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9476 / 2027
The Gaseous State

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)

01

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.

Kinetic-theory assumptions for an ideal gas
AssumptionMeaning
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 understanding

Hydrogen and oxygen are at the same temperature. Which is the same for both: molecular speed or mean kinetic energy?

  1. Speed; the heavier O2 molecules carry more energy.
  2. Mean kinetic energy; the lighter H2 molecules move faster.
  3. Both, since the two gases share the same temperature.
Show the answer

Mean kinetic energy; the lighter H2 molecules move faster.

Temperature is proportional to mean kinetic energy, (1/2)mv2. With the same energy, a smaller mass means a higher speed.

02

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.

A qualitative isotherm. The balance depends on the gas and temperature; not every real gas follows this exact curve over every pressure range.

Check your understanding

A student says, "A real gas always has a lower pressure than an ideal gas." Why is this wrong?

  1. At high density, particle volume and repulsion can make the pressure higher instead.
  2. Real gas particles have no attractions, so their pressure equals the ideal value.
  3. At low pressure, attractions make real gases deviate more, so pressure rises higher.
Show the answer

At high density, particle volume and repulsion can make the pressure higher instead.

Attractions lower the pressure compared with ideal. At very high pressure the particles' own volume matters more, raising it. Say which effect dominates under the given conditions.

03

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.

Common conversions
Given quantityConvert for SI substitution
100 kPa100000 Pa
250 cm3250 × 10-6 m3 = 2.50 × 10-4 m3
1.00 dm31.00 × 10-3 m3
25.0 degrees C298.15 K; 298 K when consistent with the precision given

Worked exampleFind 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Convert V to 2.50 × 10-4 m3 and p to 1.00 × 105 Pa.
  2. n = pV/RT = (1.00 × 105 × 2.50 × 10-4)/(8.31 × 298) = 0.0101 mol.
  3. M = m/n = 0.444/0.0101, retaining unrounded n during calculation.
Answer

M = 44.0 g mol-1 to three significant figures. The corresponding relative molecular mass Mr is 44.0 and has no units.

Try it step by step

Work it out in steps: Volume in m3, then Amount of gas, n = pV/RT, then Molar mass, M = m/n.

  1. Problem 1. A 0.320 g sample of gas occupies 250 cm3 at 100 kPa and 300 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Find its molar mass in g mol-1 to 3 significant figures.
  2. Problem 2. A 0.580 g sample of gas occupies 240 cm3 at 100 kPa and 298 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Find its molar mass in g mol-1 to 3 significant figures.
  3. Problem 3. A 1.08 g sample of gas occupies 600 cm3 at 101 kPa and 298 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Give the volume in m3, then n and M to 3 significant figures.
Show the answer

Problem 1.

Volume in m3: V = 250 x 10-6 = 2.50 x 10-4 m3, and p = 1.00 x 105 Pa.

Amount of gas, n = pV/RT: n = (1.00 x 105 x 2.50 x 10-4) / (8.31 x 300) = 0.0100 mol.

Molar mass, M = m/n: M = 0.320 / 0.0100 = 31.9 g mol-1 (keep the unrounded n), which suggests O2.

Problem 2.

Volume in m3: V = 240 x 10-6 = 2.40 x 10-4 m3, and p = 1.00 x 105 Pa.

Amount of gas, n = pV/RT: n = (1.00 x 105 x 2.40 x 10-4) / (8.31 x 298) = 0.00969 mol.

Molar mass, M = m/n: M = 0.580 / 0.00969 = 59.8 g mol-1.

Problem 3.

Volume in m3: V = 600 x 10-6 = 0.000600 m3, and p = 101 000 Pa.

Amount of gas, n = pV/RT: n = (101 000 x 0.000600) / (8.31 x 298) = 0.0245 mol.

Molar mass, M = m/n: M = 1.08 / 0.0245 = 44.1 g mol-1, which suggests CO2 (44.0).

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 understanding

A sealed rigid container of gas is warmed from 20 degrees C to 40 degrees C. Does the pressure double?

  1. No. The pressure stays the same, because the rigid container keeps the gas volume fixed.
  2. No. It rises 313/293 times, about 1.07, because p follows kelvin temperature.
  3. Yes. The temperature has doubled, so the pressure doubles as well.
Show the answer

No. It rises 313/293 times, about 1.07, because p follows kelvin temperature.

At fixed n and V, p is proportional to T in kelvin. 20 degrees C is 293 K and 40 degrees C is 313 K.

04

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 exampleSplit 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Total amount = 0.500 mol; mole fractions are 0.600 and 0.400.
  2. p(N2) = 0.600 × 250 = 150 kPa.
  3. p(O2) = 0.400 × 250 = 100 kPa; their sum checks against 250 kPa.
Answer

150 kPa nitrogen and 100 kPa oxygen. Use mole fractions, not mass fractions.

Worked exampleGas 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.

Try it first. Then show the worked stepsHide the worked steps
  1. p(dry gas) = 105 - 3.17 = 101.83 kPa.
  2. Use p = 101830 Pa and V = 185 × 10-6 m3.
  3. n = 101830 × 185 × 10-6/(8.31 × 298).
Answer

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.

Check your understanding

A mixture of 0.100 mol helium and 0.300 mol argon has a total pressure of 200 kPa. What is the partial pressure of helium?

  1. 50 kPa, from its mole fraction of 0.25
  2. 20 kPa, from 0.100 x 200 kPa
  3. 100 kPa, half the total for each gas
Show the answer

50 kPa, from its mole fraction of 0.25

x(He) = 0.100 / 0.400 = 0.250. p(He) = x(He) x p(total) = 0.250 x 200 = 50 kPa.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Gas calculations: choose the correct pressure and units
SituationRelationship or decision
Single ideal gaspV = nRT; kelvin temperature is essential.
Molar massM = mRT/pV.
Mixturepi = xi ptotal; use mole fractions.
Collected over waterSubtract water-vapour pressure before calculating dry-gas amount.
Most nearly idealLow pressure, high temperature.
DeviationAttractions and finite particle volume can have different effects.

Scope

Learning outcomes

3. The Gaseous State. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 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

    Pressure comes from collisions with the container

  2. 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

    Low pressure and high temperature favour ideal behaviour

  3. 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

    Convert the units before substituting

  4. 3(d) Use Dalton's law in gas mixtures.

    • Partial and total pressure
    • Mole fractions
    • Connection to gaseous equilibria

    Each gas contributes its own partial pressure