Will gravity create high pressure at the bottom of a container filled with a gas? if there's a large closed container, about a km in height, which is filled with a gas will there be a pressure differential created at the bottom of the container with respect to the top? like it does when the container is filled with a liquid
 A: Yes. (See Earth's atmosphere for an example). The density distribution settles to one where pressure differential between adjacent heights is just enough to balance weight of gas between those heights. 
A: There is quiete a simple model that incorporates your thoughts: the scale height. Say, you are in this gas. When you travel the scale height, the pressure increases by a factor of $e$. This can be motivated as follows:
We want the exponential relation of height and pressure, such that $P=P_0\exp(-\frac{z}{H}$). This is also the solution to a differential equation of the form $\frac{\mathrm{d}P}{\mathrm{d}z}=-\frac{P}{H}$. But on the other hand it should also be true, that the change in pressure with height is proportional to the density times the gravitational acceleration -- higher density means higher pressure, but stronger gravitational pull should also imply higher pressure, so we have something like $\frac{\mathrm{d}P}{\mathrm{d}z}=-g\rho$. Bringing both equation together we find:
\begin{equation}
\frac{P}{H}=g\rho
\end{equation}
From an ideal gas and thermodynamics we know: $PV=nRT$ and by definition we have $\rho=\frac{M}{V}$, which leads to
\begin{align}
\frac{nRT}{VH}&=g\frac{M}{V}\\
\frac{nRT}{H}&=gM\\
\frac{nRT}{gM}&=H
\end{align}
So the pressure increases with $\exp(-\frac{z}{\frac{nRT}{gM}})=\exp(-\frac{zgM}{nRT})$.
So yes, there is always a pressure difference in a gas container, and the difference even increases exponentially!
