Full chapter
The Gaseous State
Know when the ideal-gas model works, calculate gas quantities and separate the pressures in a mixture.
A-Level 9476 (2026-2027)
Pressure comes from collisions with the container
An ideal gas is a model with negligible particle volume and no intermolecular forces.
Gas particles move continually and randomly. When they collide with a wall, their momentum changes; the resulting force per unit area is the gas pressure. Increasing the number of particles per volume or their typical speed increases the rate or force of those impacts.
| Assumption | Meaning |
|---|---|
| Particles have negligible volume. | Their own volume is tiny compared with the container volume. |
| No intermolecular attractions or repulsions. | Particles move freely between collisions. |
| Continuous random straight-line motion between collisions. | There is no preferred direction in a gas at equilibrium. |
| Collisions are perfectly elastic. | Total kinetic energy is conserved in collisions; individual particles can exchange energy. |
| Mean kinetic energy depends on absolute temperature. | At the same temperature, different ideal gases have the same mean translational kinetic energy, not the same mean speed. |
Temperature must be measured in kelvin because the model relates mean kinetic energy to absolute temperature. At the same temperature lighter molecules typically move faster than heavier molecules. Heating a fixed amount in a rigid sealed container increases pressure because the particles strike the walls harder and more frequently.
Check your understandingHydrogen and oxygen are at the same temperature. Which quantity is equal: particle speed or mean kinetic energy?Think it through, then reveal the answer
Low pressure and high temperature favour ideal behaviour
Give both reasons: particle size matters less, and attractions matter less.
At low pressure the particles are far apart, so their own volume is a small fraction of the total and intermolecular attractions are less important. At high temperature their kinetic energies are large compared with the energies associated with attraction. These conditions make the ideal assumptions more reasonable.
At low temperature, attraction can significantly reduce the momentum transferred to the walls and can eventually cause condensation. At very high pressure, the particles are crowded: their finite volume reduces the free space available and short-range repulsion becomes important. Both ideal assumptions fail; a real gas is not just an ideal gas with slower particles.
Two different reasons for non-ideal behaviour
At fixed temperature a schematic real-gas curve of Z = pV/nRT begins near the ideal line Z = 1 at low pressure, dips below it when attractions dominate, and rises above it when finite particle volume and repulsion become important at high pressure. The curve is illustrative rather than data for a named gas.
Check your understandingWhy is "real-gas pressure is always lower than ideal pressure" wrong?Think it through, then reveal the answer
Convert the units before substituting
With R = 8.31 J mol-1 K-1, pressure is in Pa and volume in m3.
The ideal-gas equation is pV = nRT. It relates pressure p, gas volume V, amount n and absolute temperature T. With R = 8.31 J mol-1 K-1, use Pa and m3, since Pa m3 = J. A value of R with different units requires the matching pressure and volume units.
| Given quantity | Convert for SI substitution |
|---|---|
| 100 kPa | 100000 Pa |
| 250 cm3 | 250 × 10-6 m3 = 2.50 × 10-4 m3 |
| 1.00 dm3 | 1.00 × 10-3 m3 |
| 25.0 degrees C | 298.15 K; 298 K when consistent with the precision given |
Worked example
Find a gas molar mass
A 0.444 g sample occupies 250 cm3 at 100 kPa and 298 K. Treat it as ideal. Find its molar mass.
- Convert V to 2.50 × 10-4 m3 and p to 1.00 × 105 Pa.
- n = pV/RT = (1.00 × 105 × 2.50 × 10-4)/(8.31 × 298) = 0.0101 mol.
- M = m/n = 0.444/0.0101, retaining unrounded n during calculation.
M = 44.0 g mol-1 to three significant figures. The corresponding relative molecular mass Mr is 44.0 and has no units.
Combining n = m/M with the gas equation gives M = mRT/pV. Alternatively, with density ρ = m/V, M = ρRT/p. If density is in kg m-3 with SI R, the resulting molar mass is in kg mol-1; multiply by 1000 to obtain g mol-1.
A fixed gas sample also obeys p1V1/T1 = p2V2/T2, provided the amount is unchanged and the ideal approximation is suitable. Do not use a memorised room-temperature molar volume when the stated pressure or temperature differs.
Check your understandingA sealed rigid container is warmed from 20 to 40 degrees C. Does its gas pressure double?Think it through, then reveal the answer
Each gas contributes its own partial pressure
For an ideal mixture, partial pressure equals mole fraction times total pressure.
Dalton's law states that the total pressure of a mixture of non-reacting ideal gases is the sum of their partial pressures. The partial pressure of one component is the pressure it would exert alone at the same temperature and total volume. Since pi = niRT/V, pi = xiptotal, where xi = ni/ntotal.
Worked example
Split the total pressure
A mixture contains 0.300 mol N2 and 0.200 mol O2 at a total pressure of 250 kPa. Find the partial pressures.
- Total amount = 0.500 mol; mole fractions are 0.600 and 0.400.
- p(N2) = 0.600 × 250 = 150 kPa.
- p(O2) = 0.400 × 250 = 100 kPa; their sum checks against 250 kPa.
150 kPa nitrogen and 100 kPa oxygen. Use mole fractions, not mass fractions.
Worked example
Gas collected over water is not dry gas
185 cm3 of gas is collected over water at 298 K. Total pressure is 105 kPa and water-vapour pressure is 3.17 kPa. Find the amount of dry gas, assuming negligible dissolution.
- p(dry gas) = 105 - 3.17 = 101.83 kPa.
- Use p = 101830 Pa and V = 185 × 10-6 m3.
- n = 101830 × 185 × 10-6/(8.31 × 298).
n = 7.61 × 10-3 mol. Using the full pressure would incorrectly count water vapour as the collected gas.
In a practical determination, check for leaks, allow the sample to reach the measured temperature, and correct any pressure difference from unequal liquid levels. Water collection is unsuitable when appreciable gas dissolves or reacts with water. For a volatile-liquid molar-mass experiment, incomplete vaporisation, air remaining in the vessel or loss of vapour can invalidate the assumed gas amount.
Partial pressures also enter Kp expressions in equilibrium calculations. Adding an inert gas at fixed volume raises the total pressure but does not change the partial pressures of the existing ideal-gas components: each still has the same ni, T and V.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Situation | Relationship or decision |
|---|---|
| Single ideal gas | pV = nRT; kelvin temperature is essential. |
| Molar mass | M = mRT/pV. |
| Mixture | pi = xi ptotal; use mole fractions. |
| Collected over water | Subtract water-vapour pressure before calculating dry-gas amount. |
| Most nearly ideal | Low pressure, high temperature. |
| Deviation | Attractions and finite particle volume can have different effects. |
Scope and references
Learning outcomes and sources
3. The Gaseous State. Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
3(a) State the ideal-gas kinetic-theory assumptions.
- Negligible particle volume
- No intermolecular forces
- Random motion and elastic collisions
- Mean kinetic energy and absolute temperature
3(b) Explain when the ideal approximation succeeds or fails.
- (i) Low pressure and high temperature
- (ii) Very high pressure and very low temperature
- Particle size and intermolecular forces
3(c) Use pV = nRT, including relative molecular mass.
- Consistent pressure, volume, temperature and R units
- Amount, molar mass and density calculations
- Distinguish molar mass from dimensionless Mr
3(d) Use Dalton's law in gas mixtures.
- Partial and total pressure
- Mole fractions
- Connection to gaseous equilibria
- SEAB H2 Chemistry 9476, examination 2026
Topic 3, printed page 15; all four outcomes and both parts of 3(b) inspected. Calculations and qualitative graph are original.