worked Example 8 — Industrial-scale Haber-process feed

An ammonia plant produces $1000\text{ tonnes}$1000 tonnes of NH₃ per day. Calculate the volume of H₂ at $298\text{ K}$298 K and $2.00 \times 10^7\text{ Pa}$2.00×107 Pa (typical Haber-process pressure) required to produce this amount.

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})$N2​(g)+3H2​(g)→2NH3​(g)

• $n(\text{NH}_3) = 10^9/17.0 = 5.88 \times 10^7\text{ mol}$n(NH3​)=109/17.0=5.88×107 mol
• Ratio $3\text{H}_2 : 2\text{NH}_3$3H2​:2NH3​ → $n(\text{H}_2) = (3/2)(5.88 \times 10^7) = 8.82 \times 10^7\text{ mol}$n(H2​)=(3/2)(5.88×107)=8.82×107 mol
• $V = nRT/p = (8.82 \times 10^7)(8.314)(298)/(2.00 \times 10^7) = 2.186 \times 10^{11}/(2.00 \times 10^7) =$V=nRT/p=(8.82×107)(8.314)(298)/(2.00×107)=2.186×1011/(2.00×107)= $1.09 \times 10^4\text{ m}^3$1.09×104 m3 per day at process pressure.

Worked Example 9 — Two-point gas law

A fixed amount of gas has $V = 500\text{ cm}^3$V=500 cm3 at $27\,^{\circ}\text{C}$27∘C and $100\text{ kPa}$100 kPa. Find $V$V at $327\,^{\circ}\text{C}$327∘C and $200\text{ kPa}$200 kPa.

$\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$T1​p1​V1​​=T2​p2​V2​​ (since $n$n is fixed)

• $T_1 = 300\text{ K}$T1​=300 K; $T_2 = 600\text{ K}$T2​=600 K
• $V_2 = p_1V_1T_2/(T_1p_2) = (100)(500)(600)/[(300)(200)] = 3.0 \times 10^7/(6.0 \times 10^4) =$V2​=p1​V1​T2​/(T1​p2​)=(100)(500)(600)/[(300)(200)]=3.0×107/(6.0×104)= $500\text{ cm}^3$500 cm3

Doubling both $T$T and $p$p leaves $V$V unchanged.

Worked Example 10 — Vapour-density method for $M_r$Mr​

A volatile organic compound is vaporised in a sealed bulb at $373\text{ K}$373 K and $101\,000\text{ Pa}$101000 Pa. The bulb of internal volume $250\text{ cm}^3$250 cm3 contains $0.625\text{ g}$0.625 g of the vapour. Calculate the molar mass of the compound.

• $V = 2.50 \times 10^{-4}\text{ m}^3$V=2.50×10−4 m3
• $n = pV/RT = (101\,000)(2.50 \times 10^{-4})/[(8.314)(373)] = 25.25/3101.1 = 8.143 \times 10^{-3}\text{ mol}$n=pV/RT=(101000)(2.50×10−4)/[(8.314)(373)]=25.25/3101.1=8.143×10−3 mol
• $M = m/n = 0.625/8.143 \times 10^{-3} =$M=m/n=0.625/8.143×10−3= $76.7\text{ g mol}^{-1}$76.7 g mol−1

This matches reasonably well with CS₂ ($M_r = 76.1$Mr​=76.1) or CH₃CHO + H₂O if there's residual humidity. Vapour-density measurements were the historical method for determining the molar mass of volatile organic compounds before mass spectrometry became routine.

Worked Example 11 — Combustion-of-organic gas stoichiometry

What volume of CO₂ (at $298\text{ K}$298 K, $101\,000\text{ Pa}$101000 Pa) is produced by the complete combustion of $2.00\text{ dm}^3$2.00 dm3 of methane gas at the same conditions?

$\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O(l)}$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l)

By Avogadro's law (equal volumes contain equal moles at constant $T, p$T,p), the volume of CO₂ produced equals the volume of CH₄ consumed (1:1 in the equation). So $V(\text{CO}_2) = 2.00\text{ dm}^3$V(CO2​)=2.00 dm3.

