Yes, it's almost all very intuitive.
Let's figure out the force $F$ that the gas applies to a (flat) section of its container wall, with area $A$.
You know that $F = ma = \frac{dmv}{dt} = \frac{dp}{dt}$, that is the rate at which momentum is transferred from the wall to molecules of the gas.
This is of course proportional to the rate at which molecules collide with the wall, which is in turn proportional to the area of the wall, $A$, because that's how much is exposed to the gas, the density of gas molecules, $n/V$, and the average speed of those molecules, $|v|$, because if they go faster then they collide more often.
It's also proportional to the average momentum imparted per collision. When a particle collides, the component of its momentum normal to the wall is reversed, so the total momentum imparted is proportional to its average momentum, or $m|v|$.
Putting that all together, we have, for some constant $R$:
$$F = \frac{nRAm|v|^2}{V} \Leftrightarrow P=\frac{nRm|v|^2}{V}\Leftrightarrow PV=nRm|v|^2$$
We're almost there. Temperature, $T$, is average kinetic energy per molecule, and $m|v|^2$ certainly has the right units. But remember that $|v|$ is the average speed of each molecule, and (average $|v|$)2 is not the same as average $(|v|^2)$. However, if we assume that the shape of the distribution of velocities doesn't depend on the average speed, or the molecular mass, then these two quantities will be proportional, and we can fold the factor into our constant $R$, finally leaving $PV = nRT$.