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I have rarefied, dilute, diatomic gas (oxygen) and I have to calculate the viscosity using the Chapman-Enskog theory; however I couldn't find anywhere the formula the allows me to do such calculation. The only formula I was able to use is the one found in the Molecular Gas Dynamics and the Direct Simulation of Gas Flows book by Bird G.A. (at page 67 chapter 3) Such formula refers however to the viscosity coefficient of a monatomic gas

$$\mu = \frac{5}{16}(\frac{RT}{\pi})^{\frac{1}{2}}(\frac{m}{d^2})$$

As I said this formula is valid for a monatomic gas under the assumption of a hard sphere collisional model.

What I need is the equivalent formula for a diatomic gas or at least the reference where I can find it.

Federico Gentile
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    Viscosity is due to translational non-equilibrium -- do you think there is a difference between monatomic and diatomic translational energy modes? If you do, what are the differences? – tpg2114 Jan 02 '15 at 19:45
  • @tpg2114 I believe there should be no difference between monatomic and diatomic TRANSITIONAL energy modes; however what do you mean by saying "non-equilibrium"? And from where did you get that information from? – Federico Gentile Jan 02 '15 at 19:47
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    If all of the molecules were in translational equilibrium there would be no coherent gradients in velocity (there would be isotropically random variations according to the Maxwell distribution still) and so there would be no viscosity to speak of. They are in translational non-equilibrium because portions of the flow must exhibit different velocities and the "viscous" force is what drives the translational motion back into equilibrium. – tpg2114 Jan 02 '15 at 19:51
  • Also recall that there are two viscosities -- $\mu$ and $\lambda$, which for traditional fluids, Stokes' hypothesis is employed which says that $\lambda = -2/3 \mu$. In a gas dynamics sense, $\mu$ is a function of translational non-equilibrium and $\lambda$ is a function of rotational non-equilibrium. For a monatomic gas, $\lambda = 0$ but for a polyatomic gas that is no longer the case. – tpg2114 Jan 02 '15 at 19:54
  • Regarding where I got the information -- this is the textbook I used when I took gas dynamics: http://www.amazon.com/Introduction-Physical-Gas-Dynamics-Vincenti/dp/0882753096 and I'm just going from memory of that class. – tpg2114 Jan 02 '15 at 19:58
  • @tpg2114 What I am talking about is dynamic viscosity i.e. μ; what it bugs me is that (according to what you say) it would be the same both for a monatomic gas and a diatomic one... is that possible or not? this is strange since there should not be needed to be pointed out as i found in many text books – Federico Gentile Jan 05 '15 at 00:52
  • The equation may be the same, but the value certainly isn't. It depends on the mass of the molecule and the diameter of the approximate sphere the molecule represents. The further from spherical your molecule is, the less accurate this expression could be. So in theory it may actually work better for something like methane than oxygen since methane is "more" spherical that pure oxygen. But you should be able to work through the derivation/arguments in your book when the expression is presented to see whether diatomic molecules alter anything. – tpg2114 Jan 05 '15 at 01:20
  • Bird et al. wrote Transport Phenomena which has (p.26) $\mu=\frac{5}{16}\frac{\sqrt{\pi m\kappa T}}{\pi\sigma^2\Omega_\mu}$, where "If the gas were made up of rigid spheres of diamater $\sigma$ (instead of real molecules with attractive and repulsive forces), then $\Omega_\mu$ would be exactly unity. Hence the function $\Omega_\mu$ may be interpreted as describing the deviation from rigid-sphere behaviour". The book goes on to tabulate some $\Omega_\mu$ for some gases, eg. CO2 (varying from ~ 1-1.5 as a function of $T$) and calls Reid, Prausnitz & Poling as a standard reference for more. – alarge Jan 05 '15 at 12:24

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