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1.5 Ideal gases

Syllabus
First assessment 2025
Topic
1.5
Level
SL

Learning objectives

1.5.1Ideal gas model assumptions• Moving particles, negligible volume, no intermolecular forces• Elastic collisions1.5.2Real gases• Deviate from ideal at low temperature and high pressure1.5.3Molar volume• Constant at specific T and P• Analyse temperature, pressure, and volume graphs for a fixed mass of gas1.5.4Ideal gas equation• PV = nRT• Combined gas law: P₁V₁/T₁ = P₂V₂/T₂• Solve ideal gas problems with correct pressure, volume, and temperature units

The Ideal Gas Model

Assumption Ideal-gas statement
Particle motion Particles move continuously
Particle volume Particle volume is negligible compared with the gas volume
Intermolecular forces Forces between particles are negligible
Collisions Collisions are elastic

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.

When Real Gases Deviate

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.

pressure is vertical and volume is horizontal; the ideal-gas curve passes through (V,p) and (2V,0.5p); the real-gas curve deviates from the inverse ideal relationship; solid and dashed curve labels are unambiguous.
Ideal assumption that fails Real-gas consequence
Intermolecular forces are negligible Attractions matter when particles have low kinetic energy and are close
Particle volume is negligible Finite molecular volume matters at very high pressure

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'.

Explaining Real-Gas Behaviour

2 marks

Outline why the volume occupied by propane(g) at very high pressure is higher than the value calculated using PV=nRT.

Molar Volume and Gas Relationships

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.

Forfixedamountandtemperature:P∝1/VFor fixed amount and temperature: P ∝ 1/V

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.

Fixed amount of gas; held constant Relationship Graph/interpretation check
temperature P ∝ 1/V P–V is inverse, not a straight decreasing line
pressure V ∝ T V–T is linear only with T in kelvin
volume P ∝ T P–T is linear only with T in kelvin

State the fixed variable and use absolute temperature before interpreting a gas graph.

Applying Molar Volume

1 mark

Deduce the relationship between the pressure and volume of the sample of carbon dioxide gas.

Using the Ideal Gas Equation

PV=nRTPV = nRT

P1V1/T1=P2V2/T2P₁V₁/T₁ = P₂V₂/T₂

Worked example — amount and molar mass from gas data

The local course book gives P=101.3 kPaP=101.3\,\mathrm{kPa}, V=1.91 dm3V=1.91\,\mathrm{dm^3}, m=3.30 gm=3.30\,\mathrm{g} and T=150 ∘C=423.15 KT=150\,^\circ\mathrm{C}=423.15\,\mathrm{K}. Because kPa dm3=J\mathrm{kPa\,dm^3=J}, use R=8.31 J mol−1 K−1R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}: n=PV/(RT)=(101.3×1.91)/(8.31×423.15)=0.0550 moln=PV/(RT)=(101.3\times1.91)/(8.31\times423.15)=0.0550\,\mathrm{mol}. Then M=m/n=3.30 g/0.0550 mol=60.0 g mol−1M=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.

Solving Ideal-Gas Problems

2 marks

0.108 g of the vaporized compound was found to have a volume of 55.7 cm355.7 \mathrm{~cm}^{3} at 100∘C100^{\circ} \mathrm{C} and a pressure of 1.00×105 Pa1.00 \times 10^{5} \mathrm{~Pa}.
Calculate the amount, in moles, of the compound. Use sections 1, 2 and 4 of the data booklet.

Ideal Gases Summary

Retrieve the model: ideal particles have negligible volume and forces with elastic collisions; low temperature and high pressure expose real-gas limits; molar volume and PV=nRT then connect amount, pressure, volume, and temperature.

Before calculating, check whether the question uses STP molar volume or PV=nRT, identify the fixed conditions, convert temperature to kelvin, and align pressure and volume units.