Hard-sphere kinetic theory

If one models gas molecules as elastic hard spheres (with mass ${\displaystyle m}$ and diameter ${\displaystyle \sigma }$), then elementary kinetic theory predicts that viscosity increases with the square root of absolute temperature ${\displaystyle T}$:

${\displaystyle \mu =1.016\cdot {\frac {5}{16\sigma ^{2}}}\left({\frac {k_{\rm {B}}mT}{\pi }}\right)^{1/2}}$

where ${\displaystyle k_{\text{B}}}$ is the Boltzmann constant. While correctly predicting the increase of gaseous viscosity with temperature, the ${\displaystyle T^{1/2}}$ trend is not accurate; the viscosity of real gases increases more rapidly than this. Capturing the actual ${\displaystyle T}$ dependence requires more realistic models of molecular interactions, in particular the inclusion of attractive interactions which are present in all real gases.^[2]

Power-law force

A modest improvement over the hard-sphere model is a repulsive inverse power-law force, where the force between two molecules separated by distance ${\displaystyle r}$ is proportional to ${\displaystyle 1/r^{\nu }}$, where ${\displaystyle \nu }$ is an empirical parameter.^[3] This is not a realistic model for real-world gases (except possibly at high temperature), but provides a simple illustration of how changing intermolecular interactions affects the predicted temperature dependence of viscosity. In this case, kinetic theory predicts an increase in temperature as ${\displaystyle T^{s}}$, where ${\displaystyle s=(1/2)+2/(\nu -1)}$. More precisely, if ${\displaystyle \mu '}$ is the known viscosity at temperature ${\displaystyle T'}$, then

${\displaystyle \mu =\mu '(T/T')^{s}}$

Taking ${\displaystyle \nu \rightarrow \infty }$ recovers the hard-sphere result, ${\displaystyle s=1/2}$. For finite ${\displaystyle \nu }$, corresponding to softer repulsion, ${\displaystyle s}$ is greater than ${\displaystyle 1/2}$, which results in faster increase of viscosity compared with the hard-sphere model. Fitting to experimental data for hydrogen and helium gives predictions for ${\displaystyle s}$ and ${\displaystyle \nu }$ shown in the table. The model is modestly accurate for these two gases, but inaccurate for other gases.

Table: Inverse power law potential parameters for hydrogen and helium[3]
Gas	${\displaystyle s}$	${\displaystyle v}$	Temp. range (K)
Hydrogen	0.668	12.9	273–373
Helium	0.657	13.7	43–1073

Sutherland model

Another simple model for gaseous viscosity is the Sutherland model, which adds weak intermolecular attractions to the hard-sphere model.^[4] If the attractions are small, they can be treated perturbatively, which leads to

${\displaystyle \mu ={\frac {5}{16\sigma ^{2}}}\left({\frac {k_{\text{B}}mT}{\pi }}\right)^{1/2}\cdot \left(1+{\frac {S}{T}}\right)^{-1}}$

where ${\displaystyle S}$, called the Sutherland constant, can be expressed in terms of the parameters of the intermolecular attractive force. Equivalently, if ${\displaystyle \mu '}$ is a known viscosity at temperature ${\displaystyle T'}$, then

${\displaystyle \mu =\mu '\left({\frac {T}{T'}}\right)^{3/2}{\frac {T'+S}{T+S}}}$

Values of ${\displaystyle S}$ obtained from fitting to experimental data are shown in the table below for several gases. The model is modestly accurate for a number of gases (nitrogen, oxygen, argon, air, and others), but inaccurate for other gases like hydrogen and helium. In general, it has been argued that the Sutherland model is actually a poor model of intermolecular interactions, and is useful only as a simple interpolation formula for a restricted set of gases over a restricted range of temperatures.

Table: Sutherland constants of selected gases[4]
Gas	${\displaystyle S}$ (K)	Temp. range (K)
Dry air	113	293–373
Helium	72.9	293–373
Neon	64.1	293–373
Argon	148	293–373
Krypton	188	289–373
Xenon	252	288–373
Nitrogen	104.7	293–1098
Oxygen	125	288–1102

Lennard-Jones

Under fairly general conditions on the molecular model, the kinetic theory prediction for ${\displaystyle \mu }$ can be written in the form

${\displaystyle \mu ={\frac {5}{16\pi ^{1/2}}}{\frac {(mk_{\text{B}}T)^{1/2}}{\sigma ^{2}\Omega (T)}}}$

where ${\displaystyle \Omega }$ is called the collision integral and is a function of temperature as well as the parameters of the intermolecular interaction.^[5] It is completely determined by the kinetic theory, being expressed in terms of integrals over collisional trajectories of pairs of molecules. In general, ${\displaystyle \Omega }$ is a complicated function of both temperature and the molecular parameters; the power-law and Sutherland models are unusual in that ${\displaystyle \Omega }$ can be expressed in terms of elementary functions.

The Lennard–Jones model assumes an intermolecular pair potential of the form

${\displaystyle V(r)=4\epsilon \left[\left({\frac {\sigma }{r}}\right)^{12}-\left({\frac {\sigma }{r}}\right)^{6}\right]}$

where ${\displaystyle \epsilon }$ and ${\displaystyle \sigma }$ are parameters and ${\displaystyle r}$ is the distance separating the centers of mass of the molecules. As such, the model is designed for spherically symmetric molecules. Nevertheless, it is frequently used for non-spherically symmetric molecules provided these do not possess a large dipole moment.^[5][6]

The collisional integral ${\displaystyle \Omega }$ for the Lennard-Jones model cannot be expressed exactly in terms of elementary functions. Nevertheless, it can be calculated numerically, and the agreement with experiment is good – not only for spherically symmetric molecules such as the noble gases, but also for many polyatomic gases as well.^[6] An approximate form of ${\displaystyle \Omega }$ has also been suggested:^[7]

${\displaystyle \Omega (T)=1.16145\left(T^{*}\right)^{-0.14874}+0.52487e^{-0.77320T^{*}}+2.16178e^{-2.43787T^{*}}}$

where ${\displaystyle T^{*}\equiv k_{\text{B}}T/\epsilon }$. This equation has an average deviation of only 0.064 percent of the range ${\displaystyle 0.3<T^{*}<100}$.

