Ideal gases — OCR A-Level Physics

The kinetic theory model treats a gas as a large number of identical point molecules in continuous random motion. Their volume is negligible compared with the container, and collisions between molecules and with the walls are perfectly elastic. Between collisions no forces act, except during contact. Pressure arises from the rate of change of momentum when molecules collide with the walls. For example, doubling the absolute temperature of a fixed volume of gas increases the mean square speed of its molecules, so they strike the walls more often and harder, raising the pressure. This model explains the macroscopic gas laws in terms of molecular behaviour.

Pressure in the kinetic theory model is the force per unit area exerted by molecular collisions with the container walls. Each collision changes a molecule's momentum; the rate of change of momentum gives the force. Summing over many molecules and dividing by the wall area gives pressure. For a gas in a box, the pressure depends on the number of molecules, their mass, the mean square speed and the volume: p = (1/3)ρ⟨c²⟩, where ρ is density and ⟨c²⟩ is mean square speed. Since ⟨c²⟩ is proportional to absolute temperature, this links microscopic motion to measurable pressure.

This guided reading supports the ideal gas section. Read the specification statements for 5.1.4 carefully, noting that (d) introduces the equation of state pV = nRT and the Boltzmann constant. Use the textbook or revision guide to identify how the kinetic theory model leads to the ideal gas equation. Work through worked examples that convert between p, V, n, R and T, paying attention to units: p in Pa, V in m³, T in K, n in mol, R = 8.31 J mol⁻¹ K⁻¹. Check that you can rearrange the equation and use it with the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹.

The equation of state for an ideal gas is pV = nRT, where p is pressure in pascals, V is volume in cubic metres, n is the number of moles, R is the molar gas constant (8.31 J mol⁻¹ K⁻¹) and T is thermodynamic temperature in kelvin. It combines Boyle's law, Charles's law and Avogadro's law. For a fixed mass of gas, pV/T is constant. For example, compressing 2.0 mol of gas at 300 K into 0.050 m³ gives p = nRT/V = (2.0 × 8.31 × 300) / 0.050 = 1.0 × 10⁵ Pa. The equation assumes ideal behaviour: negligible molecular volume and no intermolecular forces.

This practical investigates two gas relationships. For Boyle’s law, trap a fixed mass of air in a sealed syringe or tube, vary its volume by moving the piston, and record the pressure with a pressure sensor while keeping temperature constant. Plot p against 1/V; a straight line through the origin supports pV = constant. For the pressure–temperature relationship, heat a sealed rigid container of fixed volume in a water bath, record pressure and temperature in kelvin, and plot p against T. Extrapolating to p = 0 estimates absolute zero. Control variables: fixed mass of gas, no leaks, and thermal equilibrium before readings. Repeat readings and average to reduce random error. Assessed by planning, data handling, graph plotting and uncertainty evaluation.

Absolute zero is estimated by measuring the pressure of a fixed mass of gas at constant volume as its temperature is varied. Temperatures must be converted to kelvin. A graph of pressure p against thermodynamic temperature T is a straight line; extrapolating it to p = 0 gives the intercept on the T-axis, which is absolute zero (0 K or −273 °C). Alternatively, plot p against temperature in degrees Celsius; the intercept on the temperature axis is −273 °C. The method assumes the gas behaves ideally and does not liquefy. In a multiple-choice question, you may be asked to identify the correct graph, the value of absolute zero, or the effect of a systematic error such as a leak.

The kinetic theory equation pV = ⅓ N m c² relates the macroscopic pressure and volume of an ideal gas to the microscopic motion of its particles. Here N is the number of particles (atoms or molecules), m is the mass of one particle, and c² is the mean square speed of the particles. The equation assumes a large number of identical particles in random motion, with negligible volume and no intermolecular forces except during elastic collisions. It can be used to calculate pressure, volume, number of particles, or mean square speed. In multiple-choice questions, you may be asked to rearrange the equation, identify the meaning of each symbol, or calculate a quantity given the others.

The mean square speed c² is the average of the squares of the speeds of all particles in a gas. The root mean square (r.m.s.) speed is the square root of the mean square speed: c_rms = √(c²). It is a representative speed for the particles, but it is not the same as the average speed. The r.m.s. speed is used in the kinetic theory equation pV = ⅓ N m c². In multiple-choice questions, you may be asked to calculate c² from given speeds, find c_rms, or compare r.m.s. speed with mean speed. Remember that c² is always positive and c_rms is also positive.

The Boltzmann constant k relates the macroscopic molar gas constant R to the number of particles in one mole, the Avogadro constant NA. Because R = 8.31 J mol⁻¹ K⁻¹ applies to a mole of particles, dividing by NA = 6.02 × 10²³ mol⁻¹ gives the gas constant per single particle: k = R / NA ≈ 1.38 × 10⁻²³ J K⁻¹. This lets you write gas equations per molecule rather than per mole, for example pV = NkT, where N counts molecules. Check units: J mol⁻¹ K⁻¹ divided by mol⁻¹ leaves J K⁻¹, the unit of k. In calculations, use k when the question gives a number of molecules or atoms, and R when it gives an amount in moles.

These two equations describe an ideal gas at the particle level. pV = NkT relates pressure p, volume V, number of molecules N, Boltzmann constant k and thermodynamic temperature T; it is the per-particle form of pV = nRT. The second, ½mc² = 3/2 kT, gives the mean translational kinetic energy of a molecule of mass m and mean-square speed c². The factor 3/2 arises because a molecule can move in three independent directions. Combining them shows that temperature alone fixes mean kinetic energy, so at a given T all gas molecules have the same average translational energy regardless of mass, while lighter molecules have higher mean-square speeds.

The internal energy of an ideal gas is the total kinetic energy of its molecules, because ideal-gas molecules are modelled as having no intermolecular potential energy. For N molecules each with mean translational kinetic energy 3/2 kT, the internal energy is U = N × 3/2 kT = 3/2 NkT. Since NkT = pV, this can also be written U = 3/2 pV. Internal energy therefore depends only on the temperature (or on the product pV) and on the number of molecules, not on the type of gas. Doubling the thermodynamic temperature of a fixed amount of ideal gas doubles its internal energy.