This is the volume-ratio shortcut: for gas-phase reactions at the same $T$T and $p$p, mole ratios from the balanced equation can be applied directly to volume ratios. No moles calculation is needed if all species are gases at the stated conditions.

Worked Example 12 — Non-STP conditions and gas mixture

A sample of $0.500\text{ mol}$0.500 mol N₂(g) and $0.300\text{ mol}$0.300 mol He(g) is placed in a $5.00\text{ dm}^3$5.00 dm3 container at $50\,^{\circ}\text{C}$50∘C. Calculate the total pressure.

• $n_\text{total} = 0.800\text{ mol}$ntotal​=0.800 mol
• $V = 5.00 \times 10^{-3}\text{ m}^3$V=5.00×10−3 m3; $T = 323\text{ K}$T=323 K
• $p = nRT/V = (0.800)(8.314)(323)/(5.00 \times 10^{-3}) = 2148.4/0.005 = 4.30 \times 10^5\text{ Pa}$p=nRT/V=(0.800)(8.314)(323)/(5.00×10−3)=2148.4/0.005=4.30×105 Pa

In SI units, $p =$p= $4.30 \times 10^5\text{ Pa}$4.30×105 Pa (430 kPa).

By Dalton's law of partial pressures, this total pressure decomposes into:

• $p(\text{N}_2) = (0.500/0.800) \times 4.30 \times 10^5 = 2.69 \times 10^5\text{ Pa}$p(N2​)=(0.500/0.800)×4.30×105=2.69×105 Pa
• $p(\text{He}) = (0.300/0.800) \times 4.30 \times 10^5 = 1.61 \times 10^5\text{ Pa}$p(He)=(0.300/0.800)×4.30×105=1.61×105 Pa

Each partial pressure is the pressure that species would exert alone in the same container at the same temperature. Partial-pressure arithmetic is the basis of $K_p$Kp​ in Module 5 equilibrium work.

Worked Example 13 — Gas-phase synthesis stoichiometry

In the Haber process, $1.00\text{ kg}$1.00 kg of nitrogen reacts with hydrogen at $773\text{ K}$773 K and $200\text{ atm}$200 atm ($2.02 \times 10^7\text{ Pa}$2.02×107 Pa). What volume of ammonia (at the same conditions, assuming 100 % conversion) is produced?

$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​

• $n(\text{N}_2) = 1000/28.0 = 35.71\text{ mol}$n(N2​)=1000/28.0=35.71 mol
• $n(\text{NH}_3) = 2 \times 35.71 = 71.43\text{ mol}$n(NH3​)=2×35.71=71.43 mol
• $V = nRT/p = (71.43)(8.314)(773)/(2.02 \times 10^7) = 459\,180/(2.02 \times 10^7) =$V=nRT/p=(71.43)(8.314)(773)/(2.02×107)=459180/(2.02×107)= $2.27 \times 10^{-2}\text{ m}^3$2.27×10−2 m3 (22.7 L)

The high industrial pressure compresses the same moles into a much smaller volume than would be needed at atmospheric pressure (about 4530 L) — the engineering trade-off driving the Haber-process design.

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The Ideal Gas Model — Assumptions and Limitations

An ideal gas is a theoretical construct satisfying:

1. Gas particles have negligible volume relative to the container.
2. No intermolecular forces act between particles.
3. All collisions are perfectly elastic (no kinetic energy lost).
4. Particles are in constant random motion.
5. Time between collisions is much greater than the duration of a collision.

Real gases deviate from ideal behaviour most at:

• High pressure — particle volume becomes a significant fraction of container volume; intermolecular attractions pull particles closer together, reducing measured pressure below the ideal-gas prediction.
• Low temperature — intermolecular forces become comparable to kinetic energies; the gas approaches liquefaction.

Gases behave most ideally at low pressure and high temperature. Helium and neon come closest to ideal because they are small, non-polar and have only very weak London dispersion forces. H₂O(g) and NH₃ deviate significantly because of hydrogen bonding.

Beyond the ideal model — van der Waals

The 1873 van der Waals equation corrects for particle volume ($b$b) and intermolecular attraction ($a$a):