Gas Viscosity, Diffusion, and Thermal Conductivity

Rub your hands together and you feel friction from air molecules colliding across a velocity gradient. That everyday drag, the slow mixing of two gases, and the way a hot poker warms the air around it are all governed by one microscopic idea. Molecules carry momentum, mass, and energy along with them as they fly between collisions. Kinetic theory converts this simple picture into precise formulas for viscosity, diffusion, and thermal conductivity. Remarkably, these formulas predict that gas viscosity and thermal conductivity are independent of pressure. This surprised nineteenth-century physicists and was later confirmed experimentally. This article derives all three transport coefficients from the mean free path framework. You will see the common derivation strategy first. Then you will see how each coefficient's pressure and temperature dependence follows from the same logic. Finally, two worked examples show the numbers behind nitrogen and helium.

Key takeaways

Viscosity, thermal conductivity, and diffusion all arise from molecules transporting momentum, energy, or mass across a gradient during free flight.
All three coefficients follow from a single "flux-carrying quantity times mean free path" argument.
Viscosity $\eta$ and thermal conductivity $\kappa$ are predicted to be independent of pressure; only diffusion $D$ depends on pressure (through density).
All three coefficients increase with temperature, roughly as $\sqrt{T}$ for hard spheres, though real gases climb faster due to softer intermolecular repulsion.
The Eucken correction improves the crude kinetic-theory thermal conductivity formula for polyatomic gases with internal degrees of freedom.
Key idea: transport coefficients measure how efficiently random thermal motion carries momentum, energy, or mass across a spatial gradient, one mean free path at a time.

The core insight

Every transport property is built from the same skeleton: a number density of carriers, an average speed at which they cross a plane, and a mean free path over which they carry their "cargo" (momentum for viscosity, kinetic energy for thermal conductivity, molecular identity for diffusion) before colliding and equilibrating to local conditions.

The physical picture

Picture a gas confined between two plates in relative motion, a hot plate facing a cold one, or two isotopes of a gas in contact. In each case, some property varies smoothly in one direction, call it $z$ : the drift velocity $v_x(z)$ , the temperature $T(z)$ , or the mole fraction of a labeled species $n_A(z)$ . Molecules do not care about the gradient directly. They simply fly in straight lines between collisions, carrying whatever momentum, energy, or identity they had at their last collision point. A molecule crossing a reference plane at $z = 0$ typically last collided one mean free path, $\lambda$ , away. If it came from $z = -\lambda$ , it carries the properties characteristic of that layer; if from $z = +\lambda$ , the properties of the other layer. Because molecules from the "hotter," "faster," or "more concentrated" side carry a systematically larger value of the transported quantity than molecules from the other side, there is a net flux of that quantity down the gradient, even though the molecular flux itself is balanced (equal numbers cross in each direction at equilibrium). This is the same transport framework introduced for mean free path and general flux laws: a property $\psi$ (momentum $m v_x$ , energy $\tfrac{3}{2}k_BT$ or $C_{V,m}T/N_A$ , or number density of tagged particles) is carried across a plane by molecules arriving isotropically from both sides, biased by the local value of $\psi$ one mean free path back.

Figure

Momentum transport between sliding gas layers gives rise to viscosity

The three transport coefficients differ only in what "cargo" the molecules carry and how the gradient is defined. Once $\eta$ , $D$ , or $\kappa$ is derived from this bookkeeping, its dependence on pressure and temperature follows immediately from how $n$ , $\bar{c}$ , and $\lambda$ separately depend on $p$ and $T$ .

Figure

How each coefficient depends on pressure and temperature

The core relations

\eta = \frac{1}{3} n m \bar{c} \lambda

where: $n$ is number density, $m$ molecular mass, $\bar{c}$ mean speed, $\lambda$ mean free path.

D = \frac{1}{3} \bar{c} \lambda

where: $D$ is the self-diffusion coefficient, same $\bar{c}$ and $\lambda$ as above.

\kappa = \frac{1}{3} n \bar{c} \lambda C_{V,m}/N_A

where: $C_{V,m}$ is the molar heat capacity at constant volume, $N_A$ Avogadro's constant, so $C_{V,m}/N_A$ is the heat capacity per molecule.

\bar{c} = \left(\frac{8k_BT}{\pi m}\right)^{1/2}, \qquad \lambda = \frac{k_BT}{\sqrt{2}\,\sigma p}

where: $k_B$ Boltzmann's constant, $\sigma$ the collision cross-section, $p$ pressure.

\eta = \frac{2}{3\pi^{1/2}\sigma}\,(m k_B T)^{1/2}, \qquad \kappa = \frac{\eta C_{V,m}}{M}

where: $M = N_A m$ is the molar mass; both expressions are pressure-independent because the $p$ in $n$ cancels the $1/p$ in $\lambda$ .

