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4.1 Gases: ideal, real and pV = nRT

Syllabus
9701–2028–2029
Topic
4.1
Level
AS

Gas pressure comes from particles colliding with container walls

Gas molecules move continuously and collide with the container walls. A molecule's momentum changes during a wall collision; the wall exerts a force on the molecule and the molecule exerts an equal and opposite force on the wall.

pressure=forcearea\mathrm{pressure=\frac{force}{area}}

The enormous number of collisions produces a steady average force. Pressure is this average force per unit wall area, so more frequent collisions or a greater momentum change per collision increase the pressure.

At constant temperature, decreasing the container volume shortens the average distance to a wall. Collisions with the walls become more frequent, so the pressure rises even though the number of molecules is unchanged.

Molecule–molecule collisions redistribute momentum, but gas pressure is measured from momentum transferred to the container wall. Pressure is a collective effect of many particles, not a property carried by one stationary molecule.

An ideal gas has zero particle volume and no intermolecular attraction

The ideal-gas model makes two required assumptions: gas particles have zero volume, and there are no intermolecular forces of attraction between them.

Ideal assumption Meaning in the model Why a real gas can deviate
zero particle volume the whole container volume is available for particle motion at high pressure, particles are close and their own volume is no longer negligible
no intermolecular attraction particles do not pull one another away from wall collisions at low temperature, attractions matter more relative to particle kinetic energy

Real gases approach ideal behaviour most closely at low pressure, where particles are far apart, and high temperature, where their kinetic energy makes attractions less significant.

An ideal gas is a simplifying model, not a special real substance. ‘Zero particle volume’ means volume is neglected in the model; it does not mean real molecules literally occupy no space.

Use pV = nRT with SI units before calculating Mᵣ

pV=nRTR=8.31 J K−1 mol−1\mathrm{pV=nRT}\qquad R=8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}

Symbol Quantity SI unit used with R = 8.31
p pressure Pa
V gas volume m³
n amount mol
T absolute temperature K

Convert before substitution: kPa × 1000 gives Pa; cm³ × 10⁻⁶ or dm³ × 10⁻³ gives m³; and T/K = temperature/°C + 273. Keep unrounded values through the calculation.

n=pVRTmolar mass=mn=mRTpV\mathrm{n=\frac{pV}{RT}\qquad molar\ mass=\frac{m}{n}=\frac{mRT}{pV}}

A 0.880 g gas sample occupies 500 cm³ at 100 kPa and 300 K. Using SI units, molar mass = (0.880 × 8.31 × 300) ÷ (100000 × 5.00 × 10⁻⁴) = 43.9 g mol⁻¹, so the numerical value of Mᵣ is 43.9.

Mᵣ is dimensionless, whereas molar mass has unit g mol⁻¹; their numerical values coincide when molar mass is expressed in g mol⁻¹. Never substitute °C, kPa, cm³ or dm³ directly with the SI value of R.