The Ideal Gas Equation

The three individual gas laws — Boyle's, Charles's, and the pressure law — each describe what happens when one variable changes while the others are held constant. The ideal gas equation combines all three into a single, powerful equation that describes the behaviour of an ideal gas under any conditions.

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The Molar Form: pV = nRT

The most commonly used form of the ideal gas equation is:

pV = nRT

where:

• p = pressure (Pa)
• V = volume (m³)
• n = number of moles of gas (mol)
• R = molar gas constant = 8.31 J mol⁻¹ K⁻¹
• T = absolute temperature (K)

This equation tells you that for a given amount of gas, the product pV is proportional to the absolute temperature. If you know any three of the four variables (p, V, n, T), you can calculate the fourth.

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The Molecular Form: pV = NkT

An alternative form uses the total number of molecules rather than moles:

pV = NkT

where:

• N = total number of molecules
• k = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹

The Boltzmann constant is the gas constant per molecule, while R is the gas constant per mole.

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Connecting the Two Forms

The link between the molar and molecular forms involves Avogadro's number (Nₐ):

Nₐ = 6.02 × 10²³ mol⁻¹

This is the number of particles (atoms or molecules) in one mole of any substance.

The relationships are:

• N = nNₐ (total molecules = moles × Avogadro's number)
• k = R/Nₐ (Boltzmann constant = molar gas constant ÷ Avogadro's number)

You can verify: k = 8.31 / (6.02 × 10²³) = 1.38 × 10⁻²³ J K⁻¹ ✓

graph LR
    A["pV = nRT\n(molar form)"] -->|"N = nNₐ\nk = R/Nₐ"| B["pV = NkT\n(molecular form)"]
    C["n = m/M"] --> A
    D["N = nNₐ"] --> B
    E["Constants"] --> F["R = 8.31 J mol⁻¹ K⁻¹"]
    E --> G["k = 1.38 × 10⁻²³ J K⁻¹"]
    E --> H["Nₐ = 6.02 × 10²³ mol⁻¹"]

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Key Constants Reference Table

Constant	Symbol	Value	Unit
Molar gas constant	R	8.31	J mol⁻¹ K⁻¹
Boltzmann constant	k	1.38 × 10⁻²³	J K⁻¹
Avogadro constant	Nₐ	6.02 × 10²³	mol⁻¹
Standard atmospheric pressure	p₀	1.01 × 10⁵	Pa
Molar volume at STP	V_m	22.4 × 10⁻³	m³ mol⁻¹

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Molar Mass and Number of Moles

The molar mass (M) of a substance is the mass of one mole, measured in kg mol⁻¹ (or g mol⁻¹ in chemistry contexts, but physics uses SI units).

n = m/M

where:

• n = number of moles
• m = mass of gas (kg)
• M = molar mass (kg mol⁻¹)

Gas	Formula	Molar mass M / kg mol⁻¹	Molar mass / g mol⁻¹
Hydrogen	H₂	0.002	2
Helium	He	0.004	4
Nitrogen	N₂	0.028	28
Oxygen	O₂	0.032	32
Neon	Ne	0.020	20
Argon	Ar	0.040	40
Carbon dioxide	CO₂	0.044	44
Water vapour	H₂O	0.018	18

For example, the molar mass of oxygen (O₂) is 32 g mol⁻¹ = 0.032 kg mol⁻¹. If you have 0.064 kg of oxygen:

n = 0.064 / 0.032 = 2.0 mol

You can also find the number of molecules: N = nNₐ = 2.0 × 6.02 × 10²³ = 1.20 × 10²⁴

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Unit Conversions

Getting the units right is essential. The ideal gas equation works in SI units:

Quantity	SI unit	Common conversions
Pressure (p)	Pa (= N m⁻²)	1 atm = 1.01 × 10⁵ Pa; 1 kPa = 10³ Pa
Volume (V)	m³	1 litre = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³; 1 mL = 1 cm³
Temperature (T)	K	T(K) = θ(°C) + 273
Mass (m)	kg	1 g = 10⁻³ kg

A very common error is to use volume in litres or cm³. Always convert to m³.

