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Derivation

The Ideal Gas Law from Kinetic Theory

D-149 Home PU-203 Threads matter · energy · force Depends on —
Statement

For a dilute gas of N identical point molecules of mass m confined to a rigid container of volume V, averaging the momentum delivered to the walls by elastic molecular collisions yields the equation of state PV = NkBT, provided one identifies the absolute temperature through ½m⟨v²⟩ = &frac32;kBT — i.e. the mean translational kinetic energy per molecule is &frac32;kBT.

Why it matters

This is the microscopic origin of the ideal gas law. It replaces the empirical constants of Boyle, Charles and Avogadro with a single mechanical picture: pressure is nothing but the time-averaged momentum flux carried by molecules striking a surface. Nothing about the gas being "ideal" is assumed at the outset — ideality emerges as the limit in which molecules are point-like and non-interacting.

Crucially, the derivation forces temperature to acquire a mechanical meaning. The quantity we measure with a thermometer turns out to be proportional to the mean kinetic energy of random molecular motion, which is the first quantitative bridge between thermodynamics and mechanics and the seed of the entire kinetic theory of matter.

Assumptions
Molecules are point particles with negligible volume.If dropped, the free volume available for motion is V − Nb rather than V, and the equation of state gains the van der Waals excluded-volume correction; pressure rises above the ideal value at high density. No intermolecular forces except during instantaneous, perfectly elastic collisions.If dropped, attractive tails reduce wall pressure (the a/V² term of van der Waals) and inelasticity would let translational energy leak into other modes, breaking the energy balance that fixes temperature. Collisions with the walls are elastic and specular, and the walls are rigid.If dropped, the wall could absorb or add momentum and net energy, so a steady mean pressure would not exist and the container would exchange work with the gas even in equilibrium. The velocity distribution is isotropic and stationary.If dropped, ⟨vx²⟩ = ⟨vy²⟩ = ⟨vz²⟩ fails, pressure becomes direction-dependent (a stress tensor rather than a scalar), and no single P can be defined. Molecular chaos: successive collisions are statistically uncorrelated (dilute limit).If dropped, one cannot replace the sum over molecules by N times a single-particle average, and correlations produce virial corrections to the pressure.
Derivation
1
Δpx = m vx − (−m vx) = 2 m vx
A single molecule with x-velocity vx striking the wall at x = L reverses its x-momentum while vy, vz are untouched; specular elastic collision. A
2
Δt = 2L / vx
Between two hits on the same wall the molecule crosses the cube of side L and returns; free flight (no interior collisions in the dilute limit). A
3
f1 = Δpx / Δt = (2 m vx) / (2L / vx) = m vx² / L
Newton's second law in impulse form: the time-averaged force from one molecule equals momentum delivered per collision divided by the interval between collisions. A
4
F = ∑i=1N m vx,i² / L = (m / L) ∑i vx,i²
Forces from independent molecules add; the wall feels the total. Valid because molecular chaos lets contributions be summed without correlation terms. B
5
∑i vx,i² = N ⟨vx²⟩
Definition of the ensemble mean: the sum of squares over N molecules is N times the average of vx². A
6
P = F / A = (m N ⟨vx²⟩ / L) / L² = m N ⟨vx²⟩ / L³ = m N ⟨vx²⟩ / V
Pressure is force per unit wall area A = L², and V = L³. A
7
⟨v²⟩ = ⟨vx²⟩ + ⟨vy²⟩ + ⟨vz²⟩ = 3⟨vx²⟩ ⇒ ⟨vx²⟩ = ⅓⟨v²⟩
Isotropy of the equilibrium velocity distribution makes the three Cartesian averages equal. This is where the stationary, isotropic-distribution assumption enters. B
8
P V = ⅓ N m ⟨v²⟩
Substituting step 7 into step 6 and multiplying by V gives the central kinetic result — still pure mechanics, no temperature yet. B
9
P V = ⅔ N · (½ m ⟨v²⟩) = ⅔ N ⟨εtr⟩
Rewrite in terms of the mean translational kinetic energy ⟨εtr⟩ = ½m⟨v²⟩ per molecule — algebraic regrouping. A
10
½ m ⟨v²⟩ ≡ &frac32; kB T  ⇒  PV = N kB T
The temperature is defined (equivalently, delivered by the equipartition theorem: ½kBT per quadratic degree of freedom, three translational here) so that the mechanical result matches the empirical gas law. This identification is what makes T a state variable. C
Result
PV = NkBT   with   ½m⟨v²⟩ = &frac32;kBT

Reading. The pressure a gas exerts is the accumulated recoil of countless molecular impacts: PV equals two-thirds of the total translational kinetic energy of the gas. Temperature is not a substance but a measure of that random kinetic energy — heating a gas literally means speeding up its molecules. The Boltzmann constant kB = 1.381×10−23 J/K is the conversion factor between the kelvin scale and joules of molecular energy. Writing N = nNA recovers the chemist's form PV = nRT with R = NAkB.

