The Ideal-Gas Law

Mechanics hands you Newton's laws; a gas hands you one extra fact, and it has to be measured. Each of the three classic experiments pins two variables and watches a third, so each one on its own is only a fragment: p ∝ 1/V at fixed n and T, p ∝ T at fixed n and V, p ∝ n at fixed V and T. The single relation that collapses to all three fragments at once is p ∝ n T/V, and turning a proportionality into an equation costs exactly one constant. The surprise is which constant. It is the same number for hydrogen, argon and carbon dioxide alike, R = 8.314 J mol⁻¹ K⁻¹, so a table of R values would have one row in it — the identity of the gas has dropped out of the bookkeeping entirely. Audit the units before trusting any of that. Pressure in pascals is N m⁻², so p V is N m⁻² × m³ = N m, a joule, and R must therefore carry J mol⁻¹ K⁻¹, which is exactly how it is quoted. That audit doubles as the warning: the SI set p in Pa, V in m³, n in mol, T in K stands or falls together, and slipping litres into the volume slot while R still expects cubic metres makes every pressure you compute a thousand times too small.

Why kelvin, and not the scale on the wall? Put a fixed amount of dilute gas in a rigid bulb and plot its pressure against the Celsius temperature. The points lie on a straight line, and extrapolating that line down to p = 0 meets the axis at −273.15 °C — the same point whether the bulb holds helium, nitrogen or air, and whatever amount of gas you started with. That common intercept is what the law is pointing at: p is proportional not to θ but to θ + 273.15. Define T = θ + 273.15 and the proportionality becomes exact, which is the whole content of the kelvin scale for this topic. The practical consequence is that every ratio must be taken in kelvin. Heating that bulb from 0 °C to 100 °C raises the pressure by a factor 373.15/273.15 = 1.366, and a Celsius ratio would have put a zero in the denominator.

n counts moles, and one mole is N_A = 6.022 × 10²³ molecules, so N = n N_A converts between the chemist's variable and the physicist's. Substitute n = N/N_A into the law and it becomes p V = N k_B T, with the per-molecule constant k_B = R/N_A = 1.381 × 10⁻²³ J K⁻¹ — the same statement counted in molecules rather than moles. Read either form carefully and notice what is missing: nothing in it names the gas. No molar mass, no molecular size, no chemical identity appears. One litre at 300 K and 1.013 × 10⁵ Pa holds n = 0.0406 mol, or 2.45 × 10²² molecules, and that count is the same for helium as for carbon dioxide. What is not the same is the mass: 0.163 g of helium against 1.79 g of carbon dioxide, a factor of eleven. Pressure is set by how many molecules there are and how hot they are; mass enters only when you ask what the sample weighs.

Three processes carry most of the problems. Isothermal: T fixed, so p V = constant and the state slides along a hyperbola on a p–V diagram — halve the volume and the pressure doubles. Isobaric: p fixed, usually because a free piston or the atmosphere sets it, so V/T = constant and V against T is a straight line through the origin. Isochoric: V fixed by a rigid wall, so p/T = constant. All three are cases of one relation, p₁V₁/T₁ = p₂V₂/T₂, which holds between any two states of a sample whose n has not changed, and is usually faster than finding n first. The isothermal case is also where the law goes under an integral: the work done by the gas is ∫p dV from V₁ to V₂, and substituting p = n R T/V at fixed T gives W = n R T ln(V₂/V₁). Compressing 1.00 mol at 300 K from 24.0 L to 12.0 L gives W = 8.314 × 300 × ln(0.500) = −1.73 × 10³ J, so 1.73 kJ has to be done on the gas.

Two pieces of bookkeeping cost more marks than the physics does. First, p in the law is absolute pressure, while most instruments read gauge pressure, the excess over the surroundings. A tyre gauge showing 2.20 × 10⁵ Pa on a day when the atmosphere is 1.01 × 10⁵ Pa means the gas is at 3.21 × 10⁵ Pa. Drive until the air warms from 15 °C to 45 °C at fixed volume and p₂ = 3.21 × 10⁵ × (318.15/288.15) = 3.54 × 10⁵ Pa, so the gauge now reads 2.53 × 10⁵ Pa. Running the same ratio on the gauge readings alone predicts 2.43 × 10⁵ Pa, and the whole error is the missing atmosphere. Second, check that n really is constant. It is not if the vessel leaks, if the gas dissolves in a liquid, if a tube is open to the room, or if a liquid in the container is evaporating into the space above it. Declare the system and look for ways gas can get in or out before writing p₁V₁/T₁ = p₂V₂/T₂.

The cleanest test of the model is Z = p V/(n R T), which the ideal law pins at exactly 1. Measure Z for real gases and it sits close to 1 while the gas is dilute and drifts far from it as the gas nears condensation. The two repairs sit in the van der Waals equation, (p + a n²/V²)(V − n b) = n R T. The term n b subtracts the volume the molecules themselves occupy, which pushes the pressure up; the term a n²/V² accounts for the attraction that slows a molecule as it approaches the wall, which pulls the pressure down. Both corrections are judged against the molar volume, so the questions are whether Vₘ is large compared with b and whether p is large compared with a/Vₘ². For air at 1.01 × 10⁵ Pa and 293 K the molar volume is Vₘ = R T/p = 2.41 × 10⁻² m³ mol⁻¹, and air's measured constants are a = 0.136 Pa m⁶ mol⁻² and b = 3.64 × 10⁻⁵ m³ mol⁻¹, so b/Vₘ = 0.15% and a/Vₘ² = 234 Pa, which is 0.23% of the pressure. The two errors partly cancel and the ideal law is good to about a tenth of a percent — which is why it serves so well in a lecture room and so badly inside a gas cylinder.