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B.3

Gas laws

Theme B · The particulate nature of matter · SL and HL

🎯What you need to be able to do

  • Use the mole, molar mass and the Avogadro constant.
  • Apply the ideal gas law \( pV = nRT \), and the combined form \( p_1V_1/T_1 = p_2V_2/T_2 \).
  • Explain gas pressure in terms of molecular collisions.
  • Use the kinetic theory results, including that average kinetic energy is proportional to absolute temperature.
  • State the assumptions of an ideal gas, and say when a real gas stops behaving like one.

📚The physics

Counting molecules. One mole contains \( N_A = 6.02 \times 10^{23} \) particles. The number of moles is \( n = N/N_A \), and also \( n = m/M \) where \(M\) is the molar mass. Most gas-law errors are not physics errors at all — they are someone putting a mass in grams where the formula wants moles.

The ideal gas law ties everything together:

\( pV = nRT \)
\( pV = NkT \)

with \( R = 8.31 \) J mol\(^{-1}\) K\(^{-1}\) and the Boltzmann constant \( k = R/N_A \). Temperature must be in kelvin — always, without exception. Pressure in pascals, volume in cubic metres.

The familiar named laws are just special cases, which is worth seeing rather than memorising. Hold \(T\) constant and \(pV\) is constant (Boyle). Hold \(p\) constant and \( V \propto T \) (Charles). Hold \(V\) constant and \( p \propto T \) (Gay-Lussac). If you can rearrange \( pV = nRT \) you never need to recall which name goes with which.

Where pressure comes from. Molecules collide with the container walls; each collision reverses the molecule’s momentum component perpendicular to the wall, and by Newton’s second law a rate of change of momentum is a force. Pressure is that force per unit area, averaged over an enormous number of collisions. Raising the temperature makes molecules faster, so collisions are both harder and more frequent, and pressure rises. Reducing the volume makes collisions more frequent for the same speeds, and pressure rises again. Both results fall out of the same picture.

Kinetic theory makes that quantitative:

\[ pV = \tfrac{1}{3}Nm\overline{c^{2}} \]

where \( \overline{c^{2}} \) is the mean square speed. Comparing this with \( pV = NkT \) gives the result that matters most:

\[ E_k = \tfrac{3}{2}kT \]

Temperature is average molecular kinetic energy, in disguise. Note that this depends only on temperature, not on which gas — at the same temperature, helium and xenon molecules have the same average kinetic energy, so the heavier xenon molecules must be moving more slowly.

The ideal gas assumptions, which you should be able to list: the molecules are identical, small compared with the separation between them, and in constant random motion; collisions between molecules and with the walls are perfectly elastic; the duration of a collision is negligible compared with the time between collisions; and there are no intermolecular forces except during collisions.

When real gases stop cooperating follows directly from those assumptions. At high pressure the molecules are close together, so their own volume is no longer negligible. At low temperature they move slowly enough for intermolecular attractions to matter. Push far enough in either direction and the gas condenses — something an ideal gas can never do, because an ideal gas has no attractive forces to condense with.

✏️Worked example

A sealed steel canister holds 0.12 m\(^{3}\) of nitrogen at \( 2.4 \times 10^{5} \) Pa and 17 °C. Take \( R = 8.31 \) J mol\(^{-1}\) K\(^{-1}\) and the molar mass of N\(_2\) as 28 g mol\(^{-1}\).

(a) How many moles? Convert the temperature first: \( T = 17 + 273 = 290 \) K. Then

\[ n = \frac{pV}{RT} = \frac{2.4 \times 10^{5} \times 0.12}{8.31 \times 290} = \frac{28800}{2410} = 12\ \text{mol} \]

(b) What mass of gas? \( m = nM = 12 \times 0.028 = 0.33 \) kg. Note the molar mass converted to kilograms.

(c) The canister is left in the sun and warms to 45 °C. What is the new pressure? The canister is sealed and rigid, so \(n\) and \(V\) are fixed and \( p \propto T \). With \( T_2 = 318 \) K, \( p_2 = 2.4 \times 10^{5} \times 318/290 = 2.6 \times 10^{5} \) Pa.

The trap. Using 17 and 45 instead of 290 and 318 gives a predicted pressure increase of a factor of 2.6 rather than 1.1 — an answer wrong by more than a factor of two, from a units slip rather than a physics one. This is the single most common error in the topic.

🔭See it happen

PhET, Gases Intro. Shrink the container at fixed temperature and watch the pressure gauge climb while the speed distribution stays put — that is Boyle’s law with the mechanism visible. Then heat at fixed volume and watch the distribution shift right instead.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the number of moles in 0.050 m\(^{3}\) of gas at \( 1.0 \times 10^{5} \) Pa and 300 K. Take \( R = 8.31 \) J mol\(^{-1}\) K\(^{-1}\).
\( n = \dfrac{pV}{RT} = \dfrac{1.0 \times 10^{5} \times 0.050}{8.31 \times 300} = \dfrac{5000}{2493} = 2.0 \) mol.
2. A gas occupying 250 cm\(^{3}\) at \( 1.0 \times 10^{5} \) Pa is compressed to 100 cm\(^{3}\) at constant temperature. Find the new pressure.
At constant \(T\), \( pV \) is constant (Boyle): \( p_2 = 1.0 \times 10^{5} \times 250/100 = 2.5 \times 10^{5} \) Pa. The volumes can stay in cm\(^{3}\) because only their ratio matters.
3. 2.0 m\(^{3}\) of gas at 27 °C is heated to 127 °C at constant pressure. Find the new volume.
Convert to kelvin first: 300 K and 400 K. At constant \(p\), \( V \propto T \), so \( V_2 = 2.0 \times 400/300 = 2.7 \) m\(^{3}\). Using 27 and 127 would give 9.4 m\(^{3}\) — wrong by more than a factor of three.
4. Find the average translational kinetic energy of a molecule at 27 °C. Take \( k = 1.38 \times 10^{-23} \) J K\(^{-1}\).
\( E_k = \tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.2 \times 10^{-21} \) J. Note this depends only on temperature, not on which gas.
5. Helium and xenon are held at the same temperature. Compare the average kinetic energies and the average speeds of their molecules.
The average kinetic energies are equal, because \( E_k = \tfrac{3}{2}kT \) depends only on temperature. Since \( E_k = \tfrac{1}{2}m\overline{c^{2}} \) and xenon molecules are far heavier, xenon must have the smaller mean square speed. The ratio of root-mean-square speeds is \( \sqrt{M_{\text{Xe}}/M_{\text{He}}} \).
6. State two conditions under which a real gas stops behaving like an ideal gas, and explain each in terms of the ideal gas assumptions.
At high pressure the molecules are close together, so their own volume is no longer negligible compared with the volume of the container — breaking the assumption that molecules are small compared with their separation. At low temperature the molecules move slowly enough for intermolecular attractive forces to have a significant effect — breaking the assumption that there are no forces except during collisions. Push far enough either way and the gas condenses, which an ideal gas can never do.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Hypertextbook — gas laws and kinetic theory
  • HyperPhysics — ideal gas law and kinetic theory of gases