Units check. [PV] = Pa·m³ = (N·m−2)(m³) = N·m = J. [NkBT] = (1)(J·K−1)(K) = J. Both sides are energies. Likewise [½m⟨v²⟩] = kg·(m/s)² = J = [&frac32;kBT]. Consistent.

Limiting cases
  • Fixed T, vary V (Boyle): P ∝ 1/V since NkBT is constant.
  • Fixed P, vary T (Charles): V ∝ T; extrapolating to V → 0 defines absolute zero.
  • T → 0: ⟨v²⟩ → 0 classically (molecules freeze); in reality quantum statistics and condensation intervene long before.
  • N → 0 (vacuum limit): P → 0; the law is exact precisely where the gas is most dilute.
  • Mixture of gases: each species contributes NikBT/V independently → Dalton's law of partial pressures.
Breaks when
  • High density / high pressure. Molecular volume and attractive forces are no longer negligible; the excluded volume and the pressure-lowering attraction give the van der Waals equation (P + aN²/V²)(V − Nb) = NkBT. Near the critical point the ideal law fails badly.
  • Low temperature / high density (quantum regime). When the thermal de Broglie wavelength λth = h/√(2πmkBT) becomes comparable to the interparticle spacing (V/N)1/3, Maxwell–Boltzmann statistics break down and Bose–Einstein or Fermi–Dirac corrections dominate (degeneracy pressure, condensation).
  • Very low density / large mean free path. When the mean free path exceeds the container size (Knudsen regime), molecules bounce wall-to-wall without intermolecular collisions; a local thermodynamic pressure is ill-defined and continuum reasoning fails.
  • Internal structure at high T. Once kBT excites rotation, vibration, dissociation or ionization, energy no longer resides only in translation; the equation of state still holds but the energy–temperature relation and heat capacities change.
Failure modes
  • Confusing ⟨v⟩ with √⟨v²⟩. The mean speed and the root-mean-square speed differ (⟨v⟩ = √(8/3π)·vrms ≈ 0.921 vrms); only vrms enters the pressure.
  • Using 2mv for the momentum change but forgetting the Δt. Students compute impulse per collision but drop the collision rate, losing a factor of vx and getting P ∝ vx instead of vx².
  • Forgetting the factor ⅓. Writing P = Nm⟨v²⟩/V without projecting onto one axis triple-counts and gives triple the correct pressure.
  • Treating &frac32;kBT as the total energy of a diatomic molecule. It is the translational energy only; rotation adds kBT more.
  • Molar/molecular mixups. Using R with N molecules, or kB with n moles — off by NA ≈ 6×1023.
  • Assuming heavier molecules hit harder, so heavier gases have higher pressure. At the same T they move slower (vrms ∝ 1/√m) in exactly compensating fashion; pressure depends only on N/V and T.
Discussion

The single most important conceptual output is not PV = NkBT itself — that relation was known empirically for a century before Maxwell and Boltzmann — but the identification of temperature with mean molecular kinetic energy. This severs temperature from any particular thermometric substance and grounds it in mechanics. The same identity, ½m⟨v²⟩ = &frac32;kBT, is what allows the ideal gas to serve as the reference for the absolute (thermodynamic) temperature scale.

Notice that mass dropped out of the equation of state entirely: P depends only on the number density N/V and the temperature, never on what the molecules are. This is Avogadro's hypothesis derived from first principles — equal volumes of any two ideal gases at the same P and T contain equal numbers of molecules. Heavier molecules move more slowly by precisely the factor needed to keep their momentum flux the same.

The result PV = ⅔Utr is the mechanical equation of state, and it generalises. The factor ⅔ is specific to a non-relativistic gas in three dimensions with a quadratic dispersion ε ∝ p². For an ultra-relativistic gas or a photon gas (ε = pc) the same momentum-flux argument gives PV = ⅓U instead. The prefactor is thus a fingerprint of the dispersion relation and the dimensionality, tying this elementary derivation to radiation pressure and to the equation of state of stellar interiors. The equipartition step (each quadratic degree of freedom carries ½kBT) is itself a theorem of classical statistical mechanics, and its breakdown — frozen-out vibrational modes, the heat capacity of hydrogen falling below &frac72;R at low T — was one of the earliest quantitative signposts pointing toward quantum mechanics.

Common misconceptions. Pressure is not caused by molecules colliding with each other — intermolecular collisions merely randomise the velocity distribution; the wall pressure comes from molecule–wall impacts and would be identical for a hypothetical collisionless gas with the same isotropic distribution. Also, temperature is a property of the distribution, not of an individual molecule: a single fast molecule has no temperature, and "hot" means the ensemble average of v² is large.