Values of ${\displaystyle \sigma }$ and ${\displaystyle \epsilon }$ estimated from experimental data are shown in the table below for several common gases.

Table: Lennard-Jones parameters of selected gases[8]
Gas	${\displaystyle \sigma }$ (angstroms)	${\displaystyle \epsilon /k_{\text{B}}}$ (K)
Dry air	3.617	97.0
Helium	2.576	10.2
Hydrogen	2.915	38.0
Argon	3.432	122.4
Nitrogen	3.667	99.8
Oxygen	3.433	113
Carbon dioxide	3.996	190
Methane	3.780	154

Two-parameter exponential

A simple and widespread empirical correlation for liquid viscosity is a two-parameter exponential:

${\displaystyle \mu =Ae^{B/T}}$

This equation was first proposed in 1913, and is commonly known as the Andrade equation (named after British physicist Edward Andrade). It accurately describes many liquids over a range of temperatures. Its form can be motivated by modeling momentum transport at the molecular level as an activated rate process,^[10] although the physical assumptions underlying such models have been called into question.^[11]

The table below gives estimated values of ${\displaystyle A}$ and ${\displaystyle B}$ for representative liquids. Comprehensive tables of these parameters for hundreds of liquids can be found in the literature.^[12]

Fitting parameters for the correlation ${\displaystyle \mu =Ae^{B/T}}$[13]
Liquid	Chemical formula	A (mPa·s)	B (K)	Temp. range (K)
Bromine	Br2	0.0445	907.6	269–302
Acetone	C3H6O	0.0177	845.6	193–333
Bromoform	CHBr3	0.0332	1195	278–363
Pentane	C5H12	0.0191	722.2	143–313
Bromobenzene	C6H5Br	0.02088	1170	273–423

Three- and four-parameter exponentials

One can also find tabulated exponentials with additional parameters, for example

${\displaystyle \mu =A\exp {\left({\frac {B}{T-C}}\right)}}$

and

${\displaystyle \mu =A\exp {\left({\frac {B}{T}}+CT+DT^{2}\right)}}$

Representative values are given in the tables below.

Fitting parameters for the correlation ${\displaystyle \mu =A\exp {\left({\frac {B}{T-C}}\right)}}$[14]
Liquid	Chemical formula	A (mPa·s)	B (K)	C (K−1)	Temp. range (K)
Mercury	Hg	0.7754	117.91	124.04	290–380
Fluorine	F2	0.09068	45.97	39.377	60–85
Lead	Pb	0.7610	421.35	266.85	600–1200
Hydrazine	N2H4	0.03625	683.29	83.603	280–450
Octane	C8H18	0.007889	1456.2	−51.44	270–400

Fitting parameters for the correlation ${\displaystyle \mu =A\exp {\left({\frac {B}{T}}+CT+DT^{2}\right)}}$[13]
Liquid	Chemical formula	A (mPa·s)	B (K)	C (K−1)	D (K−2)	Temp. range (K)
Water	H2O	1.856·10−11	4209	0.04527	−3.376·10−5	273–643
Ethanol	C2H6O	0.00201	1614	0.00618	−1.132·10−5	168–516
Benzene	C6H6	100.69	148.9	−0.02544	2.222·10−5	279–561
Cyclohexane	C6H12	0.01230	1380	−1.55·10−3	1.157·10−6	280–553
Naphthalene	C10H8	3.465·10−5	2517	0.01098	−5.867·10−6	354–748

Models for kinematic viscosity

The effect of temperature on the kinematic viscosity ${\displaystyle \nu }$ has also been described by a number of empirical equations.^[15]

The Walther formula is typically written in the form

${\displaystyle \log _{10}(\log _{10}(\nu +\lambda ))=A-B\,\log _{10}T}$

where ${\displaystyle \lambda }$ is a shift constant, and ${\displaystyle A}$ and ${\displaystyle B}$ are empirical parameters. In lubricant specifications, normally only two temperatures are specified, in which case a standard value of ${\displaystyle \lambda }$ = 0.7 is normally assumed.

The Wright model has the form

${\displaystyle \log _{10}(\log _{10}(\nu +\lambda +f(\nu )))=A-B\,\log _{10}T}$

where an additional function ${\displaystyle f(v)}$, often a polynomial fit to experimental data, has been added to the Walther formula.

The Seeton model is based on curve fitting the viscosity dependence of many liquids (refrigerants, hydrocarbons and lubricants) versus temperature and applies over a large temperature and viscosity range:

${\displaystyle \ln \left({\ln \left({\nu +0.7+e^{-\nu }K_{0}\left({\nu +1.244067}\right)}\right)}\right)=A-B\ln T}$

where ${\displaystyle T}$ is absolute temperature in kelvins, ${\displaystyle \nu }$ is the kinematic viscosity in centistokes, ${\displaystyle K_{0}}$ is the zero order modified Bessel function of the second kind, and ${\displaystyle A}$ and ${\displaystyle B}$ are empirical parameters specific to each liquid.

For liquid metal viscosity as a function of temperature, Seeton proposed:

${\displaystyle \ln \left({\ln \left({\nu +0.7+e^{-\nu }K_{0}\left({\nu +1.244067}\right)}\right)}\right)=A-{B \over T}}$