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Worked Example 1: Molar Volume at STP

Calculate the volume occupied by 1.0 mol of gas at standard conditions (p = 1.01 × 10⁵ Pa, T = 273 K).

pV = nRT

V = nRT/p = (1.0 × 8.31 × 273) / (1.01 × 10⁵)

V = 2268.6 / (1.01 × 10⁵) = 0.0225 m³ = 22.5 litres

This is the molar volume at STP — approximately 22.4 litres per mole. This value appears frequently in chemistry and physics. It applies to any ideal gas regardless of its molecular mass — an extraordinary result showing that equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules (Avogadro's law).

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Worked Example 2: Finding Volume from Mass

A cylinder contains 0.080 kg of helium (M = 0.004 kg mol⁻¹) at 27 °C. The pressure is 3.0 × 10⁵ Pa. Calculate the volume of the gas.

Step 1: Find the number of moles. n = m/M = 0.080/0.004 = 20 mol

Step 2: Convert temperature to kelvin. T = 27 + 273 = 300 K

Step 3: Use pV = nRT. V = nRT/p = (20 × 8.31 × 300) / (3.0 × 10⁵) V = 49 860 / (3.0 × 10⁵) = 0.166 m³ = 166 litres

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Worked Example 3: Molecular Form

A container of volume 0.020 m³ holds gas at 400 K and a pressure of 5.0 × 10⁵ Pa. How many molecules are in the container?

pV = NkT

N = pV/(kT) = (5.0 × 10⁵ × 0.020) / (1.38 × 10⁻²³ × 400)

N = 10 000 / (5.52 × 10⁻²¹) = 1.81 × 10²⁴ molecules

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Worked Example 4: Density of a Gas

Calculate the density of nitrogen gas (N₂, M = 0.028 kg mol⁻¹) at 25 °C and 1.01 × 10⁵ Pa.

From pV = nRT and n = m/M:

pV = (m/M)RT → m/V = pM/(RT) → ρ = pM/(RT)

ρ = (1.01 × 10⁵ × 0.028) / (8.31 × 298)

ρ = 2828 / 2476.4 = 1.14 kg m⁻³

This is a useful derived formula: ρ = pM/(RT). It shows that gas density increases with pressure and molar mass but decreases with temperature.

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Worked Example 5: Pumping Down a Container

A rigid container of volume 0.50 m³ initially contains air at 1.01 × 10⁵ Pa and 20 °C. A vacuum pump removes gas until the pressure drops to 100 Pa. How many moles of air remain? How many molecules?

T = 20 + 273 = 293 K

n = pV/(RT) = (100 × 0.50) / (8.31 × 293) = 50 / 2434.8 = 0.0205 mol ≈ 0.021 mol

N = nNₐ = 0.0205 × 6.02 × 10²³ = 1.24 × 10²² molecules

Even at this low pressure (about 1/1000 of atmospheric), the container still holds over 10²² molecules — showing how many molecules are in even a "near vacuum."

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What is an Ideal Gas?

An ideal gas is a theoretical gas that perfectly obeys pV = nRT at all pressures and temperatures. The assumptions behind this model are:

1. The gas consists of a very large number of identical molecules in random motion.
2. The volume of the molecules themselves is negligible compared to the volume of the container.
3. Collisions between molecules (and with the walls) are perfectly elastic.
4. There are no intermolecular forces between the molecules (except during collisions).
5. The time of a collision is negligible compared to the time between collisions.

Real gases approximate ideal behaviour at low pressures and high temperatures — conditions where the molecules are far apart and moving fast, so their own volume and the intermolecular forces are negligible.

