TheIBAxolotl
The ideal gas model
An ideal gas is a model built on simplifying assumptions:
- Particles are in constant, random, straight-line motion.
- The volume of the particles themselves is negligible compared with the container volume.
- There are no intermolecular forces between particles (except during collisions).
- Collisions are perfectly elastic: total kinetic energy is conserved.
- Average kinetic energy is proportional to absolute temperature.
Real versus ideal behaviour
Real gases deviate when the assumptions break down:
- High pressure: particles are crowded, so their own volume is no longer negligible.
- Low temperature: particles move slowly, so attractive forces between them become significant (and the gas may condense).
Real gases behave most ideally at high temperature and low pressure. Small, non-polar particles with weak intermolecular forces (He, H\(_2\)) are closest to ideal; polar gases such as NH\(_3\) deviate more.
Gas laws
For a fixed amount of gas:
- Boyle: \(p \propto 1/V\) at constant T
- Charles: \(V \propto T\) at constant p
- Gay-Lussac: \(p \propto T\) at constant V
These combine into the combined gas law:
\[\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\]
T must be in kelvin; p and V can be in any units as long as both sides match.
Worked example (combined gas law)
A balloon holds 2.00 dm\(^3\) of gas at 20 °C and 100 kPa. It rises to where the pressure is 50.0 kPa and the temperature is −30 °C. Find the new volume.
\[V_2 = \frac{p_1 V_1 T_2}{p_2 T_1} = \frac{100 \times 2.00 \times 243}{50.0 \times 293} = 3.32\ \text{dm}^3\]
The pressure halving doubles the volume, but cooling shrinks it a little, so 3.32 dm\(^3\) is sensible.
Molar volume
At STP (273 K and 100 kPa), one mole of any ideal gas occupies 22.7 dm\(^3\). So \(n = V / 22.7\) for gases at STP.
The ideal gas equation
\[pV = nRT\]
with \(R = 8.31\) J K\(^{-1}\) mol\(^{-1}\). In SI units: p in Pa, V in m\(^3\), T in K.
| Conversion | |
|---|---|
| 1 kPa | \(10^3\) Pa |
| 1 dm\(^3\) | \(10^{-3}\) m\(^3\) |
| 1 cm\(^3\) | \(10^{-6}\) m\(^3\) |
Worked example (molar mass of a gas)
A 0.500 g sample of gas occupies 250 cm\(^3\) at 25.0 °C and 101 kPa. Find its molar mass.
- \(p = 1.01 \times 10^{5}\) Pa, \(V = 2.50 \times 10^{-4}\) m\(^3\), \(T = 298\) K
- \(n = \dfrac{pV}{RT} = \dfrac{(1.01 \times 10^{5})(2.50 \times 10^{-4})}{8.31 \times 298} = 1.02 \times 10^{-2}\) mol
- \(M = \dfrac{m}{n} = \dfrac{0.500}{1.02 \times 10^{-2}} = 49.0\ \text{g mol}^{-1}\)
(Carrying the unrounded n, 0.010 20 mol, also gives 49.0; using T = 298.15 K would give 49.1, so keep T consistent with the data.)
You can also combine the steps: \(M = \dfrac{mRT}{pV}\).
Common mistakes
- Using °C instead of K: the most common lost mark in this topic.
- Using kPa with m\(^3\) (answer off by 1000) or dm\(^3\) with Pa.
- Saying real gases are ideal at high pressure; it is the opposite.
- Forgetting that STP here is 100 kPa, giving 22.7 dm\(^3\) mol\(^{-1}\), not the older 22.4.
Exam tips
- Write out the units of every quantity before substituting into \(pV = nRT\).
- A neat shortcut: kPa × dm\(^3\) also works with R = 8.31, because the two factors of 1000 cancel.
- When explaining deviation, link each condition to the specific assumption that fails.
- Sketch a \(pV\) against \(p\) or \(pV/nRT\) against \(p\) graph: an ideal gas gives a horizontal line.