Derivation AMomentum flux and the viscosity coefficient

Goal. Derive $\eta = \tfrac{1}{3} n m \bar c \lambda$ from molecular flux across a plane in a gas with a velocity gradient $dv_x/dz$ , then substitute kinetic-theory expressions for $\bar c$ and $\lambda$ to get $\eta$ in terms of $T$ , $m$ , and $\sigma$ alone.

Step 1: Flux of molecules crossing a plane. In an ideal gas at rest (no bulk flow), elementary kinetic theory gives the one-way molecular flux crossing any plane, from either side, as

\Phi = \frac{1}{4} n \bar c .

This factor of $1/4$ comes from averaging the speed component normal to the plane over a Maxwell–Boltzmann distribution restricted to molecules moving toward the plane.

Step 2: Assign each stream its "last-collision" property. Molecules crossing upward through $z=0$ typically had their last collision near $z = -\lambda$ ; downward-crossing molecules last collided near $z=+\lambda$ . If the fluid has a shear flow $v_x(z)$ , a molecule arriving from $z = \mp\lambda$ carries an extra x-momentum

m\,v_x(\mp\lambda) \approx m\left[v_x(0) \mp \lambda \frac{dv_x}{dz}\right],

using a first-order Taylor expansion, valid because $\lambda$ is microscopically small compared to the scale over which $v_x$ varies.

Step 3: Net momentum flux (shear stress). The net x-momentum flux across the plane, $J_p$ , is the flux from below minus the flux from above (each weighted by momentum carried):

J_p = \Phi\, m\left[v_x(0) + \lambda \frac{dv_x}{dz}\right] - \Phi\, m\left[v_x(0) - \lambda \frac{dv_x}{dz}\right] = 2\Phi m \lambda \frac{dv_x}{dz}.

Substituting $\Phi = \tfrac14 n\bar c$ gives

J_p = \frac{1}{2} n m \bar c \lambda \, \frac{dv_x}{dz}.

Step 4: Match to Newton's law of viscosity. Newtonian viscous stress is defined by $\tau = -\eta \, dv_x/dz$ , with the flux of x-momentum in the $-z$ direction equal to $\tau$ . A more careful integration over the full Maxwell–Boltzmann angular distribution (rather than the simplified two-stream argument) replaces the prefactor $1/2$ by $1/3$ , the standard result quoted in every kinetic-theory textbook:

\eta = \frac{1}{3} n m \bar c \lambda.

Step 5: Insert $\bar c$ and $\lambda$ . Using

\bar c = \left(\frac{8k_BT}{\pi m}\right)^{1/2}, \qquad \lambda = \frac{k_BT}{\sqrt2\,\sigma p},

and the ideal-gas relation $n = p/(k_BT)$ , the pressure cancels exactly:

\eta = \frac{1}{3}\cdot\frac{p}{k_BT}\cdot m \cdot \left(\frac{8k_BT}{\pi m}\right)^{1/2}\cdot \frac{k_BT}{\sqrt2\,\sigma p} = \frac{2}{3\pi^{1/2}\sigma}(m k_B T)^{1/2}.

Boxed result.

\boxed{\eta = \frac{2}{3\pi^{1/2}\sigma}\,(m k_B T)^{1/2}}

independent of pressure, and growing as $T^{1/2}$ for a hard-sphere gas with temperature-independent $\sigma$ . The identical bookkeeping, replacing $mv_x$ by molecular energy $\tfrac32 k_BT$ (or by tagged-molecule concentration), yields $\kappa$ (or $D$ ) with the same prefactor $\tfrac13 n\bar c \lambda$ times the appropriate per-molecule carried quantity.

Worked examples

Example 1 — Viscosity of nitrogen gas at 298 K. Nitrogen, $M = 28.02\text{ g mol}^{-1}$ , so $m = 28.02\times10^{-3}/6.022\times10^{23} = 4.653\times10^{-26}\text{ kg}$ . Take a typical collision cross-section $\sigma = 0.43\text{ nm}^2 = 4.3\times10^{-19}\text{ m}^2$ (consistent with a kinetic diameter near 370 pm). Mean speed at $T = 298$ K:

\bar c = \left(\frac{8(1.381\times10^{-23})(298)}{\pi (4.653\times10^{-26})}\right)^{1/2}.

Numerator: $8 \times 1.381\times10^{-23}\times298 = 3.293\times10^{-20}$ . Denominator: $\pi \times 4.653\times10^{-26} = 1.462\times10^{-25}$ . Ratio $= 2.253\times10^{5}\text{ m}^2\text{s}^{-2}$ , so

\bar c = 474.6\text{ m s}^{-1}.

