The ideal-gas equation of state

Measure pV/(nT) for any dilute gas and plot it against p. Each gas gives a slightly different curve, and every curve lands on the same intercept as p → 0: R = 8.314 J mol⁻¹ K⁻¹. That limit is the ideal gas, and pV = nRT is its equation of state, a single relation among the three variables that fix a state of a given amount, so only two are free. Check the dimensions: Pa × m³ = N m⁻² × m³ = J, so R is an energy per mole per kelvin, the first sign that temperature measures energy per molecule, which Topic 06.05 makes exact. The number also fixes the molar volume. At 0 °C and 101.325 kPa, V/n = RT/p = 8.314 × 273.15/101 325 = 2.241 × 10⁻² m³ = 22.41 L, and 22.71 L at the 1 bar standard. Boyle's, Charles's and Avogadro's laws are what the one equation says when two variables are held fixed: the equation of state is the parent, not the sum.

Chemistry counts in moles, kinetic theory counts molecules and the balance reads kilograms, and the equation of state must accept all three. Dividing by Avogadro's number, n = N/N_A, turns pV = nRT into pV = N k_B T with k_B = R/N_A = 1.381 × 10⁻²³ J K⁻¹, the same law written per molecule. Read it as a number density: N/V = p/(k_B T) = 101 325/(1.381 × 10⁻²³ × 300) = 2.45 × 10²⁵ m⁻³ at 300 K and 1 atm, so each molecule owns (V/N)¹⁄³ = 3.4 nm of room, about nine nitrogen diameters, which is the picture behind treating molecules as points. From a mass, n = m/M with the molar mass in kg mol⁻¹: 0.0290 for air, 0.0280 for N₂, 0.0020 for H₂. Eliminating n gives the density form ρ = pM/(RT): air at 20 °C and 101.325 kPa has ρ = 101 325 × 0.02897/(8.314 × 293.15) = 1.20 kg m⁻³. Which gas you are holding enters through M alone.

A process is a path on the surface p = nRT/V, and three paths recur. Hold T: pV = constant, a rectangular hyperbola on the p–V diagram, with hotter isotherms further from the origin. Hold p: V/T = constant, a horizontal line on p–V and a straight line through the origin on V–T. Hold V: p/T = constant, a vertical line on p–V and a straight line through the origin on p–T with slope nR/V. Take 0.0802 mol at state A (100 kPa, 2.00 L, 300 K). Isothermal compression to 0.500 L: p = 400 kPa, state B. Isochoric heating to 450 K: p = 400 × 1.5 = 600 kPa, state C. Isobaric expansion back to 2.00 L: T = 450 × 4 = 1800 K, state D. Now compare D with A directly: same volume, six times the pressure, so six times the temperature, 1800 K. The equation of state returns the same answer by either route because it describes states, not routes, and that comparison is the check every multi-step problem should end with.

The isochore on the p–T diagram is p = (nR/V) T, a line through the origin, and that origin is T = 0 on the absolute scale. Rewrite it in Celsius, p = (nR/V)(θ + 273.15), and the line acquires an intercept: p is proportional to T but only affine in θ. Ratios therefore work in kelvin and fail in Celsius. Heat sealed air at fixed volume from 20 °C to 40 °C and the pressure rises by 313.15/293.15 = 1.068, a 6.8% increase, not the doubling that 40/20 suggests. Differences are safe in either scale, since a rise of 20 K is a rise of 20 °C, so the derivative (∂p/∂T)_V needs no conversion while the ratio p₂/p₁ always does. Extrapolating a measured isochore to p = 0 is how the gas thermometer of Topic 06.01 located −273.15 °C in the first place; here the same line is read the other way, as the reason the equation of state is written with T and nothing else.

With two free variables, every rate of change needs a statement of what is held fixed, which is what the subscript on a partial derivative says. From V = nRT/p: (∂V/∂T)ₚ = nR/p = V/T, so the isobaric expansion coefficient is β = (1/V)(∂V/∂T)ₚ = 1/T, and (∂V/∂p)_T = −nRT/p² = −V/p, so the isothermal compressibility is κ = −(1/V)(∂V/∂p)_T = 1/p. At 300 K, β = 3.33 × 10⁻³ K⁻¹, about ninety times that of steel; at 1 atm, κ = 9.87 × 10⁻⁶ Pa⁻¹, twenty thousand times that of water. The third response, (∂p/∂T)_V = nR/V = p/T, equals β/κ, as it must for any equation of state. For air at 300 K and 101.3 kPa it is 338 Pa K⁻¹, and because nR/V is constant in a rigid sealed vessel the rise is exact, Δp = (p/T) ΔT, not a first-order estimate. Where Topic 06.02 asked what stress a blocked expansion produces, the ideal gas answers from its equation of state alone.

pV = nRT neglects two things, and each has a size you can compute. Molecules occupy volume: for nitrogen the excluded volume is b ≈ 3.9 × 10⁻⁵ m³ mol⁻¹, against a molar volume of 2.24 × 10⁻² m³ mol⁻¹ at STP, a fraction b/Vₘ = 0.17%. Molecules attract: the van der Waals a ≈ 0.137 Pa m⁶ mol⁻² makes the attraction energy per mole a/Vₘ = 6.1 J mol⁻¹, against RT = 2271 J mol⁻¹, a ratio of 0.27%. Both sit below the precision of most laboratory gauges, which is why the model works. Compress to 100 atm and Vₘ falls a hundredfold: the ratios become 17% and 27%, of opposite sign and no longer negligible, and the fact that they partly cancel is luck, not a virtue of the model. Cool the gas and the attraction ratio grows as 1/T until the molecules condense, a state the equation has no word for. The ideal gas is exact in the limit p → 0 and an approximation everywhere else; Topic 06.09 supplies the corrections.