Summary
From dimensional analysis I find that the dynamic viscosity of an ideal gas must depend on its pressure $p$, density $\rho$ and mean molecular free path $l$ in this way:
$$
\mu = C \sqrt{\rho p} l.\quad
$$
Here, $C\geq0$ is a non-dimensional constant.
However, I find it counter intuitive that the dynamic viscosity, the 'internal friction', of the fluid increases with an increasing mean free path. My intuition tells me that the internal friction is low if the molecules are widely separated.
- Have I missed some quantity that should enter the expression?
- Has my derivation failed in some other way?
- Is my intuitive picture wrong?
The derivation
In an ideal gas, molecules are interacting only through ellastic collisions. The equation of state is:
$$
p = \rho R T. \quad (1)
$$
The variables and their units are:
- $p$: Pressure [kg/(m s$^2$)]
- $\rho$: Density [kg/(m$^3$)]
- $R$: Specific gas constant [m$^2$/(s$^2$ K)]
- $T$: Temperature [K]
In general, these are field variables, so $p = p(\mathbf{x},t)$, $\rho = \rho(\mathbf{x},t)$ and $T = T(\mathbf{x},t)$. In fluid dynamics, a common assumption is that each infinitesimally small volume is in thermodynamic equilibrium, so that (1) holds at every point in the fluid. I make this assumption. I also assume that the fluid is 'Newtonian', so that the viscous stress tensor is proportional to the rate of strain. The constant of proportionality is the dynamic viscosity, $\mu$, whose unit is [kg/(m s)].
The dynamic viscosity is a 'material property'; it is independent of the motion of the fluid. In general, it is varying over space, so that $\mu = \mu(\mathbf{x},t)$. It's value is a property of the material and depends on its thermodynamic state.
It seems impossible to find how $\mu$ depends on the thermodynamic state from (1). Pressure has 'almost' the correct units, but I need to multiply the pressure by some time scale $\tau$ [s]. This time scale must depend on the microscopic properties of the material, and the only way I find it possible to construct it is by using the $l$ [m] the mean free path of the molecules in the fluid. The time scale contructed is:
$$
\tau = \sqrt{\frac{\rho}{p}} l.\quad (2)
$$
Using (2) I find that the dynamic viscosity must depend on $p$, $\rho$ and $l$ in this way:
$$
\mu = C \sqrt{\rho p} l,\quad (3)
$$
where $C\geq0$ is a non-dimensional constant.
Best Answer
Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$
Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here: We have $y-y_0=I\,l$. The mean velocity difference for such particles is therefore just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.
Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.
P.S.: Also see the derivation in the Wikipedia article on viscosity.