Now use the pressure-independent formula directly:

\eta = \frac{2}{3\pi^{1/2}\sigma}(mk_BT)^{1/2} = \frac{2}{3\sqrt\pi \,(4.3\times10^{-19})}\left[(4.653\times10^{-26})(1.381\times10^{-23})(298)\right]^{1/2}.

Inside the brackets: $4.653\times10^{-26}\times1.381\times10^{-23}\times298 = 1.915\times10^{-46}$ , so the square root is $1.384\times10^{-23}$ . Prefactor: $3\sqrt\pi = 3\times1.7725 = 5.317$ ; so $3\sqrt\pi \,\sigma = 5.317\times4.3\times10^{-19} = 2.286\times10^{-18}$ .

\eta = \frac{2\times 1.384\times10^{-23}}{2.286\times10^{-18}} = 1.211\times10^{-5}\text{ Pa s}.

This is $12.1\ \mu\text{Pa s}$ , matching the experimental value for N₂ near room temperature (about $17.9\ \mu\text{Pa s}$ measured; the kinetic-theory hard-sphere estimate typically underestimates by 30–50%, a known limitation discussed below). The important qualitative points confirmed here are the order of magnitude ( $10^{-5}\text{ Pa s}$ ) and the complete absence of $p$ in the formula: doubling the pressure at fixed $T$ leaves $\eta$ unchanged, which is exactly what is observed experimentally up to several atmospheres.

Example 2 — Thermal conductivity and mean free path of helium at 300 K and 1.00 atm, with the Eucken correction check. Helium is monatomic, $M = 4.003\times10^{-3}\text{ kg mol}^{-1}$ , so $m = 6.646\times10^{-27}\text{ kg}$ . Use $\sigma = 0.135\text{ nm}^2 = 1.35\times10^{-19}\text{ m}^2$ (helium's small kinetic diameter, about 260 pm).

Step A: mean free path. At $p = 1.013\times10^{5}$ Pa, $T = 300$ K:

\lambda = \frac{k_BT}{\sqrt2\,\sigma p} = \frac{(1.381\times10^{-23})(300)}{1.4142\times(1.35\times10^{-19})(1.013\times10^{5})}.

Numerator: $4.143\times10^{-21}$ . Denominator: $1.4142\times1.35\times10^{-19}\times1.013\times10^{5} = 1.934\times10^{-14}$ .

\lambda = 2.142\times10^{-7}\text{ m} = 214\text{ nm}.

This is about 800 times the atomic diameter, confirming the dilute-gas assumption behind the theory.

Step B: mean speed.

\bar c = \left(\frac{8(1.381\times10^{-23})(300)}{\pi(6.646\times10^{-27})}\right)^{1/2}.

Numerator: $8\times1.381\times10^{-23}\times300 = 3.314\times10^{-20}$ . Denominator: $\pi\times6.646\times10^{-27} = 2.088\times10^{-26}$ . Ratio $= 1.587\times10^{6}$ , square root gives

\bar c = 1259\text{ m s}^{-1}.

Step C: number density.

n = \frac{p}{k_BT} = \frac{1.013\times10^{5}}{(1.381\times10^{-23})(300)} = 2.445\times10^{25}\text{ m}^{-3}.

Step D: crude kinetic-theory thermal conductivity. For a monatomic ideal gas, $C_{V,m} = \tfrac32 R = 12.47\text{ J K}^{-1}\text{mol}^{-1}$ , so the per-molecule heat capacity is $C_{V,m}/N_A = 2.071\times10^{-23}\text{ J K}^{-1}$ .

\kappa = \frac13 n \bar c \lambda \, \frac{C_{V,m}}{N_A}.

Compute $n\bar c \lambda = 2.445\times10^{25}\times1259\times2.142\times10^{-7} = 6.594\times10^{21}\text{ m}^{-2}\text{s}^{-1}$ . Multiply by $C_{V,m}/N_A$ : $6.594\times10^{21}\times2.071\times10^{-23} = 0.1366$ . Divide by 3:

\kappa = 0.0455\text{ W m}^{-1}\text{K}^{-1}.

The experimental thermal conductivity of helium near 300 K is about $0.152\text{ W m}^{-1}\text{K}^{-1}$ , roughly $3\times$ higher. For a monatomic gas there are no rotational or vibrational modes to complicate the Eucken correction (it applies its enhancement mainly to polyatomic gases), so the persistent discrepancy here is due to helium's genuinely soft, non-hard-sphere repulsive potential and to the crude two-stream derivation underestimating the numerical prefactor; more refined Chapman–Enskog theory raises the coefficient from $1/3$ toward roughly $5/2 \times (k_B/m)\times$ a collision integral, closing most of the gap. Even so, the pressure-independence and $\sqrt{T}$ -type trend predicted here are exactly what is observed for real helium.

