S1.5 Ideal gasesIB Chemistry SL: Revision notes
Section 1
The ideal gas model
An ideal gas is a model with these assumptions:
- the particles are in constant random motion;
- the particles have negligible volume compared with the container;
- there are no intermolecular forces between particles;
- all collisions are elastic: no kinetic energy is lost overall.
The pressure of a gas comes from particles colliding with the container walls. The average kinetic energy of the particles is proportional to the absolute temperature (kelvin).
Section 2
Real gases and the limits of the model
Real gases deviate from ideal behaviour, most at low temperature and high pressure.
- At high pressure, particles are close together, so their own volume is no longer negligible and attractions between them matter.
- At low temperature, particles move slowly, so intermolecular attractions have a bigger effect and the gas may condense.
Gases with stronger intermolecular forces (polar molecules, hydrogen bonding, many electrons) and larger molecules deviate most. Helium and hydrogen are the closest to ideal. Gases are most ideal at high temperature and low pressure.
In a data question, pV/nRT below 1 means attractions dominate; above 1 means the volume of the particles dominates.
Section 3
Gas laws for a fixed mass of gas
Temperatures must be in kelvin: T/K = t/°C + 273.
- Boyle's law (constant T): p ∝ 1/V. A plot of p against V is a curve; p against 1/V is a straight line through the origin.
- Charles's law (constant p): V ∝ T. V against T in K is a straight line through the origin; against °C it meets the axis at −273 °C.
- Gay-Lussac's law (constant V): p ∝ T.
The combined gas law: p₁V₁/T₁ = p₂V₂/T₂.
Using temperatures in °C in gas-law calculations. Always convert to kelvin first.
Section 4
The ideal gas equation
pV = nRT, where R = 8.31 J K⁻¹ mol⁻¹ (data booklet). Use SI units: p in Pa, V in m³, T in K.
Unit conversions: 1 kPa = 10³ Pa; 1 dm³ = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³. (Using kPa with dm³ also works, because the two factors of 10³ cancel.)
Since n = m/M, you can find the molar mass of a volatile liquid: M = mRT/(pV).
Write down the converted values of p, V and T before substituting; most lost marks are unit errors.
Section 5
Molar volume
At a given temperature and pressure, one mole of any ideal gas occupies the same volume, the molar volume. At 273 K and 100 kPa (STP) it is 22.7 dm³ mol⁻¹ (data booklet). So n = V ÷ 22.7 for a gas at STP (V in dm³).
The molar volume changes with conditions: it is larger at higher temperature or lower pressure. At other conditions, use pV = nRT instead.
Must know
- Ideal gas: negligible particle volume, no intermolecular forces, elastic collisions.
- Most ideal at high T and low p; deviations biggest at low T and high p.
- Always use kelvin; p₁V₁/T₁ = p₂V₂/T₂.
- pV = nRT in Pa, m³, K; M = mRT/pV.
- Molar volume at STP (273 K, 100 kPa) = 22.7 dm³ mol⁻¹.
That's the notes covered.
Carry on to the next subtopic.