Worked examples
1
Root-mean-square speed of N2 at room temperature
Find vrms = √⟨v²⟩ for nitrogen at T = 300 K. A
2
½m⟨v²⟩ = &frac32;kBT ⇒ ⟨v²⟩ = 3kBT / m
Solve the temperature–energy relation for the mean square speed — symbols first. A
3
m = M / NA = (0.028 kg/mol) / (6.022×1023 mol−1) = 4.65×10−26 kg
Mass of one N2 molecule from molar mass 28 g/mol. A
4
⟨v²⟩ = 3(1.381×10−23)(300) / (4.65×10−26) = 2.67×105 m²/s²
Insert numbers with SI units throughout. A
vrms = √(2.67×105) ≈ 517 m/s

Reading. Air molecules at room temperature move at roughly 500 m/s — faster than the speed of sound (343 m/s), consistent with sound being carried by these same molecules. Lighter gases are faster: helium at the same T gives vrms ≈ 1370 m/s.

1
Pressure of helium from number density and speed
A vessel holds helium at number density n = N/V = 2.50×1025 m−3 with vrms = 1360 m/s. Find P. A
2
P = ⅓ (N/V) m ⟨v²⟩ = ⅓ n m vrms²
Kinetic result from step 8 of the derivation, written with number density. A
3
mHe = (0.004 kg/mol)/(6.022×1023) = 6.64×10−27 kg,  vrms² = 1.85×106 m²/s²
Molecular mass of He and the square of the given speed. A
4
P = ⅓(2.50×1025)(6.64×10−27)(1.85×106)
Substitute; units (m−3)(kg)(m²/s²) = kg·m−1·s−2 = Pa. B
P ≈ 1.02×105 Pa ≈ 1.0 atm

Reading. A monatomic gas at ordinary density and thermal speeds sits near atmospheric pressure, as it should. Cross-check via P = nkBT: with T = ⅓m vrms²/kB ≈ 297 K, P = (2.50×1025)(1.381×10−23)(297) ≈ 1.03×105 Pa. Consistent.

Problems
  1. A sealed 2.0 L rigid flask contains an ideal gas at 1.0 atm and 300 K. It is heated to 450 K. Find the new pressure.
    Solution At fixed N and V, P/T is constant: P2 = P1(T2/T1) = (1.0 atm)(450/300) = 1.5 atm = 1.52×105 Pa. The molecules move faster (vrms ∝ √T, up by √1.5 = 1.22×) and each impact carries more momentum, raising the pressure.
  2. Compute the number of molecules in 1.0 m³ of an ideal gas at STP (P = 1.013×105 Pa, T = 273 K).
    Solution N = PV/kBT = (1.013×105)(1.0)/[(1.381×10−23)(273)]. Denominator = 3.77×10−21. N = 2.69×1025 molecules (Loschmidt's number). Equivalently n = N/NA = 44.6 mol, giving molar volume 22.4 L/mol.
  3. Two ideal gases, hydrogen (M = 2 g/mol) and oxygen (M = 32 g/mol), are at the same temperature. Find the ratio of their rms speeds.
    Solution At equal T, ½m⟨v²⟩ is equal, so vrms ∝ 1/√m ∝ 1/√M. Thus vrms,H2/vrms,O2 = √(32/2) = √16 = 4. Hydrogen molecules move four times faster, which is why light gases escape planetary atmospheres more readily (Graham's law of effusion follows from the same relation).
  4. The mean translational kinetic energy of a gas molecule is measured to be 6.0×10−21 J. Find the temperature.
    Solution ⟨εtr⟩ = &frac32;kBT ⇒ T = 2⟨εtr⟩/(3kB). T = 2(6.0×10−21)/[3(1.381×10−23)] = (1.2×10−20)/(4.14×10−23) = 290 K, i.e. roughly room temperature.
  5. Estimate the density (in kg/m³) at which the ideal gas law starts to fail badly for helium at 300 K, by requiring the thermal de Broglie wavelength to reach the interparticle spacing. Take mHe = 6.64×10−27 kg, h = 6.63×10−34 J·s.
    Solution Thermal wavelength λth = h/√(2πm kBT). Denominator: 2π(6.64×10−27)(1.381×10−23)(300) = 1.73×10−46; square root = 1.31×10−23. So λth = (6.63×10−34)/(1.31×10−23) = 5.05×10−11 m. Degeneracy sets in when spacing (V/N)1/3 ≈ λth, i.e. N/V ≈ λth−3 = (5.05×10−11)−3 = 7.8×1030 m−3. Mass density ρ = (N/V)mHe = (7.8×1030)(6.64×10−27) ≈ 5×104 kg/m³. This is far denser than liquid helium (125 kg/m³), confirming that at 300 K helium is comfortably classical — quantum degeneracy only matters at cryogenic temperatures where λth is much larger.