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Common Exam Mistakes

Mistake	Consequence	Correction
Volume in litres or cm³	Answer wrong by factor of 10³ or 10⁶	1 L = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³
Temperature in °C	Completely wrong answer	T(K) = θ(°C) + 273
Molar mass in g mol⁻¹	Answer wrong by factor of 10³	Convert to kg mol⁻¹
Confusing n (moles) and N (molecules)	Wrong by factor of ~10²³	n = N/Nₐ; check which form you need
Using R when you need k (or vice versa)	Wrong by factor of ~10²³	R goes with n (moles); k goes with N (molecules)

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Summary

• pV = nRT: n in moles, R = 8.31 J mol⁻¹ K⁻¹.
• pV = NkT: N in molecules, k = 1.38 × 10⁻²³ J K⁻¹.
• Nₐ = 6.02 × 10²³ mol⁻¹ links moles to molecules: N = nNₐ, k = R/Nₐ.
• n = m/M links mass to moles; ρ = pM/(RT) gives gas density.
• All values must be in SI units (Pa, m³, K, kg mol⁻¹).
• Ideal gas behaviour is best approximated at low pressure and high temperature.

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A-Level Deep Dive: Ideal Gas Equation

Spec mapping

Edexcel 9PH0 specification Topic 9 — Thermodynamics covers the ideal gas equation $pV = nRT$pV=nRT alongside the alternative molecular form $pV = NkT$pV=NkT, where students are expected to use these relations together with the gas laws (Boyle's law, Charles' law and the pressure law) and the kinetic-theory model (refer to the official specification document for exact wording). Topic 9 follows Topic 5 (materials) and feeds forward into Topic 13 (oscillations) where energy distributions reappear, and into Topic 14 (astrophysics) where stellar atmospheres are modelled as ideal gases at extremes of pressure and temperature. The Edexcel 9PH0 formula booklet provides $pV = NkT$pV=NkT and the molar gas constant $R$R along with Avogadro's number $N_A$NA​ and the Boltzmann constant $k$k, so candidates need not memorise the constants — but they must know which equation pairs with which constant, and they must convert between mass, moles and number of molecules without prompting.

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Worked example with full mark scheme

Question (8 marks):

A sealed rigid cylinder of volume $2.50 \times 10^{-3}$2.50×10−3 m³ contains nitrogen gas (molar mass $M = 28.0 \times 10^{-3}$M=28.0×10−3 kg mol⁻¹) at a pressure of $1.20 \times 10^{5}$1.20×105 Pa and temperature $295$295 K.

(a) Calculate the amount of gas in the cylinder, in moles. (3)

(b) The cylinder is heated until the pressure rises to $1.80 \times 10^{5}$1.80×105 Pa. The volume of the cylinder is unchanged. Calculate the new temperature of the gas. (3)

(c) Calculate the mass of nitrogen in the cylinder. (2)

Solution with mark scheme:

(a) Step 1 — rearrange the ideal gas equation.

From $pV = nRT$pV=nRT, rearrange for $n$n:

$n = \dfrac{pV}{RT}$n=RTpV​

M1 — correct rearrangement of the ideal gas equation in terms of $n$n. Common error: writing $n = pV \cdot RT$n=pV⋅RT or $n = RT/(pV)$n=RT/(pV). Either inversion loses both M1 and the substitution mark.

Step 2 — substitute SI values.

$n = \dfrac{(1.20 \times 10^{5})(2.50 \times 10^{-3})}{(8.31)(295)}$n=(8.31)(295)(1.20×105)(2.50×10−3)​

M1 — correct substitution with all quantities in SI units (Pa, m³, K, J mol⁻¹ K⁻¹). Examiners deduct here when candidates leave volume in cm³ or temperature in °C.

Step 3 — evaluate.

Numerator: $1.20 \times 10^{5} \times 2.50 \times 10^{-3} = 300$1.20×105×2.50×10−3=300 Pa m³ (i.e. 300 J).

Denominator: $8.31 \times 295 = 2451.45$8.31×295=2451.45 J mol⁻¹.

$n = \dfrac{300}{2451.45} = 0.1224 \approx 0.122 \text{ mol}$n=2451.45300​=0.1224≈0.122 mol

A1 — correct numerical answer to an appropriate number of significant figures (3 s.f. matches data).

(b) Step 1 — choose the correct law.

Constant $V$V, constant $n$n implies $\dfrac{p}{T} = \text{constant}$Tp​=constant (the pressure law, a special case of the combined gas law).

M1 — recognising that $V$V and $n$n are fixed and applying $p_1/T_1 = p_2/T_2$p1​/T1​=p2​/T2​.

