Pressure dependence of viscosity Does viscosity depend on pressure? If depends, is there any equation by which I can calculate viscosity at a specific pressure?
 A: In the ideal gas limit of low pressures, viscosity is a function only of temperature.  But at higher pressures, above the ideal gas region, viscosity also depends on pressure (and increases with pressure).  Bird et al, Transport Phenomena, present a corresponding states graph of reduced viscosity as a function of reduced pressure and reduced temperature showing all this.
For liquids, the effect of pressure on viscosity is typically negligible.
A: Viscosity does not depend directly on pressure. However it does depend on temperature. In liquids it usually decreases with increasing temperature, whereas in gases viscosity increases with increasing temperature. The relationship in the case of gases is that
$\eta \approx \sqrt{T} $
where $\eta$ is the viscosity and $T$ is the temperature.
A: No, the viscosity of gases does not depend pressure. Viscosity of liquids may depend on pressure at extreme pressures, but the dependence is very weak.
Why? The classical easy derivation for basic courses is to take the kinetic theory of gases and simply assume how the chaotic movement of molecules with different velocities transfers momentum between those layers. The momentum flux is the viscous friction force.
Simply put, when you have two layers of fluid, one faster and one slower, and their distance is equal to the mean free path, the molecules from the slower layer bring their smaller momentum to the faster layer and the molecules from the faster layer bring their higher momentum to the slower layer. We just have to be careful not to mix the mean translational velocity of the fluid layer and the chaotic velocities of the individual molecules. Those are different and both are important.
You can check the kinetic theory calculation for a simple dilute gas on Wikipedia https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Viscosity_and_kinetic_momentum The momentum flux is linearly dependent on the velocity gradient and the constant of the proportionality is the (dynamic) viscosity and for this simple model it is proportional to $\sqrt{MRT}$ where $M$ is the molar mass, $R$ is the gas constant and $T$ is the temperature.

Notice, that the pressure does not appear in the starting points for the derivation. You can insert it through the mean free path
$$l = \frac{k_B T}{\sqrt{2} \pi d^2 p}$$
but then you have
$$\mu = \frac{1}{3} \bar{v} n m l = \frac{1}{3} \bar{v} \rho l =  \frac{1}{3} \bar{v} n m l = \frac{1}{3} \bar{v} \frac{k_B }{\sqrt{2} \pi d^2} \frac{\rho T}{p}.$$
And according to the ideal gas law
$$ p = \rho R T$$
you get
$$ \frac{\rho T}{p} = \frac{1}{R}$$ which is just a constant so the pressure disappears from the equation for the dynamic viscosity $\mu$.
And the mean velocity of the chaotic motion of the molecules depends just on the temperature.

For a more realistic model of a non-dilute gas one should use the more refined Chapman-Enskog theory, which accounts for the finite size of the molecules and the forces between them, when they spend a lot of time near each other. You can again see that $\mu$ is proportional to $\sqrt{T}$ with some corrections depending on the exact model for the inter-molecular forces.
For liquids there is no general theory like the Chapman-Enskog above, even though they are in some sense limit of the gas models when the molecules are very close to each other all the time. But in that case the real behaviour is extremely dependent on the exact details of the inter-molecular forces. The pressure dependence will probably appear there but the molecules are already close to each other even at lower pressures and many liquids are only compressible very weakly. The distance of the molecules only gets lower at very high pressures.
The gases are definitely a simpler system to look at.