Limiting cases & common traps

Situation	What happens	Why
Increase $p$ at fixed $T$	$\eta$ , $\kappa$ unchanged; $D$ decreases as $1/p$	$n \propto p$ and $\lambda \propto 1/p$ cancel in $\eta,\kappa$ but $D \propto \lambda$ alone
Increase $T$ at fixed $n$	All three coefficients increase	$\bar c \propto T^{1/2}$ ; for hard spheres $\sigma$ is $T$ -independent, so all scale near $T^{1/2}$
Very low pressure (below ~ $10^{-3}$ atm in small vessels)	Kinetic-theory formulas above break down	$\lambda$ becomes comparable to or larger than the container size (Knudsen regime)
Polyatomic gas	Simple $\kappa$ formula underestimates conductivity	Rotational/vibrational energy is also transported; Eucken correction adds a factor accounting for internal modes
Real (non-hard-sphere) molecules	$\sigma$ effectively decreases with $T$	Softer repulsive potential lets faster molecules approach more closely, so $\eta,\kappa$ rise faster than $T^{1/2}$

The chemist behind it

James Clerk Maxwell derived the first quantitative kinetic-theory expression for gas viscosity in 1860, and he was startled by his own prediction that $\eta$ should be independent of pressure. This flatly contradicted the intuition, common at the time, that a rarer gas should offer less resistance to flow. Maxwell tested the prediction himself with a torsion-pendulum viscometer, varying gas pressure over a wide range, and confirmed that viscosity indeed stayed constant down to quite low pressures, a triumph that helped establish the kinetic theory of gases as physically real rather than a convenient mathematical fiction. Rudolf Clausius had already introduced the mean free path concept a few years earlier, giving Maxwell the tool he needed to connect molecular speeds and collision frequency directly to macroscopic transport.

Origins · 1 minJames Clerk Maxwell and how this was discovered

James Clerk Maxwell

1831–1879 · Derived the kinetic-theory viscosity formula and experimentally confirmed its startling pressure-independence

Ask the tutor: How does the Chapman–Enskog collision integral refine these hard-sphere estimates for real gases like helium and nitrogen?

Why are gas viscosity and thermal conductivity independent of pressure, but liquid viscosity is not?

In the kinetic-theory derivation, viscosity contains the product $n\lambda$ , and since $n \propto p$ while $\lambda \propto 1/p$ , the pressure cancels exactly, leaving $\eta$ a function of $T$ alone (for a given gas). This cancellation is a real, experimentally confirmed feature of dilute gases, verified from millimeter to atmospheric pressures. Liquids have no comparable mean-free-path picture: molecules are in constant contact, and viscosity instead reflects the strength of intermolecular cohesive forces, which do respond to compression, so liquid viscosity depends on pressure at high compressions.

Why do gas viscosity, diffusion, and thermal conductivity all increase with temperature?

All three coefficients contain the mean molecular speed $\bar c \propto T^{1/2}$ , which sets how fast molecules can ferry momentum, energy, or identity across a gradient. For an idealized hard-sphere gas with a temperature-independent cross-section $\sigma$ , all three transport coefficients scale close to $T^{1/2}$ . Real gases climb somewhat faster than $T^{1/2}$ because the effective collision cross-section shrinks slightly as molecules move faster and probe less of each other's soft repulsive potential, an effect captured quantitatively by more advanced Chapman–Enskog transport theory.

How accurate is the simple kinetic-theory formula for viscosity in practice?

The functional form, pressure-independence and roughly $T^{1/2}$ growth, is confirmed extremely well. The absolute numerical prefactor from the crude two-stream, hard-sphere derivation typically underestimates measured viscosities by 30 to 50 percent, as seen for nitrogen in the worked example above. Chapman–Enskog theory replaces the simple mean free path average with a proper collision integral over a realistic intermolecular potential (such as the Lennard-Jones potential), which brings predictions within a few percent of experiment across a wide temperature range.

Does the ideal kinetic-theory picture apply at very low or very high pressure?

At very low pressure, or in narrow tubes and pores, the mean free path can become comparable to or exceed the size of the container, entering the Knudsen regime, where the kinetic-theory transport formulas above no longer apply and transport instead depends on the container geometry itself. At very high pressure, the gas is dense enough that molecular volume and intermolecular attractions violate the dilute-gas assumptions behind $\lambda$ , and real-gas corrections, or a liquid-like transport picture, become necessary.