1.5 Ideal gases

An ideal gas is a simplified particle model. Use its assumptions as a checklist before deciding whether PV=nRT is a suitable description of a real sample.

Use the assumptions to make predictions: compressing a gas until particle volume is no longer negligible weakens the model, while raising temperature usually reduces the relative importance of attractions. Ideal particles still move and collide; only the collisions are treated as elastic.

Gas pressure arises from particle collisions with the container walls and the associated momentum transfer. At a higher Kelvin temperature, the greater average kinetic energy established in Structure 1.1.3 changes the collision behaviour; this does not replace the separate assumptions that particle volume and intermolecular attractions are negligible.

Real gases deviate most from ideal behaviour at low temperature and high pressure. Low temperature reduces particle kinetic energy, while high pressure brings particles close together.

A strong explanation names the condition, identifies the failed ideal assumption, and links it to the observed deviation.

Diagnose the cause from the condition. Cooling makes attractive forces more important because particle kinetic energy is lower; strong compression exposes both attractions and finite particle volume. Name the failed ideal assumption rather than stating only that the gas is 'non-ideal'.

Molar volume is the volume occupied by one mole of gas at a specified temperature and pressure. At STP as used by the current IB data context (273.15 K and 100 kPa), Vₘ is approximately 22.7 dm³ mol⁻¹; a different condition requires its own value.

Read pressure–volume graphs as an inverse relationship, not simply as one quantity increasing while the other decreases. At STP, use the stated molar volume with the balanced-equation mole ratio.

Attach conditions to every molar volume. Once Vₘ is valid for the stated temperature and pressure, convert gas volume to moles, apply the balanced-equation ratio, then convert back if needed. Do not carry one tabulated Vₘ into a different set of conditions.

Worked example — amount and molar mass from gas data

The local course book gives $P=101.3\,\mathrm{kPa}$, $V=1.91\,\mathrm{dm^3}$, $m=3.30\,\mathrm{g}$ and $T=150\,^\circ\mathrm{C}=423.15\,\mathrm{K}$. Because $\mathrm{kPa\,dm^3=J}$, use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$: $n=PV/(RT)=(101.3\times1.91)/(8.31\times423.15)=0.0550\,\mathrm{mol}$. Then $M=m/n=3.30\,\mathrm{g}/0.0550\,\mathrm{mol}=60.0\,\mathrm{g\,mol^{-1}}$. The final value is the mass of one mole of the vaporized compound under the stated ideal-gas model.

Convert Celsius to kelvin before substitution, and make pressure and volume units consistent with the chosen gas constant. Rearrange the equation only after the known quantities and units are identified.

Before solving, write a unit line beside P, V and T. With R = 8.31 J mol⁻¹ K⁻¹, use pressure in Pa, volume in m³ and temperature in K; using kPa with dm³ is also consistent because kPa·dm³ equals J. Judge model suitability before trusting the numerical result.