Step 2 — substitute.

$T_2 = T_1 \cdot \dfrac{p_2}{p_1} = 295 \cdot \dfrac{1.80 \times 10^{5}}{1.20 \times 10^{5}}$T2​=T1​⋅p1​p2​​=295⋅1.20×1051.80×105​

M1 — correct substitution with temperatures in kelvin (not celsius — this is where most marks are lost).

Step 3 — evaluate.

$T_2 = 295 \cdot 1.50 = 442.5 \approx 443 \text{ K}$T2​=295⋅1.50=442.5≈443 K

A1 — correct value with units. Equivalent answer $170$170 °C is acceptable only if the candidate states the conversion explicitly.

(c) Step 1 — relate moles to mass.

$m = nM$m=nM, where $M$M is the molar mass.

M1 — using the relation $m = nM$m=nM rather than confusing molar mass with the relative molecular mass.

Step 2 — evaluate.

$m = 0.1224 \times 28.0 \times 10^{-3} = 3.43 \times 10^{-3} \text{ kg} = 3.43 \text{ g}$m=0.1224×28.0×10−3=3.43×10−3 kg=3.43 g

A1 — correct mass with units.

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A weather balloon contains $120$120 moles of helium at ground level, where the pressure is $1.01 \times 10^{5}$1.01×105 Pa and the temperature is $288$288 K. The balloon rises until the external pressure is $2.50 \times 10^{4}$2.50×104 Pa and the temperature is $220$220 K. Assume the helium remains in equilibrium with the surroundings.

(a) Calculate the volume of the balloon at ground level. (2)

(b) Calculate the volume of the balloon at altitude. (2)

(c) Comment on the assumption that helium behaves as an ideal gas under these conditions. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO2) — correct rearrangement and substitution into $pV = nRT$pV=nRT giving $V_1 = nRT_1/p_1$V1​=nRT1​/p1​.
• A1 (AO2) — $V_1 = (120)(8.31)(288)/(1.01 \times 10^{5}) = 2.84$V1​=(120)(8.31)(288)/(1.01×105)=2.84 m³ (3 s.f.).

(b)

• M1 (AO2) — recognising $n$n is constant and using $p_1V_1/T_1 = p_2V_2/T_2$p1​V1​/T1​=p2​V2​/T2​ (or recomputing $V_2 = nRT_2/p_2$V2​=nRT2​/p2​). Either method earns the M1.
• A1 (AO2) — $V_2 = 8.77$V2​=8.77 m³ (3 s.f.).

(c)

• M1 (AO3) — identifies that helium is monatomic with weak intermolecular forces and small molecular volume, conditions under which the ideal gas approximation is excellent.
• A1 (AO3) — extends the comment: at $220$220 K and $2.5 \times 10^{4}$2.5×104 Pa the gas remains far from its boiling point ($4.2$4.2 K), so the ideal gas equation is a strong fit.

Total: 6 marks split AO2 = 4, AO3 = 2. Paper 2 frequently couples a numerical pV calculation with a short evaluative comment — the AO3 marks reward physics judgement, not arithmetic.

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Synoptic links

Connects to:

1. Kinetic theory of gases (Topic 9): the ideal gas equation $pV = nRT$pV=nRT is identical in form to $pV = \tfrac{1}{3}Nm\langle c^2 \rangle$pV=31​Nm⟨c2⟩ derived from kinetic theory, allowing direct identification $\tfrac{1}{2}m\langle c^2 \rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT. The mean translational kinetic energy of a gas particle is $\tfrac{3}{2}kT$23​kT — temperature is, mechanically, a measure of mean kinetic energy.

2. Boyle's, Charles' and pressure laws (Topic 9): each historical gas law is a special case of $pV = nRT$pV=nRT. Boyle's law ($pV = \text{const}$pV=const) holds at fixed $n, T$n,T; Charles' law ($V/T = \text{const}$V/T=const) at fixed $n, p$n,p; the pressure law ($p/T = \text{const}$p/T=const) at fixed $n, V$n,V. Recognising which constraint applies in a problem is the AO2 reasoning Edexcel rewards.